Mathematical Thinking Questions \u2014 Interactive<\/title>\n\n <script>\n window.MathJax = {\n tex: {\n inlineMath: [['$', '$'], ['\\\$', '\\\$']],\n displayMath: [['\\\\[', '\\\\]']]\n },\n svg: { fontCache: 'global' }\n };\n <\/script>\n <script src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-svg.js\" async><\/script>\n\n <style>\n :root {\n --bg: #f5f7fb;\n --card: #ffffff;\n --text: #1f2937;\n --muted: #6b7280;\n --border: #d6dbe6;\n --accent: #0f766e;\n --accent-dark: #115e59;\n --accent-light: #e6f4f1;\n --hint-bg: #fff7ed;\n --hint-border: #fed7aa;\n --answer-bg: #eef8ee;\n --answer-border: #bbf7d0;\n --shadow: 0 10px 25px rgba(15, 23, 42, 0.08);\n }\n\n * { box-sizing: border-box; }\n\n body {\n margin: 0;\n font-family: system-ui, -apple-system, BlinkMacSystemFont, \"Segoe UI\", sans-serif;\n background: var(--bg);\n color: var(--text);\n line-height: 1.55;\n }\n\n header {\n background: linear-gradient(135deg, var(--accent), var(--accent-dark));\n color: white;\n 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.nav-btn.secondary {\n background: #e5e7eb;\n color: #1f2937;\n }\n\n .panel {\n display: none;\n margin-top: 0.75rem;\n padding: 0.9rem;\n border-radius: 13px;\n }\n\n .panel.hint {\n background: var(--hint-bg);\n border: 1px solid var(--hint-border);\n }\n\n .panel.answer {\n background: var(--answer-bg);\n border: 1px solid var(--answer-border);\n }\n\n .nav-row {\n display: flex;\n justify-content: space-between;\n align-items: center;\n gap: 0.8rem;\n margin-top: 1.2rem;\n flex-wrap: wrap;\n }\n\n .letter-list, .inner-list {\n margin-top: 0.5rem;\n }\n\n .letter-list {\n list-style: none;\n counter-reset: proof-item;\n padding-left: 2.1rem;\n }\n\n .letter-list > li {\n counter-increment: proof-item;\n position: relative;\n margin: 0.35rem 0;\n }\n\n .letter-list > li::before {\n content: \"(\" counter(proof-item, lower-alpha) \")\";\n position: absolute;\n left: -2.1rem;\n font-weight: 700;\n }\n\n\n .step-support-reminder {\n margin: 1.25rem 0 0;\n padding-top: 0.65rem;\n border-top: 1px solid var(--border);\n color: var(--muted);\n font-size: 0.84rem;\n font-style: italic;\n }\n\n .step-support-reminder a {\n color: var(--accent-dark);\n font-weight: 700;\n text-decoration: none;\n }\n\n .step-support-reminder a:hover {\n text-decoration: underline;\n }\n\n .nav-btn.definition-btn {\n background: #eef2ff;\n color: #3730a3;\n border: 1px solid #c7d2fe;\n }\n\n .footer {\n text-align: center;\n color: var(--muted);\n font-size: 0.9rem;\n margin-top: 1.5rem;\n }\n\n @media (max-width: 640px) {\n header { padding-bottom: 2.8rem; }\n main { margin-top: -1.5rem; }\n .step-header { display: block; }\n .step-count { margin-top: 0.4rem; }\n }\n <\/style>\n<\/head>\n\n<body class=\"hint-mode-none answer-mode-none\">\n <header>\n <h1>Mathematical Thinking Questions<\/h1>\n <p>Interactive self-assessment guide for applicants to Mathematics with Computer Science at GTIIT.<\/p>\n <\/header>\n\n <main>\n <section class=\"card note\">\n <h2>A note to the reader<\/h2>\n <p>\n The purpose of these problems is not simply to obtain the answers.\n Finding the answer is easy; one can ask a search engine, or even an AI system.\n But that is not the point.\n <\/p>\n <p>\n The only thing that really matters is the mathematical work that you do by yourself.\n Mathematics is not mainly about collecting correct answers. It is about understanding ideas,\n recognizing patterns, making definitions precise, finding examples, discovering counterexamples,\n and building arguments.\n <\/p>\n <p>\n Do not rush. Take your time. Try examples. Make mistakes. Start again.\n Write down your reasoning.\n <\/p>\n <p>\n You are not expected to solve all problems listed here.\n <\/p><div class=\"section-nav\">\n <button class=\"section-btn\" onclick=\"goToSection('eventually')\">1. Eventually and frequently equal to a value<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('intersections')\">2. Intersections of sets<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('countable')\">3. Countable sets<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('pigeonhole')\">4. The Pigeonhole Principle<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('proof')\">5. Build the proof<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('sequences')\">6. Sequences and formulas<\/button>\n <\/div>\n\n\n <div class=\"support-box\" id=\"supportBox\">\n <p class=\"support-note\">\n <span class=\"lead\">Need a little help?<\/span>\n You may choose to show hint buttons or answer buttons below.\n You can show support only for some later items, or for all items.\n <\/p>\n <div class=\"support-controls\">\n <button id=\"hintSomeToggle\" class=\"toggle-support-btn hints-toggle\" onclick=\"setHintMode('some')\">Some hint buttons<\/button>\n <button id=\"hintAllToggle\" class=\"toggle-support-btn hints-toggle\" onclick=\"setHintMode('all')\">All hint buttons<\/button>\n <button id=\"answerSomeToggle\" class=\"toggle-support-btn answers-toggle\" onclick=\"setAnswerMode('some')\">Some answer buttons<\/button>\n <button id=\"answerAllToggle\" class=\"toggle-support-btn answers-toggle\" onclick=\"setAnswerMode('all')\">All answer buttons<\/button>\n <\/div>\n <\/div>\n <\/section>\n\n <section class=\"card\" id=\"stepCard\">\n \n <article class=\"step visible\" data-step=\"1\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Definitions<\/h2>\n <\/div>\n <div class=\"step-count\">Step 1 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nSequences often exhibit patterns. Some patterns are temporary, some repeat infinitely many times,\nand some eventually stabilize. In this section we introduce two simple but useful notions.\n<\/p>\n<p>Let $(a_n)$ be a sequence, and let $c$ be a number.<\/p>\n\n<div class=\"definition-box\">\n <p>\n We say that $(a_n)$ is <strong>eventually equal to $c$<\/strong> if, from some point on,\n all its terms are equal to $c$. In other words, there exists a natural number $N$ such that\n <\/p>\n \\[\n a_n=c\n \\]\n <p>for every $n\\geq N$.<\/p>\n<\/div>\n\n<div class=\"definition-box\">\n <p>\n We say that $(a_n)$ is <strong>frequently equal to $c$<\/strong> if the value $c$\n appears infinitely many times in the sequence. Equivalently, for every natural number $N$,\n there exists some $n\\geq N$ such that\n <\/p>\n \\[\n a_n=c.\n \\]\n<\/div>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"2\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Warm-up problems<\/h2>\n <\/div>\n <div class=\"step-count\">Step 2 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n1,1,1,1,1,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint1')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer1')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint1\">\n <p>Every term of the sequence is equal to $1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer1\">\n <p>Yes. It is eventually equal to $1$, for example with $N=1$. It is also frequently equal to $1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n1,2,1,2,1,2,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$? Is it frequently equal to $2$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint2')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer2')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint2\">\n <p>The values $1$ and $2$ alternate forever.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer2\">\n <p>It is not eventually equal to $1$, because $2$ keeps appearing. It is frequently equal to both $1$ and $2$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"3\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Understanding the distinction<\/h2>\n <\/div>\n <div class=\"step-count\">Step 3 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n5,4,3,2,1,1,1,1,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$? Is it frequently equal to another number?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint3')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer3')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint3\">\n <p>Look at what happens from the fifth term onward.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer3\">\n <p>It is eventually equal to $1$ and frequently equal to $1$. It is not frequently equal to $2$, $3$, $4$, or $5$, since each appears only once.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n1,2,1,1,2,1,1,1,2,1,1,1,1,2,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$? Is it frequently equal to $2$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint4')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer4')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint4\">\n <p>The blocks of $1$'s get longer, but the value $2$ still appears again and again.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer4\">\n <p>It is not eventually equal to $1$, because $2$ appears infinitely many times. It is frequently equal to $1$ and frequently equal to $2$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Can a sequence be frequently equal to $c$, but not eventually equal to $c$? Give an example or explain why this is impossible.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint5')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer5')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint5\">\n <p>Try alternating $c$ with another value.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer5\">\n <p>Yes. For example, if $d\\neq c$, then $c,d,c,d,c,d,\\ldots$ is frequently equal to $c$ but not eventually equal to $c$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Can a sequence be eventually equal to $c$, but not frequently equal to $c$? Give an example or explain why this is impossible.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint6')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer6')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint6\">\n <p>If from some point on all terms are $c$, how many times does $c$ appear?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer6\">\n <p>No. Eventually equal to $c$ always implies frequently equal to $c$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"4\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Formula examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 4 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\na_n=\\left\\lfloor \\frac{2026}{n}\\right\\rfloor+2026.\n\\]\n<p>Compute several terms of the sequence. Is $(a_n)$ eventually equal to $2026$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint7')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer7')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint7\">\n <p>Ask when $\\left\\lfloor 2026\/n\\right\\rfloor=0$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer7\">\n <p>For $n>2026$, we have $0<2026\/n<1$, so $\\left\\lfloor 2026\/n\\right\\rfloor=0$. Hence $a_n=2026$ for all $n\\geq 2027$. The sequence is eventually equal to $2026$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\nb_n=\\frac{1}{n}.\n\\]\n<p>Is $(b_n)$ eventually equal to $0$? Is it frequently equal to $0$?\nDoes there exist $c>0$ and $d<0$ such that $b_n\\geq c$ or $b_n\\leq d$, for all $n\\in\\mathbb{N}$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint8')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer8')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint8\">\n <p>The terms get closer and closer to $0$, but no term is actually equal to $0$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer8\">\n <p>\nIt is not eventually equal to $0$ and not frequently equal to $0$, since $1\/n\\neq 0$ for every $n$.\nThere is no fixed $c>0$ such that $1\/n\\geq c$ for every $n$.\nIndeed, if $c>0$, then $1\/c>0$, and we can choose a natural number $n>1\/c$.\nIt follows that\n<\/p>\n\\[\n0<\\frac{1}{n}<c.\n\\]\n<p>\nAlso, since $1\/n>0$ for every $n$, there is no $d<0$ such that $1\/n\\leq d$ for all $n$.\n<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"5\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Conceptual questions<\/h2>\n <\/div>\n <div class=\"step-count\">Step 5 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Can a sequence be eventually equal to two different numbers $c$ and $d$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint9')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer9')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint9\">\n <p>If a late term must be both $c$ and $d$, what follows?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer9\">\n <p>Only if $c=d$. If $c\\neq d$, this is impossible.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Can a sequence be frequently equal to two different numbers $c$ and $d$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint10')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer10')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint10\">\n <p>Think of alternating values.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer10\">\n <p>Yes. If $c\\neq d$, the sequence $c,d,c,d,c,d,\\ldots$ is frequently equal to both $c$ and $d$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"6\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Challenge<\/h2>\n <\/div>\n <div class=\"step-count\">Step 6 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct, if possible, a sequence that is frequently equal to $n$, for every natural number $n$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint11')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer11')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint11\">\n <p>Try writing blocks: $1$, then $1,2$, then $1,2,3$, then $1,2,3,4$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer11\">\n \n<p>One construction is<\/p>\n\\[\n1,\\quad 1,2,\\quad 1,2,3,\\quad 1,2,3,4,\\quad \\ldots\n\\]\n<p>written as a single sequence:<\/p>\n\\[\n1,1,2,1,2,3,1,2,3,4,\\ldots\n\\]\n<p>For each fixed $n$, the value $n$ appears in every block $1,2,\\ldots,k$ with $k\\geq n$, hence infinitely many times.<\/p>\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"7\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Definitions<\/h2>\n <\/div>\n <div class=\"step-count\">Step 7 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nSets allow us to collect objects and study them together. If $A$ and $B$ are sets,\ntheir <strong>intersection<\/strong>, denoted by $A\\cap B$, is the set of all elements\nthat belong to both $A$ and $B$.\n<\/p>\n<p>Thus, $x\\in A\\cap B$ means that $x\\in A$ and $x\\in B$.\nIf two sets have no elements in common, then $A\\cap B=\\emptyset$.<\/p>\n<p>For three sets $A$, $B$, and $C$, the intersection $A\\cap B\\cap C$\nis the set of all elements that belong to all three sets at the same time.<\/p>\n<p>If $A_1,A_2,A_3,\\ldots$ is a family of sets, then<\/p>\n\\[\n\\bigcap_{n=1}^{\\infty} A_n\n\\]\n<p>is the set of all elements that belong to every one of $A_1,A_2,A_3,\\ldots$.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"8\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Computing intersections<\/h2>\n <\/div>\n <div class=\"step-count\">Step 8 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find $A\\cap B$, for<\/p>\n\\[\nA=\\{1,2,3,4\\},\\qquad B=\\{3,4,5,6\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint12')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer12')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint12\">\n <p>Look for the elements that appear in both sets.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer12\">\n \\[A\\cap B=\\{3,4\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find $A\\cap B$, for<\/p>\n\\[\nA=\\{1,3,5,7\\},\\qquad B=\\{2,4,6,8\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint13')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer13')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint13\">\n <p>One set contains odd numbers and the other contains even numbers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer13\">\n \\[A\\cap B=\\emptyset.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find $A\\cap B$, $A\\cap C$, $B\\cap C$ and $A\\cap B\\cap C$, for<\/p>\n\\[\nA=\\{1,2,3,4,5\\},\\quad B=\\{2,4,6,8\\},\\quad C=\\{4,5,6,7\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint14')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer14')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint14\">\n <p>First compare two sets at a time. Then look for elements common to all three.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer14\">\n \n\\[\nA\\cap B=\\{2,4\\},\\quad A\\cap C=\\{4,5\\},\\quad B\\cap C=\\{4,6\\},\n\\]\n\\[\nA\\cap B\\cap C=\\{4\\}.\n\\]\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"9\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Constructing sets with prescribed intersections<\/h2>\n <\/div>\n <div class=\"step-count\">Step 9 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Construct two different finite sets $A$ and $B$ such that<\/p>\\[A\\cap B=\\{3,5\\}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint15')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer15')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint15\">\n <p>Put $3$ and $5$ in both sets. Then add different extra elements to each set.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer15\">\n <p>For example, $A=\\{1,3,5\\}$ and $B=\\{3,5,7\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\\[A\\cap B\\cap C=\\{1,2\\}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint16')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer16')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint16\">\n <p>Put $1$ and $2$ in all three sets. Add some different extra elements.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer16\">\n <p>For example, $A=\\{1,2,3\\}$, $B=\\{1,2,4\\}$, $C=\\{1,2,5\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\\[A\\cap B=\\{1,2\\}\\quad\\text{and}\\quad A\\cap B\\cap C=\\{1\\}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint17')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer17')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint17\">\n <p>Make $1$ and $2$ belong to both $A$ and $B$, but put only $1$ in $C$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer17\">\n <p>For example, $A=\\{1,2,3\\}$, $B=\\{1,2,4\\}$, $C=\\{1,5\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Construct three different finite sets $A$, $B$, and $C$ such that $A\\cap B\\cap C=\\emptyset$, but none of $A$, $B$, and $C$ is empty.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint18')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer18')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint18\">\n <p>This is possible even if the sets are very small.