Bijective Function: f: [0,1] --> [0,1]

[CarmineCortez]

Is this function bijective ?

f: [0,1] --> [0,1] f(x) = x if x E [0,1] intersection Q
f(x) = 1-x if x E [0,1]\Q

 

[HallsofIvy]

The question then is whether it is possible to get f(x1)= f(x2) for two different values of x1, x2 in [0, 1]. If both are rational, then that says x1= x2 so that case cannot happen. If both are irrational, then 1-x1= 1-x2 which also leads to x1= x2. If x1 is rational and x2 irrational, then x1= 1- x2 or if x1 is irrational and x2 rational, 1-x1= x2.

Do you see that in both cases one of f(x1), f(x2) is rational and the other irrational?