Properties of the Absolute Value

Just wanted to say hi before I start my post!

As you may know there is a property of the absolute value that states; for $$a, b \in R$$;

$$|ab| = |a||b|$$

Well, my friend asked me if I knew a proof for this... but I don't know...
How can we prove this statement/property? I know there is a proof for the triangle inequality but for this I really don't know but I'm curious.

I'd appreciate it if anyone could show me any kind of proof or send me some links etc. Thanks!

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Gold Member Science Advisor Staff Emeritus You prove that |ab|= |a||b| the same way you prove any such elementary statement: use the definitions. The simplest definition of |x| (there are several equivalent definitions) is that |x|= x if x is positive or 0, -x if x is negative. Now break it into "cases": case 1: x and y are both positive: |x|= x and |y|= y. xy is also positive so |xy|= xy= |x||y|. case 2: x is positive while y is negative: |x|= x and |y|= -y. xy is negative so |xy|= -xy= x(-y)= |x||y|. case 3: x is negative while y is positive: |x|= -x and |y|= y. xy is negative so |xy|= -xy= (-x)y= |x||y|. case 4: x and y are both negative: |x|= -x and |y|= -y. xy is positive so |xy|= xy= (-x)(-y)= |x||y|. case 5: x= 0 and y is positive: |x|= 0 and |y|= y. xy= 0 so |xy|= 0= 0(y)= |x||y|. case 6: x= 0 and y is negative: |x|= 0 and |y|= -y. xy= 0 so |xy|= 0= (0)(-y)= |x||y|. case 7: x is positive and y is 0: |x|= x and |y|= 0. xy= 0 so |xy|= 0= x(0)= |x||y|. case 8: x is negative and y is 0: |x|= -x and |y|= 0. xy= 0 so |xy|= 0 = (-x)(0)= |x||y|. case 9: both x and y are 0: |x|= 0 and |y|= 0. xy= 0 so |xy|= 0= (0)(0)= |x||y|. There are simpler ways to prove that but I thought this would be conceptually clearest.
 First, we have to understand that the absolute value is a function defined by: $$|x| = \begin{cases} x & \text{if } x\geq 0 \\ -x & \text{if } x<0 \end{cases}$$ So, $$|ab| = \begin{cases} ab & \text{if } ab\geq 0 \\ -ab & \text{if } ab<0 \end{cases}$$ Now, let's see what |a||b| is: $$|a||b| = \begin{cases} ab & \text{if } a\geq 0 \wedge b\geq0 \\ (-a)(-b) & \text{if } a\leq 0 \wedge b\leq0 \\ (-a)b & \text{if } a> 0 \wedge b<0 \\ a(-b) & \text{if } a<0 \wedge b<0 \end{cases} \Leftrightarrow$$ $$|a||b| = \begin{cases} ab & \text{if } a\geq 0 \wedge b\geq0 \\ ab & \text{if } a\leq 0 \wedge b\leq0 \\ -ab & \text{if } a> 0 \wedge b<0 \\ -ab & \text{if } a<0 \wedge b<0 \end{cases} \Leftrightarrow$$ Notice that you have ab if a and b have the same sign and that you use -ab otherwise. Now, if a and b have the same sign, $$ab\geq0$$. If they have opposite signs (and are different than zero), $$ab<0$$. Using this, $$|a||b| = \begin{cases} ab & \text{if } ab \geq 0 \\ -ab & \text{if } ab<0 \end{cases} = |ab|$$ Quod erat demonstrandum

Properties of the Absolute Value

let a = mcis(kpi) where m => 0, k is integer
and b = ncis(hpi) where n => 0, h integer
(clearly a,b are real)

|a||b|= mn
|ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn

as required

Recognitions:
Gold Member
Staff Emeritus
 Quote by quark1005 let a = mcis(kpi) where m => 0, k is integer and b = ncis(hpi) where n => 0, h integer (clearly a,b are real) |a||b|= mn |ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn as required
This would make more sense if you had said that "mcis(hpi)" is
$$m (cos(h\pi)+ i sin(h\pi))$$
That is much more an "engineering notation" than mathematics.

If you really want to go to complex numbers, why not
if
$$x= r_xe^{i\theta_x}$$
and
$$y= r_ye^{i\theta_y}$$, then
$$|xy|= |r_xe^{i\theta_x}r_ye^{i\theta_y}|= |(r_xr_y)e^{i(\theta_x+\theta_y)}|$$

But for any $z= re^{i\theta}$, |z|= r, so
$$|xy|= r_x r_y= |x||y|$$

 $$\Huge |ab|=\sqrt{(ab)^2}=\sqrt{a^2b^2}=\sqrt{a^2}\sqrt{b^2}=|a||b|$$ and $$\large |a+b|=\sqrt{(a+b)^2}=\sqrt{a^2+2ab+b^2}\leq \sqrt{a^2+|2ab|+b^2}=\sqrt{(|a|+|b|)^2}=||a|+|b||$$ where I utilize the following fact: |2ab|=|2(ab)|=|2||ab|=2|a||b| ----- À bientôt ?;-D