Question

Let \(x\) be an irrational number. Show that for some positive integer \(j\) not exceeding the positive integer \(n\), the absolute value of the difference between \(jx\) and the nearest integer to \(jx\) is less than \(1/n\).

Short answer

The resultant answer is that setting \(j = (a - b)\) and \(k = \left\lfloor {ax} \right\rfloor - \left\lfloor {bx} \right\rfloor \) gives us \(|jx - k| < \frac{1}{n}\).

Step-by-step solution

Given data

The given data \(x\) is an irrational number.

Concept of Pigeonhole principle

If the number of pigeons exceeds the number of pigeonholes, at least one hole will hold at least two pigeons, according to the pigeon hole principle.

Contradict and simplify the expression

Consider \(\{ x\} = x - \left\lfloor x \right\rfloor \), the fractional part of the irrational number \(x\), obviously \(0 \le \{ x\} < 1\)

Consider the \(n + 1\) numbers \(\{ ax\} \), where \(1 \le a \le n + 1\). If those irrational numbers are put into the \(n\) pigeonholes \(\left( {0,\frac{1}{n}} \right),\left( {\frac{1}{n},\frac{2}{n}} \right), \ldots ,\left( {1 - \frac{1}{n},1} \right)\), then one pigeonhole must contain at least two of them. Suppose \(\{ ax\} \) and \(\{ bx\} \) belong to the same pigeonhole where \(a > b\), which means \(|\{ ax\} - \{ bx\} | < \frac{1}{n}\) with \((a - b) \le n\). Setting \(j = (a - b)\) and \(k = \left\lfloor {ax} \right\rfloor - \left\lfloor {bx} \right\rfloor \) gives us \(|jx - k| < \frac{1}{n}\).