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:
\(\text {(1)}: \quad\) | \(\ds x\) | \(\ge\) | \(\ds 0\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds x\) | \(<\) | \(\ds 0\) |
Suppose $x \ge 0$.
Then:
\(\ds y_1 = \map f x\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds y_2 = \map f x\) | \(=\) | \(\ds x\) |
That is:
- $y_2 = y_1 = x$
Suppose $x < 0$.
Then:
\(\ds y_1 = \map f {x_1}\) | \(=\) | \(\ds -x\) | ||||||||||||
\(\ds y_2 = \map f {x_2}\) | \(=\) | \(\ds -x\) |
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.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 4$. Relations; functional relations; mappings: Exercise $1$