I have tried this problem five times but I keep getting stuck. I keep following the proof for $\sqrt{2}$. I know that I have to prove that the set is nonempty. Which I do by induction.
$2^1 > 1$
assume $2^n > n$ and prove $2^{n+1} > n+1$
$2^n + 2^n > n+n > n$
Then I think I have to prove that it has an upper bound. Also, I think it has something to do with prime numbers?
We need to show that if $n^{1/2}$ is a rational number, then $n^{1/2}$ is an integer.
Suppose that $n^{1/2}=\frac{a}{b}$, where $a$ and $b$ are integers. We may without loss of generality assume that $a$ and $b$ are relatively prime, and that $b\ge 1$. We show that $b=1$.
Suppose to the contrary that $b\gt 1$. Then some prime $p$ divides $b$.
From $n^{1/2}=\frac{a}{b}$, we get that $b^2n=a^2$. Since $p$ divides $b$, we conclude that $p$ must divide $a^2$, so $p$ divides $a$. This contradicts the fact that $a$ and $b$ are relatively prime.
So $n^{1/2}=\frac{a}{1}$, and therefore $n^{1/2}$ is an integer.