n-th root 2

Number Theory Level 2

Is there an integer $n > 1$ such that $\sqrt[n]{2}$ is rational?

Yes No

1 solution

João Areias
Nov 18, 2017

Relevant wiki: Fermat's Last Theorem

Let's assume there is such a number. This means that:

$\frac{a}{b} = \sqrt[n]{2}$

$(\frac{a}{b})^n = 2$

$a^n = 2b^n$

$a^n = b^n + b^n$

Which can only be true if Fermat's last theorem was false or if $n = 2$ , since we know that $\sqrt{2}$ is irrational and that Fermat's last theorem is true, $\sqrt[n]{2}$ will never be rational.

If n=1/2, 2^(1/(1/2))= 2^2 = 4, right?

Yan Pereira - 3 years, 6 months ago

Dammit, I'm sorry, the problem is for integers n, didn't realize that mistake, I'm gonna edit it but you are absolutely right

João Areias - 3 years, 6 months ago

Thanks. I see that this problem has been edited. Those who previously answered "Yes" has been marked correct.

In the future, if you have concerns about a problem's wording/clarity/etc., you can report the problem. See how here .

Brilliant Mathematics Staff - 3 years, 6 months ago

n-th root 2

1 solution

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