Permutation/Examples/Addition of Constant on Integers
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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
\(\ds \forall x, y \in \Z: \, \) | \(\ds \map f x\) | \(=\) | \(\ds \map f y\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x + a\) | \(=\) | \(\ds y + a\) | Definition of $f$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) |
demonstrating that $f$ is injective.
Then:
\(\ds \forall y \in \Z: \, \) | \(\ds y\) | \(=\) | \(\ds \paren {y - a} + a\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f {y - a}\) | 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 $55$