for every positive real number X prove that there exists an irrational number satisfying zero greater than greater than x
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Answer: See the answer in the attachment and mark as a brainliest
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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.
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