# 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:

\(\text {(1)}: \quad\) | \(\displaystyle x\) | \(\ge\) | \(\displaystyle 0\) | ||||||||||

\(\text {(2)}: \quad\) | \(\displaystyle x\) | \(<\) | \(\displaystyle 0\) |

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

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 4$. Relations; functional relations; mappings: Exercise $1$