Proof that $\sqrt {n+1} - \sqrt n ≤ 1$ I have some difficulties proving the following statement: $\sqrt {n+1} - \sqrt n ≤ 1$
Thus far I have the following:
The root of a positive integer is always smaller than or equal to the integer, since $a^2 ≥ a$ where a is a positive integer.
Therefore $\sqrt {n+1} ≤ \sqrt n + 1$ and thus $\sqrt {n+1} - \sqrt n ≤ 1$
However I feel like I am jumping to a conclusion and I'd like to be more precise.
This form would be precise if I'd want to prove any of the following:
$\sqrt {n+1} ≤ n + 1$
or
$\sqrt n + \sqrt 1 ≤ \sqrt n + 1$
But I don't think it is precise enough for what I actually want to prove. Any ideas?
I understand the statement is true. Adding an integer $x$ to an integer $n$ and then taking the square root of that will always be less than the squre root of $n$ + $x$, I just don't know how to prove it.

$\sqrt {n+1} - \sqrt n ≤ 1$
$(\sqrt {n+1} - \sqrt n) (\sqrt {n+1} + \sqrt n)≤\sqrt {n+1} + \sqrt n $
$ 1 \leq \sqrt {n+1} + \sqrt n=f(n)$
$f(n)$ increase on [0,$\infty$]...with minimum at $n=0$ then $f(n)=1$
