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.

0) For $n=0:$ $(1)^{1/2}-0 =1$.
1) Consider $n \ne 0$.
$f(x) = x^{1/2}$ is continuos in $[n,n+1]$, differentiable in $(n,n+1)$.
Mean value Theorem:
$\dfrac{f(n+1) -f(n)}{1} = f'(t)$, 
where $t \in (n,n+1)$.
$(n+1)^{1/2} - (n)^{1/2} =$
$ (1/2)(t)^{-1/2} \lt 1$,
where $t \in (n,n+1)$.
