Bijection between Integers and Even Integers

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Theorem

Let $\Z$ be the set of integers.

Let $2 \Z$ be the set of even integers.


Then there exists a bijection $f: \Z \to 2 \Z$ between the two.


Proof

Let $f: \Z \to 2 \Z$ be the mapping defined as:

$\forall n \in \Z: \map f n = 2 n$

Let $m, n \in \Z$ such that $\map f m = \map f n$.

\(\displaystyle \map f m\) \(=\) \(\displaystyle \map f n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 m\) \(=\) \(\displaystyle 2 n\) Definition of $f$
\(\displaystyle \leadsto \ \ \) \(\displaystyle m\) \(=\) \(\displaystyle n\) dividing both sides by $2$

Thus $f$ is an injection.


Let $n \in 2 Z$.

Then by definition:

$n = 2 r$

for some $r \in \Z$

That is:

$n = \map f r$

and so $f$ is a surjection.


So $f$ is both an injection and a surjection.

Hence by definition, $f$ is a bijection.

$\blacksquare$