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer18\">\n <p>For example, $A=\\{1\\}$, $B=\\{2\\}$, $C=\\{3\\}$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"10\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Pairwise intersections and full intersection<\/h2>\n <\/div>\n <div class=\"step-count\">Step 10 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\n\\[\nA\\cap B\\neq\\emptyset,\\qquad A\\cap C\\neq\\emptyset,\n\\]\n<p>but<\/p>\n\\[\nB\\cap C=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint19')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer19')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint19\">\n <p>Let $A$ share one element with $B$ and a different element with $C$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer19\">\n <p>For example, $A=\\{1,2\\}$, $B=\\{1\\}$, $C=\\{2\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\n\\[\nA\\cap B\\neq\\emptyset,\\qquad A\\cap C\\neq\\emptyset,\\qquad B\\cap C\\neq\\emptyset,\n\\]\n<p>but<\/p>\n\\[\nA\\cap B\\cap C=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint20')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer20')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint20\">\n <p>Try to make each pair share a different element.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer20\">\n <p>For example, $A=\\{1,2\\}$, $B=\\{2,3\\}$, $C=\\{1,3\\}$. Then every pair intersects, but no element belongs to all three sets.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"11\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Infinite families of sets<\/h2>\n <\/div>\n <div class=\"step-count\">Step 11 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>For every natural number $n$, let $A_n=\\{1,2,\\ldots,n\\}$. Notice that $A_n\\subseteq A_{n+1}$ for every $n$.<\/p>\n<p>Compute<\/p>\n\\[\n\\bigcap_{n=1}^{\\infty} A_n.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint21')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer21')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint21\">\n <p>An element must belong to $A_1$ and also to every later set.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer21\">\n \\[\\bigcap_{n=1}^{\\infty} A_n=\\{1\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>With the same sets $A_n=\\{1,2,\\ldots,n\\}$, compute<\/p>\n\\[\n\\bigcap_{n=100}^{\\infty} A_n.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint22')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer22')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint22\">\n <p>The sets are increasing. The smallest set in this tail is $A_{100}$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer22\">\n \\[\\bigcap_{n=100}^{\\infty} A_n=A_{100}=\\{1,2,\\ldots,100\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>More generally, for a given $N\\in\\mathbb{N}$, compute<\/p>\n\\[\n\\bigcap_{n=N}^{\\infty} A_n.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint23')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer23')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint23\">\n <p>Again, the smallest set in the family is $A_N$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer23\">\n \\[\\bigcap_{n=N}^{\\infty} A_n=A_N=\\{1,2,\\ldots,N\\}.\\]\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"12\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Challenges<\/h2>\n <\/div>\n <div class=\"step-count\">Step 12 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct four finite sets $A$, $B$, $C$, and $D$ such that the intersection of any three of them is nonempty, but<\/p>\n\\[\nA\\cap B\\cap C\\cap D=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint24')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer24')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint24\">\n <p>Try using the set $\\{1,2,3,4\\}$, and let each set miss exactly one element.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer24\">\n \n<p>One example is<\/p>\n\\[\nA=\\{2,3,4\\},\\quad B=\\{1,3,4\\},\\quad C=\\{1,2,4\\},\\quad D=\\{1,2,3\\}.\n\\]\n<p>The intersection of all four is empty, but the intersection of any three contains the element missing from the fourth set.<\/p>\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Can you generalize the previous construction? For a given natural number $n$, can you construct $n$ finite sets such that the intersection of any $n-1$ of them is nonempty, but the intersection of all $n$ sets is empty?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint25')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer25')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint25\">\n <p>Use the set $\\{1,2,\\ldots,n\\}$, and make the $i$-th set miss only $i$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer25\">\n \n<p>Let<\/p>\n\\[\nA_i=\\{1,2,\\ldots,n\\}\\setminus\\{i\\},\\qquad i=1,\\ldots,n.\n\\]\n<p>The intersection of all $n$ sets is empty. But if we omit $A_i$, then the intersection of the remaining $n-1$ sets contains $i$.<\/p>\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct a family of sets<\/p>\n\\[\nA_1\\supseteq A_2\\supseteq A_3\\supseteq\\cdots\n\\]\n<p>such that every $A_n$ is nonempty, but<\/p>\n\\[\n\\bigcap_{n=1}^{\\infty} A_n=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint26')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer26')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint26\">\n <p>Try removing the first few natural numbers step by step.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer26\">\n \n<p>Take<\/p>\n\\[\nA_n=\\{n,n+1,n+2,\\ldots\\}.\n\\]\n<p>Then $A_1\\supseteq A_2\\supseteq A_3\\supseteq\\cdots$, and every $A_n$ is nonempty. But no natural number belongs to every $A_n$, so the full intersection is empty.<\/p>\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"13\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Definition and first idea<\/h2>\n <\/div>\n <div class=\"step-count\">Step 13 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nSome infinite sets can be listed one element after another, in a single infinite row.\nFor example, the positive even integers can be listed as\n<\/p>\n\\[\n2,4,6,8,10,\\ldots\n\\]\n<p>We say that a set $A$ is <strong>countable<\/strong> if its elements can be listed in a sequence<\/p>\n\\[\na_1,a_2,a_3,\\ldots\n\\]\n<p>so that every element of $A$ appears somewhere in the list.<\/p>\n<p>\nEquivalently, the elements of $A$ can be paired with the natural numbers.\nFor the positive even integers, the pairing is $n\\longleftrightarrow 2n$.\n<\/p>\n<p>The order of the list is not important. What matters is that every element appears at some finite position.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"14\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>First examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 14 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set $\\{10,20,30,40,\\ldots\\}$. Which element is paired with the natural number $n$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint27')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer27')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint27\">\n <p>Look at the common factor.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer27\">\n <p>The list is $10,20,30,\\ldots$. The number $n$ is paired with $10n$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set of positive odd integers. Can you exhibit an explicit pairing?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint28')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer28')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint28\">\n <p>The positive odd integers are $1,3,5,\\ldots$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer28\">\n <p>The list is $1,3,5,\\ldots$. The number $n$ is paired with $2n-1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set $\\{1,4,9,16,25,\\ldots\\}$. Which element is paired with the natural number $n$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint29')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer29')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint29\">\n <p>These are squares.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer29\">\n <p>The list is $1,4,9,16,\\ldots$. The number $n$ is paired with $n^2$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"15\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>More challenging examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 15 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Is it possible to list all the negative integers?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint30')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer30')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint30\">\n <p>Start with $-1$, then $-2$, then $-3$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer30\">\n <p>Yes: $-1,-2,-3,-4,\\ldots$ lists all negative integers.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Is it possible to list all the integers? [Hint: Yes, it is possible.]<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint31')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer31')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint31\">\n <p>Try alternating positive and negative integers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer31\">\n <p>Yes. One list is $0,1,-1,2,-2,3,-3,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set of all integer multiples of $3$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint32')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer32')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint32\">\n <p>Use the list of all integers and multiply by $3$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer32\">\n <p>One list is $0,3,-3,6,-6,9,-9,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"16\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Combining countable sets<\/h2>\n <\/div>\n <div class=\"step-count\">Step 16 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Suppose that the elements of a set $A$ can be listed as $a_1,a_2,a_3,\\ldots$\nand the elements of a set $B$ can be listed as $b_1,b_2,b_3,\\ldots$.\nConstruct a list that contains all elements of $A\\cup B$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint33')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer33')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint33\">\n <p>Alternate between the two lists.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer33\">\n <p>One possible list is $a_1,b_1,a_2,b_2,a_3,b_3,\\ldots$, skipping repeated elements if necessary.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Use the previous idea to list all numbers in<\/p>\n\\[\n\\{1,3,5,7,\\ldots\\}\\cup \\{2,4,6,8,\\ldots\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint34')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer34')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint34\">\n <p>This union is the set of all positive integers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer34\">\n <p>One list is $1,2,3,4,5,6,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Suppose $A$ and $B$ are countable sets. Is $A\\cup B$ countable? Explain.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint35')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer35')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint35\">\n <p>Use the alternating list idea.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer35\">\n <p>Yes. If $A$ and $B$ can be listed, then $A\\cup B$ can be listed by alternating between the two lists and ignoring repetitions.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Suppose $A_1,A_2,A_3,\\ldots$ are countable sets. Do you think the union<\/p>\n\\[\nA_1\\cup A_2\\cup A_3\\cup\\cdots\n\\]\n<p>must be countable?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint36')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer36')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint36\">\n <p>Imagine arranging the lists in rows and moving along diagonals.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer36\">\n <p>Yes. A countable union of countable sets is countable. One can arrange the elements in an infinite table and list them by diagonals.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"17\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Pairs of natural numbers<\/h2>\n <\/div>\n <div class=\"step-count\">Step 17 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Consider the set $\\mathbb{N}\\times\\mathbb{N}$ of all pairs $(m,n)$ of natural numbers. Try to list its elements.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint37')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer37')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint37\">\n <p>A row-by-row list fails. Diagonals work better.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer37\">\n <p>One beginning is $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>A first attempt might be to list the pairs row by row:<\/p>\n\\[\n(1,1),(1,2),(1,3),(1,4),\\ldots\n\\]\n<p>Why does this attempt fail to list all pairs?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint38')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer38')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint38\">\n <p>Do we ever leave the first row?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer38\">\n <p>It never reaches pairs like $(2,1)$ or $(3,1)$, because it stays forever in the first row.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Try instead to list the pairs by diagonals:<\/p>\n\\[\nDiag_i=\\{(n,m): n+m=i+1\\}.\n\\]\n<ol class=\"inner-list\">\n<li>Identify $Diag_1$, $Diag_2$ and $Diag_3$.<\/li>\n<li>Put, in a finite list, all the elements of $Diag_1\\cup Diag_2\\cup Diag_3$.<\/li>\n<li>List $\\mathbb{N}\\times\\mathbb{N}$.<\/li>\n<li>Explain why every pair $(m,n)$ eventually appears.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint39')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer39')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint39\">\n <p>A pair $(m,n)$ lies on the diagonal determined by $m+n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer39\">\n \n<p>We have<\/p>\n\\[\nDiag_1=\\{(1,1)\\},\\quad Diag_2=\\{(1,2),(2,1)\\},\\quad Diag_3=\\{(1,3),(2,2),(3,1)\\}.\n\\]\n<p>Every pair $(m,n)$ appears on the diagonal with index $m+n-1$.<\/p>\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Conclude that $\\mathbb{N}\\times\\mathbb{N}$ is countable.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint40')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer40')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint40\">\n <p>If every pair appears in the diagonal list, then the pairs can be listed.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer40\">\n <p>Yes, $\\mathbb{N}\\times\\mathbb{N}$ is countable.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"18\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Rational numbers<\/h2>\n <\/div>\n <div class=\"step-count\">Step 18 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Every positive rational number can be written as $\\dfrac{p}{q}$, where $p$ and $q$ are natural numbers.\nThis representation is not unique. Is there a way to represent them in a unique way?\nExplain why this relates the positive rational numbers to $\\mathbb{N}\\times\\mathbb{N}$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint41')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer41')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint41\">\n <p>Each fraction gives a pair $(p,q)$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer41\">\n <p>Every positive rational number comes from some pair $(p,q)\\in\\mathbb{N}\\times\\mathbb{N}$. A unique representation is obtained by requiring $\\gcd(p,q)=1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Use the previous subsection to argue that the set of positive rational numbers is countable.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint42')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer42')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint42\">\n <p>List all pairs $(p,q)$ and then read them as fractions $p\/q$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer42\">\n <p>Since $\\mathbb{N}\\times\\mathbb{N}$ is countable, the fractions $p\/q$ can be listed. Repetitions do not matter, so the positive rational numbers are countable.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Do you think the set of all rational numbers is countable? Explain how one might list them.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint43')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer43')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint43\">\n <p>Combine positive rationals, negative rationals, and zero.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer43\">\n <p>Yes. One can list $0$, then alternate positive and negative rational numbers: $q_1,-q_1,q_2,-q_2,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"19\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>A tough challenge<\/h2>\n <\/div>\n <div class=\"step-count\">Step 19 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Is every infinite set countable? In other words, can the elements of every infinite set be listed in a single infinite row?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint44')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer44')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint44\">\n <p>The real numbers are the standard place where this question becomes surprising.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer44\">\n <p>No. There are infinite sets that are not countable, such as the real numbers between $0$ and $1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the set $A$ of real numbers between $0$ and $1$ whose decimal expression only contains $0$'s and $1$'s, for example<\/p>\n\\[\n0.001101010111\\ldots\n\\quad\\text{or}\\quad\n0.1001010000111111\\ldots\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint45')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer45')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint45\">\n <p>This set is much larger than it may first appear.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer45\">\n <p>This is a good test set for diagonal thinking: each number corresponds to an infinite sequence of $0$'s and $1$'s.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Try to list all the elements of $A$. What difficulties appear?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint46')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer46')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint46\">\n <p>Any proposed list can be challenged digit by digit.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer46\">\n <p>The difficulty is that a list has rows, and each row has infinitely many digits. One can construct a new number by changing the diagonal digits.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Suppose someone claims to have listed all the elements of $A$:<\/p>\n\\[\nr_1,r_2,r_3,\\ldots\n\\]\n<p>Can you imagine a way to construct a new real number between $0$ and $1$ that is different from every number on the list?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint47')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer47')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint47\">\n <p>Make the new number differ from $r_1$ in the first digit, from $r_2$ in the second digit, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer47\">\n \n<p>This is Cantor's diagonal idea. If the $n$-th digit of $r_n$ is $0$, choose the $n$-th digit of the new number to be $1$; if it is $1$, choose $0$. The new number differs from $r_n$ in the $n$-th digit, so it is not on the list.