Linear Function on Real Numbers is Bijection
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Theorem
Let $a, b \in \R$ be real numbers.
Let $f: \R \to \R$ be the real function defined as:
- $\forall x \in \R: \map f x = a x + b$
Then $f$ is a bijection if and only if $a \ne 0$.
Proof
Let $a \ne 0$.
Let $y = \map f x$.
\(\ds y\) | \(=\) | \(\ds \map f x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a x + b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac {y - b} a\) |
and so:
- $\forall y \in \R: \exists x \in \R; y = \map f x$
demonstrating that $f$ is surjective.
Then:
\(\ds \map f {x_1}\) | \(=\) | \(\ds \map f {x_2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a x_1 + b\) | \(=\) | \(\ds a x_2 + b\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a x_1\) | \(=\) | \(\ds a x_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1\) | \(=\) | \(\ds x_2\) | as $a \ne 0$ |
demonstrating that $f$ is injective.
Thus $f$ is a bijection by definition.
$\Box$
Let $a = 0$.
Then:
- $\forall x \in \R: \map f x = b$
Thus for example:
- $\map f 1 = \map f 2$
and $f$ is trivially not injective.
Also:
- $\forall y \in \R: y \ne b \implies \nexists x \in \R: \map f x = y$
and $f$ is equally trivially not surjective either.
Thus $f$ is not a bijection by definition.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $3$. Mappings: Exercise $4$