For any positive real number x, prove that there exist an irrational number y such that 0
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For any positive real number x, prove that there exist an irrational number y such that 0
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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 [tex]\sf y = \dfrac{x}{2}[/tex] is also an irrational number such that 0 < y < x.
★ If x is rational,
Then [tex]\sf y = \dfrac{x}{ \sqrt{2}}[/tex] is an irrational such that
[tex]\sf y = \dfrac{x}{ \sqrt{2}}\qquad\bigg\lgroup\bf as\; \sqrt{2} > 1\bigg\rgroup[/tex]
Hence, for any positive real number x, there exists an irrational number y such that 0 < y < x.
[tex]\rule{200}3[/tex]