Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was thinking about this. My intuition is that there is a counterexample. Suppose $f:[0,1]\longrightarrow \mathbb{R}$ is continuous. Also suppose

  1. If $q\in\mathbb{Q}$, then $f(q)\in\mathbb{Q}$.
  2. $f(0)<0$
  3. $f(1)>1$

By the Intermediate Value Theorem, for any rational number $r$ with $0<r<1$ there is an $x$ with $0<x<1$ such that $f(x)=r$. Is this true of the restriction $f:\mathbb{Q}\longrightarrow\mathbb{Q}$.

share|improve this question
    
I can't figure out why enumerate won't work. –  Joe Johnson 126 Dec 8 '12 at 0:23
    
I don't get whats in the box –  Amr Dec 8 '12 at 0:23
    
@JoeJohnson126: edited. –  Sigur Dec 8 '12 at 0:29
1  
Your intuition is correct. For an ordered field, the Intermediate Value Property is equivalent to completeness. –  lhf Dec 8 '12 at 1:13
1  
@Sigur Thanks for the edit. –  Joe Johnson 126 Dec 8 '12 at 14:38

1 Answer 1

up vote 8 down vote accepted

$$ f(q) = 3 q^2 - 1 {}{}{}{}{}{}{} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.