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

`f(x)= x`


`text { is } x in Q 1-x`


` text { if } x notin Q text { for } x in Q`


`text { for } x notin Q f(x)=x`


` f(x)=1-x f circ f(x)=f[f(x)] `


`f circ f(x)=f[f(x)] =f(x) `


` =f[1-x] =x `


` =1-(1-x) `


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