The absolute value is a multipath function defined by:

$|x| = \begin{cases} x > 0, \quad x \\ x = 0, \quad 0 \\ x < 0, \quad -x \end{cases} $ The following section describes useful theorems arising from the definition.

1) $\forall x \in \R$, $-|x| \leq x \leq |x|$

2) $\forall x,y \in \R$, $|x| < y \leftrightarrow -y < x < y$

3) $\forall x,y \in \R$, $|xy| = |x||y|$

4) The triangle inequality: $\forall x,y \in R$, $|x + y| \leq |x| + |y|$.



Proof:

1) The method of proof is by cases:
$ x \geq 0$, $-|x| = -x \leq x$ (since $2x \geq 0)$ and $|x| = x \geq x$.

$ x < 0$, $-|x| = -(-x) = x \leq x$ and $ |x| = -x \geq x$, since $-x > 0$, and so $-2x \geq 0$.



2) Again the bidirectional implication is established by cases:

Assume that $|x| < y$. If $x \geq 0$, then $x < y$.

If $x < 0 $, then $ -x < y / \quad \cdot (-1) \rightarrow x > -y $.

Combining both results covers $\forall x \in \R$ and gives $-y < x < y$.



3) This is proved by the same proof by cases technique.
(Using the Lemma $\forall x \in \R$, $|-x| = |x|$).



4) Again, by cases,

a. Assume $a + b \geq 0$, then, $|a + b| = a + b \leq |a| + |b|$, using the property (1).

b. Assume $a + b < 0 $, then, $|a + b| = -(a + b) = -a-b \leq |-a| + |-b| = |a| + |b|$.

The cases cover $\forall a+b \in \R$, where $|a + b| \leq |a| + |b|$ is satisfied.