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I am trying to show that if $x$ is irrational, then so is $\sqrt[n]{x}$ via contrapositive; that is, I want to assume that if $\sqrt[n]{x}$ is rational, then so is $x$ and this is how I proceeded with my attempt:

if $\sqrt[n]{x}$ is rational, then we write $\sqrt[n]{x}=\frac{u}{v}$ with $u\in\mathbb{Z^{+}}$ and $v\in\mathbb{Z}$ and $\gcd(u,v)=1$. We then have $x=\frac{u^{n}}{v^{n}}$ which implies that $u^{n}=xv^{n}$.

From this point, how can I continue, or should I abandon this method?

Bill Dubuque
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    You method is correct. Notice that $u^n \in \mathbb Z^+$ and $v^n \in \mathbb Z$. – player3236 Oct 02 '20 at 14:43
  • However, this needs proof, how can I show what you stated? should I prove that $\gcd(u^{n},v^{n})=1$? if so then how? and thank you sir. –  Oct 02 '20 at 14:44
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    It doesn't actually matter that $\gcd(u,v)=1$ and it doesn't matter that $\gcd(u^n,v^n)=1$. So long as something can be written as the ratio of two integers, regardless whether it was in simplest form or not, that is enough to say it is rational. It can sometimes be helpful to insist on simplest form for a fraction to help our argument (e.g. for contradiction by noting that such a simplest form cannot exist as there is always simpler possible) but it is not needed here. – JMoravitz Oct 02 '20 at 14:45
  • The $\gcd$ plays no role in determining whether a number is rational. To prove $u^n$ is a positive integer just note that a product of positive integers is always a positive integer. – player3236 Oct 02 '20 at 14:46
  • thank you very much for your comments, never the less, I must show that $u^{n}$ and $v^{n}$ are both integers using axioms of $Z$. –  Oct 02 '20 at 14:48
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    Because an integer times an integer is always another integer you have that $u$ an integer implies that $u\times u$ an integer. $u$ and $u^k$ an integer implies that $u^{k+1}=u^k\times u$ an integer. Together by induction you have $u$ an integer implies $u^n$ an integer for all natural $n\geq 1$ – JMoravitz Oct 02 '20 at 14:48

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Supposed that $n \in \mathbb{N}$, then you've already finished the proof: check that you reached $x = \frac{u^n}{v^n}$, which is the ratio of two integers, hence $x$ is rational.

Miguel González
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You want to show that if $x$ is irrational, then so is $x^{\frac{1}{n}}$. Well suppose $x$ is irrational and $x^{\frac{1}{n}}$ is rational.

Let $$x^{\frac{1}{n}}=\frac{u}{v}$$

Where $u$ and $v$ are coprime integers. Then, $$x=\frac{u^n}{v^n}$$

So $x$ is rational, which contradicts our claim. Thus $x^{\frac{1}{n}}$ is irrational.