Prove Function is Constant on [a,b]

[chief12]

Homework Statement

for function f:[a,b]--> Q is continuous on [a,b],
prove that f is constant on [a,b]

Homework Equations

The Attempt at a Solution

proof by contradiction,
suppose there is a k, such that a<k<b,
since f is continuous on [a,b], there must be a j, such that f(k)=j.

Since on any fixed interval there is an uncountable number of irrational numbers, then there must be a k, such that f(k) = j and j is not a rational number.

therefore is f:[a,b] --> Q is continuous, f must be constant on [a,b]



I was told that my suggestion is "plausible" but not a proof. Any help? thanks

 

[gb7nash]

A contradiction proof sounds like the right way to start.

So assume: f:[a,b]--> Q is continuous on [a,b] and that that f is not constant on [a,b].

So there exists c,d in [a,b] such that f(c) != f(d). (WLOG, f(c)<f(d)). Between any two rational numbers there exists an irrational number, so there exists e such that f(c) < e < f(d). Using the intermediate value theorem, what do we know?

Just realized, do you "know" intermediate value theorem yet? If not you might have to take another approach.

 

[Stephen Tashi]

The mean value theorem won't help. It isn't given that the function is differentiable.

 

[gb7nash]

The mean value theorem won't help. It isn't given that the function is differentiable.

Yeah, I fixed it. Was getting them confused.