Would my proof that $ \mathbb{Q} \sim\mathbb{N} $ count?

Question

So I saw that $ \mathbb{Q}\sim\mathbb{N} $ and I tried to prove it.

The official proof was a function using prime factors.

I'm learning and in doing so I try to prove each theorem and corollary or example before I read the one in the script or book. But mine was a little funky as I tried long and hard to think of something, but unfortunately couldn't come up with anything better.

Would this work:

Every number in $ \mathbb{Q}$ can be interpreted as two integers $a$ and $b$. Then we can devise a function $f: \mathbb{Q} \rightarrow \mathbb{N} $ where $a$ is the first digit and $b$ is appended.

For example $f(5/4) = 54$ or $f(1/1) = 11$.

Is such a function even allowed? Would it be injective and bijective?

Also, if this is an allowed function. Would $0/4$ and $0/1$ count as two different rational numbers or the same? ( I was struggling finding a definition for the zero numbers if they count as different).

Sorry if this is complete nonsense!

Answer 1

As you probably know by now, your $f$ is not injective (if it were a function), but most importantly, it is actually not a function as it is not well defined.

That is, as you pointed out, $0=\frac{0}{1}=\frac{0}{4}=\frac{0}{n}$. Thus, $f(0)$ could actually be any integer, so $f$ is not even a function.

If you want a explicit bijection you can have a look here. It shows a bijection from rationals to naturals, but it is easy to modify it to get integers of course.

Answer 2

I like Cantor's pairing function. You can look it up and get a formula. But it is utterly trivial from a geometric point of view. Roughly speaking, arrange the rationals in an array, and then just zig-zag your way from the upper left hand corner to put them in a list.

Answer 3

Here is another route. It is quite common when trying to prove that two sets $A$ and $B$ have the same cardinality that you can fairly easily find a function $f:A \rightarrow B$ which is an injection but not a surjection and another $g:B \rightarrow A$ which is also an injection but not a surjection.

Intuitively, this says that $|A| \le |B|$ and $|B| \le |A|$ so $|A| = |B|$.

The Schröder–Bernstein theorem deals with this case: you can deduce that the sets have the same cardinality. It is worth studying that and then you can stop once you find such a pair of injections.

Even in a fairly simple case such as yours, an explicit bijection can be rather messy.