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

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The expression $\sqrt{6}-\sqrt{2}$ is not equal to $2\sqrt{2-\sqrt{3}}$. Numerical approximations show $\sqrt{6}-\sqrt{2} \approx 1.03528$ while $2\sqrt{2-\sqrt{3}}$ approximates to $1.09638$. The initial confusion arose from a misinterpretation of the notation, as the original post incorrectly referenced $2\sqrt{2-\sqrt{3}}$ instead of $2\sqrt{2}-\sqrt{3}$. The correct simplification shows that $\sqrt{6}-\sqrt{2}$ can be expressed as $\sqrt{2}(\sqrt{3}-1)$, leading to the conclusion that it can be rewritten in terms of $2\sqrt{2-\sqrt{3}}$. Therefore, the two expressions are not 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 $2 \sqrt{2} - \sqrt{3}$.
 
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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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