Inequality to prove:
$|a+b|\leq |a| + |b|$
Proof:
$-|a| \leq a \leq |a|$
$-|b| \leq b \leq |b|$
Add 1 and 2 together to get:
$-(|a|+|b|)\leq a+b\leq|a|+|b|$
$|a+b|\leq|a|+|b|$
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?
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 $-(a + b) = -a-b < |a| + |b|$.
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 $x+p\leq y+q$.
$\mathbb{R}$ 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”