# 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 $x$, then the absolute value of $x$ which can be written as $|x|$ is equal to

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

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

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

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

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

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

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

If the $-x$ part is a bit confusing, consider take for example $\sqrt{(-8)^2 } = -(-8) = 8$.

From the discussion above we can conclude that $|x| = \sqrt{x^2}$.

Using this fact we are going to prove to arguments in this post.

If $a$ and $b$ are two real numbers,

(1) the absolute value of their product is equal to the product of their absolute values or $|ab| = |a||b|$

and

(2) the absolute value of their quotient is equal to the quotient of their absolute values or $\left |\displaystyle \frac{a}{b} \right | = \displaystyle \frac{|a|}{|b|}$

Proof of (1)

From above, we know that $|x| = \sqrt{x^2}$, so $|ab| = \sqrt{(ab)^2} = \sqrt{a^2b^2} = \sqrt{a^2} \sqrt{b^2}$

But from the definition above, $\sqrt{a^2} \sqrt{b^2} = |a||b|$

Therefore, $|ab| = |a||b|$.

Proof of (2) $\left |\displaystyle \frac{a}{b} \right | = \sqrt {\left ( \displaystyle \frac{a}{b} \right ) ^2} = \displaystyle \frac{\sqrt{a^2}}{\sqrt{b^2}}$

Since $\sqrt{x^2} = |x|$ $\displaystyle \displaystyle \frac{\sqrt{a^2}}{\sqrt{b^2}} =\frac{|a|}{|b|}$

Therefore, $\left |\displaystyle \frac{a}{b} \right | = \displaystyle \frac{|a|}{|b|}$

That’s all for now, see you in the next post.