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I know the classic argument for countability of $\mathbb{Q}$ is the zig-zagging traversal, but could I also prove that the rationals are countable using prime factorization of the natural numbers? For instance, defining an injective function $f: \mathbb{Q} \to \mathbb{N}$ where $f(\frac{a}{b}) = 2^{a}3^{b}$. This should work, unless I'm missing something.

Adam B.
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  • Well, $\frac{-1}{2}$ for example wouldn't work, since $2^{-1}3^2\not\in\Bbb N$. Salvagable, but should be noted –  May 03 '17 at 20:12
  • You need to add something like $f(-\frac ab)=5^a7^b$ for $a,b\in \mathbb N$ – lulu May 03 '17 at 20:14
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    You're tantalizingly close to proving that the Cartesian product of a finite number of countable sets is countable. – Austin Mohr May 03 '17 at 20:16
  • See my answer to an old question to see how to create an explicit bijection between the naturals and the rationals using prime factorization: https://math.stackexchange.com/a/1067928/7933 – Thomas Andrews May 03 '17 at 20:26
  • lulu. Or we could do $f(\pm a/b) = 2^a3^b5^{1/2 \pm 1/2}$ i.e. multiply by five if negative. Or $f(|a/b|) = 2^{2|a|}3^{|b|}$ and $f(-|a/b|) = 2^{2|a| + 1}3^{|b|}$. – fleablood May 03 '17 at 20:27

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This almost works. You have to worry a little about the sign of $\frac{a}{b}$: as it stands your function doesn't work on negative rationals. But you could make some definition like $g(c\frac{a}{b})=2^a3^b5^{1+c}$, where $a$ is nonnegative, $b$ is positive, $\gcd(a,b)=1$, and $c \in \{-1,1\}$.

Micah
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Just to give a single, simple expression that handles all cases, if $q=a/b$ with $\gcd(a,b)=1$ and $b\ge1$, then

$$f(q)=2^{|a|}\cdot3^b\cdot5^{|a|+a}$$

will be an injection from $\mathbb{Q}$ to $\mathbb{N}$.

Barry Cipra
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Need to 1) specify that $\gcd(a,b) = 1$ (but we can assume that is a given)

Need to state $0 = 0/1$ (we can probably take that as a given--- oh, wait, of course we can. $\gcd(0,a) = a$ so $a/b =0; \gcd(a,b)=1$ can only but $b =1$).

Need to take negative numbers into account. We can declare a rational number has a unique representation as $\pm \frac ab$ where $a \ge 0$, $b > 0$ and $\gcd(a,b) = 1$. Then we can have $f(+a/b) = 2^a3^b$ and $f(-a/b) = 2^a3^b*5$.

If for some reason we want to keep just the two prime factors we can declare that positives and zero go to even powers of $2$ and negatives to odd powers. i.e. $f(\pm \frac ab= (-1)^n \frac ab; n = \{0, 1\}) = 2^{2a+n}3^b$.

So, yes, these are perfect well-defined injective functions into $\mathbb N$ and so $\mathbb Q$ are countable.

fleablood
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