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Prove that rational numbers (not just positive) are countable without using axiom of choice(since it is controversial).

I have seen proofs that use the fact that union of countable sets is countable, which is proved using axiom of choice (if it is not, can you provide a proof showing that). I have also seen many proofs that showing that positive rational numbers are countable, but not both positive and negative rational numbers. I dislike the listing all the rational numbers and assigning a one-one correspondence proof as well (e.g. Cantor's proof) because it feels like cheating to me.

However, I can't find a good proof myself. Hence, I really hope that someone can provide me with a nice proof on this, nice being explicit bijection. Thanks.

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    $\begingroup$ Once we have a bijection between the positive rationals and the natural numbers, it is easy to produce a bijection between all rationals and the natural numbers. $\endgroup$ – André Nicolas Jul 11 '15 at 16:31
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You don't need the axiom of choice for the following statement:

If $X$ is countable, and $f$ is a function whose domain is $X$, then the range of $f$ is countable.

You also don't need the axiom of choice for the following statement:

$\Bbb{N\times Z}$ is countable.

Finally, define $f(n,m)=\frac nm$ or $0$ if $m=0$, and show that this is a surjection onto the rational numbers.

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Explicit bijections are rather tedious to come up with. However, the Schroeder-Bernstein theorem implies that a subset of a countable set is countable (or finite, if your definition of countable excludes finite sets) and its proof does not require AC (see https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem). This provides an alternative to the nice observations in Asaf Karaglia's answer: it is easy to identify $\mathbb{Q}$ with an (infinite) subset of $\mathbb{N} \times \mathbb{Z}$ and then use the countability of the latter. (You can reconstruct a more or less explicit bijection from the usual proofs of the S-B theoem, if you really want one.)

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Every positive rational number can be written as a finite continued fraction$^*$, and every finite continued fraction can be associated with a string over the alphabet $\Sigma=\{0,1\}$. For instance: $$ \frac{89}{13}=[6;1,5,2] \longrightarrow 11111101111100, $$ $$ \frac{101}{47}=[2;6,1,2,2]\longrightarrow 1100000010011, $$ so, by reading that string as the binary representation of an integer number, we have the existence of an injective map from $\mathbb{Q}^+$ to $\mathbb{N}^+$. So $\mathbb{Q}^+$ is countable. Moreover, it is not difficult to modify a bit the above contruction in order to get a bijective map from $\mathbb{Q}$ to $\mathbb{N}$. The key idea is just to identify any (positive) rational number as a path in the Stern-Brocot tree.

$^*$: the representation is unique if we require that the last element of the continued fraction is not one.
So the canonical representation of $\frac{89}{13}$ is $[6;1,5,2]$, not $[6;1,5,1,1]$.

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This is a proof I came up recently: First, note that to prove: $$(1)\text{ If } f: A \to B \text{ is injective and } B \text{ is countable, then } A \text{ is countable.} $$ it is sufficient to prove that for every $A \subseteq \mathbb{N}$, $A$ is countable - and you can prove it without AC. Then, from $(1)$, we can conclude that $\mathbb{N} \times \mathbb{N}$ if countable by defining the function $f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$, with $f(n, m) = 2^n 3^m $, and proving that it is injective. Hence the cartesian product of countable sets is countable.

Now, define $g: \mathbb{N} \times \mathbb{Z} \to \mathbb{Q}$ by $g(n, m) = \frac{n}{m}$. Obviously, $g$ is surjective. From the hypothesis, $\mathbb{N} \times \mathbb{Z} = A$ is countable, then there is a enumeration $\phi: \mathbb{N} \to A$.

Now, set $h = g \circ \phi$ (which is surjective) and define the following relation in $\mathbb{N}$: $$n \sim m \text{ iff } h(n) = h(m)$$ $\sim$ is an equivalence relation.

For each $X \in \mathbb{N}/\sim$, let $n_X$ be its least element and define $H: \mathbb{N}/\sim \to \mathbb{Q}$ by setting $H(X) = h(n_X)$. Then H is a bijection.

Beacause $\sim$ is an equivalence relation, the function that takes $X \in \mathbb{N}/\sim$ to $n_X$ is an injection, hence $\mathbb{N}/\sim$ is countable.

It follows that $\mathbb{Q}$ is countable. $\square$

I guess I may be too late.

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Let $S = \{q \in \mathbb Q \, | \, 0 \le q \lt 1\}$.

Let $\mathbb N = \text{ the positive integers}$.

Claim a function

$\tag 1 f: \mathbb N \to S$

can be defined using recursion that is a bijective correspondence.

There is a demonstration of the validity of this claim in the next section. But for now just assume we have such an explicit bijection $f$.

Now it is easy to see that $S \times \mathbb Z$ can be naturally put into a bijective correspondence with the set of all rational numbers $\mathbb Q$ (the coordinate $(p, m)$ corresponding to $p + m$). But since $S \times \mathbb Z$ is now countable (we have $f$ and see also Asaf Karagila's answer), the set $\mathbb Q$ is countable.

The reader is invited to edit the next section to get an explicit bijection between $\mathbb N$ and $\mathbb Q$.


The implementation of this function $f$ is described by the following computer program using Python:

#--------*---------*---------*---------*---------*---------*---------*---------*
# Desc: Implement a bijective corresponce between the integers {n | n > 0} and
#       the set of rational numbers {q | 0 <= q < 1} 
#--------*---------*---------*---------*---------*---------*---------*---------*

from fractions import Fraction
import sys

def daFunctionGraph(curCoordinate):
    nextInteger = curCoordinate[0] + 1
    p = curCoordinate[1]
    tickPrecision = curCoordinate[1].denominator
    if p + Fraction(1, tickPrecision) == 1:
        return [nextInteger, Fraction(1, tickPrecision + 1)]
    else:
        while True:
            q = p + Fraction(1, tickPrecision)
            if q.denominator == tickPrecision:
                return [nextInteger, q]
            else:
                p = p + Fraction(1, tickPrecision)


#--------*---------*---------*---------*---------*---------*---------*---------#
while True:#                       M A I N L I N E                             #
#--------*---------*---------*---------*---------*---------*---------*---------#
    current = [1, Fraction(0, 1)]
    while True:
        print(current[1], end=', ')
        if current[0] == 100:
            break
        current = daFunctionGraph(current)

    sys.exit() # END PROGRAM

The first 100 outputs from $f$ are enumerated by this program; you can use the 'slider' to see all the outputs:

0, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 3/8, 5/8, 7/8, 1/9, 2/9, 4/9, 5/9, 7/9, 8/9, 1/10, 3/10, 7/10, 9/10, 1/11, 2/11, 3/11, 4/11, 5/11, 6/11, 7/11, 8/11, 9/11, 10/11, 1/12, 5/12, 7/12, 11/12, 1/13, 2/13, 3/13, 4/13, 5/13, 6/13, 7/13, 8/13, 9/13, 10/13, 11/13, 12/13, 1/14, 3/14, 5/14, 9/14, 11/14, 13/14, 1/15, 2/15, 4/15, 7/15, 8/15, 11/15, 13/15, 14/15, 1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16, 1/17, 2/17, 3/17, 4/17, 5/17, 6/17, 7/17, 8/17, 9/17, 10/17, 11/17, 12/17, 13/17, 14/17, 15/17, 16/17, 1/18, 5/18, 7/18, 11/18, 
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