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.

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