Absolute Value Function is Even Function

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.

$\blacksquare$