Is it possible? If so, provide an example. If not, prove it!

  • #1

Homework Statement


A continuous function g: Q x Q --> R such that g(0,0)=0 and g(1,1)=1, but there does not exist any x,y\in Q such that g(x,y)=1/2


Homework Equations


Mean value theorem?


The Attempt at a Solution


I want to say no, because I'm sure there's something going on because the domain is not R x R....but I can't put my finger on it. Any advice?
 

Answers and Replies

  • #2
557
1
I don't know if I say something meaningful here, but first of all:
- may we first think of a [tex]\mathbb{Q} \rightarrow \mathbb{R}[/tex] application ? It looks simpler and it could provide a useful mind training.

So that, if I let [tex]g(x)=(x+ \pi) [/tex], there's no way to make [tex]g(x)= {1 \over 2}[/tex].

[tex] \pi [/tex] is irrational, it has infinite figures, so there's no way to make a rational x to cancel out all the decimals of [tex] \pi [/tex].
 
  • #3
160
2
I think the whole two-variables thing is a bit of a red herring, so to make things a bit more transparent, rephrase the question by letting g(x,y)=f((x+y)/2).

We're after f:Q-->R with f(0)=0 and f(1)=1, and no x with f(x)=1/2. We can actually do much better and have function Q-->Q. This is an excercise in showing that rational numbers are rubbish for analysis: it's a counterexample to the intermediate value theorem for Q.

Here's a clue: what's the square root of 1/2?
 
  • #4
557
1
I think the whole two-variables thing is a bit of a red herring, so to make things a bit more transparent, rephrase the question by letting g(x,y)=f((x+y)/2).

We're after f:Q-->R with f(0)=0 and f(1)=1, and no x with f(x)=1/2. We can actually do much better and have function Q-->Q. This is an excercise in showing that rational numbers are rubbish for analysis: it's a counterexample to the intermediate value theorem for Q.

Here's a clue: what's the square root of 1/2?
[tex]1 \over \sqrt2[/tex]

Let me say:[tex] f(x) = x^2[/tex]

[tex] f(0) = 0 [/tex]
[tex] f(1) = 1 [/tex]
[tex] f(x) = 1/2 , x [/tex] is irrational
 
  • #5
Is there a way to prove it beyond just giving a counterexample?
 
  • #6
HallsofIvy
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A counterexample is how you prove a general statement is NOT true.
 
  • #7
But it's not a general statement. It says to give an example if it is possible....so not for all, just there exists...don't I have to do more to prove it isn't possible at all?
 
  • #8
Nevermind, I read it wrong. Thanks so much for your help!
 
  • #9
Redbelly98
Staff Emeritus
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Thread locked.
 
Last edited:

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