Prove peicewise function is or is not onto.

[teroenza]

Homework Statement

I am told to prove or disprove whether the following function is 1:1 and onto.

f(x)= x if x is rational
x^2 if x is not rational

The Domain and Range are both given as [0,1]

I have provided a counter example to show it is not 1:1. x=1/2 and sqrt(0.5) both map to 0.5.

I am stuck on onto. If I am to disprove it being onto, I need to find/show
there is a b in the range
such that there is no a in the range
for which (a,b) is an element of f(a). I am confused as to where to begin to formulate this.

 

[glb_lub]

I am stuck on onto. If I am to disprove it being onto, I need to find/show
there is a b in the range
such that there is no a in the range
for which (a,b) is an element of f(a). I am confused as to where to begin to formulate this.

Why are you trying to disprove it , when you can prove it.
Take any 'b' in the co-domain and show that there is an 'a' in the domain such that f(a) = b.By the way this thread does not belong to the Advanced Physics sub-forum. You should have posted it here - https://www.physicsforums.com/forumdisplay.php?f=156 , i.e in the Calculus and Beyond forum.

 

[teroenza]

Thank you.

Apologies for the mislabeling.