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