Question

If $f\left(x\right)$ is defined on $[0,1]$ by the rule $f\left(x\right)=\{\begin{array}{cccc}x& ,& if& x\in Q\\ 1-x& ,& if& x\notin Q\end{array}$
Then, for all $x\in [0,1]$ prove that $f\left(f\right(x\left)\right)=x$.

Answer 1

We have,
$f\left(x\right)=\{\begin{array}{cccc}x& ,& if& x\in Q\\ 1-x& ,& if& x\notin Q\end{array}$
$\therefore f\left(f\right(x\left)\right)=\{\begin{array}{ccc}f\left(x\right)& ,& x\in Q\\ f(1-x)& ,& x\notin Q\end{array}$
$=\{\begin{array}{ccc}f\left(x\right)& ,& x\notin Q\\ 1-(1-x)& ,& x\notin Q\end{array}$ $\left[\begin{array}{c}\therefore x\in Q\Rightarrow 1-x\notin Q\\ \because f(1-x)=1-(1-x)\end{array}\right]$
$=\{\begin{array}{c}x,x\notin Q\\ x,x\notin Q\end{array}$
$=$ x for all $x\in [0,1]$.