If f (x) is defined on [0,1] as f (x)={x if x belongs to Q and 1-x if x does not belongs to Q } where f( f (x) )=

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Solution

Q denotes a rational number
All decimal numbers are rational
so if x is a rational number then f(x) =x is also a rational number
if x is irrational
then f(x)=1-x is also an irrational
so f(f(x)) has same property of f(x) in the question

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Q. If

f (x)

is defined on

[0, 1]

by the rule

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Then, for all

x \in [0, 1]

prove that

f (f (x)) = x

Let f(x) be a function defined on [0, 1] such that $f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}$
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f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}

Then, for all

x ϵ [0, 1], f (f (x)) =

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Then, for all $x ϵ [0, 1], f (f (x)) =$

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If f (x) is defined on [0,1] as f (x)={x if x belongs to Q and 1-x if x does not belongs to Q } where f( f (x) )=

Q denotes a rational number All decimal numbers are rational so if x is a rational number then f(x) =x is also a rational number if x is irrational then f(x)=1-x is also an irrational so f(f(x)) has same property of f(x) in the question

Q denotes a rational number
All decimal numbers are rational
so if x is a rational number then f(x) =x is also a rational number
if x is irrational
then f(x)=1-x is also an irrational
so f(f(x)) has same property of f(x) in the question