<\/p>\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"20\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>The principle<\/h2>\n <\/div>\n <div class=\"step-count\">Step 20 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nThe <strong>Pigeonhole Principle<\/strong> is a simple combinatorial idea with many surprising consequences.\nIf more than $n$ objects are placed into $n$ boxes, then at least one box contains at least two objects.\n<\/p>\n<p>\nFor example, if $8$ students are in a room, then at least two of them were born on the same day of the week,\nbecause there are only $7$ days of the week.\n<\/p>\n<p>The main difficulty is often not applying the principle, but deciding what the boxes should be.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"21\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>First examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 21 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>There are $13$ students in a room. Show that at least two of them were born in the same month.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint48')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer48')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint48\">\n <p>Use the $12$ months as boxes.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer48\">\n <p>There are $13$ students and only $12$ months, so two students must share a birth month.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>There are $27$ students in a room. Show that at least three of them were born in the same month.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint49')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer49')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint49\">\n <p>If each month had at most two students, how many students could there be?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer49\">\n <p>If every month had at most two students, there would be at most $24$ students. Since there are $27$, some month has at least three.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>A drawer contains only black socks and white socks. How many socks must you take to be sure that you have two socks of the same color? Explain your answer as a consequence of the Pigeonhole Principle. Which are the boxes?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint50')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer50')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint50\">\n <p>The boxes are the two colors.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer50\">\n <p>You need $3$ socks. With $3$ objects and $2$ boxes, one color must contain at least two socks.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"22\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Remainders as boxes<\/h2>\n <\/div>\n <div class=\"step-count\">Step 22 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $6$ integers. Show that at least two of them have the same remainder when divided by $5$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint51')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer51')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint51\">\n <p>The possible remainders are $0,1,2,3,4$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer51\">\n <p>There are $6$ integers and only $5$ possible remainders, so two have the same remainder modulo $5$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $11$ integers. Show that at least two of them differ by a multiple of $10$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint52')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer52')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint52\">\n <p>Two integers differ by a multiple of $10$ exactly when they have the same remainder modulo $10$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer52\">\n <p>There are $10$ possible remainders modulo $10$ and $11$ integers. Two share a remainder, hence their difference is divisible by $10$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ integers. Show that there are two of them such that their difference is divisible by $n$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint53')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer53')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint53\">\n <p>Use remainders modulo $n$ as boxes.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer53\">\n <p>There are $n$ possible remainders modulo $n$, and $n+1$ integers. Two have the same remainder, so their difference is divisible by $n$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"23\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Choosing the boxes<\/h2>\n <\/div>\n <div class=\"step-count\">Step 23 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Choose any $6$ numbers from<\/p>\n\\[\n\\{1,2,3,4,5,6,7,8,9,10\\}.\n\\]\n<p>Show that at least two of the chosen numbers have sum $11$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint54')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer54')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint54\">\n <p>Group the numbers into pairs with sum $11$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer54\">\n <p>Use the boxes $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$, $\\{5,6\\}$. Choosing $6$ numbers from $5$ boxes forces two chosen numbers to be in the same box, hence to sum to $11$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why the boxes $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$, $\\{5,6\\}$ prove the previous result.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint55')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer55')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint55\">\n <p>Each box contains exactly two numbers with the desired property.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer55\">\n <p>Since there are $5$ boxes and $6$ chosen numbers, one box contains two chosen numbers. The two numbers in any one box have sum $11$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Choose any $7$ numbers from<\/p>\n\\[\n\\{1,2,3,\\ldots,12\\}.\n\\]\n<p>Show that at least two of the chosen numbers are consecutive.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint56')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer56')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint56\">\n <p>Use boxes $\\{1,2\\}$, $\\{3,4\\}$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer56\">\n <p>There are $6$ boxes: $\\{1,2\\},\\{3,4\\},\\ldots,\\{11,12\\}$. Choosing $7$ numbers forces two in the same box, and those two are consecutive.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>In the previous problem, what are the useful boxes?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint57')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer57')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint57\">\n <p>Pair consecutive numbers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer57\">\n \\[\\{1,2\\},\\{3,4\\},\\{5,6\\},\\{7,8\\},\\{9,10\\},\\{11,12\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n \n<p>Choose any $7$ numbers from<\/p>\n\\[\n\\{1,2,3,\\ldots,13\\}.\n\\]\n<p>Is it necessarily true that two of the chosen numbers are consecutive? Give a proof or a counterexample.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint58')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer58')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint58\">\n <p>Try choosing every other number.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer58\">\n <p>No. The set $\\{1,3,5,7,9,11,13\\}$ has $7$ numbers and no two are consecutive.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"24\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Geometric examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 24 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $5$ points inside a square of side length $2$. Show that at least two points are at distance at most $\\sqrt{2}$ from each other.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint59')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer59')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint59\">\n <p>Divide the large square into four unit squares.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer59\">\n <p>Four unit squares are the boxes. With $5$ points, two lie in the same unit square. The maximum distance between two points in a unit square is its diagonal, $\\sqrt{2}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why dividing the square into four unit squares is useful in the previous problem.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint60')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer60')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint60\">\n <p>The diameter of each small square is exactly $\\sqrt{2}$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer60\">\n <p>It creates $4$ boxes, each with the property that any two points inside the same box are at distance at most $\\sqrt{2}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $10$ points inside an equilateral triangle of side length $3$. Show that at least two points are at distance at most $1$ from each other.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint61')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer61')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint61\">\n <p>Divide the triangle into $9$ equilateral triangles of side length $1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer61\">\n <p>The $9$ small triangles are the boxes. With $10$ points, two lie in the same small triangle, whose diameter is $1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Suppose every point in the plane is colored either red or blue. Show that there exist two points of the same color at distance exactly $1$ from each other.<\/p><p><em>Hint:<\/em> consider the vertices of an equilateral triangle of side length $1$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint62')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer62')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint62\">\n <p>There are $3$ vertices but only $2$ colors.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer62\">\n <p>Among the three vertices of an equilateral triangle of side $1$, two must have the same color. Those two vertices are at distance $1$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"25\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Challenges<\/h2>\n <\/div>\n <div class=\"step-count\">Step 25 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ numbers from $\\{1,2,3,\\ldots,2n\\}$. Show that at least two have sum $2n+1$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint63')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer63')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint63\">\n <p>Pair $1$ with $2n$, $2$ with $2n-1$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer63\">\n <p>Use boxes $\\{1,2n\\}$, $\\{2,2n-1\\}$, ..., $\\{n,n+1\\}$. There are $n$ boxes and $n+1$ chosen numbers.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ numbers from $\\{1,2,3,\\ldots,2n\\}$. Show that at least two are relatively prime.<\/p><p><em>Hint:<\/em> think about consecutive integers.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint64')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer64')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint64\">\n <p>Use boxes of consecutive pairs.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer64\">\n <p>Pair the numbers as $\\{1,2\\}$, $\\{3,4\\}$, ..., $\\{2n-1,2n\\}$. Two chosen numbers must lie in the same pair, and consecutive integers are relatively prime.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ numbers from $\\{1,2,3,\\ldots,2n\\}$. Show that at least one of the chosen numbers divides another.<\/p><p><em>Hint:<\/em> write each number in the form $2^k m$, where $m$ is odd.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint65')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer65')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint65\">\n <p>The odd part $m$ can only be one of $1,3,5,\\ldots,2n-1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer65\">\n <p>There are $n$ possible odd parts. With $n+1$ chosen numbers, two have the same odd part: $2^a m$ and $2^b m$. The one with the smaller power of $2$ divides the other.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"26\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>What is the task?<\/h2>\n <\/div>\n <div class=\"step-count\">Step 26 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nA proof is not only a collection of true statements. The statements must also appear in the correct logical order.\nSome statements depend on previous ones; some introduce definitions; some use the hypothesis; and some give the final conclusion.\n<\/p>\n<p>In this section, each problem gives a mathematical claim and a list of statements in the wrong order. Rearrange the statements to obtain a correct proof.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"27\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Warm-up proofs<\/h2>\n <\/div>\n <div class=\"step-count\">Step 27 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 1<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> If $a$ and $b$ are even integers, then $a+b$ is even.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, $a+b$ is even.<\/li>\n<li>Since $a$ and $b$ are even, there exist integers $m,n$ such that $a=2m$ and $b=2n$.<\/li>\n<li>Then $a+b=2m+2n=2(m+n)$.<\/li>\n<li>Since $m+n$ is an integer, $a+b$ is divisible by $2$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint66')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer66')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint66\">\n <p>Start by using the meaning of \u201ceven\u201d.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer66\">\n <p>Correct order: (b), (c), (d), (a).<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 2<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> If $a$ and $b$ are odd integers, then $ab$ is odd.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Since $m,n$ are integers, $2mn+m+n$ is an integer.<\/li>\n<li>Then $ab=(2m+1)(2n+1)$.<\/li>\n<li>Therefore, $ab$ is odd.<\/li>\n<li>Since $a$ and $b$ are odd, there exist integers $m,n$ such that $a=2m+1$ and $b=2n+1$.<\/li>\n<li>Expanding, $ab=4mn+2m+2n+1=2(2mn+m+n)+1$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint67')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer67')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint67\">\n <p>Begin with the definition of odd integers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer67\">\n <p>Correct order: (d), (b), (e), (a), (c).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"28\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Proofs by contradiction<\/h2>\n <\/div>\n <div class=\"step-count\">Step 28 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 3<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> If $n^2$ is even, then $n$ is even.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, if $n^2$ is even, then $n$ must be even.<\/li>\n<li>Suppose, for contradiction, that $n$ is odd.<\/li>\n<li>Hence $n^2$ is odd.<\/li>\n<li>Then there exists an integer $k$ such that $n=2k+1$.<\/li>\n<li>Thus $n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1$.<\/li>\n<li>This contradicts the assumption that $n^2$ is even.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint68')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer68')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint68\">\n <p>Because this is a contradiction proof, begin by assuming the opposite of what you want to prove.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer68\">\n <p>Correct order: (b), (d), (e), (c), (f), (a).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"29\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Proofs with sets<\/h2>\n <\/div>\n <div class=\"step-count\">Step 29 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 4<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> Prove the inclusion<\/p>\n\\[\nA\\cap(B\\cup C)\\subseteq (A\\cap B)\\cup(A\\cap C).\n\\]\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, $x\\in (A\\cap B)\\cup(A\\cap C)$.<\/li>\n<li>This proves the inclusion.<\/li>\n<li>Suppose $x\\in A\\cap(B\\cup C)$.<\/li>\n<li>Hence $x\\in A$ and $x\\in B\\cup C$.<\/li>\n<li>Therefore $x\\in B$ or $x\\in C$.<\/li>\n<li>If $x\\in B$, then $x\\in A\\cap B$.<\/li>\n<li>If $x\\in C$, then $x\\in A\\cap C$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint69')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer69')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint69\">\n <p>Start by taking an arbitrary element of the left-hand side.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer69\">\n <p>Correct order: (c), (d), (e), (f) and (g), (a), (b). Statements (f) and (g) are the two cases.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 5<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> Prove the reverse inclusion<\/p>\n\\[\n(A\\cap B)\\cup(A\\cap C)\\subseteq A\\cap(B\\cup C).\n\\]\n<p>Write your own proof. Try to imitate the structure of the previous problem.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint70')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer70')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint70\">\n <p>Start with $x\\in (A\\cap B)\\cup(A\\cap C)$ and split into cases.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer70\">\n <p>If $x\\in A\\cap B$, then $x\\in A$ and $x\\in B$, so $x\\in B\\cup C$, hence $x\\in A\\cap(B\\cup C)$. Similarly, if $x\\in A\\cap C$, then $x\\in A\\cap(B\\cup C)$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"30\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Divisibility<\/h2>\n <\/div>\n <div class=\"step-count\">Step 30 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 6<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> For every integer $n$, the number $n^3-n$ is divisible by $3$.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, one of $n-1,n,n+1$ is divisible by $3$.<\/li>\n<li>Hence $n^3-n$ is divisible by $3$.<\/li>\n<li>We can factor: $n^3-n=n(n-1)(n+1)$.<\/li>\n<li>Among any three consecutive integers, exactly one is divisible by $3$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint71')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer71')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint71\">\n <p>Factor first.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer71\">\n <p>Correct order: (c), (d), (a), (b).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"31\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>A classical proof<\/h2>\n <\/div>\n <div class=\"step-count\">Step 31 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 7<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> There are infinitely many prime numbers.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, $q$ is a prime not among $p_1,p_2,\\ldots,p_k$.