Show that if $|a-5| < \frac{1}{2}$ and $|b-8| < \frac{1}{2}$, then $|(a+b) - 13| < 1$

Question

Show that if $|a-5| < \frac{1}{2}$ and $|b-8| < \frac{1}{2}$, then $|(a+b) - 13| < 1$

It looks like you just "add" the inequalities together, but what are the rules for doing that?

For reference, this is from Rogowski Calculus ET 3e, Section 1.1 #29.

Answer 1

$|(a+b)-13|=|(a-5)+(b-8)| \le |a-5|+|b-8| < \frac{1}{2}+\frac{1}{2}=1$

(triangle inequality !)

Answer 2

Use $|x+y|\le |x|+|y|$ with $x:=a-5$ and $y:=b-8$.

Answer 3

In general if you have $| x-y | < z $ this is equivalent to saying $ -z < x-y < z $ so by applying to both $| a-5 |<\frac{1}{2} $ and $ | b-8 |<\frac{1}{2} $ we have

$$ -\frac{1}{2} < a-5 <\frac{1}{2} $$ $$ -\frac{1}{2} < b-8 <\frac{1}{2} $$

If we add these inequalities together we have

$$ -1 < a+b-13 < 1 $$

Then using the rule above in reverse we have $|a+b-13|<1$. Hope this gives some clarity