Absolute Value is Many-to-One

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Theorem

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.


Proof

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:

\((1):\quad\) \(\displaystyle x\) \(\ge\) \(\displaystyle 0\)
\((2):\quad\) \(\displaystyle x\) \(<\) \(\displaystyle 0\)


Suppose $x \ge 0$.

Then:

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

That is:

$y_2 = y_1 = x$


Suppose $x < 0$.

Then:

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

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.

$\blacksquare$


Sources