Prove equality

• Oct 16th 2011, 12:21 PM
Fabio010
[RESOLVED]Prove equality
$a$ and $b$ are real numbers.

demonstrate the following equality.
$|ab|=|a||b|$

I have no idea, how to prove that.

I just know $|a+b|\leq|a|+|b|$ so i dont know how to start.
• Oct 16th 2011, 12:31 PM
Plato
Re: Prove equality
Quote:

Originally Posted by Fabio010
$a$ and $b$ are real numbers. demonstrate the following equality.
$|ab|=|a||b|$

Just bite the bullet.
There are four cases:
$\begin{gathered} a \geqslant 0\;\& \,b \geqslant 0 \hfill \\ a \geqslant 0\;\& \,b < 0 \hfill \\ a < 0\;\& \,b \geqslant 0 \hfill \\ a < 0\;\& \,b < 0 \hfill \\ \end{gathered}$

Take the last case:
$(-a)(-b)=ab>0$ so $ab=|ab|$.
But $-a=|a|~\&~-b=|b|$ so $|a||b|=|ab|$
• Oct 16th 2011, 12:52 PM
Fabio010
Re: Prove equality
Quote:

Originally Posted by Plato
Just bite the bullet.
There are four cases:
$\begin{gathered} a \geqslant 0\;\& \,b \geqslant 0 \hfill \\ a \geqslant 0\;\& \,b < 0 \hfill \\ a < 0\;\& \,b \geqslant 0 \hfill \\ a < 0\;\& \,b < 0 \hfill \\ \end{gathered}$

Take the last case:
$(-a)(-b)=ab>0$ so $ab=|ab|$.
But $-a=|a|~\&~-b=|b|$ so $|a||b|=|ab|$

OK thanks a lot for the help
but

even if $ab<0$ isnt $ab= |ab|$ ??
• Oct 16th 2011, 12:56 PM
Plato
Re: Prove equality
Quote:

Originally Posted by Fabio010
OK thanks a lot for the help
but even if $ab<0$ isnt $ab= |ab|$ ??

No one said it was. Nowhere was that ever even mentioned.
This is true: If $ab<0$ then $-ab=|ab|$
• Oct 16th 2011, 01:02 PM
Fabio010
Re: Prove equality
ok

so i can prove with all cases :/???
because if $-ab = |ab|$
$-a = |a|~\&~b = |b|~so |a||b| = |ab|$

sorry if i am being confusing. :)
• Oct 17th 2011, 09:42 AM
Fabio010
Re: Prove equality
??can i prove with all cases?
• Oct 17th 2011, 10:11 AM
Plato
Re: Prove equality
Quote:

Originally Posted by Fabio010
??can i prove with all cases?

What does that mean?
• Oct 17th 2011, 10:46 AM
Fabio010
Re: Prove equality
Sorry i was confusing.
Now i understood it completely.

Thanks for the help.