# Proving |a+b| = |a|+|b|

• Jun 5th 2010, 04:38 AM
rowe
Proving |a+b| = |a|+|b|
Hi guys, I'm doing a case by case proof of

$\displaystyle |a+b| \leq |a|+|b|$

from Spivak, and he says:

When $\displaystyle a \geq 0$ and $\displaystyle b \leq 0$, we must prove that:

$\displaystyle |a+b| \leq a-b$

I'm a bit stuck on his line of reasoning, could someone explain why we have to prove the above?
• Jun 5th 2010, 05:05 AM
Plato
Quote:

Originally Posted by rowe
I'm doing a case by case proof of
$\displaystyle |a+b| = |a|+|b|$ from Spivak, and he says

You cannot prove $\displaystyle |a+b|=|a|+|b|$ because it is not true.
So exactly what is the correct statement of this problem?
• Jun 5th 2010, 05:28 AM
rowe
Typo, sorry. Fixed original post
• Jun 5th 2010, 08:53 AM
running-gag
Quote:

Originally Posted by rowe
Hi guys, I'm doing a case by case proof of

$\displaystyle |a+b| \leq |a|+|b|$

from Spivak, and he says:

When $\displaystyle a \geq 0$ and $\displaystyle b \leq 0$, we must prove that:

$\displaystyle |a+b| \leq a-b$

I'm a bit stuck on his line of reasoning, could someone explain why we have to prove the above?

Hi

If $\displaystyle a \geq 0$ then $\displaystyle |a| = a$
and if $\displaystyle b \leq 0$ then $\displaystyle |b| = -b$

Therefore when $\displaystyle a \geq 0$ and $\displaystyle b \leq 0$, $\displaystyle |a|+|b| = a-b$