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Fie \( A=\left\{ \left.\sqrt{\frac{1}{a}+\frac{1}{b}}\right|a,b\in\mathbb{N}^{*},\ a\neq b\right\} \) si \( B=\left\{ \left.\sqrt{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\right|x,\ y,\ z\in\mathbb{N}^{*},\ x>y>z\right\} \). Demonstati ca \( A \cap B \) contine o infinitate de numere rationale si o infinitate de numere irationale.

Cristina si Claudiu Andone, lista scurta, 2006

pentru \( a=42c \) si \( b=42 \cdot 6c \) obtinem
\( \sqrt{\frac{1}{a}+\frac{1}{b}}=\frac{1}{6\sqrt c} \)

pentru \( z=66c, y=66\cdot 2c, z=66\cdot 3c \) obtinem
\( \sqrt{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{1}{6\sqrt{c}} \)

deci \( \frac{1}{6\sqrt{c}} \subset A \cap B \forall c\in \mathbb{N}^{*} \)
daca c e patrat perfect gasim o infinitate de numere rationale, iar daca c nu e patrat perfect, gasim o infinitate de numere irationale.