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Does anybody know of proof that the rationals are countable using an array that starts with 1/1 2/1 1/2 3/1 2/3 3/2 1/3 etc. where you generate from p/q the next two fractions to p+q/q and p/p+q

Why are all the rationals in the array and why are they all in reduced form?

Luis Shein
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  • All the rationals are in the array so we can enumerate them. They are reduced because, for example, $\frac{1}{2}=\frac{2}{4}$. By the way, you can leave them non-reduced and show that the natural numbers are in one-to-one correspondence with a super set of the rationals. That, combined with the fact that the naturals are a proper subset of the rationals gives you that the rationals are countable. – John Douma Dec 27 '21 at 18:30
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    It's better to think of it as a binary tree than an array. – DanielV Dec 27 '21 at 18:30
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Dec 27 '21 at 18:46

2 Answers2

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why are they all in reduced form?

You only have 2 construction rules. So to construct $a/b$ you can only come from $(a-b)/b$ or $a/(b-a)$ . And since we are restricted to positive rationals, there really is only 1 place a rational can come from, depending on whether $a > b$. So try constructing something like $2/2$ or $3/9$ by working backwards and you'll see why it is impossible.

Why are all the rationals in the array?

First figure out how to construct all the rationals of the form $a/1$ and $1/b$. Then note that when you do a backwards transform like $a/b \leftarrow c/d$ (like $5/7 \leftarrow 5/2$) at least one number is made smaller, but not made into zero. That gives you a proof by descent that every positive rational is constructed.

DanielV
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Define $f:\mathbb Q^+\to \mathbb Q_{\ge 0}$ as follows: $$ f(x) = \begin{cases} x-1 & x \ge 1 \\ \frac1{(1/x)-1} & 0<x<1 \end{cases} $$

Lemma: For all $x\in \mathbb Q^+$, there exists $n\in \mathbb N$ such that $f^{(n)}(x)=1$, where $f^{(n)}$ denotes iterated composition. That is, $f^{(0)}(x)=x$, while $f^{(n+1)}(x)=f(f^{(n)}(x))$ for all $n\ge 0$.

Proof idea: Write $x=p/q$ in reduced form, and prove the statement by induction on $p+q$.

We have shown that for any rational number $x$, there is a chain $$ x\longrightarrow f^{(1)}(x) \longrightarrow f^{(2)}(x) \longrightarrow \dots \longrightarrow f^{(n)}(x)=1 \tag{$*$} $$

Note that if $p$ and $q$ are coprime positive integers, then $$ f\left(\frac{p+q}{q}\right)=\frac pq,\\ f\left(\frac{p}{p+q}\right)=\frac pq. $$ Therefore, if follow the chain in $(*)$ in reverse, you get a way to produce starting from $1/1$, and at each step moving from $f^{(k+1)}(x)=p/q$ to either $(p+q)/q$ or $p/(p+q)$, depending on whether $f^{(k)}(x)$ is greater than $1$. That is, the two maps $(p/q)\mapsto (p+q)/q$ and $(p/q)\mapsto p/(p+q)$ generate all positive rational numbers, as desired.

Mike Earnest
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