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This seems like a rather obvious fact, but I can't figure out how to prove (or disprove) it.

Suppose $a, b, \in \mathbb{R}$, and $a^2 = b^2$. If I take square roots, I get $a = |b|$ and $b = |a|$. I want to conclude that $|a| = |b|$, and this seems to be rather obviously true, but I can't seem to get it via substitution. Perhaps the solution is to consider cases and prove that $|a| - |b| \geq 0$ and $|b| - |a| \geq 0$.

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    $\begingroup$ Note: $||a||=|a|$ (idempotence) $\endgroup$ – J. W. Tanner Mar 22 '20 at 4:30
  • $\begingroup$ I am ok with that, but since taking absolute values doesn't preserve equalities, I don't know how to prove that. (Or does it? Perhaps I am thinking of inequalities.) $\endgroup$ – John P. Mar 22 '20 at 4:32
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    $\begingroup$ $|a| = \sqrt{a^2}$, so taking the square root on both sides of $a^2=b^2$ implies $|a|=|b|$. $\endgroup$ – Hayden Mar 22 '20 at 4:34
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    $\begingroup$ John: Doesn’t $a^2=b^2$ imply $\sqrt{a^2}=\sqrt{b^2}$? $\endgroup$ – J. W. Tanner Mar 22 '20 at 4:35
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    $\begingroup$ I should have realized it was that easy. Thanks. $\endgroup$ – John P. Mar 22 '20 at 4:36
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$$a^2=b^2$$

$$a^2-b^2=0$$

$$(a-b)(a+b)=0$$

$$a=b \textrm{ or } a=-b$$

In either case: $$|a|=|b|$$

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$a^2=b^2\implies\sqrt{a^2}=\sqrt{b^2}\implies|a|=|b|$

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