Prove that for all real numbers $a$ and $b$, $|a|\leq b$ iff $-b\leq a\leq b$.

Question

Not a duplicate of

If $a\leq b$ and $-a\leq b$, then $|a|\leq b$.

if $-a\leq b\leq a$, then $|b|\leq a$

Is my proof of $|a| \leq b \iff -b \leq a \leq b$ correct?

Prove that for all real numbers $a$ and $b$, $|a| \leq b$ iff $-b \leq a \leq b$

This is exercise $3.5.12.a$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Prove that for all real numbers $a$ and $b$, $|a|\leq b$ iff $-b\leq a\leq b$.

I am familiar with the routine proof of the above theorem but I was wondering whether we could prove the right-to-left direction of the above theorem in the following simple way:

Suppose $-b\leq a\leq b$. Since $a\leq b$ then $a\leq b$ or $-a\leq b$ and thus by definition $|a|\leq b$. Therefore if $-b\leq a\leq b$ then $|a|\leq b$. $Q.E.D.$

I am suspicious of my proof! Is it correct$?$ If not, then why$?$

Thanks for your attention.

Answer 1

When you said "Since $a\le b$ then $a\le b$ or $-a\le b$",

you should have said since $a\le b$ then $a\le b$ and since $-b\le a$ then $b\ge -a$.