# Permutation/Examples/Addition of Constant on Integers

Jump to navigation
Jump to search

## Examples of Permutations

Let $\Z$ denote the set of integers.

Let $a \in \Z$.

Let $f: \Z \to \Z$ denote the mapping defined as:

- $\forall x \in \Z: \map f x = x + a$

Then $f$ is a permutation on $\Z$.

## Proof

\(\, \displaystyle \forall x, y \in \Z: \, \) | \(\displaystyle \map f x\) | \(=\) | \(\displaystyle \map f y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x + a\) | \(=\) | \(\displaystyle y + a\) | Definition of $f$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) |

demonstrating that $f$ is injective.

Then:

\(\, \displaystyle \forall y \in \Z: \, \) | \(\displaystyle y\) | \(=\) | \(\displaystyle \paren {y - a} + a\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map f {y - a}\) | Definition of $f$ |

As $y - a \in \Z$ it follows that $f$ is a surjection

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

By definition, then, $f$ is a bijection.

As the domain and codomain of $f$ is the same, $f$ is by definition a permutation/.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations: Example $55$