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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 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...

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I think you mean $x \in \mathbb{R} \setminus \mathbb{Q}$, not $x \in \mathbb{R}/\mathbb{Q}$. – ShreevatsaR Jan 30 '11 at 17:30
@shreevatsaR: Yes, does it make any difference – anonymous Jan 30 '11 at 17:31
@shreevatsaR: Ok anyhow i have edited it. – anonymous Jan 30 '11 at 17:31
+1 that is a nice question! – milcak Jan 30 '11 at 17:41
possible duplicate of Decimal representation series – Aryabhata Jan 30 '11 at 20:29
up vote 10 down vote accepted

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.

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As to your edit, denseness does not follow from "since it has no smallest positive element". Any sequence of positive numbers converging to zero has this property. Not all of them are dense. – Aryabhata Jan 31 '11 at 21:29
A subgroup of $\mathbb{R}$ containing a positive sequence converging to zero is dense. – Plop Jan 31 '11 at 23:50

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?)

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