Bijection/Examples/ax+b on Real Numbers
Jump to navigation
Jump to search
Example of Bijection
Let $a, b \in \R$ such that $a \ne 0$.
Let $f_{a, b}: \R \to \R$ be the mapping defined on the set of real numbers as:
- $\forall x \in \R: \map f x = a x + b$
Then $f$ is a bijection.
Proof
Let $x_1$ and $x_2$ be real numbers.
Then:
\(\ds \map f {x_1}\) | \(=\) | \(\ds \map f {x_2}\) | by supposition | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a x_1 + b\) | \(=\) | \(\ds a x_2 + b\) | Definition of $f$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1\) | \(=\) | \(\ds x_2\) |
Hence by definition $f$ is an injection.
$\Box$
Let $y \in \R$.
Let $x = \dfrac {y - b} a$.
We have that:
- $x \in \R$
- $\map f x = y$
Hence by definition $f$ is a surjection.
$\Box$
Thus $f$ is both an injection and a surjection, and so a bijection.
$\blacksquare$