# How to prove $-|a|\le a\le |a|$ based on the fact that $|a|\le b ⇔ -b\le a\le b$

I tried doing this:
If $$a\ge 0$$, then it's obvious that $$-a\le |a|$$
If $$a\le 0$$, then it's obvious that $$a\le |a|$$
Combining two cases we can see, $$-|a|\le a\le |a|$$
However, I am definitely sure that this proof is a bit tacky, and also doesn't utilize the fact provided in the question. This is a rephrasing of a question in Spivak Calculus Chapter $$1$$ Question no. $$14$$(iii). How can I resolve this?

• Just put $b=|a|$ and you are done. Jun 4 '20 at 8:59
• put b = |a|. QED Jun 4 '20 at 9:00
• @KaviRamaMurthy is there anything wrong with the way I proved it? Jun 4 '20 at 9:03
• Richie: You solution is perfectly fine but it is unncesessrily long while Kavi solution is nicer, more elegant... but such results are trivial enough that dont require proofs to be honest Jun 4 '20 at 9:06
• Yes, there is something wrong. You did not use what you were supposed to use. Jun 4 '20 at 9:06