×

# Decimal parts of multiples

For a real number $$x$$, let $$f(x)$$ be a function such that $$f(x) = x - \lfloor x \rfloor$$; and let $$A(x)$$ be a set such that $$A(x) = \{f(xn) : n \in \mathbb{N} \}$$.

Prove the following:

(1) $$x$$ is rational if and only if $$A(x)$$ is finite;

(2) $$x$$ is irrational if and only if $$A(x)$$ is dense in the interval $$[0,1)$$.

Have fun :)

Note by Ariel Gershon
2 years, 4 months ago

Sort by:

Suppose $$A(x)$$ is finite. Then by the Pigeonhole Principle, there are two distinct integers $$m$$ and $$n$$ such that $$f(xm) = f(xn)$$. So $$xm-xn = k$$ is an integer. Then $$x = \frac{k}{m-n}$$ is rational.

Suppose $$A(x)$$ is dense in $$[0,1)$$. Then it is certainly infinite. So $$x$$ can't be rational, by (1).

What remains is to show that if $$x$$ is irrational, then $$A(x)$$ is dense in $$[0,1)$$.

- 2 years, 4 months ago

You can do this by the Pigeonhole Principle as well:

http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset

- 2 years, 4 months ago

We see that $$f(x) = \{x\}$$, i.e. it is the fractional part of $$x$$.

(1) $$\longrightarrow$$ In this direction we assume that $$x$$ is rational and try to prove $$A(x)$$ is finite. Since $$x$$ is rational, we can write it as $$x = \frac{p}{q}$$ where $$p$$ and $$q$$ are coprime integers and $$q$$ is not zero.

Claim: $$A(x)$$ will have exactly $$q$$ elements.

The elements in $$A(x)$$ will be $$\big \{\frac{p}{q} \big \}, \big \{\frac{2p}{q} \big \}, \big \{\frac{3p}{q} \big \}, \big \{\frac{4p}{q} \big \}, \ldots$$.

Since $$p$$ and $$q$$ are integers, these terms can be written as $$\frac{p ~\bmod~ q}{q}, \frac{2p ~\bmod ~ q}{q}, \frac{3p ~\bmod ~ q}{q}, \frac{4p ~\bmod~ q}{q}, \ldots$$. There can be only $$q$$ such terms, viz $$\frac{0}{q}, \frac{1}{q}, \frac{2}{q}, \ldots \frac{q-1}{q}$$. Since $$q$$ is a finite number, $$A(x)$$ also has a finite number of elements.

- 2 years, 4 months ago