Prove that if $n \geq 2$, then $\sqrt[n]{n}$ is irrational. Hint, show that if $n \geq 2$, then $2^{n} > n$.

So, my thought process was that I could show that $2^{n} > n$ using induction, but I'm not sure how that helps to solve the original problem. Showing $2^{n} > n$ means I could take the $n^{th}$ root of each side, giving me $2 > \sqrt[n]{n}$, but that's not quite showing that it's irrational.

My other thought was proof by contradiction, I could start assuming that it's rational -- ie that there exists some integers $r$ and $s$ such that $\sqrt[n]{n} = r/s$. Then I could say that $n = r^{n} / s^{n}$. But I have no way of showing that this is impossible for any $n \geq 2$.

Any hints as to what a good direction to start in would be appreciated.

  • 2
    $\begingroup$ What's the smallest number, greater than one, whose $n^\text{th}$ root is a positive integer? Does this place an upper bound on $\sqrt[n]{n}$? $\endgroup$ – Eric Towers Jul 6 '14 at 23:13
  • $\begingroup$ Though it isn't stated, you intend that $n$ is integral. By continuity, it is patently false otherwise. $\endgroup$ – MPW Jul 7 '14 at 0:45

Suppose there is a rational number $a/b$ such that gcd$(a,b)=1$ and $a^n/b^n=n.$ Then $a^n=b^nn.$ So $a^n$ divides $b^nn$. Since gcd$(a,b)=1 \implies $ gcd$(a^n,b^n)=1$, we know $a^n$ must divide $n$. Thus, $n \geq a^n$. But note that the integer $a$ must be at least $2$ since $(1/b)^n$ clearly cannot be $n$. Combining with the hint gives a contradiction.

  • $\begingroup$ Wait, I'm slightly confused. How did you decide that $a$ must be at least 2? Sure, $(1/b)^{n}$ cannot be $n$, that's totally okay, but I'm unsure where you're pulling the fact that one of them must be equal to $n$. $\endgroup$ – Katlyn Edwards Jul 7 '14 at 0:29
  • $\begingroup$ I am doing a proof by contradiction--I assume that $n^{1/n}$ IS rational, and show this implies a contradiction. So, by assumption $a$ and $b$ are positive integers, and $a^n/b^n=n$. This implies $a$ cannot be $1$. So it is at least $2$. I did not understand what you mean by "one of them must be equal to $n$"; could you please clarify? I didn't say $a$ or $b$ must be $n$. $\endgroup$ – StrangerLoop Jul 7 '14 at 0:43
  • $\begingroup$ Ohhh shoot, I see, I misunderstood what you were doing in the last step then. I see. Thanks a bunch! $\endgroup$ – Katlyn Edwards Jul 7 '14 at 0:48
  • $\begingroup$ You're welcome :) $\endgroup$ – StrangerLoop Jul 7 '14 at 0:49

If $$\sqrt[n]{n}=\frac{a}{b}$$then $nb^n=a^n$ and if $a$ and $b$ are in lowest terms this implies that $b=1$. So it must be an integer, now its easy to show this cannot happen.

  • $\begingroup$ How do you deduce that $b=1$? $\endgroup$ – DanZimm Jul 6 '14 at 23:30
  • $\begingroup$ Agreed, I don't think having $a$ and $b$ in lowest terms means that $b = 1$ $\endgroup$ – Katlyn Edwards Jul 6 '14 at 23:31
  • $\begingroup$ If $p|b$ then $p|a^n$ and so $p|a$, thus $b$ has no prime divisors. $\endgroup$ – Rene Schipperus Jul 6 '14 at 23:33

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