<\/li>\n<li>Suppose, for contradiction, that there are only finitely many primes: $p_1,p_2,\\ldots,p_k$.<\/li>\n<li>Let $N=p_1p_2\\cdots p_k+1$.<\/li>\n<li>Hence there must be infinitely many primes.<\/li>\n<li>Let $q$ be a prime divisor of $N$.<\/li>\n<li>None of the primes $p_1,p_2,\\ldots,p_k$ divides $N$, since each leaves remainder $1$.<\/li>\n<li>This contradicts the assumption that the list contained all primes.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint72')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer72')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint72\">\n <p>Start by assuming there are finitely many primes, then construct a new number.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer72\">\n <p>Correct order: (b), (c), (e), (f), (a), (g), (d).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"32\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Challenge<\/h2>\n <\/div>\n <div class=\"step-count\">Step 32 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 8<\/div>\n <div class=\"problem-statement\">\n \n<p>Write a proof, by contradiction, that $\\sqrt{2}$ is irrational.<\/p>\n<p>You may assume as starting point that<\/p>\n\\[\n\\sqrt{2}=\\frac{a}{b},\n\\]\n<p>where $a$ and $b$ are positive integers with no common factor. Then write an integral equation that $a$ and $b$ should satisfy.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint73')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer73')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint73\">\n <p>Square both sides and clear denominators.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer73\">\n <p>Squaring gives $2=a^2\/b^2$, hence $a^2=2b^2$. Thus $a^2$ is even, so $a$ is even. Write $a=2k$. Then $4k^2=2b^2$, so $b^2=2k^2$, hence $b$ is even. This contradicts that $a$ and $b$ have no common factor.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"33\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Definition and first idea<\/h2>\n <\/div>\n <div class=\"step-count\">Step 33 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>A <strong>sequence<\/strong> is a list of numbers written in a definite order:<\/p>\n\\[\na_1,a_2,a_3,a_4,\\ldots\n\\]\n<p>The number $a_n$ is called the $n$-th term of the sequence.<\/p>\n<p>\nSometimes a sequence is described by giving its first few terms. Sometimes it is described by a formula.\nA useful skill is to move between the first terms of a sequence and a formula for its $n$-th term.\n<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"34\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Computing terms from a formula<\/h2>\n <\/div>\n <div class=\"step-count\">Step 34 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $a_n=(-1)^n$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint74')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer74')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint74\">\n <p>Substitute $n=1,2,3,4,5,6$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer74\">\n \\[-1,1,-1,1,-1,1.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $b_n=(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint75')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer75')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint75\">\n <p>The exponent is one more than before.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer75\">\n \\[1,-1,1,-1,1,-1.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Compare the two sequences above. How are they similar? How are they different?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint76')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer76')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint76\">\n <p>Both alternate between $1$ and $-1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer76\">\n <p>They have the same two values, but start with different signs. One is the negative of the other.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $c_n=2(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint77')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer77')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint77\">\n <p>Use the sequence from part (b) and multiply by $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer77\">\n \\[2,-2,2,-2,2,-2.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $d_n=2+2(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint78')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer78')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint78\">\n <p>Start from $2(-1)^{n+1}$ and add $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer78\">\n \\[4,0,4,0,4,0.\\]\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"35\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Finding formulas<\/h2>\n <\/div>\n <div class=\"step-count\">Step 35 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $1,-1,1,-1,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint79')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer79')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint79\">\n <p>Compare with $(-1)^n$ and $(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer79\">\n \\[a_n=(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $2,-2,2,-2,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint80')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer80')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint80\">\n <p>Multiply the previous sequence by $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer80\">\n \\[a_n=2(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $4,0,4,0,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint81')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer81')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint81\">\n <p>Think of it as a constant part plus an alternating part.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer81\">\n \\[a_n=2+2(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $3,7,3,7,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint82')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer82')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint82\">\n <p>The average is $5$, and the deviation is $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer82\">\n \\[a_n=5-2(-1)^{n+1}\\quad\\text{or}\\quad a_n=5+2(-1)^n.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $10,20,10,20,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint83')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer83')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint83\">\n <p>The average is $15$, and the deviation is $5$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer83\">\n \\[a_n=15-5(-1)^{n+1}\\quad\\text{or}\\quad a_n=15+5(-1)^n.\\]\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"36\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Alternating between two values<\/h2>\n <\/div>\n <div class=\"step-count\">Step 36 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Let $a$ and $b$ be two given numbers. Find a formula for $a,b,a,b,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint84')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer84')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint84\">\n <p>Use the average $(a+b)\/2$ and half-difference $(a-b)\/2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer84\">\n \\[x_n=\\frac{a+b}{2}+\\frac{a-b}{2}(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Check your formula when $a=1$ and $b=-1$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint85')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer85')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint85\">\n <p>The average is $0$ and the half-difference is $1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer85\">\n \\[x_n=(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Check your formula when $a=4$ and $b=0$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint86')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer86')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint86\">\n <p>The average is $2$ and the half-difference is $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer86\">\n \\[x_n=2+2(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why $a,b,a,b,\\ldots$ can be thought of as a constant part plus an alternating part.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint87')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer87')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint87\">\n <p>The midpoint between $a$ and $b$ is constant.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer87\">\n <p>The sequence oscillates around the average $(a+b)\/2$, moving alternately by $+(a-b)\/2$ and $-(a-b)\/2$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n <p>Show that one possible formula is<\/p>\\[x_n=\\frac{a+b}{2}+\\frac{a-b}{2}(-1)^{n+1}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint88')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer88')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint88\">\n <p>Evaluate separately for odd $n$ and even $n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer88\">\n <p>If $n$ is odd, then $(-1)^{n+1}=1$, so $x_n=a$. If $n$ is even, then $(-1)^{n+1}=-1$, so $x_n=b$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"37\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Looking for structure<\/h2>\n <\/div>\n <div class=\"step-count\">Step 37 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>In the formula $x_n=\\frac{a+b}{2}+\\frac{a-b}{2}(-1)^{n+1}$, what is the role of $\\frac{a+b}{2}$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint89')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer89')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint89\">\n <p>It is the midpoint between $a$ and $b$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer89\">\n <p>It is the constant part, or average, around which the sequence alternates.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>What is the role of $\\frac{a-b}{2}(-1)^{n+1}$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint90')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer90')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint90\">\n <p>It changes sign depending on the parity of $n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer90\">\n <p>It is the alternating part that moves the value from the average to either $a$ or $b$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why the formula gives $a$ when $n$ is odd and $b$ when $n$ is even.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint91')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer91')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint91\">\n <p>Use $(-1)^{n+1}=1$ for odd $n$ and $-1$ for even $n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer91\">\n <p>Odd $n$ gives $\\frac{a+b}{2}+\\frac{a-b}{2}=a$. Even $n$ gives $\\frac{a+b}{2}-\\frac{a-b}{2}=b$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find another way to describe the same sequence using cases:<\/p>\n\\[\nx_n=\n\\begin{cases}\n?, & n \\text{ odd},\\\\\n?, & n \\text{ even}.\n\\end{cases}\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint92')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer92')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint92\">\n <p>Read the sequence directly: first term $a$, second term $b$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer92\">\n \n\\[\nx_n=\n\\begin{cases}\na, & n \\text{ odd},\\\\\nb, & n \\text{ even}.\n\\end{cases}\n\\]\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"38\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Challenges<\/h2>\n <\/div>\n <div class=\"step-count\">Step 38 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $a,b,c,a,b,c,\\ldots$, where $a$, $b$, and $c$ are given numbers.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint93')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer93')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint93\">\n <p>Using cases modulo $3$ is natural.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer93\">\n \n<p>One clear rule is<\/p>\n\\[\nx_n=\n\\begin{cases}\na, & n\\equiv 1\\pmod 3,\\\\\nb, & n\\equiv 2\\pmod 3,\\\\\nc, & n\\equiv 0\\pmod 3.\n\\end{cases}\n\\]\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula, or a clear rule, for $1,2,3,1,2,3,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint94')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer94')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint94\">\n <p>Again use the remainder of $n$ modulo $3$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer94\">\n \\[x_n=((n-1)\\bmod 3)+1.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Can the same sequence be described by two different formulas? Give an example or explain why not.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint95')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer95')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint95\">\n <p>Think of the alternating sequence.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer95\">\n <p>Yes. For example, $(-1)^n$ and $\\cos(n\\pi)$ describe the same sequence for $n=1,2,3,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n\n <div class=\"nav-row\">\n <button class=\"nav-btn secondary\" onclick=\"previousStep()\">Previous step<\/button>\n <button id=\"definitionNavButton\" class=\"nav-btn definition-btn\" onclick=\"goToDefinition()\">Definition<\/button>\n <button class=\"nav-btn\" onclick=\"nextStep()\">Next step<\/button>\n <\/div>\n <\/section>\n\n <div class=\"footer\">\n Interactive prototype with six sections. Each step opens progressively; optional hints and answers can be enabled above.\n <\/div>\n <\/main>\n\n <script>\n let currentStep = 1;\n const totalSteps = 38;\n\n\n function prepareProblemFocus(stepEl) {\n if (!stepEl) return;\n const problems = Array.from(stepEl.querySelectorAll('.problem'));\n problems.forEach((problem, index) => {\n let openButton = problem.querySelector('.open-item-btn');\n if (!openButton) {\n const title = problem.querySelector('.problem-title');\n const label = title ? title.textContent.trim() : `item ${index + 1}`;\n openButton = document.createElement('button');\n openButton.className = 'open-item-btn';\n openButton.type = 'button';\n openButton.textContent = index === 0 ? `Open first item` : `Open item ${label}`;\n openButton.addEventListener('click', () => expandProblem(problem));\n problem.insertBefore(openButton, problem.firstChild);\n }\n\n problem.classList.toggle('collapsed', index !== 0);\n problem.querySelectorAll('.panel').forEach(panel => {\n panel.style.display = 'none';\n });\n });\n }\n\n function expandProblem(problem) {\n const stepEl = problem.closest('.step');\n if (!stepEl) return;\n\n \/\/ Once an item is opened, keep it open. This lets the reader see previous\n \/\/ items while working on later ones.\n problem.classList.remove('collapsed');\n problem.scrollIntoView({ behavior: 'smooth', block: 'start' });\n\n if (window.MathJax) {\n MathJax.typesetPromise([problem]);\n }\n }\n\n function showStep(step) {\n currentStep = Math.max(1, Math.min(totalSteps, step));\n document.querySelectorAll('.step').forEach(el => {\n el.classList.remove('visible');\n });\n const active = document.querySelector(`[data-step=\"${currentStep}\"]`);\n active.classList.add('visible');\n prepareProblemFocus(active);\n\n const sectionSlug = active.getAttribute('data-section');\n const firstStepOfSection = document.querySelector(`.step[data-section=\"${sectionSlug}\"]`);\n const definitionButton = document.getElementById('definitionNavButton');\n\n if (definitionButton) {\n definitionButton.style.display =\n active === firstStepOfSection ? 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'none' : 'block';\n if (window.MathJax) {\n MathJax.typesetPromise([panel]);\n }\n }\n markSomeSupportButtons();\n prepareProblemFocus(document.querySelector('.step.visible'));\n const initialDefinitionButton = document.getElementById('definitionNavButton');\n if (initialDefinitionButton) {\n initialDefinitionButton.style.display = 'none';\n }\n <\/script>\n<\/body>\n<\/html>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"<p>Mathematical Thinking Questions \u2014 Interactive Mathemati …<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/sites.gtiit.edu.cn\/admissions\/math-test-text\/\"> <span class=\"screen-reader-text\">\u6570\u5b66\u6d4b\u8bd5\u9898<\/span> \u67e5\u770b\u5168\u6587 »<\/a><\/p>\n","protected":false},"author":314,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"elementor_header_footer","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts\/46822"}],"collection":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/users\/314"}],"replies":[{"embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/comments?post=46822"}],"version-history":[{"count":4,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts\/46822\/revisions"}],"predecessor-version":[{"id":47013,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts\/46822\/revisions\/47013"}],"wp:attachment":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/media?parent=46822"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/categories?post=46822"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/tags?post=46822"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

