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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. |
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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 $y<2$. |
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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. |
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