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:

\(\displaystyle x\) \(>\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle -x\) \(<\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size {-x}\) \(=\) \(\displaystyle x\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle \size x\) Definition of Absolute Value

$\Box$


\(\displaystyle x\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle -x\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size {-x}\) \(=\) \(\displaystyle 0\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle \size x\) Definition of Absolute Value

$\Box$


\(\displaystyle x\) \(<\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle -x\) \(>\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size {-x}\) \(=\) \(\displaystyle -x\) Definition of Absolute Value: as $-x > 0$
\(\displaystyle \) \(=\) \(\displaystyle \size x\) Definition of Absolute Value: as $x < 0$

$\Box$


Hence the result by definition of even function.

$\blacksquare$