Is $\mathbb{Z}= \{\dots -3, -2, -1, 0 ,1 ,2 , 3, \dots \}$ countable? 
Question:
Is $\mathbb{Z}= \{\dots, -3, -2, -1, 0 ,1 ,2 , 3, \dots \}$ countable?

My attemp so far:
Let us create the following one-to-one correspondence between $\mathbb{Z}$ and $\mathbb{N}$.
$$\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & \dots \\ \updownarrow & \updownarrow & \updownarrow & \updownarrow & \updownarrow & \updownarrow & \updownarrow & \dots\\0 & 1 & -1 & 2 & -2 & 3 & -3 & \dots\end{matrix}$$
In order for $\mathbb{Z}$ to be countable we must define a function $f:\mathbb{N} \to \mathbb{Z}$ so that $\mathbb{Z} \sim \mathbb{N}$.
$$\displaystyle f(n) = \begin{cases} \frac{n}{2} & n \text{ is even} \\ -\frac{n-1}{2} & n\text{ is odd} \end{cases}$$
To show that $\mathbb{Z} \sim \mathbb{N}$ we require $f$ to be bijective.
From the picture above, we can clearly see that $f$ is surjective since $\forall z \in \mathbb{Z}\  \exists n \in \mathbb{N}$ such that $f(n) = z$. Hence we must now show that $f$ is injective.
We need to consider three cases:

*

*$n_1, n_2$ are odd

*$n_1, n_2$ are even

*$n_1$ is odd and $n_2$ is even

Case 1:
\begin{align}f(n_1) &= f(n_2) \\ \implies -\frac{n_1 -1}{2} &= -\frac{n_2 -1}{2} \\ \implies n_1 -1 & = n_2 -1 \\ \implies n_1 &= n_2\end{align}
Case 2:
\begin{align}f(n_1) &= f(n_2) \\ \implies \frac{n_1}{2} &= \frac{n_2}{2} \\ \implies n_1 &= n_2\end{align}
I am, however, experiencing some difficulty showing the injective property for the $3^{rd}$ case. Can anyone please give me some assistance with that specific case?
 A: The third case is fairly easy to see intuitively, given that even numbers map to positive numbers and odd numbers map to non-positive ones. However, if you want to solve it, just suppose that, for even $n_1$ and odd $n_2$:
$$\frac{n_1}2=-\frac{n_2-1}2$$
we can solve this to
$$n_1=-n_2-1$$
$$n_1+n_2=1$$
However, there can be no pair of an even and an odd natural number summing to $1$, thus if $n_1$ and $n_2$ have different parity, it cannot be that $f(n_1)=f(n_2)$.
By the way, your proof gets the right point, but it isn't quite written correctly; you should write, for instance, that
$$\begin{align}f(n_1) &= f(n_2) \\ \implies -\frac{n_1 -1}{2} &=-\frac{n_2 -1}{2} \\ \implies n_1 -1 & = n_2 -1 \\ \implies n_1 &= n_2\end{align}$$
where equality is used rather than inequality; $f(n_1)\neq f(n_2)$ implies $n_1\neq n_2$ for all functions - the inverse of the statement is what defines injectivity. (It also looks like you have a typo in your definition of $f$ and switched even and odd cases)
A: No need to check the third case cause you can see that when $n$ is odd then function is negative and when it's even it's positive so the sides can never be equal in 3rd case.By this I mean $f(n_1)\leq 0,f(n_2)>0$