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\n\t\t\t\n\n\n \n \n Mathematical Thinking Questions \u2014 Interactive<\/title>\n\n <script>\n window.MathJax = {\n tex: {\n inlineMath: [['$', '$'], ['\\\$', '\\\$']],\n displayMath: [['\\\\[', '\\\\]']]\n },\n svg: { fontCache: 'global' }\n };\n <\/script>\n <script src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-svg.js\" async><\/script>\n\n <style>\n :root {\n --bg: #f5f7fb;\n --card: #ffffff;\n --text: #1f2937;\n --muted: #6b7280;\n --border: #d6dbe6;\n --accent: #0f766e;\n --accent-dark: #115e59;\n --accent-light: #e6f4f1;\n --hint-bg: #fff7ed;\n --hint-border: #fed7aa;\n --answer-bg: #eef8ee;\n --answer-border: #bbf7d0;\n --shadow: 0 10px 25px rgba(15, 23, 42, 0.08);\n }\n\n * { box-sizing: border-box; }\n\n body {\n margin: 0;\n font-family: system-ui, -apple-system, BlinkMacSystemFont, \"Segoe UI\", sans-serif;\n background: var(--bg);\n color: var(--text);\n line-height: 1.55;\n }\n\n header {\n background: linear-gradient(135deg, var(--accent), 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padding-top: 0.65rem;\n border-top: 1px solid var(--border);\n color: var(--muted);\n font-size: 0.84rem;\n font-style: italic;\n }\n\n .step-support-reminder a {\n color: var(--accent-dark);\n font-weight: 700;\n text-decoration: none;\n }\n\n .step-support-reminder a:hover {\n text-decoration: underline;\n }\n\n .nav-btn.definition-btn {\n background: #eef2ff;\n color: #3730a3;\n border: 1px solid #c7d2fe;\n }\n\n .footer {\n text-align: center;\n color: var(--muted);\n font-size: 0.9rem;\n margin-top: 1.5rem;\n }\n\n @media (max-width: 640px) {\n header { padding-bottom: 2.8rem; }\n main { margin-top: -1.5rem; }\n .step-header { display: block; }\n .step-count { margin-top: 0.4rem; }\n }\n <\/style>\n<\/head>\n\n<body class=\"hint-mode-none answer-mode-none\">\n <header>\n <h1>Mathematical Thinking Questions<\/h1>\n <p>Interactive self-assessment guide for applicants to Mathematics with Computer Science at GTIIT.<\/p>\n <\/header>\n\n <main>\n <section class=\"card note\">\n <h2>A note to the reader<\/h2>\n <p>\n The purpose of these problems is not simply to obtain the answers.\n Finding the answer is easy; one can ask a search engine, or even an AI system.\n But that is not the point.\n <\/p>\n <p>\n The only thing that really matters is the mathematical work that you do by yourself.\n Mathematics is not mainly about collecting correct answers. It is about understanding ideas,\n recognizing patterns, making definitions precise, finding examples, discovering counterexamples,\n and building arguments.\n <\/p>\n <p>\n Do not rush. Take your time. Try examples. Make mistakes. Start again.\n Write down your reasoning.\n <\/p>\n <p>\n You are not expected to solve all problems listed here.\n <\/p><div class=\"section-nav\">\n <button class=\"section-btn\" onclick=\"goToSection('eventually')\">1. Eventually and frequently equal to a value<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('intersections')\">2. Intersections of sets<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('countable')\">3. Countable sets<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('pigeonhole')\">4. The Pigeonhole Principle<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('proof')\">5. Build the proof<\/button>\n<button class=\"section-btn\" onclick=\"goToSection('sequences')\">6. Sequences and formulas<\/button>\n <\/div>\n\n\n <div class=\"support-box\" id=\"supportBox\">\n <p class=\"support-note\">\n <span class=\"lead\">Need a little help?<\/span>\n You may choose to show hint buttons or answer buttons below.\n You can show support only for some later items, or for all items.\n <\/p>\n <div class=\"support-controls\">\n <button id=\"hintSomeToggle\" class=\"toggle-support-btn hints-toggle\" onclick=\"setHintMode('some')\">Some hint buttons<\/button>\n <button id=\"hintAllToggle\" class=\"toggle-support-btn hints-toggle\" onclick=\"setHintMode('all')\">All hint buttons<\/button>\n <button id=\"answerSomeToggle\" class=\"toggle-support-btn answers-toggle\" onclick=\"setAnswerMode('some')\">Some answer buttons<\/button>\n <button id=\"answerAllToggle\" class=\"toggle-support-btn answers-toggle\" onclick=\"setAnswerMode('all')\">All answer buttons<\/button>\n <\/div>\n <\/div>\n <\/section>\n\n <section class=\"card\" id=\"stepCard\">\n \n <article class=\"step visible\" data-step=\"1\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Definitions<\/h2>\n <\/div>\n <div class=\"step-count\">Step 1 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nSequences often exhibit patterns. Some patterns are temporary, some repeat infinitely many times,\nand some eventually stabilize. In this section we introduce two simple but useful notions.\n<\/p>\n<p>Let $(a_n)$ be a sequence, and let $c$ be a number.<\/p>\n\n<div class=\"definition-box\">\n <p>\n We say that $(a_n)$ is <strong>eventually equal to $c$<\/strong> if, from some point on,\n all its terms are equal to $c$. In other words, there exists a natural number $N$ such that\n <\/p>\n \\[\n a_n=c\n \\]\n <p>for every $n\\geq N$.<\/p>\n<\/div>\n\n<div class=\"definition-box\">\n <p>\n We say that $(a_n)$ is <strong>frequently equal to $c$<\/strong> if the value $c$\n appears infinitely many times in the sequence. Equivalently, for every natural number $N$,\n there exists some $n\\geq N$ such that\n <\/p>\n \\[\n a_n=c.\n \\]\n<\/div>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"2\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Warm-up problems<\/h2>\n <\/div>\n <div class=\"step-count\">Step 2 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n1,1,1,1,1,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint1')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer1')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint1\">\n <p>Every term of the sequence is equal to $1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer1\">\n <p>Yes. It is eventually equal to $1$, for example with $N=1$. It is also frequently equal to $1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n1,2,1,2,1,2,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$? Is it frequently equal to $2$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint2')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer2')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint2\">\n <p>The values $1$ and $2$ alternate forever.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer2\">\n <p>It is not eventually equal to $1$, because $2$ keeps appearing. It is frequently equal to both $1$ and $2$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"3\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Understanding the distinction<\/h2>\n <\/div>\n <div class=\"step-count\">Step 3 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n5,4,3,2,1,1,1,1,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$? Is it frequently equal to another number?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint3')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer3')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint3\">\n <p>Look at what happens from the fifth term onward.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer3\">\n <p>It is eventually equal to $1$ and frequently equal to $1$. It is not frequently equal to $2$, $3$, $4$, or $5$, since each appears only once.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\n1,2,1,1,2,1,1,1,2,1,1,1,1,2,\\ldots\n\\]\n<p>Is this sequence eventually equal to $1$? Is it frequently equal to $1$? Is it frequently equal to $2$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint4')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer4')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint4\">\n <p>The blocks of $1$'s get longer, but the value $2$ still appears again and again.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer4\">\n <p>It is not eventually equal to $1$, because $2$ appears infinitely many times. It is frequently equal to $1$ and frequently equal to $2$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Can a sequence be frequently equal to $c$, but not eventually equal to $c$? Give an example or explain why this is impossible.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint5')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer5')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint5\">\n <p>Try alternating $c$ with another value.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer5\">\n <p>Yes. For example, if $d\\neq c$, then $c,d,c,d,c,d,\\ldots$ is frequently equal to $c$ but not eventually equal to $c$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Can a sequence be eventually equal to $c$, but not frequently equal to $c$? Give an example or explain why this is impossible.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint6')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer6')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint6\">\n <p>If from some point on all terms are $c$, how many times does $c$ appear?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer6\">\n <p>No. Eventually equal to $c$ always implies frequently equal to $c$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"4\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Formula examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 4 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\na_n=\\left\\lfloor \\frac{2026}{n}\\right\\rfloor+2026.\n\\]\n<p>Compute several terms of the sequence. Is $(a_n)$ eventually equal to $2026$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint7')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer7')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint7\">\n <p>Ask when $\\left\\lfloor 2026\/n\\right\\rfloor=0$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer7\">\n <p>For $n>2026$, we have $0<2026\/n<1$, so $\\left\\lfloor 2026\/n\\right\\rfloor=0$. Hence $a_n=2026$ for all $n\\geq 2027$. The sequence is eventually equal to $2026$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the sequence<\/p>\n\\[\nb_n=\\frac{1}{n}.\n\\]\n<p>Is $(b_n)$ eventually equal to $0$? Is it frequently equal to $0$?\nDoes there exist $c>0$ and $d<0$ such that $b_n\\geq c$ or $b_n\\leq d$, for all $n\\in\\mathbb{N}$?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint8')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer8')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint8\">\n <p>The terms get closer and closer to $0$, but no term is actually equal to $0$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer8\">\n <p>\nIt is not eventually equal to $0$ and not frequently equal to $0$, since $1\/n\\neq 0$ for every $n$.\nThere is no fixed $c>0$ such that $1\/n\\geq c$ for every $n$.\nIndeed, if $c>0$, then $1\/c>0$, and we can choose a natural number $n>1\/c$.\nIt follows that\n<\/p>\n\\[\n0<\\frac{1}{n}<c.\n\\]\n<p>\nAlso, since $1\/n>0$ for every $n$, there is no $d<0$ such that $1\/n\\leq d$ for all $n$.\n<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"5\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Conceptual questions<\/h2>\n <\/div>\n <div class=\"step-count\">Step 5 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Can a sequence be eventually equal to two different numbers $c$ and $d$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint9')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer9')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint9\">\n <p>If a late term must be both $c$ and $d$, what follows?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer9\">\n <p>Only if $c=d$. If $c\\neq d$, this is impossible.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Can a sequence be frequently equal to two different numbers $c$ and $d$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint10')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer10')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint10\">\n <p>Think of alternating values.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer10\">\n <p>Yes. If $c\\neq d$, the sequence $c,d,c,d,c,d,\\ldots$ is frequently equal to both $c$ and $d$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"6\" data-section=\"eventually\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">1. Eventually and frequently equal to a value<\/div>\n <h2>Challenge<\/h2>\n <\/div>\n <div class=\"step-count\">Step 6 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct, if possible, a sequence that is frequently equal to $n$, for every natural number $n$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint11')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer11')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint11\">\n <p>Try writing blocks: $1$, then $1,2$, then $1,2,3$, then $1,2,3,4$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer11\">\n \n<p>One construction is<\/p>\n\\[\n1,\\quad 1,2,\\quad 1,2,3,\\quad 1,2,3,4,\\quad \\ldots\n\\]\n<p>written as a single sequence:<\/p>\n\\[\n1,1,2,1,2,3,1,2,3,4,\\ldots\n\\]\n<p>For each fixed $n$, the value $n$ appears in every block $1,2,\\ldots,k$ with $k\\geq n$, hence infinitely many times.<\/p>\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"7\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Definitions<\/h2>\n <\/div>\n <div class=\"step-count\">Step 7 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nSets allow us to collect objects and study them together. If $A$ and $B$ are sets,\ntheir <strong>intersection<\/strong>, denoted by $A\\cap B$, is the set of all elements\nthat belong to both $A$ and $B$.\n<\/p>\n<p>Thus, $x\\in A\\cap B$ means that $x\\in A$ and $x\\in B$.\nIf two sets have no elements in common, then $A\\cap B=\\emptyset$.<\/p>\n<p>For three sets $A$, $B$, and $C$, the intersection $A\\cap B\\cap C$\nis the set of all elements that belong to all three sets at the same time.<\/p>\n<p>If $A_1,A_2,A_3,\\ldots$ is a family of sets, then<\/p>\n\\[\n\\bigcap_{n=1}^{\\infty} A_n\n\\]\n<p>is the set of all elements that belong to every one of $A_1,A_2,A_3,\\ldots$.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"8\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Computing intersections<\/h2>\n <\/div>\n <div class=\"step-count\">Step 8 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find $A\\cap B$, for<\/p>\n\\[\nA=\\{1,2,3,4\\},\\qquad B=\\{3,4,5,6\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint12')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer12')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint12\">\n <p>Look for the elements that appear in both sets.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer12\">\n \\[A\\cap B=\\{3,4\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find $A\\cap B$, for<\/p>\n\\[\nA=\\{1,3,5,7\\},\\qquad B=\\{2,4,6,8\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint13')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer13')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint13\">\n <p>One set contains odd numbers and the other contains even numbers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer13\">\n \\[A\\cap B=\\emptyset.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find $A\\cap B$, $A\\cap C$, $B\\cap C$ and $A\\cap B\\cap C$, for<\/p>\n\\[\nA=\\{1,2,3,4,5\\},\\quad B=\\{2,4,6,8\\},\\quad C=\\{4,5,6,7\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint14')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer14')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint14\">\n <p>First compare two sets at a time. Then look for elements common to all three.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer14\">\n \n\\[\nA\\cap B=\\{2,4\\},\\quad A\\cap C=\\{4,5\\},\\quad B\\cap C=\\{4,6\\},\n\\]\n\\[\nA\\cap B\\cap C=\\{4\\}.\n\\]\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"9\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Constructing sets with prescribed intersections<\/h2>\n <\/div>\n <div class=\"step-count\">Step 9 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Construct two different finite sets $A$ and $B$ such that<\/p>\\[A\\cap B=\\{3,5\\}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint15')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer15')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint15\">\n <p>Put $3$ and $5$ in both sets. Then add different extra elements to each set.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer15\">\n <p>For example, $A=\\{1,3,5\\}$ and $B=\\{3,5,7\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\\[A\\cap B\\cap C=\\{1,2\\}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint16')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer16')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint16\">\n <p>Put $1$ and $2$ in all three sets. Add some different extra elements.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer16\">\n <p>For example, $A=\\{1,2,3\\}$, $B=\\{1,2,4\\}$, $C=\\{1,2,5\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\\[A\\cap B=\\{1,2\\}\\quad\\text{and}\\quad A\\cap B\\cap C=\\{1\\}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint17')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer17')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint17\">\n <p>Make $1$ and $2$ belong to both $A$ and $B$, but put only $1$ in $C$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer17\">\n <p>For example, $A=\\{1,2,3\\}$, $B=\\{1,2,4\\}$, $C=\\{1,5\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Construct three different finite sets $A$, $B$, and $C$ such that $A\\cap B\\cap C=\\emptyset$, but none of $A$, $B$, and $C$ is empty.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint18')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer18')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint18\">\n <p>This is possible even if the sets are very small.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer18\">\n <p>For example, $A=\\{1\\}$, $B=\\{2\\}$, $C=\\{3\\}$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"10\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Pairwise intersections and full intersection<\/h2>\n <\/div>\n <div class=\"step-count\">Step 10 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\n\\[\nA\\cap B\\neq\\emptyset,\\qquad A\\cap C\\neq\\emptyset,\n\\]\n<p>but<\/p>\n\\[\nB\\cap C=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint19')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer19')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint19\">\n <p>Let $A$ share one element with $B$ and a different element with $C$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer19\">\n <p>For example, $A=\\{1,2\\}$, $B=\\{1\\}$, $C=\\{2\\}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct three different finite sets $A$, $B$, and $C$ such that<\/p>\n\\[\nA\\cap B\\neq\\emptyset,\\qquad A\\cap C\\neq\\emptyset,\\qquad B\\cap C\\neq\\emptyset,\n\\]\n<p>but<\/p>\n\\[\nA\\cap B\\cap C=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint20')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer20')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint20\">\n <p>Try to make each pair share a different element.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer20\">\n <p>For example, $A=\\{1,2\\}$, $B=\\{2,3\\}$, $C=\\{1,3\\}$. Then every pair intersects, but no element belongs to all three sets.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"11\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Infinite families of sets<\/h2>\n <\/div>\n <div class=\"step-count\">Step 11 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>For every natural number $n$, let $A_n=\\{1,2,\\ldots,n\\}$. Notice that $A_n\\subseteq A_{n+1}$ for every $n$.<\/p>\n<p>Compute<\/p>\n\\[\n\\bigcap_{n=1}^{\\infty} A_n.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint21')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer21')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint21\">\n <p>An element must belong to $A_1$ and also to every later set.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer21\">\n \\[\\bigcap_{n=1}^{\\infty} A_n=\\{1\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>With the same sets $A_n=\\{1,2,\\ldots,n\\}$, compute<\/p>\n\\[\n\\bigcap_{n=100}^{\\infty} A_n.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint22')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer22')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint22\">\n <p>The sets are increasing. The smallest set in this tail is $A_{100}$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer22\">\n \\[\\bigcap_{n=100}^{\\infty} A_n=A_{100}=\\{1,2,\\ldots,100\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>More generally, for a given $N\\in\\mathbb{N}$, compute<\/p>\n\\[\n\\bigcap_{n=N}^{\\infty} A_n.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint23')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer23')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint23\">\n <p>Again, the smallest set in the family is $A_N$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer23\">\n \\[\\bigcap_{n=N}^{\\infty} A_n=A_N=\\{1,2,\\ldots,N\\}.