Hi I am having trouble with a proof, was wondering if someone could give me some insight...
Here is the problem:
Suppose f:[a,b] --> Q is continuous on [a,b]. Prove that f is constant on [a,b].
I did a proof by contradiction, but realized that nowhere in my proof was there anything about Q, which Im sure is the crux of the issue.
Any insight into how to prove this using contradiction?
My contradiction was to assume that for some x,y existing in [a,b], f(x) does not equal f(y)....
I just want to make sure that the above proof by contradiction is based on the assumption that is NOT a constant function .
Then the density of irrational numbers in the set of rational numbers would provide the contradiction as needed.