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Prove: $\sqrt[4]{4}$ is irrational

I know that $\sqrt{p}$ is irrational where $p$ is a prime number.

So $\sqrt[4]{4}=\sqrt[4]{2*2}=16*\sqrt[4]{2}=16*2^{\frac{1}{2}^\frac{1}{2}}$

What can I say about $2^{\frac{1}{2}^\frac{1}{2}}$?

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    $\begingroup$ Sorry if I say that but there are wrong things here: 1) the main issues is that $\sqrt[4]4=\sqrt[4]2\times\sqrt[4]2\neq 16\times\sqrt[4]2$ because that would imply that $\sqrt[4]2=16$ so you have confused exponentiation with roots... $\endgroup$ – Renato Faraone Oct 19 '15 at 18:39
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There are a few misunderstandings here:

1)The main issue is that $\sqrt[4]4=\sqrt[4]2\times\sqrt[4]2\neq 16\times\sqrt[4]2$ because that would imply that $\sqrt[4]2=16$ (you have confused exponentiation with roots).

2)While it is true that $\sqrt[4]2=(2^{\frac 12})^{\frac 12}$ it isn't that $\sqrt[4]2=2^{{\frac 12}^{\frac 12}}$ because $2^{{\frac 12}^{\frac 12}}=2^{\sqrt {\frac 12}}=2^{\frac {\sqrt2}2}=\sqrt{2^{\sqrt2}}$.

You should simply write:

$$\sqrt[4]4=(2^2)^{\frac 14}=2^{\frac 12}=\sqrt2$$

That is irrational (if you haven't prove that you can find a lot of demonstration on this online).

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  • $\begingroup$ Thanks, I have mixed up, I need to take the root and not the exponent as you said. for example: $\sqrt{32}=\sqrt{8*4}=2\sqrt{8}$ $\endgroup$ – gbox Oct 19 '15 at 19:33
  • $\begingroup$ Yes! My advice is to always think that as a fractional exponent like: $\sqrt{32}=2^{\frac 52}=2^{2+\frac 12}=4\times\sqrt2$ @gbox $\endgroup$ – Renato Faraone Oct 19 '15 at 19:47
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That is wrong, it is $$\sqrt[4]{4}=\sqrt[4]{2^2}=\sqrt{2}$$ so it is irrational.

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