Absolute Value is Many-to-One

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Let $f: \R \to \R$ be the absolute value function:

$\forall x \in \R: \map f x = \begin{cases}

x & : x \ge 0 \\ -x & : x < 0 \end{cases}$

Then $f$ is a many-to-one relation.


Aiming for a contradiction, suppose $f$ is not a many-to-one relation.

Then there exists $y_1 \in \R$ and $y_2 \in \R$ where $y_1 \ne y_2$ such that:

$\exists x \in \R: \map f x = y_1, \map f x = y_2$

There are two possibilities:

\(\text {(1)}: \quad\) \(\ds x\) \(\ge\) \(\ds 0\)
\(\text {(2)}: \quad\) \(\ds x\) \(<\) \(\ds 0\)

Suppose $x \ge 0$.


\(\ds y_1 = \map f x\) \(=\) \(\ds x\)
\(\ds y_2 = \map f x\) \(=\) \(\ds x\)

That is:

$y_2 = y_1 = x$

Suppose $x < 0$.


\(\ds y_1 = \map f {x_1}\) \(=\) \(\ds -x\)
\(\ds y_2 = \map f {x_2}\) \(=\) \(\ds -x\)

That is:

$y_2 = y_1 = -x$

So by Proof by Cases we have that $y_1 = y_2$ whatever $x$ may be.

This contradicts our assertion that $y_1 \ne y_2$.

Hence it follows by Proof by Contradiction that $f$ is many-to-one.