Question

For any positive real number X, prove that there exists an irrational number y such that 0

Answer 1

that is a irrational number 0

Answer 2

For any positive number X. There exists an irrational number y such that 0 < y < x

Stepwise explanation is given below:

- If x is irrational, then y = x/2 is also an irrational number such that 0 < y < x , is greater than 0 and is less than X.

- If x is rational, then y = x/√2 is an irrational number such that it is less than X and is greater than 0.

y=x/√2 <x (as √2>1)

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