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I have proved this for the square root of n, but not for any root of n. To prove this where k=2, I showed that:

If n is a perfect square then $\sqrt n$ is an integer and not irrational.

If $n$ is not a perfect square, assume to the contrary that $\sqrt n$ is rational and can be written as $\frac{a}{b}$.

$\sqrt n$ = $\frac{a}{b}$ where $a,b$ are integers with $GCD(a,b) = 1$

then

$ n = \frac{a^2}{b^2}$ where $b \neq 1$

But $\frac{a^2}{b^2}$ is not an integer so there is a contradiction. So $ \sqrt n \neq\frac{a}{b}$ and $\sqrt n$ must be irrational.

How would I expand this to show that it works for any value of $k$ where $k$ is the $kth$ root of $n$?

mathstudent288
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    Your claim that $\frac{a^2}{b^2}$ is not an integer is true, but it needs some justification. And then you can use essentially the same argument to show that $\frac{a^k}{b^k}$ is not an integer if $\gcd(a,b)=1$ and $b\ne 1$. – TonyK Oct 08 '20 at 23:57

1 Answers1

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In fact, the $k$-th root of $n$ satisfies the equation $x^k - n = 0$.

Therefore it is an algebraic integer. If it is also rational, then it is actually an integer, as $\Bbb Z$ is integrally closed.

The simple proof of that: if $x = \frac a b$ is rational, with $a, b$ coprime integers, then we have $a^k - n b^k = 0$, which leads to $b \mid a^k$.

But $a, b$ are coprime, so it must follow that $b = 1$, and thus $x = a$ is an integer.

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