Question

For any positive real number x, prove that there exists an irrational number y such that 0 < y < x.

Answer 1

If $x$ is irrational, then $y=\frac{x}{2}$ is also an irrational number such that $0<y<x$. If $x$ is irrational, then $\frac{\sqrt{2}}{x}$ is an irrational number such that $\frac{\sqrt{x}}{2}<x$ as $\sqrt{2}>1$.

hence , for any positive real number x, there exists an irrational number y such that $0<y<x$.