# 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.

$\blacksquare$