Will be happy if you guys check my proof.
I have to prove that:
$$ |x| < m \iff -m < x < m$$
so I'm going back to the basic definition of the absolute value:
$$ x < 0 \implies |x| = -x$$
$$ x > 0 \implies |x| = x$$
considering $x$ in both cases and createing to inequalities:
$$-x < m \implies x > -m $$
$$ x < m$$
now from the two inequalities above, concluding about the range of $x$ :
$$-m
Trait proof with absolute value
1
$\begingroup$
calculus
algebra-precalculus
absolute-value
1 Answers
0
Your proof is essentially correct. I would suggest just a little adjustment.
1) For $x<0$ you got $x>-m$ so you have $-m 2) For $x\ge 0$ you got $x Putting both together you get: $$-m
-
0@Ozk: is it clear? – 2017-01-16