I came across this observation in an exam today, and thought that this might be useful in making certain algorithms run faster, but first I want a way to prove that this is true. How can I do this? I'm fairly new to proofs, outside of geometry and trig identities.

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    $\begingroup$ $a=\sqrt{ a^2} ,\sqrt{ab}=\sqrt a \sqrt b$ if $a\geq 0$. $\endgroup$ – kingW3 Mar 22 at 3:08
  • $\begingroup$ Do you mean $\sqrt x/n$? $\endgroup$ – Lord Shark the Unknown Mar 22 at 4:26
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    $\begingroup$ $\displaystyle\mathrm{sgn}\left(n\right)\,\sqrt{\, x\,}\,$. $\endgroup$ – Felix Marin Mar 22 at 5:45
  • $\begingroup$ @LordSharktheUnknown no. $\sqrt{20} = \sqrt{4 \cdot 5} = \sqrt{2 \cdot 2 \cdot 5} = 2\sqrt{5}$ and $2 \sqrt{\frac{20}{4}} = 2\sqrt{5}$ . Thus $\sqrt{20} = 2\sqrt{\frac{20}{4}}$. $sqrt{20} \neq \frac{\sqrt{20}}{2}$ $\endgroup$ – jstowell Mar 22 at 5:56

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