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 :)

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TopNewestSuppose \( A(x) \) 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 \( [0,1) \). – Patrick Corn · 2 years, 1 month ago

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http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset – Patrick Corn · 2 years, 1 month ago

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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. – Pranshu Gaba · 2 years, 1 month ago

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