fractional part of the square of natural number

Question

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 1

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 $n<x$.

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 $n<x$, which is a contradiction.

Answer 2

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?