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

[rollinthedeep]

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?

 

[Quinzio]

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

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

$$\pi$$ is irrational, it has infinite figures, so there's no way to make a rational x to cancel out all the decimals of $$\pi$$.

 

[henry_m]

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?

 

[Quinzio]

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?

$$1 \over \sqrt2$$

Let me say:$$f(x) = x^2$$

$$f(0) = 0$$
$$f(1) = 1$$
$$f(x) = 1/2 , x$$ is irrational

 

[rollinthedeep]

Is there a way to prove it beyond just giving a counterexample?

 

[HallsofIvy]

A counterexample is how you prove a general statement is NOT true.

 

[rollinthedeep]

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?

 

[rollinthedeep]

Nevermind, I read it wrong. Thanks so much for your help!

 

[Redbelly98]

Thread locked.