Question 984120
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The square root of 32 is a positive real number and thus has a square root itself.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{\sqrt{32}}\ =\ \sqrt{4\sqrt{2}}\ =\ 2\sqrt{\sqrt{2}}]


Another way to look at this is to recall that *[tex \Large \sqrt[n]{x}\ =\ x^{\frac{1}{n}].


So *[tex \Large \sqrt{2}\ =\ 2^{\frac{1}{2}}], and then *[tex \Large \sqrt{\sqrt{2}}\ =\ \left(2^{\frac{1}{2}}\right)^{\frac{1}{2}}\ =\ 2^{\frac{1}{4}}\ =\ \sqrt[4]{2}], and so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{\sqrt{32}}\ =\ \sqrt{4\sqrt{2}}\ =\ 2\sqrt{\sqrt{2}}\ =\ 2\sqrt[4]{2}]


Roughly 2.378414230005442133434999941121


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \