Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite

Inequality to prove:

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

Proof:

  1. $-|a| \leq a \leq |a|$

  2. $-|b| \leq b \leq |b|$

Add 1 and 2 together to get:

$-(|a|+|b|)\leq a+b\leq|a|+|b|$

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

The reason I'm asking is because this looks like the simplest proof of all proofs I've seen but it's rarely used. I am wondering why more "complicated" proofs are being used. Is there something wrong with this proof?

share|improve this question
The complicated proofs are for $a,b\in R^n$. For $n=1$ everything is simple. – Fabian Jan 5 at 13:01
Well, the second-to-last line of mathematics in your question actually means that $$|a+b|\leq \left|\,|a|+|b|\,\right|\ldots$$ This is a tad short of what you actually want, but you've made almost all. – DonAntonio Jan 5 at 13:02
Yours is perfect. Now replace $a$ by $a-b$ in the last inequality to find $|a-b|\geq|a|-|b|$ – Babak S. Jan 5 at 13:02
Note that $-y\le x\le y$ implies $x\le y\land -x\le y$ and hence $|x|=\max\{x,-x\}\le y$. – Hagen von Eitzen Jan 5 at 13:03
This proof is hard to generalize by another cases, For example, $\mathbb{R}^n$ and $\mathbb{C}$. – tetori Jan 5 at 13:11

1 Answer

up vote 1 down vote

The proof is indeed simple for $x, y \in \mathbb{R}$. Things get more complicated for $\mathbb{R}^n$ where you cannot compare $x$ with $|x|$. This is also true for metric spaces in general.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.