Proving a function to be constant

Halen
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The question states:
Suppose that f:R--->R is continuous and that f(x) in the set Q (f(x)eQ) for all x in the Reals (xeR)
Prove that f is constant.

How would you go about this question? Any help is appreciated!

i know we have to prove that f(x) is equal to some constant (p/q) ; q not equal to 0 for all xeR but how would you prove this?
 
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Do you know anything about topology? If so, what do you know about R? And what do you know about what f(R)? What does f(R) have that Q doesn't?
 
no, unforunately, i don't do topology..

but i can try answering the question..

Has it anything to do with f(Q) being countable?
 
I never said anything about f(Q). I said about f(R). But if you don't know some basic topology, then it'd be harder to explain. I think we can side-step it if with a certain equivalent.

Since f is continuous, it satisfies the intermediate value theorem. Why is this a problem?
 
thank you! that is great help indeed!

i have started the proof.. am i going on the right track?

suppose there exists a,b e R such that f(a), f(b) e Q.
WLOG, f(a)<f(b)
Between any two rational numbers, there is an irrational number, say e such that f(a)<e<f(b)
by IVT, there exists ce(a,b) such that f(c)=e

now the remaining part is to prove f(c)=e.. is that right?
 
Halen said:
now the remaining part is to prove f(c)=e.. is that right?

You've already proved that f(c) = e from the IVT. You're trying to reach a contradiction since this is a proof by contradiction. What's the contradiction from f(c) = e?
 
ohh.. the contradiction would be that f(c) is rational but e is not..
 
Halen said:
ohh.. the contradiction would be that f(c) is rational but e is not..

no

f(c) = e is irrational. What do we know about the function f from the hypothesis?
 
  • #10
the function f is continuous and has a domain and range in the reals.
 
  • #11
Halen said:
the function f is continuous and has a domain and range in the reals.

Read it again. You typed it in the first post! The domain is the real numbers, but what is the range?
 
  • #12
i really hope I'm right now.. :) is it f(x)eQ?
 
  • #13
Halen said:
i really hope I'm right now.. :) is it f(x)eQ?

Yes, so what's the contradiction here?
 
  • #14
so there does not exist a point such that f(c)=e because it is irrational but we are told f(x)eQ? and so f is constant
 
  • #15
Halen said:
so there does not exist a point such that f(c)=e because it is irrational but we are told f(x)eQ? and so f is constant

You got it. :approve:
 
  • #16
thank you! finally! :)
 
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