$\displaystyle a$ and $\displaystyle b$ are real numbers.

demonstrate the following equality.

$\displaystyle |ab|=|a||b|$

I have no idea, how to prove that.

I just know $\displaystyle |a+b|\leq|a|+|b|$ so i dont know how to start.

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- Oct 16th 2011, 12:21 PMFabio010[RESOLVED]Prove equality
$\displaystyle a$ and $\displaystyle b$ are real numbers.

demonstrate the following equality.

$\displaystyle |ab|=|a||b|$

I have no idea, how to prove that.

I just know $\displaystyle |a+b|\leq|a|+|b|$ so i dont know how to start. - Oct 16th 2011, 12:31 PMPlatoRe: Prove equality
Just bite the bullet.

There are four cases:

$\displaystyle \begin{gathered} a \geqslant 0\;\& \,b \geqslant 0 \hfill \\ a \geqslant 0\;\& \,b < 0 \hfill \\ a < 0\;\& \,b \geqslant 0 \hfill \\ a < 0\;\& \,b < 0 \hfill \\ \end{gathered} $

Take the last case:

$\displaystyle (-a)(-b)=ab>0$ so $\displaystyle ab=|ab|$.

But $\displaystyle -a=|a|~\&~-b=|b|$ so $\displaystyle |a||b|=|ab|$ - Oct 16th 2011, 12:52 PMFabio010Re: Prove equality
- Oct 16th 2011, 12:56 PMPlatoRe: Prove equality
- Oct 16th 2011, 01:02 PMFabio010Re: Prove equality
ok

so i can prove with all cases :/???

because if $\displaystyle -ab = |ab|$

$\displaystyle -a = |a|~\&~b = |b|~so |a||b| = |ab|$

sorry if i am being confusing. :) - Oct 17th 2011, 09:42 AMFabio010Re: Prove equality
??can i prove with all cases?

- Oct 17th 2011, 10:11 AMPlatoRe: Prove equality
- Oct 17th 2011, 10:46 AMFabio010Re: Prove equality
Sorry i was confusing.

Now i understood it completely.

Thanks for the help.