Question

For any positive real number x, prove that there exist an irrational number y such that 0

Answer 1

Correct Question:-

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

AnswEr:-

★ If x is irrational,

Then $\sf y = \dfrac{x}{2}$ is also an irrational number such that 0 < y < x.

★ If x is rational,

Then $\sf y = \dfrac{x}{ \sqrt{2}}$ is an irrational such that

$\sf y = \dfrac{x}{ \sqrt{2}}\qquad\bigg\lgroup\bf as\; \sqrt{2} > 1\bigg\rgroup$

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

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