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This comes from one of the exercises in How to Prove It by Daniel Velleman. In this proof I invoke the theorem from the previous exercise: “for all real numbers $a$ and $b$, $|a| \leq b$ iff $-b \leq a \leq b$.” I won’t include that proof here, as I am virtually certain it is valid. Anyway, here is my proof:

Theorem. For any real number $x$, $-|x| \leq x \leq |x|$.

Proof. Let $x$ be an arbitrary real number. We can of course write $x = x$, as this is a tautology. By taking the absolute value of both sides of this equation, it follows that $|x| = |x|$. Given this, by the theorem in the previous problem we can conclude that $-|x| \leq x \leq |x|$. Since $x$ was arbitrary, this is true for all real numbers $x$.

I apologize if this seems basic, but I’m really doubting this proof, and I couldn’t find anything similar in my searches.

i-cant-name
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    What is invalid? If the previous exercise is true, let $a=x$ and $b=|x|$, and then $|x|\le|x|$ if and only if $-|x|\le x \le |x|$. (It would help in your proof to say what you are letting $a$ and $b$ equal in your use of the previous exercise.) – Steve Kass Mar 14 '19 at 03:54
  • @JavaMan Did you mean to say the proof is valid? – saulspatz Mar 14 '19 at 04:03
  • Your proof is valid. However it would improve readability by a great deal if you explicitly wrote down the choices of $a$ and $b$. Now it's up to the reader to see or indeed not see your intention. – maxmilgram Mar 14 '19 at 06:18
  • @maxmilgram I'm still getting the hang of which details to include and which to leave out. But now that you point it out I see that you're absolutely correct. I should have specified my choices for $a$ and $b$. Thanks for the feedback. – i-cant-name Mar 14 '19 at 06:31

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