Absolute Value Function is Even Function
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Theorem
Let $\size {\, \cdot \,} : \R \to \R$ denote the absolute value function on $\R$:
Then $\size {\, \cdot \,}$ is an even function.
Proof
Recall the definition of the absolute value function:
- $\size x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$
Testing the $3$ cases in turn:
\(\ds x\) | \(>\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(<\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {-x}\) | \(=\) | \(\ds x\) | Definition of Absolute Value | ||||||||||
\(\ds \) | \(=\) | \(\ds \size x\) | Definition of Absolute Value |
$\Box$
\(\ds x\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {-x}\) | \(=\) | \(\ds 0\) | Definition of Absolute Value | ||||||||||
\(\ds \) | \(=\) | \(\ds \size x\) | Definition of Absolute Value |
$\Box$
\(\ds x\) | \(<\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(>\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {-x}\) | \(=\) | \(\ds -x\) | Definition of Absolute Value: as $-x > 0$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \size x\) | Definition of Absolute Value: as $x < 0$ |
$\Box$
Hence the result by definition of even function.
$\blacksquare$