Proof of Multiplicative Property of Absolute Values

[coverband]

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

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

 

[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?

 

[coverband]

Thanks halls!

 

[danago]

My book uses the following proof:

<br /> \left| {ab} \right| = \sqrt {(ab)^2 } = \sqrt {a^2 b^2 } = \sqrt {a^2 } \sqrt {b^2 } = \left| a \right|\left| b \right|<br />

 

[HallsofIvy]

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

 

[coverband]

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

 

[Big-T]

When a<0, -a is positive.

 

[HallsofIvy]

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

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

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

 

[coverband]

When a<0, -a is positive.

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