Integers are countable

Proof

We have the injection (actually bijection) $\mathbb{Z}\to \mathbb{N}$ given by

Let $m,n\in \mathbb{Z}$ such that $f(n)=f(m)$. If $m,n\ge 0$ then $2n=2m$, in which case $m=n$. If $m,n<0$, then $2|n|-1=2|m|-1$ so $|n|=|m|$ but this means $n=m$ since $m$ and $n$ have the same sign. These cases are exhaustive, since $n<0$ and $m\ge 0$ means that $f(n)$ is odd and $f(m)$ is even. Thus $f(m)\ne f(n)$, a contradiction. The same is true for $m<0$ and $n\ge 0$.