Let `A={x in R :0lt=xlt=1}dot` If `f: AvecA` is defined by `f(x)={x ,ifx Q1-x ,ifx Q` then prove that `fof(x)=x` for all `x in Adot`

Explanation

To proof:

`fof(x)=x`

Given that `f:[0,1]->[0,1]` be defined by

`f(x)={ x, ` if x is rational
`1-x`, if x is irrational

For x is rational

`(fof)x=f(f(x))=f(x)=x`

For x is irrational

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