Proving supremum is $1$: $A = \{ √ n − [ √ n] : n ∈ \mathbb{N}\}$

Question

I need to prove that the supremum of this set is $1$: $A = \{ √ n − [ √ n] : n ∈ \mathbb{N}\}$

thanks!

Answer 1

Hint: Try $n = m^2 - 1$.

Answer 2

Since $(m+1)^2 =m^2+2m+1 $, if $M = \sqrt{m^2+2m}$ then $M = m\sqrt{1+2/m} < m+1 $ and

$\begin{array}\\ M-m &=\sqrt{m^2+2m}-m\\ &=(\sqrt{m^2+2m}-m)\frac{\sqrt{m^2+2m}+m}{\sqrt{m^2+2m}+m}\\ &=\frac{m^2+2m-m^2}{\sqrt{m^2+2m}+m}\\ &=\frac{2m}{\sqrt{m^2+2m}+m}\\ &=\frac{2}{\sqrt{1+2/m}+1}\\ &\approx\frac{2}{1+1/m+1}\\ &=\frac{1}{1+1/(2m)}\\ &=1-1/(2m) \qquad\text{for large }m\\ \end{array} $

so $M-\lfloor M \rfloor \to 1 $ as $m \to \infty$.