calculus - fractional part of the square of natural number

How can if prove that the sequence : $$a_n\:=\left\{\sqrt{n}\right\}\left(fractional\:part\:of\:\sqrt{n}\right)\:=\:\:\sqrt{n}\:-\:\left[\sqrt{n}\right]$$
is bounded from above by 1?
So far i try induction but its nothing that the assumption can help me for the "step" of the induction so i kind of stuck here.
tnx!

*($[x]$ - the floor function of x)

Answer

The fractional part of a number is, by definition, between $0$ and $1$. This is because $[x]$, the integer part of $x$, is defined as

The largest integer $n\in\mathbb Z$ such that $n

Therefore, if $x-[x] > 1$, then $[x]+1$ is:

• smaller than $x$ (because $x-[x]>1$ can be rearanged to $x>[x]+1$)


• larger than $[x]$ (by definition, it is larger by $1$.

meaning that $[x]$ is not the largest integer satisfying $n