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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:

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

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

  1. What is the meaning of adding inequalities 1 and 2? How does adding these inequalities rely on the order axioms? Could someone break down this addition into its elementary meaning? I think adding inequalities must be some "short-cut" or "trick" what is the principle behind this?

What I don't understand is how to do this without "adding inequalities". Suppose I use the order axioms and add $-|b|$ to inequality 1. We get:

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

This gives the very left inequality of the final result, but how can we use the order axioms to get the final result?

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3 Answers 3

up vote 2 down vote accepted

We have to prove $a+b\leq |a| + |b|$ and $-(a+b)\leq |a| + |b|$.

By property of mod $a \leq |a|$. Add $b$ to both sides, so by translation invariance

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

By property of mod $b \leq |b|$. Add $|a|$ to both sides, so by translation invariance

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

So by transitivity $a + b < |a| + |b|$.

Similarly $-a \leq |a|$. Add $-b$ to both sides, so by translation invariance

$-a - b \leq |a| - b.$

Similarly $-b \leq |b|$. Add $|a|$ to both sides, so by translation invariance

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

So by transitivity $-(a + b) = -a-b < |a| + |b|$.

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Adding inequalities means this. If $x\leq y$ and $p\leq q$, then $x+p\leq y+p$ and $y+p\leq y+q$, so that $x+p\leq y+q$.

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$\mathbb{R}$ is an Ordered Field.

Then:

(1) $\forall a \forall b \forall c (a\leq b \Rightarrow a+c\leq b+c)$

If $a\leq b$ and $c\leq d$, by (1):

$a+c\leq b+c$

$c+b\leq d+b$

$\Rightarrow a+c \leq b+d$, by transitivity.

This is "adding inequalities"

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