# 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.

• multiplying equation by $\sqrt{n+1}+\sqrt n$ since this expression is always positive... – Aryadeva Sep 23 '17 at 21:19

write $$\sqrt{n+1}\le 1+\sqrt{n}$$ after squaring (all summands are positive) we get $$0\le \sqrt{n}$$ which is true.

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.

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 $n \ge 1$, $\sqrt{n+1} \lt \sqrt{n}+\frac12$.

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.

Not exactly number-theoretic, but for $n=0$ the inequality is obvious, while for $n\ge1$ we have

$$\sqrt{n+1}-\sqrt n=\int_n^{n+1}{2dx\over\sqrt x}\le\int_n^{n+1}{dx\over2\sqrt n}={1\over2\sqrt n}\le1$$

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 $t \in (n,n+1)$.

$\sqrt {n+1} - \sqrt n ≤ 1$

$(\sqrt {n+1} - \sqrt n) (\sqrt {n+1} + \sqrt n)≤\sqrt {n+1} + \sqrt n$

$1 \leq \sqrt {n+1} + \sqrt n=f(n)$

$f(n)$ increase on [0,$\infty$]...with minimum at $n=0$ then $f(n)=1$