Proving a function to be constant

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Discussion Overview

The discussion revolves around proving that a continuous function f: R → R, which takes only rational values for all real inputs, must be constant. Participants explore various approaches to the proof, including the application of the intermediate value theorem (IVT) and the implications of the function's continuity.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the problem and seeks assistance in proving that f is constant given that f(x) is rational for all x in R.
  • Another participant questions the relationship between the topology of R and the properties of f(R) compared to Q.
  • There is a suggestion that the countability of f(Q) might be relevant to the proof.
  • Participants discuss the implications of the intermediate value theorem (IVT) and how it leads to a contradiction if f is not constant.
  • A participant outlines a proof strategy involving the selection of two rational outputs and the existence of an irrational number between them, leading to a contradiction.
  • Clarifications are made regarding the nature of the contradiction arising from the assumption that f(c) could equal an irrational number while being constrained to rational outputs.

Areas of Agreement / Disagreement

Participants generally agree on the use of the intermediate value theorem as a critical component of the proof. However, there is some uncertainty regarding the specifics of the contradiction and the proof's progression, indicating that the discussion remains somewhat unresolved.

Contextual Notes

There are limitations regarding the participants' familiarity with topology, which may affect the depth of the discussion. The proof relies on the properties of continuous functions and the nature of rational and irrational numbers, but these concepts are not fully explored by all participants.

Who May Find This Useful

This discussion may be useful for students or individuals interested in mathematical proofs, particularly those involving continuity, rationality, and the intermediate value theorem.

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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