Proving $n > \sqrt{n}$ for every $n \geq 2$. I'm trying to prove, as a lemma for a larger result, that when $n \geq 2$, $n > \sqrt{n}$.
I first tried induction. I don't think there's a rigorous way to check the base case other than to note that $2 > \sqrt{2} \approx 1.4142 \ldots$. If this result holds for some $k \geq 2$, then because $k > \sqrt{k}$, $k + 1 > \sqrt{k} + 1$. I need to show that $k + 1 > \sqrt{k+1}$, but I can't figure out how to change the right-hand side from a sum of squar roots (with $1 = \sqrt{1}$) instead a square root of sums without assuming the conclusion.
I also tried defining $f: \mathbb{N} \to \mathbb{N}$ by $f(n) = n - \sqrt{n}$ and showed that for a general $n$, $f(n) > 0$ if and only if $n \geq 2$. The steps are not however reversible, because I cannot assert that $n - \sqrt{n} > 0$ if and only if $(n - \sqrt{n})^2 > 0$. The forward direction certain holds, but there's no guarantee that the backward direction holds: that would require assuming the conclusion.
With that, I'm stuck. Some guidance and hints would be appreciated.
 A: Suppose $n\in\mathbb{N}_{\geq 2}.\ $ Then, $\sqrt{n}^2 = n\geq 2.\ $ If $\sqrt{n}\leq 1,$ then $n = \sqrt{n}\sqrt{n}\leq 1,\ $ contradicting $n\geq 2.$ Therefore, $\sqrt{n}>1.\ $ Therefore $n=\sqrt{n}\sqrt{n} > \sqrt{n}.$
A: Here is another way to approach it for the sake of curiosity (as suggested by @JohnOmielan):
\begin{align*}
n - \sqrt{n} = \sqrt{n}(\sqrt{n} - 1) = \frac{\sqrt{n}(n - 1)}{\sqrt{n} + 1} > 0
\end{align*}
whenever $n\geq 2$.
Hopefully this helps!
A: Try proving that $n > \sqrt{n}$ if and only if $n^2 > n$.  Then the question becomes to prove that $n^2 > n$ whenever $n > 1$, which is much easier.
A: Let's assume that $\sqrt {k+1} > \sqrt{k} + 1 $
By squaring both sides, we get:
$$k + 1> k + 2\sqrt{k} + 1$$
or, $$ 0 > 2\sqrt{k}$$
This is a contradiction. Therefore, $\sqrt k +1 \ge \sqrt{k+1}$. But, if  we equate them, we will get $k=0$ which is also a  contradiction. Therefore, there will be a strict inequality. Hence,  $\sqrt k +1 > \sqrt{k+1}$
So, $$k+1 > \sqrt{k} + 1 > \sqrt{k+1}$$
