Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm interested in finding the range of possible values for the $n$th root of a natural number $n \in \mathbb{N}$. Right now, my intuition is telling me that $\forall n \in \mathbb{N}, 1 \leq \sqrt[n]{n} < 2$.

Do you guys think this is correct and if so how would you go about proving it? I guess you could prove that 1 is the infinum and 2 is an upper bound of a set containing all $n$th roots?

Thanks in advance for your help.

share|cite|improve this question
Hint: $\sqrt[n]{n}=e^{\ln n/n}$. – 1015 Feb 2 '13 at 16:49
Can you prove it using just the field, positivity, and completeness axioms for R? – user39898 Feb 2 '13 at 16:50
I see, you are not allowed to study the function $\ln(x)/x$ and you are not even allowed to talk about $\exp$ and $\ln$. – 1015 Feb 2 '13 at 16:53
True, Avatar. Could you show that 2 is an upper bound, then? And yes, that's basically correct, Julian... – user39898 Feb 2 '13 at 16:54
In fact, the supremeum is $e^{(1/e)}\approx 1.445$, so there is a lot of margin on the top end. – Ross Millikan Feb 2 '13 at 16:59
up vote 1 down vote accepted

Fix $n\geq 1$.

By basic axioms of $\mathbb{R}$, you can prove there is a unique $y>0$ such that $y^n=n$.

Note that the function $y\longmapsto y^n$ is increasing.

Prove by induction that $2^m>m$ for all $m\geq 1$. In particular, $2^n>n$.

Conclude that $y<2$.

share|cite|improve this answer
Clear enough, thanks julien – user39898 Feb 2 '13 at 17:06

As $n\in\Bbb N\implies n\geq1$

Taking $n_{th}$ root on both sides gives, $n^{\frac{1}{n}}\geq 1$

Now, using the fact that $n\lt 2^n\implies n^{\frac{1}{n}}\lt2$

Thus, $1\leq n^{\frac{1}{n}}\lt2$

There might be some tighter bounds.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.