$|\mathbb{Z}|=|\mathbb{N}|$ 
Hello all, if you could give some critique on my proof I would be grateful.

Show that $|\mathbb{Z}|=|\mathbb{N}|$ for,
$$\displaystyle f(n) = \begin{cases} 2|n|+1, \text{if} & n\leq0 \\ 2n, \text{if} & n>0\end{cases}$$
By the definition of cardinality (def. 2.10b) we note that if a function is bijective, cardinality of the sets equal to one another. So, we want to show that $f(n)$ is bijective. To prove that $f:\mathbb{Z}\rightarrow \mathbb{N}$ injective, we must show $\forall n_1,n_2\in \mathbb{Z}$, if $f(n_1)=f(n_2)$ then $n_1=n_2$ for both conditions of $f(n)$, so,
$f(n)=2|n|+1 $, if $ n\leq 0$
$2|n_1|+1 = 2|n_2|+1$
$2|n_1|=2|n_2|$
$|n_1|=|n_2|$
and
$f(n)=2n$, if $n>0$
$2n_1=2n_2$
$n_1=n_2$
Thus we see that $f(n)$ is clearly injective. To prove $f:\mathbb{Z}\rightarrow \mathbb{N}$ is surjective, meaning $\exists n\in \mathbb{Z},\forall r\in \mathbb{N}$ s.t. $f(n)=r$, again for both conditions,
$f(n)=2|n|+1 $, if $ n\leq 0$ 
$r=2|n|+1$
$r-1=2|n|$ 
$|n|=\frac{r-1}{2}$
and 
$f(n)=2n$, if $n>0$
$r=2n$
$n=\frac{r}{2}$
Thus we see that for both conditions $f(n)=r$, and function $f:\mathbb{Z}\rightarrow \mathbb{N}$ is surjective. Hence, we prove $f:\mathbb{Z}\rightarrow \mathbb{N}$ is bijective as it is injective and surjective. Therefore, by definition of cardinality, we say that cardinality of $\mathbb{N}$ is equal to $\mathbb{Z}$. $\blacksquare$

Using common sense:


*

*send even naturals to positives,

*send odd naturals to negatives.
This is clearly reversible and establishes a bijection. Some care is required in the vicinity of $0$, but if necessary you can adjust by translation.
