# Supremum and Infimum of the set of non-integer parts of the roots of naturals

First analysis course and first question here, so excuse me if I'm missing something.

The question is: what is the supremum and infimum of the set of points $${ \sqrt n - floor(\sqrt n)}$$ with n natural.

It is trivial to prove that 0 and 1 are lower and upper bounds, but I have had great difficulty showing they are the infimum and supremum.

Thank you!

Let $$f(n) = \sqrt n - \lfloor \sqrt n \rfloor$$. Consider $$\lim_{n \to \infty} f(n^2 + 1)$$ and $$\lim_{n \to \infty} f(n^2 - 1)$$
It's easy to see that the first limit is $$0$$ and the second limit is $$1$$, thus demonstrating the value of the infinum and the supremum.
To see the first limit, note that $$f(n^2 + 1) = \sqrt {n^2 + 1} - n = \frac {(\sqrt {n^2 + 1} - n)(\sqrt {n^2 + 1} + n)}{\sqrt {n^2 + 1} + n} = \frac {1}{\sqrt {n^2 + 1} + n}$$
Taking limits throughout, we easily see that the right-hand side approaches $$0$$ so the left hand side also approaches $$0$$. Using a similar proof we can easily see that $$\sqrt{n^2-1}$$ approaches $$n$$ for large $$n$$, so $$f(n^2-1)$$ approaches $$1$$.