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I've seen this image posted: https://i.imgur.com/tDSX24E.jpg

I actually agree with it - that is not to say that $\mathbb{Q}$ isn't countable.

The definition of bijection is a useful starting point, and it is easy to see that $\mathbb{Z}$ is countable. I shall denote this as $\mathbb{Z}\sim\mathbb{N}$ for simplicity.

I then take: $$f:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Q}\text{ by }\ f:(a,b)\mapsto\left\{\begin{array}{lr}\frac{a}{b} & \text{if }b\ne 0\\ 1 & \text{otherwise}\end{array}\right.$$

There are other choices (like $\mathbb{Z}\times\mathbb{N} say) but this matters not. This mapping is surjective but not injective.

From this we know now: $|\mathbb{Q}|\le|\mathbb{N}|$

I got my set theory book out to check this, and it exhibits a similar map to mine and considers it proven that $\mathbb{Q}$ is countable, this is not enough as a function that maps numbers to even and odd has a finite range. It would prove that it is at most countable though? Can I use that?

I am not sure how to go from here, I was looking for a theorem along the lines of:

If $A\subseteq B$ then $|A|\le |B|$

We would then know that $|\mathbb{Q}|\le|\mathbb{N}|$ and $|\mathbb{Q}|\ge|\mathbb{N}|$


  1. Can I do that with infinities? Are they ordered? Does $\le$ make sense?
  2. I know that exhibiting a bijection$\implies$ countability. As this is a definition it can be taken as $\iff$ (if we have a countable set, there must be a bijection with $\mathbb{N}$ - if there isn't it violates that it is countable). Thus for the statement in the image to be true we cannot have countability of $\mathbb{Q}$ ($^*$)

($^*$) - damn, that shortens the point of this post.


marked as duplicate by marty cohen, user223391, Claude Leibovici, user91500, Julian Kuelshammer Sep 9 '15 at 11:24

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  • $\begingroup$ By the way! It also just occurred to me that a bijection could exist, it'd just mean poking a lot of holes in $\mathbb{Z}\times\mathbb{Z}$ such that each quotient only came up once - this'd be a lot of work though! It also suggests that my definition, exhibiting a bijection, is wrong. It is that there must be a bijection. I do feel quite silly now. (~2mins after post) $\endgroup$ – Alec Teal Sep 8 '15 at 20:57
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    $\begingroup$ Well, $ZF$ proves that there is a bijection $\mathbb N \rightarrow \mathbb Q$ and the Theorem of Schröder-Bernstein is not only famous for its (many different) wrong name(s), but also very useful to prove this claim. $\endgroup$ – Stefan Mesken Sep 8 '15 at 21:30
  • $\begingroup$ There are plenty of explicit bijections. $\endgroup$ – André Nicolas Sep 8 '15 at 21:48

The premise of your question is wrong. A bijection between $\mathbb Q$ and $\mathbb Z$ can be explicitly specified.

You seem to agree that there is a bijection between $\mathbb N$ and $\mathbb Z$. Now enumerate $\mathbb Q$ by considering pairs $(p,q)\in\mathbb Z\times\mathbb Z\setminus\{0\}$ with increasing values of $m=|p|+|q|$. For each value of $m$, there are finitely many pairs $(p,q)$. Order them, say, lexicographically. For each pair, associate the next element of $\mathbb N$ with $p/q\in\mathbb Q$ iff this number hasn't previously been associated with an element of $\mathbb N$.

This establishes a bijection between $\mathbb Q$ and $\mathbb N$, and thus between $\mathbb Q$ and $\mathbb Z$.

  • $\begingroup$ This is inline with the comment and something I realised when typing the second question. Before I tick though, I am uncomfortable with putting a partial ordering on cardinals (but have been assured it's okay to do so) and would love to be able to complete my proof (I was missing a $A\subseteq B\implies|A|\le|B|$ theorem, or something else I could use) $\endgroup$ – Alec Teal Sep 8 '15 at 21:47
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    $\begingroup$ @AlecTeal: I don't know what you mean by "a partial ordering of the cardinals". If you want to complete your proof, I'd suggest that you post a new question, without all the stuff about there not being a bijection between $\mathbb Z$ and $\mathbb Q$, focusing on the proof you want to complete. $\endgroup$ – joriki Sep 8 '15 at 21:50

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