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:I just asked another question in here If $(a,b)=1$ and $(\frac{a}{b})^m=n$ then show $b=1$

Which we can get If $(a,b)=1$ and $(\frac{a}{b})^m=n$ then $b=1 or -1$

Here is another question:

If n isn’t mth power of any natural number then prove $n^{\frac{1}{m}}$ is irrational.

So we use contradiction and assume :$\sqrt[m]{n}=\frac{a}{b} ,(a,b)=1$ so we want to show there is no a and b that (a,b)=1 But according to the previous question of mine we know b is 1 or -1 So (a,b) would be actually 1 And it shouldn’t be to $n^{\frac{1}{m}}$ to be irrational

Am I misunderstanding something here?

Negar Rezaei Nejad
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1 Answers1

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If $b=\pm 1$, then $n=(\pm a)^m$ and hence a $m$-th power. Contradiction

Dr. Mathva
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  • thank you so much for your answer but can you tell me what’s the flaw in my deduction? – Negar Rezaei Nejad Oct 20 '20 at 19:59
  • There is no flaw in your deduction; you've simply stopped too soon. The contradiction you are looking for is not that there exist no $a,b$ with $(alb)=1$ such that $\sqrt[m]{n}=\frac ab$. The contradiction is that this would lead to $b=\pm 1$ and, hence, $n$ would be a $m$-th power, yet we assumed that $n$ is no $m-$th power. Contradiction! – Dr. Mathva Oct 20 '20 at 20:07
  • thank you so much – Negar Rezaei Nejad Oct 20 '20 at 20:25