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I'm learning math, and found some interesting example. $$ \sqrt {2-\sqrt 3} = \frac{1}{\sqrt {2+\sqrt 3}} $$

Can you please explain me, why this happend. I guess that it depend on limits, but I still don't understand

mxxnseat
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  • multiply the LHS by $\frac{\sqrt{2+\sqrt 3}}{\sqrt{2+\sqrt 3}}$. – Surb Oct 05 '22 at 09:19
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    This uses $(a+b)(a-b)=a^2-b^2$ and is often used to remove square roots from denominators (e.g. in going from right to left in your expression) – Henry Oct 05 '22 at 09:21
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    What makes you think that this is in any way connected to limits? – José Carlos Santos Oct 05 '22 at 09:22
  • @JoséCarlosSantos Cuz, I found on wikipedia that radical from 3, can be written as $$ -1 + 2\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2} + ...}} $$ and I think that I can use limits to prove this – mxxnseat Oct 05 '22 at 09:26
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    @Surb and Henry Hah, it's was so easy, I don't think about it -_-. Thank you – mxxnseat Oct 05 '22 at 09:27
  • Note that $\sqrt{2-\sqrt3}\times\sqrt{2+\sqrt3}=1$. And $ab=1$ imply $a=\frac1b$ and $b=\frac1a$. – Etemon Oct 05 '22 at 10:04

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