{ localUrl: '../page/rationals_are_countable.html', arbitalUrl: 'https://arbital.com/p/rationals_are_countable', rawJsonUrl: '../raw/511.json', likeableId: '2933', likeableType: 'page', myLikeValue: '0', likeCount: '1', dislikeCount: '0', likeScore: '1', individualLikes: [ 'EricBruylant' ], pageId: 'rationals_are_countable', edit: '1', editSummary: '', prevEdit: '0', currentEdit: '1', wasPublished: 'true', type: 'wiki', title: 'The set of rational numbers is countable', clickbait: 'Although there are "lots and lots" of rational numbers, there are still only countably many of them.', textLength: '1504', alias: 'rationals_are_countable', externalUrl: '', sortChildrenBy: 'likes', hasVote: 'false', voteType: '', votesAnonymous: 'false', editCreatorId: 'PatrickStevens', editCreatedAt: '2016-07-03 09:47:06', pageCreatorId: 'PatrickStevens', pageCreatedAt: '2016-07-03 09:47:07', seeDomainId: '0', editDomainId: 'AlexeiAndreev', submitToDomainId: '0', isAutosave: 'false', isSnapshot: 'false', isLiveEdit: 'true', isMinorEdit: 'false', indirectTeacher: 'false', todoCount: '0', isEditorComment: 'false', isApprovedComment: 'true', isResolved: 'false', snapshotText: '', anchorContext: '', anchorText: '', anchorOffset: '0', mergedInto: '', isDeleted: 'false', viewCount: '27', text: 'The set $\\mathbb{Q}$ of [4zq rational numbers] is countable: that is, there is a [499 bijection] between $\\mathbb{Q}$ and the set $\\mathbb{N}$ of [45h natural numbers].\n\n# Proof\n\nBy the [cantor_schroeder_bernstein_theorem Schröder-Bernstein theorem], it is enough to find an [4b7 injection] $\\mathbb{N} \\to \\mathbb{Q}$ and an injection $\\mathbb{Q} \\to \\mathbb{N}$.\n\nThe former is easy, because $\\mathbb{N}$ is a subset of $\\mathbb{Q}$ so the identity injection $n \\mapsto \\frac{n}{1}$ works.\n\nFor the latter, we may define a function $\\mathbb{Q} \\to \\mathbb{N}$ as follows.\nTake any rational in its lowest terms, as $\\frac{p}{q}$, say. %%note:That is, the [greatest_common_divisor GCD] of the numerator $p$ and denominator $q$ is $1$.%%\nAt most one of $p$ and $q$ is negative (if both are negative, we may just cancel $-1$ from the top and bottom of the fraction); by multiplying by $\\frac{-1}{-1}$ if necessary, assume without loss of generality that $q$ is positive.\nIf $p = 0$ then take $q = 1$.\n\nDefine $s$ to be $1$ if $p$ is positive, and $2$ if $p$ is negative.\n\nThen produce the natural number $2^p 3^q 5^s$.\n\nThe function $f: \\frac{p}{q} \\mapsto 2^p 3^q 5^s$ is injective, because [fundamental_theorem_of_arithmetic prime factorisations are unique] so if $f\\left(\\frac{p}{q}\\right) = f \\left(\\frac{a}{b} \\right)$ (with both fractions in their lowest terms, and $q$ positive) then $|p| = |a|, q=b$ and the sign of $p$ is equal to the sign of $a$.\nHence the two fractions were the same after all.', metaText: '', isTextLoaded: 'true', isSubscribedToDiscussion: 'false', isSubscribedToUser: 'false', isSubscribedAsMaintainer: 'false', 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