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.
 A: write $$\sqrt{n+1}\le 1+\sqrt{n}$$ after squaring (all summands are positive) we get
$$0\le \sqrt{n}$$ which is true.
A: Hint:
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}$$
Use $(a+b)(a-b)=(a^2-b^2)$ for the numerator.
A: Another way,
which does not use the standard
$a^2-b^2
= (a-b)(a+b)$:
Since
$(\sqrt{n}+1/2)^2
= n+\sqrt{n}+\frac14
\gt n+1
$
for $n \ge 1$,
$\sqrt{n+1}
\lt \sqrt{n}+\frac12$.
A: Another method:
Show that $\sqrt{x+1}-\sqrt{x} \le 1$ for all non-negative real numbers ... which of course includes all the non-negative integers.
One way to do this is to show that the function $f(x)=\sqrt{x+1}-\sqrt{x}$ equals $1$ for $x=0$ and that for all $x$: $f'(x) <0$, i.e. $f(x)$ is a monotonically decreasing function.
A: Not exactly number-theoretic, but for $n=0$ the inequality is obvious, while for $n\ge1$ we have
$$\sqrt{n+1}-\sqrt n=\int_n^{n+1}{2dx\over\sqrt x}\le\int_n^{n+1}{dx\over2\sqrt n}={1\over2\sqrt n}\le1$$
A: 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 $t \in (n,n+1)$.
A: $\sqrt {n+1} - \sqrt n ≤ 1$
$(\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 $f(n)=1$
