Supremum of $\sqrt{n} -\left\lfloor\sqrt{n}\right\rfloor $ I've got the following question:
$A = \{\sqrt{n} - \left\lfloor\sqrt{n}\right\rfloor : n \in \mathbb{N}\}$
Find inf A, sup A

inf A is no big deal, but I have a hard time proving sup A is 1
I've tried to prove that:
$\sqrt{n^2 - 1} -\left\lfloor\sqrt{n^2-1}\right\rfloor > \sqrt{n} -\left\lfloor\sqrt{n}\right\rfloor $
but couldn't find a way to do that
 A: For $n=m^2-1$, we have $\lfloor\sqrt{n}\rfloor=m-1$, and
$$1-\left(\sqrt{n}-\lfloor\sqrt{n}\rfloor\right)=m-\sqrt{m^2-1}=\frac1{m+\sqrt{m^2-1}}.$$
This expression has the limit $0$.
A: Obviously
$$\sqrt n-\lfloor\sqrt n\rfloor<1.$$
Then for any $\epsilon>0$, we can find $n=m^2-1^*$ such that
$$1-\epsilon<\sqrt n-\lfloor\sqrt n\rfloor=\sqrt{m^2-1}-(m-1).$$
This occurs when
$$\sqrt{m^2-1}>m-\epsilon$$ or
$$2m>\frac1\epsilon+\epsilon$$ which is always possible.
$^*$Chosen to be the closest to a perfect square.

For example, with $\epsilon=0.0001$, $m=5001>5000.00005$ yields
$$\sqrt{25010000}-\lfloor\sqrt{25010000}\rfloor\approx0.999900019995>0.9999.$$
A: We will choose to consider integers of the form $n^2-1$. Because
$$\sqrt {n^2-1}/n\to 1$$
we can make $\sqrt {n^2-1}$ as close to $n $ as we like provided we make $n $ large enough. We also know that $\sqrt {n^2-1}<n $ and we see that we can make
$$\sqrt {n^2-1}-\lfloor\sqrt {n^2-1}\rfloor$$
as close to one as we like. It is also clear that it can't be greater than one.
