Question

for every positive real number X prove that there exists an irrational number satisfying zero greater than greater than x​

Answer 1

Answer: See the answer in the attachment and mark as a brainliest

Step-by-step explanation:

Answer 2

Step-by-step explanation:

If x is irrational, then y = x/2 is also an irrational

number such that 0 <y<x

If x is rational, then y = x/V2 is an irrational

number such that

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.