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?

  • 8
    $\begingroup$ Just put $b=|a|$ and you are done. $\endgroup$ Jun 4 '20 at 8:59
  • 2
    $\begingroup$ put b = |a|. QED $\endgroup$
    – ILoveMath
    Jun 4 '20 at 9:00
  • $\begingroup$ @KaviRamaMurthy is there anything wrong with the way I proved it? $\endgroup$
    – Richard
    Jun 4 '20 at 9:03
  • $\begingroup$ 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 $\endgroup$
    – ILoveMath
    Jun 4 '20 at 9:06
  • 1
    $\begingroup$ Yes, there is something wrong. You did not use what you were supposed to use. $\endgroup$ Jun 4 '20 at 9:06

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