Properties of the Absolute Value

[roam]

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!

 

[HallsofIvy]

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.

 

[sagardipak]

First, we have to understand that the absolute value is a function defined by:

$$|x| = \begin{cases}<br /> x & \text{if } x\geq 0 \\<br /> -x & \text{if } x<0<br /> \end{cases}$$

So,

$$|ab| = \begin{cases}<br /> ab & \text{if } ab\geq 0 \\<br /> -ab & \text{if } ab<0<br /> \end{cases}$$

Now, let's see what |a||b| is:

$$|a||b| = \begin{cases}<br /> ab & \text{if } a\geq 0 \wedge b\geq0 \\<br /> (-a)(-b) & \text{if } a\leq 0 \wedge b\leq0 \\<br /> (-a)b & \text{if } a> 0 \wedge b<0 \\<br /> a(-b) & \text{if } a<0 \wedge b<0<br /> \end{cases} \Leftrightarrow$$

$$|a||b| = \begin{cases}<br /> ab & \text{if } a\geq 0 \wedge b\geq0 \\<br /> ab & \text{if } a\leq 0 \wedge b\leq0 \\<br /> -ab & \text{if } a> 0 \wedge b<0 \\<br /> -ab & \text{if } a<0 \wedge b<0<br /> \end{cases} \Leftrightarrow$$

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,

$$|a||b| = \begin{cases}<br /> ab & \text{if } ab \geq 0 \\<br /> -ab & \text{if } ab<0<br /> \end{cases} = |ab|$$

Quod erat demonstrandum

 

[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

 

[HallsofIvy]

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
$$|xy|= r_x r_y= |x||y|$$

 

[Vincent Mazzo]

$$\Huge |ab|=\sqrt{(ab)^2}=\sqrt{a^2b^2}=\sqrt{a^2}\sqrt{b^2}=|a||b|$$

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|

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À bientôt
?;-D