Prove |ab|=|a||b|

1. Feb 14, 2008

coverband

Hi does anyone know a proof for the multiplicative propery of absolute values

i.e. Prove |ab|=|a||b|

2. Feb 14, 2008

HallsofIvy

How about doing exactly what you always do with absolute values: break it into cases.

1) If $a\ge 0$ and $b\ge 0$
Then $ab\ge 0$ so |ab|= ab while |a|= a, |b|= b. ab= (a)(b) so |ab|= |a||b|.

2) If $a\ge 0$ while b< 0
The $ab\le 0$ so |ab|= -ab while |a|= a, |b|= -b. -ab= (a)(-b) so |ab|= |a||b|.

Can you do the other two cases?

3. Feb 14, 2008

coverband

Thanks halls!

4. Feb 15, 2008

danago

My book uses the following proof:

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

5. Feb 15, 2008

HallsofIvy

Well, if you want to do it the easy way!

6. Feb 15, 2008

coverband

I still find |a|=-a when a<0 weird! Surely if a = -a, |-a| = a

7. Feb 15, 2008

Big-T

When a<0, -a is positive.

8. Feb 15, 2008

HallsofIvy

Yes that's true. Because if a= -a, then a= 0!

Are you sure that's what you meant to say?

9. Feb 15, 2008

coverband

Yeah I think when you look at the graph of y=|x| it becomes clear (as mud)!