# Proof that for all real numbers a, |-a| = |a|

We can think of the absolute value of a number as its distance from 0. So, the absolute value of a, which is denoted by $latex |a|$ is always greater than 0. In this post, we are going to prove that for all real numbers a, |-a| = |a|.

There are two possible cases: $latex a \geq 0$ and $latex b < 0$. (i) For $latex a \geq 0$, $latex since - a \leq 0$ $latex |-a| = -(-a) = a$ (since a is negative, we negate it to make it positive) $latex |a| = a$ (ii) For $latex a < 0$, since $latex - a > 0$,

$latex |-a|$ = $latex -a$ (since a is negative, we negate it to make it positive)
$latex |a| = -a$

By (i) and (ii), for any real number a,

$latex |-a| = |a|$.

# Absolute Value Proofs: Products and Quotients

The absolute value of a number is its distance from 0. Thus, all absolute values are either positive or 0. That is, if we have a number $latex x$, then the absolute value of $latex x$ which can be written as $latex |x|$ is equal to

(a) $latex x$ if $latex x$ is positive

(b) $latex -x$ if $latex x$ is negative

(c) $latex 0$ if $latex x$ is 0.

Also, for any real number, $latex x^2$ is positive or 0 (if $latex x = 0$).  Therefore, $latex \sqrt{x^2}$ is

(a) $latex x$ if $latex x$ is positive

(b) $latex -x$ if $latex x$ is negative

(c) $latex 0$ if $latex x$ is 0.

If the $latex -x$ part is a bit confusing, consider take for example  Continue reading