prove that the $\sqrt{n}$ is unbounded

Question

I wanted to check if my answer to proving that the $\sqrt{n}$ is unbounded works.

If the $\sqrt{n}$ is bounded then there exists a $K$ s.t. $|\sqrt{n}|< K$, for all n.

therefore

$|\sqrt{n}| < K \Rightarrow -K < \sqrt{n} < K \Rightarrow (-K)^2 < (\sqrt{n})^2 < (K)^2 \Rightarrow K^2 < n < K^2 $

and since this is impossible, the sequence $\sqrt{n}$ is unbounded.

Answer 1

Not a valid proof.

From $$-K < \sqrt{n} < K$$

one cannot conclude that

$$K^2 < n < K^2$$

Example: $-3<1<3$ is true but $9<1<9$ is not true because $9<1$ is not true.

Answer 2

Let $K\in I\!R$ be an bound to $\sqrt{n}$. So $|\sqrt{n}|< K \forall n$.

But take $n'=ceil(K^2)+1$, where $n'$ is the smallest integer greater than $K^2$ ($K\in I\!R$, so $K^2$ is not necessarily is an integer) plus one. Then $\sqrt{n'}>\sqrt{K^2}=K$, so $K$ can't be a bound to $\sqrt{n}$.

We showed this without specifying $K$, so it is valid for any bound, and therefore no bound exists.