Problem
ALG-B1-M06-P025 One irrational value
\(2028\) pairwise distinct nonzero irrational numbers are written around a circle. For each pair of neighboring numbers \(u\) and \(v\), the value \(\frac{uv}{u-v}\) is computed. Can exactly one of the \(2028\) obtained values be irrational?
Look at the reciprocal of \(\frac{uv}{u-v}\).
Answer: no.
For neighboring numbers \(u\neq v\), \[\left(\frac{uv}{u-v}\right)^{-1}=\frac{u-v}{uv}=\frac{1}{v}-\frac{1}{u}.\] Hence \(\frac{uv}{u-v}\) is rational if and only if the difference \(\frac{1}{v}-\frac{1}{u}\) is rational.
Assume that all values are rational except possibly the value for the pair \(a_{2028},a_1\). Then the differences \(\frac{1}{a_2}-\frac{1}{a_1}\), \(\frac{1}{a_3}-\frac{1}{a_2}\), ..., \(\frac{1}{a_{2028}}-\frac{1}{a_{2027}}\) are rational. Their sum is \(\frac{1}{a_{2028}}-\frac{1}{a_1}\), so it is rational as well.
Therefore \(\frac{a_{2028}a_1}{a_{2028}-a_1}\) is rational. This contradicts the assumption that exactly one value is irrational.
Source analysis: the objects are numbers on a cycle and rationality of a neighboring expression; the first approach is to work with the fraction directly; the hidden observation is that the reciprocal becomes a difference of reciprocals; the key ideas are substitution and a telescoping sum along the chain. The cycle size is changed and the statement is rewritten.