Prove equality

**Fabio010** · October 16th 2011, 12:21 PM

$a$ 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.

**Plato** · October 16th 2011, 12:31 PM

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 $|a||b|=|ab|$

**Fabio010** · October 16th 2011, 12:52 PM

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 $ab= |ab|$ ??

**Plato** · October 16th 2011, 12:56 PM

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 $-ab=|ab|$

**Fabio010** · October 16th 2011, 01:02 PM

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.

**Fabio010** · October 17th 2011, 09:42 AM

??can i prove with all cases?

**Plato** · October 17th 2011, 10:11 AM

Originally Posted by Fabio010

??can i prove with all cases?

What does that mean?

**Fabio010** · October 17th 2011, 10:46 AM

Sorry i was confusing.
Now i understood it completely.

Thanks for the help.

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