# Multiples of an irrational number forming a dense subset

Say you picked your favorite irrational number $q$ and looking at $S = \{nq: n\in \mathbb{Z} \}$ in $\mathbb{R}$, you chopped off everything but the decimal of $nq$, leaving you with a number in $[0,1]$. Is this new set dense in $[0,1]$? If so, why? (Basically looking at the $\mathbb{Z}$-orbit of a fixed irrational number in $\mathbb{R}/\mathbb{Z}$ where we mean the quotient by the group action of $\mathbb{Z}$.)

Thanks!

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Not that it matters terribly, but usually $q$ is used for rational numbers... – copper.hat Jan 8 '13 at 2:03
Here is a (an almost) duplicate of your question. – David Mitra Jan 8 '13 at 2:07
There is a short proof here; it’s a fairly simple application of the pigeonhole principle. – Brian M. Scott Jan 8 '13 at 14:09

Notation: For each real number $r$, let

• $\lfloor r \rfloor$ denote the largest integer $\leq r$ and
• $\{ r \}$ denote the fractional part of $r$.

Notice that $\{ r \} = r - \lfloor r \rfloor$. Hence, $\{ r \}$ is the ‘chopped-off decimal part’ of $r$ that you speak of.

Most proofs begin with the Pigeonhole Principle, but we can introduce a slightly topological flavor by using the Bolzano-Weierstrass Theorem. Full detail will be provided.

Let $\alpha$ be an irrational number. Then for distinct $i,j \in \mathbb{Z}$, we must have $\{ i \alpha \} \neq \{ j \alpha \}$. If this were not true, then $$i \alpha - \lfloor i \alpha \rfloor = \{ i \alpha \} = \{ j \alpha \} = j \alpha - \lfloor j \alpha \rfloor,$$ which yields the false statement $\alpha = \dfrac{\lfloor i \alpha \rfloor - \lfloor j \alpha \rfloor}{i - j} \in \mathbb{Q}$. Hence, $$S := \{ \{ i \alpha \} \mid i \in \mathbb{Z} \}$$ is an infinite subset of $[0,1]$. By the Bolzano-Weierstrass Theorem, $S$ has a limit point in $[0,1]$. One can thus find pairs of elements of $S$ that are arbitrarily close.

Now, fix an $n \in \mathbb{N}$. By the previous paragraph, there exist distinct $i,j \in \mathbb{Z}$ such that $$0 < |\{ i \alpha \} - \{ j \alpha \}| < \frac{1}{n}.$$ WLOG, it may be assumed that $0 < \{ i \alpha \} - \{ j \alpha \} < \dfrac{1}{n}$. Let $M$ be the largest positive integer such that $M (\{ i \alpha \} - \{ j \alpha \}) \leq 1$. The irrationality of $\alpha$ then yields $$(\spadesuit) \quad M (\{ i \alpha \} - \{ j \alpha \}) < 1.$$ Next, observe that for any $m \in \{ 0,\ldots,n - 1 \}$, we can find a $k \in \{ 1,\ldots,M \}$ such that $$k (\{ i \alpha \} - \{ j \alpha \}) \in \! \left[ \frac{m}{n},\frac{m + 1}{n} \right].$$ This is because

• the length of the interval $\left[ \dfrac{m}{n},\dfrac{m + 1}{n} \right]$ equals $\dfrac{1}{n}$, while
• the distance between $l (\{ i \alpha \} - \{ j \alpha \})$ and $(l + 1) (\{ i \alpha \} - \{ j \alpha \})$ is $< \dfrac{1}{n}$ for all $l \in \mathbb{N}$.

On the other hand, there is another expression for $k (\{ i \alpha \} - \{ j \alpha \})$: \begin{align} k (\{ i \alpha \} - \{ j \alpha \}) & = \{ k (\{ i \alpha \} - \{ j \alpha \}) \} \quad (\text{As $0 < k (\{ i \alpha \} - \{ j \alpha \}) < 1$; see ($\spadesuit$).}) \\ & = \{ k [(i \alpha - \lfloor i \alpha \rfloor) - (j \alpha - \lfloor j \alpha \rfloor)] \} \\ & = \{ k (i - j) \alpha + k (\lfloor j \alpha \rfloor - \lfloor i \alpha \rfloor) \} \\ & = \{ k (i - j) \alpha \}. \quad (\text{The $\{ \cdot \}$ function discards any integer part.}) \end{align} Hence, $$\{ k (i - j) \alpha \} \in \! \left[ \dfrac{m}{n},\dfrac{m + 1}{n} \right] \cap S.$$ As $n$ is arbitrary, every non-degenerate sub-interval of $[0,1]$, no matter how small, must contain an element of $S$.

(Note: A non-degenerate interval is an interval whose endpoints are not the same.)

Conclusion: $S$ is dense in $[0,1]$.

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Hint: Let $\{ z\}$ denote the fractional part of the number $z$. If $x$ is an irrational number, then for any given $n$, then there exists $1 \leq i \in \mathbb{N}$, $i \leq n+1$ such that $0 < \{ ix \} < \frac {1}{n}$

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