\\]\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"12\" data-section=\"intersections\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">2. Intersections of sets<\/div>\n <h2>Challenges<\/h2>\n <\/div>\n <div class=\"step-count\">Step 12 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct four finite sets $A$, $B$, $C$, and $D$ such that the intersection of any three of them is nonempty, but<\/p>\n\\[\nA\\cap B\\cap C\\cap D=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint24')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer24')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint24\">\n <p>Try using the set $\\{1,2,3,4\\}$, and let each set miss exactly one element.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer24\">\n \n<p>One example is<\/p>\n\\[\nA=\\{2,3,4\\},\\quad B=\\{1,3,4\\},\\quad C=\\{1,2,4\\},\\quad D=\\{1,2,3\\}.\n\\]\n<p>The intersection of all four is empty, but the intersection of any three contains the element missing from the fourth set.<\/p>\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Can you generalize the previous construction? For a given natural number $n$, can you construct $n$ finite sets such that the intersection of any $n-1$ of them is nonempty, but the intersection of all $n$ sets is empty?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint25')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer25')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint25\">\n <p>Use the set $\\{1,2,\\ldots,n\\}$, and make the $i$-th set miss only $i$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer25\">\n \n<p>Let<\/p>\n\\[\nA_i=\\{1,2,\\ldots,n\\}\\setminus\\{i\\},\\qquad i=1,\\ldots,n.\n\\]\n<p>The intersection of all $n$ sets is empty. But if we omit $A_i$, then the intersection of the remaining $n-1$ sets contains $i$.<\/p>\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Construct a family of sets<\/p>\n\\[\nA_1\\supseteq A_2\\supseteq A_3\\supseteq\\cdots\n\\]\n<p>such that every $A_n$ is nonempty, but<\/p>\n\\[\n\\bigcap_{n=1}^{\\infty} A_n=\\emptyset.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint26')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer26')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint26\">\n <p>Try removing the first few natural numbers step by step.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer26\">\n \n<p>Take<\/p>\n\\[\nA_n=\\{n,n+1,n+2,\\ldots\\}.\n\\]\n<p>Then $A_1\\supseteq A_2\\supseteq A_3\\supseteq\\cdots$, and every $A_n$ is nonempty. But no natural number belongs to every $A_n$, so the full intersection is empty.<\/p>\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"13\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Definition and first idea<\/h2>\n <\/div>\n <div class=\"step-count\">Step 13 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nSome infinite sets can be listed one element after another, in a single infinite row.\nFor example, the positive even integers can be listed as\n<\/p>\n\\[\n2,4,6,8,10,\\ldots\n\\]\n<p>We say that a set $A$ is <strong>countable<\/strong> if its elements can be listed in a sequence<\/p>\n\\[\na_1,a_2,a_3,\\ldots\n\\]\n<p>so that every element of $A$ appears somewhere in the list.<\/p>\n<p>\nEquivalently, the elements of $A$ can be paired with the natural numbers.\nFor the positive even integers, the pairing is $n\\longleftrightarrow 2n$.\n<\/p>\n<p>The order of the list is not important. What matters is that every element appears at some finite position.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"14\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>First examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 14 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set $\\{10,20,30,40,\\ldots\\}$. Which element is paired with the natural number $n$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint27')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer27')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint27\">\n <p>Look at the common factor.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer27\">\n <p>The list is $10,20,30,\\ldots$. The number $n$ is paired with $10n$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set of positive odd integers. Can you exhibit an explicit pairing?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint28')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer28')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint28\">\n <p>The positive odd integers are $1,3,5,\\ldots$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer28\">\n <p>The list is $1,3,5,\\ldots$. The number $n$ is paired with $2n-1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set $\\{1,4,9,16,25,\\ldots\\}$. Which element is paired with the natural number $n$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint29')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer29')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint29\">\n <p>These are squares.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer29\">\n <p>The list is $1,4,9,16,\\ldots$. The number $n$ is paired with $n^2$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"15\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>More challenging examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 15 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Is it possible to list all the negative integers?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint30')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer30')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint30\">\n <p>Start with $-1$, then $-2$, then $-3$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer30\">\n <p>Yes: $-1,-2,-3,-4,\\ldots$ lists all negative integers.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Is it possible to list all the integers? [Hint: Yes, it is possible.]<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint31')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer31')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint31\">\n <p>Try alternating positive and negative integers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer31\">\n <p>Yes. One list is $0,1,-1,2,-2,3,-3,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>List the elements of the set of all integer multiples of $3$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint32')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer32')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint32\">\n <p>Use the list of all integers and multiply by $3$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer32\">\n <p>One list is $0,3,-3,6,-6,9,-9,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"16\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Combining countable sets<\/h2>\n <\/div>\n <div class=\"step-count\">Step 16 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Suppose that the elements of a set $A$ can be listed as $a_1,a_2,a_3,\\ldots$\nand the elements of a set $B$ can be listed as $b_1,b_2,b_3,\\ldots$.\nConstruct a list that contains all elements of $A\\cup B$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint33')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer33')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint33\">\n <p>Alternate between the two lists.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer33\">\n <p>One possible list is $a_1,b_1,a_2,b_2,a_3,b_3,\\ldots$, skipping repeated elements if necessary.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Use the previous idea to list all numbers in<\/p>\n\\[\n\\{1,3,5,7,\\ldots\\}\\cup \\{2,4,6,8,\\ldots\\}.\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint34')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer34')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint34\">\n <p>This union is the set of all positive integers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer34\">\n <p>One list is $1,2,3,4,5,6,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Suppose $A$ and $B$ are countable sets. Is $A\\cup B$ countable? Explain.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint35')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer35')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint35\">\n <p>Use the alternating list idea.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer35\">\n <p>Yes. If $A$ and $B$ can be listed, then $A\\cup B$ can be listed by alternating between the two lists and ignoring repetitions.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Suppose $A_1,A_2,A_3,\\ldots$ are countable sets. Do you think the union<\/p>\n\\[\nA_1\\cup A_2\\cup A_3\\cup\\cdots\n\\]\n<p>must be countable?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint36')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer36')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint36\">\n <p>Imagine arranging the lists in rows and moving along diagonals.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer36\">\n <p>Yes. A countable union of countable sets is countable. One can arrange the elements in an infinite table and list them by diagonals.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"17\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Pairs of natural numbers<\/h2>\n <\/div>\n <div class=\"step-count\">Step 17 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Consider the set $\\mathbb{N}\\times\\mathbb{N}$ of all pairs $(m,n)$ of natural numbers. Try to list its elements.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint37')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer37')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint37\">\n <p>A row-by-row list fails. Diagonals work better.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer37\">\n <p>One beginning is $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>A first attempt might be to list the pairs row by row:<\/p>\n\\[\n(1,1),(1,2),(1,3),(1,4),\\ldots\n\\]\n<p>Why does this attempt fail to list all pairs?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint38')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer38')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint38\">\n <p>Do we ever leave the first row?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer38\">\n <p>It never reaches pairs like $(2,1)$ or $(3,1)$, because it stays forever in the first row.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Try instead to list the pairs by diagonals:<\/p>\n\\[\nDiag_i=\\{(n,m): n+m=i+1\\}.\n\\]\n<ol class=\"inner-list\">\n<li>Identify $Diag_1$, $Diag_2$ and $Diag_3$.<\/li>\n<li>Put, in a finite list, all the elements of $Diag_1\\cup Diag_2\\cup Diag_3$.<\/li>\n<li>List $\\mathbb{N}\\times\\mathbb{N}$.<\/li>\n<li>Explain why every pair $(m,n)$ eventually appears.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint39')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer39')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint39\">\n <p>A pair $(m,n)$ lies on the diagonal determined by $m+n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer39\">\n \n<p>We have<\/p>\n\\[\nDiag_1=\\{(1,1)\\},\\quad Diag_2=\\{(1,2),(2,1)\\},\\quad Diag_3=\\{(1,3),(2,2),(3,1)\\}.\n\\]\n<p>Every pair $(m,n)$ appears on the diagonal with index $m+n-1$.<\/p>\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Conclude that $\\mathbb{N}\\times\\mathbb{N}$ is countable.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint40')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer40')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint40\">\n <p>If every pair appears in the diagonal list, then the pairs can be listed.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer40\">\n <p>Yes, $\\mathbb{N}\\times\\mathbb{N}$ is countable.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"18\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>Rational numbers<\/h2>\n <\/div>\n <div class=\"step-count\">Step 18 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Every positive rational number can be written as $\\dfrac{p}{q}$, where $p$ and $q$ are natural numbers.\nThis representation is not unique. Is there a way to represent them in a unique way?\nExplain why this relates the positive rational numbers to $\\mathbb{N}\\times\\mathbb{N}$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint41')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer41')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint41\">\n <p>Each fraction gives a pair $(p,q)$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer41\">\n <p>Every positive rational number comes from some pair $(p,q)\\in\\mathbb{N}\\times\\mathbb{N}$. A unique representation is obtained by requiring $\\gcd(p,q)=1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Use the previous subsection to argue that the set of positive rational numbers is countable.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint42')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer42')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint42\">\n <p>List all pairs $(p,q)$ and then read them as fractions $p\/q$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer42\">\n <p>Since $\\mathbb{N}\\times\\mathbb{N}$ is countable, the fractions $p\/q$ can be listed. Repetitions do not matter, so the positive rational numbers are countable.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Do you think the set of all rational numbers is countable? Explain how one might list them.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint43')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer43')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint43\">\n <p>Combine positive rationals, negative rationals, and zero.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer43\">\n <p>Yes. One can list $0$, then alternate positive and negative rational numbers: $q_1,-q_1,q_2,-q_2,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"19\" data-section=\"countable\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">3. Countable sets<\/div>\n <h2>A tough challenge<\/h2>\n <\/div>\n <div class=\"step-count\">Step 19 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Is every infinite set countable? In other words, can the elements of every infinite set be listed in a single infinite row?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint44')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer44')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint44\">\n <p>The real numbers are the standard place where this question becomes surprising.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer44\">\n <p>No. There are infinite sets that are not countable, such as the real numbers between $0$ and $1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n \n<p>Consider the set $A$ of real numbers between $0$ and $1$ whose decimal expression only contains $0$'s and $1$'s, for example<\/p>\n\\[\n0.001101010111\\ldots\n\\quad\\text{or}\\quad\n0.1001010000111111\\ldots\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint45')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer45')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint45\">\n <p>This set is much larger than it may first appear.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer45\">\n <p>This is a good test set for diagonal thinking: each number corresponds to an infinite sequence of $0$'s and $1$'s.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Try to list all the elements of $A$. What difficulties appear?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint46')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer46')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint46\">\n <p>Any proposed list can be challenged digit by digit.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer46\">\n <p>The difficulty is that a list has rows, and each row has infinitely many digits. One can construct a new number by changing the diagonal digits.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Suppose someone claims to have listed all the elements of $A$:<\/p>\n\\[\nr_1,r_2,r_3,\\ldots\n\\]\n<p>Can you imagine a way to construct a new real number between $0$ and $1$ that is different from every number on the list?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint47')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer47')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint47\">\n <p>Make the new number differ from $r_1$ in the first digit, from $r_2$ in the second digit, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer47\">\n \n<p>This is Cantor's diagonal idea. If the $n$-th digit of $r_n$ is $0$, choose the $n$-th digit of the new number to be $1$; if it is $1$, choose $0$. The new number differs from $r_n$ in the $n$-th digit, so it is not on the list.<\/p>\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"20\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>The principle<\/h2>\n <\/div>\n <div class=\"step-count\">Step 20 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nThe <strong>Pigeonhole Principle<\/strong> is a simple combinatorial idea with many surprising consequences.\nIf more than $n$ objects are placed into $n$ boxes, then at least one box contains at least two objects.\n<\/p>\n<p>\nFor example, if $8$ students are in a room, then at least two of them were born on the same day of the week,\nbecause there are only $7$ days of the week.\n<\/p>\n<p>The main difficulty is often not applying the principle, but deciding what the boxes should be.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"21\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>First examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 21 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>There are $13$ students in a room. Show that at least two of them were born in the same month.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint48')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer48')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint48\">\n <p>Use the $12$ months as boxes.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer48\">\n <p>There are $13$ students and only $12$ months, so two students must share a birth month.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>There are $27$ students in a room. Show that at least three of them were born in the same month.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint49')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer49')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint49\">\n <p>If each month had at most two students, how many students could there be?<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer49\">\n <p>If every month had at most two students, there would be at most $24$ students. Since there are $27$, some month has at least three.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>A drawer contains only black socks and white socks. How many socks must you take to be sure that you have two socks of the same color? Explain your answer as a consequence of the Pigeonhole Principle. Which are the boxes?