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  4. If f (x) is defined on [0, 1] by the rule f(x) = begincases x &: textx text is rational 1 - x &: textx text is irrational endcases then for all x ∈ R, f (f(x)) is

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Q. If f (x) is defined on [0, 1] by the rule $f(x) = \begin{cases} x & : \text{x } \text{ is rational}\\ 1 - x & : \text{x} \text{ is irrational} \end{cases}$
then for all x $\in$ R, f (f(x)) is

Relations and Functions - Part 2

Solution:

f (f (x)) = f (x) = x if x is rational
= f (1 - x) = 1 - (1 - x) = x if x is irrational.
Hence f (f (x)) = x $\forall$ x $\in$ R