Prove that f is a constant function

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Homework Help Overview

The problem involves proving that a continuous function \( f: \mathbb{R} \to \mathbb{R} \) is constant, given that \( f\left( r + \frac{1}{n} \right) = f(r) \) for any rational number \( r \) and natural number \( n \). The discussion centers around the implications of continuity and the behavior of the function at rational and irrational points.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss using mathematical induction for part (i) and explore the implications of continuity for part (ii). There is a suggestion to assume the negation of the desired result and consider sequences of rationals converging to a real number.

Discussion Status

Some participants have offered insights into the role of continuity in the proof, while others are questioning the assumptions and exploring the necessary theorems that may apply. The discussion is active, with multiple lines of reasoning being examined.

Contextual Notes

Participants note the importance of continuity and the density of rational numbers in the real numbers, which are crucial to the arguments being made. There is an acknowledgment of the need for a rigorous approach to the proof without assuming the conclusion.

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Homework Statement


Suppose that the function f:ℝ→ℝ is continuous on ℝ and [itex]f\left( {r + \frac{1}{n}} \right) = f(r)[/itex] for any rational number r and natural number n.
(i) Prove that for any rational r and natural numbers n and m, [itex]f\left( {r + \frac{m}{n}} \right) = f(r)[/itex].
(ii) Prove that f is a constant function




Homework Equations





The Attempt at a Solution



For part (i), I think we can just use induction on m:
Let P(m) be the statement that [itex]f\left( {r + \frac{m}{n}} \right) = f(r)[/itex], for any rational r and natural numbers n and m
The case when m=1 is given as a hypothesis, so no need to prove it.
Assume P(k) holds, i.e. [itex]f\left( {r + \frac{k}{n}} \right) = f(r)[/itex], then
[itex]f\left( {r + \frac{k+1}{n}} \right) = f\left( {r + \frac{k}{n} + \frac{1}{n}} \right) = f\left( {r + \frac{k}{n}} \right) = f(r)[/itex] => P(k+1) also holds. By principle of mathematical induction, P(m) holds.

Any hint on part (ii) is much appreciated, as I don't even know how to get started, thanks!
 
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You'll need to use continuity, so how about supposing the negative (f(r) not equal to f(0) for some real r) and introducing a sequence of rationals converging to r?
 
haruspex said:
You'll need to use continuity, so how about supposing the negative (f(r) not equal to f(0) for some real r) and introducing a sequence of rationals converging to r?

Thank you.
Oh continuity, that's exactly what our professor gave us as a hint :) But how do you know the proof of constant function involves continuity? Also, what theorem(s) are we going to use here?
 
drawar said:
Oh continuity, that's exactly what our professor gave us as a hint
It's also a given fact about the function, which is a hint in itself.
But how do you know the proof of constant function involves continuity?
Without a condition such as continuity, there doesn't need to be any relationship between the values a function takes at different points. Since the rationals are dense in the set of reals, continuity implies strong constraints on the relationship between the values at rational and irrational points.
Also, what theorem(s) are we going to use here?
Use the definition of continuity, obviously. Start by assuming the result is not true, i.e. there exists some real r for which f(r) ≠ c, where c is the value of f at all rational points. You then need to find a sequence of rationals converging to r - that's the interesting part. There may be some theorem you've been taught that helps with that, but it's not hard from first principles. Then apply the definition of continuity to this sequence and the result should drop out.
 
Well finally something pops up in my mind.
By part (i), f(x)=c for some constant c and all rationals x. Suppose f is not a constant function, then there exists a real number r s.t f(r)≠c. By the Density Theorem, there exists a sequence (r_n) of rationals converging to r. Now since f is continuous on R,
[itex]f(r) = f(\mathop {\lim }\limits_{n \to \infty } {r_n}) = \mathop {\lim }\limits_{n \to \infty } f({r_n}) = \mathop {\lim }\limits_{n \to \infty } c = c[/itex]
, which is a contradiction!. Hence f is a constant function.
 
Bingo.
 

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