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)

Just note that by definition fractional parts should always be less then $1$. 
To get an idea of why this happens just note that any real number $x$ can be written in the form $$x=I.a_1a_2\ldots$$
for example $\dfrac{1}{3}=0.3333\ldots$ and so here $I=0$ and $a_i=3$ for all $i\in \mathbb{N}$. 
Hence $\left [x\right]=I$
Therefore $$x-\left [x\right]=I.a_1a_2\ldots-I=0.a_1a_2\ldots=\{x\}$$
Can we now conclude that $0\le\{x\}<1$? When does the equality arise?
