How is $\sqrt{6}-\sqrt{2}$ equal to $2\sqrt{2-\sqrt{3}}$?

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SUMMARY

The expression $\sqrt{6}-\sqrt{2}$ is equal to $2\sqrt{2-\sqrt{3}}$ through algebraic manipulation. The initial misunderstanding arose from the misinterpretation of the notation, where $2\sqrt{2-\sqrt{3}}$ was confused with $2\sqrt{2}-\sqrt{3}$. The correct derivation shows that $\sqrt{6}-\sqrt{2}$ simplifies to $\sqrt{2}(\sqrt{3}-1)$, which can be rewritten as $\sqrt{2(\sqrt{3}-1)^2}$, leading to the conclusion that both expressions are indeed equivalent.

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how is $\sqrt{6}-\sqrt{2}$ equal to $2\sqrt{2-\sqrt{3}}$

please explain. Thanks!
 
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It does not. $\sqrt 6 - \sqrt 2 \approx 1.03528$, and $2 \sqrt 2 - \sqrt 3 \approx 1.09638$.
 
magneto said:
It does not. $\sqrt 6 - \sqrt 2 \approx 1.03528$, and $2 \sqrt 2 - \sqrt 3 \approx 1.09638$.

Notice OP wrote $2 \sqrt{2 - \sqrt{3}}$, not $2 \sqrt{2} - \sqrt{3}$.
 
Oops. The font rendering in Safari is messing up.

Then, $\sqrt 6 - \sqrt 2 = \sqrt 2 (\sqrt 3 - 1)$. Since the number are positive, use $a = \sqrt{a^2}$. $ \sqrt 2 (\sqrt 3 - 1) = \sqrt{2 (\sqrt{3}-1)^2}$, and deduce from there $2\sqrt{2-\sqrt 3}$.
 

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