Proving that for all $x\in\mathbb R:\; -a < x < a$ if and only if $|x| < a$.

Question

I'm in an intro to proof-writing class, studying for a final exam. I've been struggling with this problem for quite some time now and am able to prove it one way and not the other. I have so far completed the proof that if $|x| < a$, then $-a < x < a$. I wrote it as follows:

Suppose that $|x| < a$. Then, we have two cases.

For the first case, suppose that $x \geq 0$. Then by definition of absolute value, $x < a$. Because $x \geq 0$, it is the case that $a > 0$. Then $-a < 0$. Thus, $-a < x < a$ when $x > 0$.

Secondly, suppose that $x < 0$. Then by definition of absolute value, $-x < a$. This implies that $x > -a$. Then because $x < 0$, $-a < 0$. Thus $a > 0$. Therefore $-a < x < a$ when $x < 0$.

Thus, if $|x| < a$, then $-a < x < a$.

However, I cannot figure out the proof that if $-a < x < a$, then $|x| < a$. Any suggestions?

Answer 1

Just definitions and reverse

If $-a < x < a$ then... well to begin with $a > 0$.... if $a = 0$ or $a < 0$ then $-a < a$ would be impossible... so

If $-a < x < a$ then either $-a < x < 0$ or $0 \le x < a$.

If $-a < x < 0$ then $x < 0$ and $|x| = -x$ and $x = -|x|$ so $-a < -|x| < 0$ so $0 < |x| < a$.

And if $0 \le x < a$ then $x \ge 0$ and $|x| = x$ so $0 \le |x| < a$.

So either way $|x| < a$.

Answer 2

If $x>0 \Rightarrow x=|x|$

$-a<x<a$ implies $-a<|x|<a$ which means $|x|<a$

If $x<0 \Rightarrow x=-|x|$ which implies $-a<-|x|$ which means $|x|<a$

I will leave $x=0$ for the reader to prove :)

P.S. I am using as definition of absolute value of $x$ $$\begin{cases} |x|=x,\ x>=0 \\ |x|=-x,\ x<0 \\ \end{cases}$$

Answer 3

Below an attempt for --> direction with emphasis on the logical structure of the reasoning