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For any positive integer x prove that there exists an irrational no. y such that 0<y<x

Posted by Vansh Girotra 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

If x is irrational, then {tex}y = \frac { x } { 2 }{/tex} is also an irrational number such that 0 < y < x.
If x is rational, then {tex}\frac { x } { \sqrt { 2 } }{/tex} is an irrational number such that {tex}\frac { x } { \sqrt { 2 } } < x{/tex} as {tex}\sqrt { 2 } > 1{/tex}.
{tex}\therefore \quad y = \frac { x } { \sqrt { 2 } }{/tex} is an irrational number such that 0 < y < x.
Hence, for any positive real number x, there exists an irrational number y such that 0 < y < x.

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CBSE > Class 10 > Mathematics