If I=[0,1] and f: I->R is defined by
f(x)= {x if x is rational, 1-x if x is irrational.
What point of I is f continuous and why?
Also, is f monotone and/or is f one-to-one on I?
Yay! Sorry I'm just excited that I managed to pull something out of my brain however what about I being one-to-one? I had to look up this definiton becaus e I couldn't remember it but it says that a 1-1 is one that assumes each value in its image exactly once. I wanna say that it is NOT one-to-one but I don't know why and I feel like I'm wrong :/
Oh no now I'm confused. So based on what you said it's making me think yes. However, when i look at the picture of the graph it cannot be that for every point and I thought that was the point of one-to-one. ...also could you possibly explain why for the first part where f is continuous at .5 I thought isaw it but i don't understand it.
If $\displaystyle f(a)=f(b)$ both are rational then $\displaystyle a=b$.
If both are irrational then $\displaystyle 1-a=1-b$ or again $\displaystyle a=b$.
If a is rational and b is irrational then that means $\displaystyle a=1-b$ or $\displaystyle b=1-a$.
But $\displaystyle 1-a$ is rational. Thus that case is impossible.
The function is one-to-one.