<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint50')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer50')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint50\">\n <p>The boxes are the two colors.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer50\">\n <p>You need $3$ socks. With $3$ objects and $2$ boxes, one color must contain at least two socks.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"22\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Remainders as boxes<\/h2>\n <\/div>\n <div class=\"step-count\">Step 22 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $6$ integers. Show that at least two of them have the same remainder when divided by $5$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint51')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer51')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint51\">\n <p>The possible remainders are $0,1,2,3,4$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer51\">\n <p>There are $6$ integers and only $5$ possible remainders, so two have the same remainder modulo $5$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $11$ integers. Show that at least two of them differ by a multiple of $10$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint52')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer52')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint52\">\n <p>Two integers differ by a multiple of $10$ exactly when they have the same remainder modulo $10$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer52\">\n <p>There are $10$ possible remainders modulo $10$ and $11$ integers. Two share a remainder, hence their difference is divisible by $10$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ integers. Show that there are two of them such that their difference is divisible by $n$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint53')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer53')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint53\">\n <p>Use remainders modulo $n$ as boxes.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer53\">\n <p>There are $n$ possible remainders modulo $n$, and $n+1$ integers. Two have the same remainder, so their difference is divisible by $n$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"23\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Choosing the boxes<\/h2>\n <\/div>\n <div class=\"step-count\">Step 23 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n \n<p>Choose any $6$ numbers from<\/p>\n\\[\n\\{1,2,3,4,5,6,7,8,9,10\\}.\n\\]\n<p>Show that at least two of the chosen numbers have sum $11$.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint54')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer54')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint54\">\n <p>Group the numbers into pairs with sum $11$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer54\">\n <p>Use the boxes $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$, $\\{5,6\\}$. Choosing $6$ numbers from $5$ boxes forces two chosen numbers to be in the same box, hence to sum to $11$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why the boxes $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$, $\\{5,6\\}$ prove the previous result.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint55')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer55')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint55\">\n <p>Each box contains exactly two numbers with the desired property.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer55\">\n <p>Since there are $5$ boxes and $6$ chosen numbers, one box contains two chosen numbers. The two numbers in any one box have sum $11$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n \n<p>Choose any $7$ numbers from<\/p>\n\\[\n\\{1,2,3,\\ldots,12\\}.\n\\]\n<p>Show that at least two of the chosen numbers are consecutive.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint56')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer56')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint56\">\n <p>Use boxes $\\{1,2\\}$, $\\{3,4\\}$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer56\">\n <p>There are $6$ boxes: $\\{1,2\\},\\{3,4\\},\\ldots,\\{11,12\\}$. Choosing $7$ numbers forces two in the same box, and those two are consecutive.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>In the previous problem, what are the useful boxes?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint57')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer57')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint57\">\n <p>Pair consecutive numbers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer57\">\n \\[\\{1,2\\},\\{3,4\\},\\{5,6\\},\\{7,8\\},\\{9,10\\},\\{11,12\\}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n \n<p>Choose any $7$ numbers from<\/p>\n\\[\n\\{1,2,3,\\ldots,13\\}.\n\\]\n<p>Is it necessarily true that two of the chosen numbers are consecutive? Give a proof or a counterexample.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint58')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer58')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint58\">\n <p>Try choosing every other number.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer58\">\n <p>No. The set $\\{1,3,5,7,9,11,13\\}$ has $7$ numbers and no two are consecutive.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"24\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Geometric examples<\/h2>\n <\/div>\n <div class=\"step-count\">Step 24 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $5$ points inside a square of side length $2$. Show that at least two points are at distance at most $\\sqrt{2}$ from each other.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint59')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer59')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint59\">\n <p>Divide the large square into four unit squares.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer59\">\n <p>Four unit squares are the boxes. With $5$ points, two lie in the same unit square. The maximum distance between two points in a unit square is its diagonal, $\\sqrt{2}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why dividing the square into four unit squares is useful in the previous problem.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint60')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer60')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint60\">\n <p>The diameter of each small square is exactly $\\sqrt{2}$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer60\">\n <p>It creates $4$ boxes, each with the property that any two points inside the same box are at distance at most $\\sqrt{2}$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $10$ points inside an equilateral triangle of side length $3$. Show that at least two points are at distance at most $1$ from each other.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint61')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer61')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint61\">\n <p>Divide the triangle into $9$ equilateral triangles of side length $1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer61\">\n <p>The $9$ small triangles are the boxes. With $10$ points, two lie in the same small triangle, whose diameter is $1$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Suppose every point in the plane is colored either red or blue. Show that there exist two points of the same color at distance exactly $1$ from each other.<\/p><p><em>Hint:<\/em> consider the vertices of an equilateral triangle of side length $1$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint62')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer62')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint62\">\n <p>There are $3$ vertices but only $2$ colors.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer62\">\n <p>Among the three vertices of an equilateral triangle of side $1$, two must have the same color. Those two vertices are at distance $1$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"25\" data-section=\"pigeonhole\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">4. The Pigeonhole Principle<\/div>\n <h2>Challenges<\/h2>\n <\/div>\n <div class=\"step-count\">Step 25 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ numbers from $\\{1,2,3,\\ldots,2n\\}$. Show that at least two have sum $2n+1$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint63')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer63')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint63\">\n <p>Pair $1$ with $2n$, $2$ with $2n-1$, and so on.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer63\">\n <p>Use boxes $\\{1,2n\\}$, $\\{2,2n-1\\}$, ..., $\\{n,n+1\\}$. There are $n$ boxes and $n+1$ chosen numbers.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ numbers from $\\{1,2,3,\\ldots,2n\\}$. Show that at least two are relatively prime.<\/p><p><em>Hint:<\/em> think about consecutive integers.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint64')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer64')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint64\">\n <p>Use boxes of consecutive pairs.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer64\">\n <p>Pair the numbers as $\\{1,2\\}$, $\\{3,4\\}$, ..., $\\{2n-1,2n\\}$. Two chosen numbers must lie in the same pair, and consecutive integers are relatively prime.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Choose any $n+1$ numbers from $\\{1,2,3,\\ldots,2n\\}$. Show that at least one of the chosen numbers divides another.<\/p><p><em>Hint:<\/em> write each number in the form $2^k m$, where $m$ is odd.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint65')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer65')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint65\">\n <p>The odd part $m$ can only be one of $1,3,5,\\ldots,2n-1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer65\">\n <p>There are $n$ possible odd parts. With $n+1$ chosen numbers, two have the same odd part: $2^a m$ and $2^b m$. The one with the smaller power of $2$ divides the other.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"26\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>What is the task?<\/h2>\n <\/div>\n <div class=\"step-count\">Step 26 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>\nA proof is not only a collection of true statements. The statements must also appear in the correct logical order.\nSome statements depend on previous ones; some introduce definitions; some use the hypothesis; and some give the final conclusion.\n<\/p>\n<p>In this section, each problem gives a mathematical claim and a list of statements in the wrong order. Rearrange the statements to obtain a correct proof.<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"27\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Warm-up proofs<\/h2>\n <\/div>\n <div class=\"step-count\">Step 27 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 1<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> If $a$ and $b$ are even integers, then $a+b$ is even.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, $a+b$ is even.<\/li>\n<li>Since $a$ and $b$ are even, there exist integers $m,n$ such that $a=2m$ and $b=2n$.<\/li>\n<li>Then $a+b=2m+2n=2(m+n)$.<\/li>\n<li>Since $m+n$ is an integer, $a+b$ is divisible by $2$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint66')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer66')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint66\">\n <p>Start by using the meaning of \u201ceven\u201d.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer66\">\n <p>Correct order: (b), (c), (d), (a).<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 2<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> If $a$ and $b$ are odd integers, then $ab$ is odd.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Since $m,n$ are integers, $2mn+m+n$ is an integer.<\/li>\n<li>Then $ab=(2m+1)(2n+1)$.<\/li>\n<li>Therefore, $ab$ is odd.<\/li>\n<li>Since $a$ and $b$ are odd, there exist integers $m,n$ such that $a=2m+1$ and $b=2n+1$.<\/li>\n<li>Expanding, $ab=4mn+2m+2n+1=2(2mn+m+n)+1$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint67')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer67')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint67\">\n <p>Begin with the definition of odd integers.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer67\">\n <p>Correct order: (d), (b), (e), (a), (c).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"28\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Proofs by contradiction<\/h2>\n <\/div>\n <div class=\"step-count\">Step 28 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 3<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> If $n^2$ is even, then $n$ is even.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, if $n^2$ is even, then $n$ must be even.<\/li>\n<li>Suppose, for contradiction, that $n$ is odd.<\/li>\n<li>Hence $n^2$ is odd.<\/li>\n<li>Then there exists an integer $k$ such that $n=2k+1$.<\/li>\n<li>Thus $n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1$.<\/li>\n<li>This contradicts the assumption that $n^2$ is even.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint68')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer68')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint68\">\n <p>Because this is a contradiction proof, begin by assuming the opposite of what you want to prove.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer68\">\n <p>Correct order: (b), (d), (e), (c), (f), (a).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"29\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Proofs with sets<\/h2>\n <\/div>\n <div class=\"step-count\">Step 29 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 4<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> Prove the inclusion<\/p>\n\\[\nA\\cap(B\\cup C)\\subseteq (A\\cap B)\\cup(A\\cap C).\n\\]\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, $x\\in (A\\cap B)\\cup(A\\cap C)$.<\/li>\n<li>This proves the inclusion.<\/li>\n<li>Suppose $x\\in A\\cap(B\\cup C)$.<\/li>\n<li>Hence $x\\in A$ and $x\\in B\\cup C$.<\/li>\n<li>Therefore $x\\in B$ or $x\\in C$.<\/li>\n<li>If $x\\in B$, then $x\\in A\\cap B$.<\/li>\n<li>If $x\\in C$, then $x\\in A\\cap C$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint69')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer69')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint69\">\n <p>Start by taking an arbitrary element of the left-hand side.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer69\">\n <p>Correct order: (c), (d), (e), (f) and (g), (a), (b). Statements (f) and (g) are the two cases.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 5<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> Prove the reverse inclusion<\/p>\n\\[\n(A\\cap B)\\cup(A\\cap C)\\subseteq A\\cap(B\\cup C).\n\\]\n<p>Write your own proof. Try to imitate the structure of the previous problem.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint70')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer70')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint70\">\n <p>Start with $x\\in (A\\cap B)\\cup(A\\cap C)$ and split into cases.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer70\">\n <p>If $x\\in A\\cap B$, then $x\\in A$ and $x\\in B$, so $x\\in B\\cup C$, hence $x\\in A\\cap(B\\cup C)$. Similarly, if $x\\in A\\cap C$, then $x\\in A\\cap(B\\cup C)$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"30\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Divisibility<\/h2>\n <\/div>\n <div class=\"step-count\">Step 30 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 6<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> For every integer $n$, the number $n^3-n$ is divisible by $3$.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, one of $n-1,n,n+1$ is divisible by $3$.<\/li>\n<li>Hence $n^3-n$ is divisible by $3$.<\/li>\n<li>We can factor: $n^3-n=n(n-1)(n+1)$.<\/li>\n<li>Among any three consecutive integers, exactly one is divisible by $3$.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint71')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer71')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint71\">\n <p>Factor first.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer71\">\n <p>Correct order: (c), (d), (a), (b).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"31\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>A classical proof<\/h2>\n <\/div>\n <div class=\"step-count\">Step 31 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 7<\/div>\n <div class=\"problem-statement\">\n \n<p><strong>Claim.<\/strong> There are infinitely many prime numbers.<\/p>\n<p>Rearrange:<\/p>\n<ol class=\"letter-list\">\n<li>Therefore, $q$ is a prime not among $p_1,p_2,\\ldots,p_k$.<\/li>\n<li>Suppose, for contradiction, that there are only finitely many primes: $p_1,p_2,\\ldots,p_k$.<\/li>\n<li>Let $N=p_1p_2\\cdots p_k+1$.<\/li>\n<li>Hence there must be infinitely many primes.<\/li>\n<li>Let $q$ be a prime divisor of $N$.<\/li>\n<li>None of the primes $p_1,p_2,\\ldots,p_k$ divides $N$, since each leaves remainder $1$.<\/li>\n<li>This contradicts the assumption that the list contained all primes.<\/li>\n<\/ol>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint72')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer72')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint72\">\n <p>Start by assuming there are finitely many primes, then construct a new number.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer72\">\n <p>Correct order: (b), (c), (e), (f), (a), (g), (d).<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"32\" data-section=\"proof\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">5. Build the proof<\/div>\n <h2>Challenge<\/h2>\n <\/div>\n <div class=\"step-count\">Step 32 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">Problem 8<\/div>\n <div class=\"problem-statement\">\n \n<p>Write a proof, by contradiction, that $\\sqrt{2}$ is irrational.<\/p>\n<p>You may assume as starting point that<\/p>\n\\[\n\\sqrt{2}=\\frac{a}{b},\n\\]\n<p>where $a$ and $b$ are positive integers with no common factor. Then write an integral equation that $a$ and $b$ should satisfy.<\/p>\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint73')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer73')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint73\">\n <p>Square both sides and clear denominators.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer73\">\n <p>Squaring gives $2=a^2\/b^2$, hence $a^2=2b^2$. Thus $a^2$ is even, so $a$ is even. Write $a=2k$. Then $4k^2=2b^2$, so $b^2=2k^2$, hence $b$ is even. This contradicts that $a$ and $b$ have no common factor.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"33\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Definition and first idea<\/h2>\n <\/div>\n <div class=\"step-count\">Step 33 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n<p>A <strong>sequence<\/strong> is a list of numbers written in a definite order:<\/p>\n\\[\na_1,a_2,a_3,a_4,\\ldots\n\\]\n<p>The number $a_n$ is called the $n$-th term of the sequence.