Supremum of $\sqrt{n} -\left\lfloor\sqrt{n}\right\rfloor $

Question

I've got the following question:

$A = \{\sqrt{n} - \left\lfloor\sqrt{n}\right\rfloor : n \in \mathbb{N}\}$

Find inf A, sup A

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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

Answer 1

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.

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For example, with $\epsilon=0.0001$, $m=5001>5000.00005$ yields

$$\sqrt{25010000}-\lfloor\sqrt{25010000}\rfloor\approx0.999900019995>0.9999.$$

Answer 2

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.

Answer 3

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 $0$.