Proving that the terms of the sequence $(nx-\lfloor nx \rfloor)$ is dense in $[0,1]$. I have been doing a basic math course on Real analysis...I encountered with a problem which follows as " Prove that $na \pmod1$ is dense in $(0,1)$..where $a$ is an Irrational number , $n\ge1$...
I tried to prove it using only basic principles...first of all I proved that above defined sequence is infinite..and also it is bounded...so by Bolzano-Weierstrass theorem it has a limit point in $(0,1)$..but to prove denseness I need to prove that for any given $(a,b)$ a subset of $(0,1)$ there is at least one element of the sequence...I am not getting how to figure out and link that limit point to that interval $(a,b)$..can any one help me in this..?...It would be of great help...
 A: A hint on one way to prove it (not necessarily shortest) without using much theory: what is the difference between successive terms in the sequence?
(More explicit hint: Let $\{x\} = x - \lfloor x \rfloor$ be the fractional part of $x$. Try proving that $$(n+1)x-\lfloor(n+1)x\rfloor \equiv (nx - \lfloor nx \rfloor) + \{x\} \mod 1 .$$
Can you take it from here, using the fact that $\{x\}$ is not rational?)
A: It is better than dense, it is equidistributed (in $\mathbb{R}/\mathbb{Z}$), meaning that if $f: \mathbb{R}/\mathbb{Z} \rightarrow \mathbb{C}$ is continuous (it is equivalent to give $f$ or a $1$-periodic continuous function from $\mathbb{R}$ to $\mathbb{C}$), $\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^n f(kx)= \int_0^1 f$.
This can be easily proven using the fact that the polynomials in $t \mapsto e^{2 i \pi t}$ and $t \mapsto e^{2 i \pi t}$ (so, sums of the form $\sum_{l=-n}^n \lambda_k e^{2 i \pi l t}$) are dense (for the supremum norm) in our space of $1$-periodic continuous functions (Weierstrass theorem), and it is easy to compute $\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^n e^{2 i \pi l k x} = \delta_{l,0}$.
The density of your sequence follows: if an open subset of $\mathbb{R}/\mathbb{Z}$ (or $]0,1[$) did not contain any element of the sequence, using a "test function" $f$, non-negative, not identically zero and whose support is contained in our given open subset, would contradict the equidistribution.
EDIT: if you don't you the density theorem of Weierstrass above, you could also use the fact that $x \mathbb{Z} + \mathbb{Z}$ is dense in $\mathbb{R}$ (since it has no smallest positive element, otherwise $x$ would be rational), but this way you only get density, not equidistribution.
