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

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$$?$$

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 $$b\ge -a$$.
• I know that but my conclusion is of the form $P$ or $Q$. I have $P$. Am I not allowed to conclude without justification that $P$ or $Q?$ – Khashayar Baghizadeh Jul 17 '20 at 12:05
• To prove $|a|\le b$, you really need $P$ and $Q$ – J. W. Tanner Jul 17 '20 at 12:09
• So why when we prove the left-to-right direction, we break it up into cases as if it is $P$ or $Q?$ – Khashayar Baghizadeh Jul 17 '20 at 12:11
• $P$ and $Q$ implies $P$ or $Q$, but not the reverse – J. W. Tanner Jul 17 '20 at 12:40