Q:

Is the map $\mathbb{Z}\times (\mathbb{Z}_{>0})\to \mathbb{Q}$, given by $(m,n)\mapsto \frac{m}{n}$ injective, surjective or bijective?

Is the following mapping $$\mathbb{Z}\times (\mathbb{Z}_{>0})\to \mathbb{Q},(m,n)\mapsto \frac{m}{n}$$ injective, surjective or bijective?

I have been working on this problem for a couple of hours and I think it is surjective, and here is my proof:
$\cfrac{m}{n}$ multiply the numerator and the denominator
by any positive integer is the same value
despite the fact that there are different numbers.
for instance: $\cfrac{4}{5}$ is the same as $\cfrac{8}{10}$ and $\cfrac{12}{15}$
But I think this might be a little weak and might be wrong, and I would be much grateful if anyone is able to point out the correct answer and suitable explanation for me, thank you

A:

Is it injective?  No.  To show this, find two elements of $\Bbb{Z} \times \Bbb{Z}_{>0}$ with the same image under your map.
\begin{align*}
(1,2) &\mapsto 1/2  \text{ and} \\
(2,4) &\mapsto 2/4 = 1/2  \text{, } 
\end{align*}
where we have used equality in the rationals.  Therefore, this map is not injective.  Consequently, it also is not bijective.
Is it surjective?  Let's find out.  We want to know if every element of the codomain is the image of some element of the domain.  So, let $m/n \in \Bbb{Q}$ with $n > 0$.  (Notice we don't even require that this is in lowest terms, which we could do by writing "with $\gcd(m,n) = 1$".)  Our choice of $n$ requires $m < 0$ if $m/n < 0$.  Then $m \in \Bbb{Z}$ and $n \in \Bbb{Z}_{>0}$ and also $(m,n) \mapsto m/n$, so the map is surjective.
Aside:  It is always a good idea to give maps names.  This makes it easier to refer to them, "so $f$ is a ..." rather than "so the map is a ...".

