Help explain proof that $\mathbb{Q}$ is denumerable 
A set $X$ is said to be denumerable if there is a bijection $\mathbb{Z}^+\rightarrow X$

Statement to prove:

The of rationals $\mathbb{Q}$ is denumerable.

Proof:

Define $f:\mathbb{Q} \rightarrow \mathbb{Z}\times \mathbb{Z}^+$ by $f(q)=(m,n)$ where $\frac{m}{n}$ is teh fraction representing $q$ in its lowest terms (i.e. $q=\frac{m}{n}, n>0$ and $\text{gcd}(m,n)=1$. This is clearly an injection. Its image is an infinite subset of the denumerable set $\mathbb{Z}\times\mathbb{Z}^+$ and so is denumerable. Hence, since $f$ defines a bijection onto its image, $\mathbb{Q}$ is denumerable.

Questions:


*

*I can see that $\mathbb{Z}\times \mathbb{Z}^+$ is denumerable. But from the way that $f$ is defined, I had a feeling that the image of $f$ is $\mathbb{Z}\times \mathbb{Z}^+$. I am unable to find a counterexample to show that this is false. Please help me come up with a counterexample and show that the image of $f$ is actually an infinite subset of $\mathbb{Z}\times \mathbb{Z}^+$.

*I'm abit unclear on the last sentence, "...$f$ defines a bijection onto its image". Does this mean that the bijection being referred to, say $g$, is:
$$g:\mathbb{Q}\rightarrow \text{Im}(f)$$
where:
$$g(q)=f(q)$$
and since $\text{Im}(f)$ is also denumerable, there is a bijection, $i:\text{Im}(f)\rightarrow \mathbb{Z}^+$.
Hence, we can construct a bijection $f\circ g:\mathbb{Q}\rightarrow \text{Im}(f) \rightarrow \mathbb{Z}^+$ which proves that $Q$ is denumerable too.
Is this what the text meant?

Question 1: The image of $f$ is not $\mathbb{Z} \times \mathbb{Z}^+$ because $(4,2) \in \mathbb{Z} \times \mathbb{Z}^+$ is not in the image. The image of $f$ is an infinite subset of $\mathbb{Z} \times \mathbb{Z}^+$ because the domain is infinite and $f$ is injective.
Question 2: Yes and yes.