<\/p>\n<p>\nSometimes a sequence is described by giving its first few terms. Sometimes it is described by a formula.\nA useful skill is to move between the first terms of a sequence and a formula for its $n$-th term.\n<\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"34\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Computing terms from a formula<\/h2>\n <\/div>\n <div class=\"step-count\">Step 34 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $a_n=(-1)^n$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint74')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer74')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint74\">\n <p>Substitute $n=1,2,3,4,5,6$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer74\">\n \\[-1,1,-1,1,-1,1.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $b_n=(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint75')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer75')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint75\">\n <p>The exponent is one more than before.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer75\">\n \\[1,-1,1,-1,1,-1.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Compare the two sequences above. How are they similar? How are they different?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint76')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer76')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint76\">\n <p>Both alternate between $1$ and $-1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer76\">\n <p>They have the same two values, but start with different signs. One is the negative of the other.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $c_n=2(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint77')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer77')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint77\">\n <p>Use the sequence from part (b) and multiply by $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer77\">\n \\[2,-2,2,-2,2,-2.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n <p>Compute the first six terms of $d_n=2+2(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint78')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer78')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint78\">\n <p>Start from $2(-1)^{n+1}$ and add $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer78\">\n \\[4,0,4,0,4,0.\\]\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"35\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Finding formulas<\/h2>\n <\/div>\n <div class=\"step-count\">Step 35 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $1,-1,1,-1,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint79')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer79')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint79\">\n <p>Compare with $(-1)^n$ and $(-1)^{n+1}$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer79\">\n \\[a_n=(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $2,-2,2,-2,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint80')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer80')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint80\">\n <p>Multiply the previous sequence by $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer80\">\n \\[a_n=2(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $4,0,4,0,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint81')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer81')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint81\">\n <p>Think of it as a constant part plus an alternating part.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer81\">\n \\[a_n=2+2(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $3,7,3,7,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint82')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer82')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint82\">\n <p>The average is $5$, and the deviation is $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer82\">\n \\[a_n=5-2(-1)^{n+1}\\quad\\text{or}\\quad a_n=5+2(-1)^n.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $10,20,10,20,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint83')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer83')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint83\">\n <p>The average is $15$, and the deviation is $5$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer83\">\n \\[a_n=15-5(-1)^{n+1}\\quad\\text{or}\\quad a_n=15+5(-1)^n.\\]\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"36\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Alternating between two values<\/h2>\n <\/div>\n <div class=\"step-count\">Step 36 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Let $a$ and $b$ be two given numbers. Find a formula for $a,b,a,b,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint84')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer84')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint84\">\n <p>Use the average $(a+b)\/2$ and half-difference $(a-b)\/2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer84\">\n \\[x_n=\\frac{a+b}{2}+\\frac{a-b}{2}(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Check your formula when $a=1$ and $b=-1$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint85')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer85')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint85\">\n <p>The average is $0$ and the half-difference is $1$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer85\">\n \\[x_n=(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Check your formula when $a=4$ and $b=0$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint86')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer86')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint86\">\n <p>The average is $2$ and the half-difference is $2$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer86\">\n \\[x_n=2+2(-1)^{n+1}.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why $a,b,a,b,\\ldots$ can be thought of as a constant part plus an alternating part.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint87')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer87')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint87\">\n <p>The midpoint between $a$ and $b$ is constant.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer87\">\n <p>The sequence oscillates around the average $(a+b)\/2$, moving alternately by $+(a-b)\/2$ and $-(a-b)\/2$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(e)<\/div>\n <div class=\"problem-statement\">\n <p>Show that one possible formula is<\/p>\\[x_n=\\frac{a+b}{2}+\\frac{a-b}{2}(-1)^{n+1}.\\]\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint88')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer88')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint88\">\n <p>Evaluate separately for odd $n$ and even $n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer88\">\n <p>If $n$ is odd, then $(-1)^{n+1}=1$, so $x_n=a$. If $n$ is even, then $(-1)^{n+1}=-1$, so $x_n=b$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"37\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Looking for structure<\/h2>\n <\/div>\n <div class=\"step-count\">Step 37 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>In the formula $x_n=\\frac{a+b}{2}+\\frac{a-b}{2}(-1)^{n+1}$, what is the role of $\\frac{a+b}{2}$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint89')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer89')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint89\">\n <p>It is the midpoint between $a$ and $b$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer89\">\n <p>It is the constant part, or average, around which the sequence alternates.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>What is the role of $\\frac{a-b}{2}(-1)^{n+1}$?<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint90')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer90')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint90\">\n <p>It changes sign depending on the parity of $n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer90\">\n <p>It is the alternating part that moves the value from the average to either $a$ or $b$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Explain why the formula gives $a$ when $n$ is odd and $b$ when $n$ is even.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint91')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer91')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint91\">\n <p>Use $(-1)^{n+1}=1$ for odd $n$ and $-1$ for even $n$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer91\">\n <p>Odd $n$ gives $\\frac{a+b}{2}+\\frac{a-b}{2}=a$. Even $n$ gives $\\frac{a+b}{2}-\\frac{a-b}{2}=b$.<\/p>\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(d)<\/div>\n <div class=\"problem-statement\">\n \n<p>Find another way to describe the same sequence using cases:<\/p>\n\\[\nx_n=\n\\begin{cases}\n?, & n \\text{ odd},\\\\\n?, & n \\text{ even}.\n\\end{cases}\n\\]\n\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint92')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer92')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint92\">\n <p>Read the sequence directly: first term $a$, second term $b$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer92\">\n \n\\[\nx_n=\n\\begin{cases}\na, & n \\text{ odd},\\\\\nb, & n \\text{ even}.\n\\end{cases}\n\\]\n\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n <article class=\"step\" data-step=\"38\" data-section=\"sequences\">\n <div class=\"step-header\">\n <div>\n <div class=\"section-kicker\">6. Sequences and formulas<\/div>\n <h2>Challenges<\/h2>\n <\/div>\n <div class=\"step-count\">Step 38 \/ 38<\/div>\n <\/div>\n <div class=\"step-body\">\n <p class=\"problem-focus-note\">The first item is shown. Open later items when you are ready; previous items will remain visible.<\/p>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(a)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula for $a,b,c,a,b,c,\\ldots$, where $a$, $b$, and $c$ are given numbers.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint93')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer93')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint93\">\n <p>Using cases modulo $3$ is natural.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer93\">\n \n<p>One clear rule is<\/p>\n\\[\nx_n=\n\\begin{cases}\na, & n\\equiv 1\\pmod 3,\\\\\nb, & n\\equiv 2\\pmod 3,\\\\\nc, & n\\equiv 0\\pmod 3.\n\\end{cases}\n\\]\n\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(b)<\/div>\n <div class=\"problem-statement\">\n <p>Find a formula, or a clear rule, for $1,2,3,1,2,3,\\ldots$.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint94')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer94')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint94\">\n <p>Again use the remainder of $n$ modulo $3$.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer94\">\n \\[x_n=((n-1)\\bmod 3)+1.\\]\n <\/div>\n <\/div>\n \n <div class=\"problem\">\n <div class=\"problem-title\">(c)<\/div>\n <div class=\"problem-statement\">\n <p>Can the same sequence be described by two different formulas? Give an example or explain why not.<\/p>\n <\/div>\n <div class=\"actions\">\n <button class=\"hint-btn\" onclick=\"togglePanel('hint95')\">Show hint<\/button>\n <button class=\"answer-btn\" onclick=\"togglePanel('answer95')\">Show answer<\/button>\n <\/div>\n <div class=\"panel hint\" id=\"hint95\">\n <p>Think of the alternating sequence.<\/p>\n <\/div>\n <div class=\"panel answer\" id=\"answer95\">\n <p>Yes. For example, $(-1)^n$ and $\\cos(n\\pi)$ describe the same sequence for $n=1,2,3,\\ldots$.<\/p>\n <\/div>\n <\/div>\n \n <p class=\"step-support-reminder\">\n Need a hint or an answer?\n <a href=\"#supportBox\">Return to the support buttons above.<\/a>\n <\/p>\n\n <\/div>\n <\/article>\n \n\n <div class=\"nav-row\">\n <button class=\"nav-btn secondary\" onclick=\"previousStep()\">Previous step<\/button>\n <button id=\"definitionNavButton\" class=\"nav-btn definition-btn\" onclick=\"goToDefinition()\">Definition<\/button>\n <button class=\"nav-btn\" onclick=\"nextStep()\">Next step<\/button>\n <\/div>\n <\/section>\n\n <div class=\"footer\">\n Interactive prototype with six sections. Each step opens progressively; optional hints and answers can be enabled above.\n <\/div>\n <\/main>\n\n <script>\n let currentStep = 1;\n const totalSteps = 38;\n\n\n function prepareProblemFocus(stepEl) {\n if (!stepEl) return;\n const problems = Array.from(stepEl.querySelectorAll('.problem'));\n problems.forEach((problem, index) => {\n let openButton = problem.querySelector('.open-item-btn');\n if (!openButton) {\n const title = problem.querySelector('.problem-title');\n const label = title ? title.textContent.trim() : `item ${index + 1}`;\n openButton = document.createElement('button');\n openButton.className = 'open-item-btn';\n openButton.type = 'button';\n openButton.textContent = index === 0 ? `Open first item` : `Open item ${label}`;\n openButton.addEventListener('click', () => expandProblem(problem));\n problem.insertBefore(openButton, problem.firstChild);\n }\n\n problem.classList.toggle('collapsed', index !== 0);\n problem.querySelectorAll('.panel').forEach(panel => {\n panel.style.display = 'none';\n });\n });\n }\n\n function expandProblem(problem) {\n const stepEl = problem.closest('.step');\n if (!stepEl) return;\n\n \/\/ Once an item is opened, keep it open. This lets the reader see previous\n \/\/ items while working on later ones.\n problem.classList.remove('collapsed');\n problem.scrollIntoView({ behavior: 'smooth', block: 'start' });\n\n if (window.MathJax) {\n MathJax.typesetPromise([problem]);\n }\n }\n\n function showStep(step) {\n currentStep = Math.max(1, Math.min(totalSteps, step));\n document.querySelectorAll('.step').forEach(el => {\n el.classList.remove('visible');\n });\n const active = document.querySelector(`[data-step=\"${currentStep}\"]`);\n active.classList.add('visible');\n prepareProblemFocus(active);\n\n const sectionSlug = active.getAttribute('data-section');\n const firstStepOfSection = document.querySelector(`.step[data-section=\"${sectionSlug}\"]`);\n const definitionButton = document.getElementById('definitionNavButton');\n\n if (definitionButton) {\n definitionButton.style.display =\n active === firstStepOfSection ? 'none' : 'inline-block';\n }\n\n document.getElementById('stepCard').scrollIntoView({ behavior: 'smooth', block: 'start' });\n if (window.MathJax) {\n MathJax.typesetPromise([active]);\n }\n }\n\n function nextStep() {\n showStep(currentStep + 1);\n }\n\n function previousStep() {\n showStep(currentStep - 1);\n }\n\n\n function goToDefinition() {\n const active = document.querySelector('.step.visible');\n if (!active) return;\n\n const sectionSlug = active.getAttribute('data-section');\n const definitionStep = document.querySelector(`.step[data-section=\"${sectionSlug}\"]`);\n if (!definitionStep) return;\n\n const step = parseInt(definitionStep.getAttribute('data-step'), 10);\n showStep(step);\n }\n\n function goToSection(slug) {\n const target = document.querySelector(`.step[data-section=\"${slug}\"]`);\n if (!target) return;\n const step = parseInt(target.getAttribute('data-step'), 10);\n showStep(step);\n }\n\n function markSomeSupportButtons() {\n document.querySelectorAll('.step').forEach(step => {\n const problems = Array.from(step.querySelectorAll('.problem'));\n const total = problems.length;\n if (total === 0) return;\n\n \/\/ \"Some\" means the last floor(total\/2) items in this step.\n \/\/ Examples:\n \/\/ 2 items -> item 2\n \/\/ 3 items -> item 3\n \/\/ 4 items -> items 3 and 4\n \/\/ 5 items -> items 4 and 5\n const firstSupportedIndex = total - Math.floor(total \/ 2);\n\n problems.forEach((problem, index) => {\n problem.querySelectorAll('.hint-btn, .answer-btn').forEach(button => {\n button.classList.remove('support-some');\n if (index >= firstSupportedIndex) {\n button.classList.add('support-some');\n }\n });\n });\n });\n }\n\n function clearPanels(kind) {\n document.querySelectorAll(`.panel.${kind}`).forEach(panel => {\n panel.style.display = 'none';\n });\n }\n\n function setHintMode(mode) {\n const body = document.body;\n const current = body.classList.contains(`hint-mode-${mode}`);\n body.classList.remove('hint-mode-none', 'hint-mode-some', 'hint-mode-all');\n\n if (current) {\n body.classList.add('hint-mode-none');\n mode = 'none';\n } else {\n body.classList.add(`hint-mode-${mode}`);\n }\n\n document.getElementById('hintSomeToggle').classList.toggle('active', mode === 'some');\n document.getElementById('hintAllToggle').classList.toggle('active', mode === 'all');\n clearPanels('hint');\n }\n\n function setAnswerMode(mode) {\n const body = document.body;\n const current = body.classList.contains(`answer-mode-${mode}`);\n body.classList.remove('answer-mode-none', 'answer-mode-some', 'answer-mode-all');\n\n if (current) {\n body.classList.add('answer-mode-none');\n mode = 'none';\n } else {\n body.classList.add(`answer-mode-${mode}`);\n }\n\n document.getElementById('answerSomeToggle').classList.toggle('active', mode === 'some');\n document.getElementById('answerAllToggle').classList.toggle('active', mode === 'all');\n clearPanels('answer');\n }\n\n\n function togglePanel(id) {\n const panel = document.getElementById(id);\n if (!panel) return;\n panel.style.display = panel.style.display === 'block' ? 'none' : 'block';\n if (window.MathJax) {\n MathJax.typesetPromise([panel]);\n }\n }\n markSomeSupportButtons();\n prepareProblemFocus(document.querySelector('.step.visible'));\n const initialDefinitionButton = document.getElementById('definitionNavButton');\n if (initialDefinitionButton) {\n initialDefinitionButton.style.display = 'none';\n }\n <\/script>\n<\/body>\n<\/html>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"<p>Mathematical Thinking Questions \u2014 Interactive Mathemati …<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/sites.gtiit.edu.cn\/admissions\/math-test-text\/\"> <span class=\"screen-reader-text\">\u6570\u5b66\u6d4b\u8bd5\u9898<\/span> \u67e5\u770b\u5168\u6587 »<\/a><\/p>\n","protected":false},"author":314,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"elementor_header_footer","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts\/46822"}],"collection":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/users\/314"}],"replies":[{"embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/comments?post=46822"}],"version-history":[{"count":4,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts\/46822\/revisions"}],"predecessor-version":[{"id":47013,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/posts\/46822\/revisions\/47013"}],"wp:attachment":[{"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/media?parent=46822"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/categories?post=46822"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sites.gtiit.edu.cn\/admissions\/wp-json\/wp\/v2\/tags?post=46822"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}