How to show this sequence is dense in [0 ,1]?

Question

The sequence frac(sqrt(n)), n = 1, 2, 3....., where frac(u) means the fractionary part of u.

Thank you

Answer 1

First, observe that we have frac(sqrt(m)) > 0 for some positive integer m when m is between 2 perfect squares. So, in such cases, m is of the form m = n^2 + k, for some positive integer n and some k in {1,2, ....2n} (note that n^2 + 2n + 1 = (n+1)^2). Let's keep n n fixed and partition [n^2, (n+1)^2] with the partition points n^2, n^2 +1, ...n^2 + 2n, (n +1)^2.

The numbers frac(n^2 + k) = sqrt(n^2 + k) - n, k =1, 2... 2n are increasing and the distances from each of them to the next are decreasing (this is consequence of the fact that f(x) = sqrt(x+1) - sqrt(x) is decreasing for x >0. Actually, f'(x) = (1/2) [1/sqrt(x+1) - 1/sqrt(x)] <0 for x > 0). So, for each n, the norm of our partition (lenght of the largest interval) is sqrt(n^2 + 1) - n

Since sqrt(n^2 + 1) - n --> 0 as n --> oo, the norm of our partition goes to 0 as n --> oo and the partitions get finer as n increases. This means that, given any r in [0,1], it's possible to find increasing values of n such that r is enclosed in one of the intervals of the partitions defined by n. In other words, we get a subsequence frac(sqrt(n_j)) such that r is enclosed in one of the intervals of the partition defined by n_k.

Since, as we have seen, the length of such intervals goes to 0, it follows frac(sqrt(n_k)) --> r. And since this holds for each r in [0,1], we conclude frac(sqrt(n)) is dense in [0, 1].

Maybe this not a completely rigorous proof, we could formalize a bit more. I leave this to you, don't have time now. But these are the basic ideas.