# Thread:

1. ## Intermediate Value Theorem

Hey guys,

Having a little difficulty trying to understand what this question is trying to ask me.

The questions says: "Suppose that f is continuous on [a,b] and f takes only rational values. What can be concluded about f?"

I know that the question is relating to the intermediate value theorem, as it is in that section of the text book, however I'm not sure what can be concluded from this.

My thinking so far has been this:
- f is on a closed, bounded interval, which implies a whole load of things such as continuity etc.
- I'm not sure if the intermediate value theorem works if the function only takes rational numbers, as the reals are dense with irrationals. But they are also dense with the rationals so maybe not?

I don't know how to go about the solution as I'm sure than it is more than "there exists a value c such that a<c<b and f(c) is rational"

Any idea what the question might be asking for???

Thanks in advance,

Mark

2. ## Re: Intermediate Value Theorem

Originally Posted by MarkJacob
Hey guys,

Having a little difficulty trying to understand what this question is trying to ask me.

The questions says: "Suppose that f is continuous on [a,b] and f takes only rational values. What can be concluded about f?"

I know that the question is relating to the intermediate value theorem, as it is in that section of the text book, however I'm not sure what can be concluded from this.

My thinking so far has been this:
- f is on a closed, bounded interval, which implies a whole load of things such as continuity etc.
- I'm not sure if the intermediate value theorem works if the function only takes rational numbers, as the reals are dense with irrationals. But they are also dense with the rationals so maybe not?

I don't know how to go about the solution as I'm sure than it is more than "there exists a value c such that a<c<b and f(c) is rational"

Any idea what the question might be asking for???

Thanks in advance,

Mark
How could it possibly be continuous when it only takes on rational values? There are always irrationals in between any two rational numbers...

3. ## Re: Intermediate Value Theorem

Sorry, my bad.
Silly mistake.

4. ## Re: Intermediate Value Theorem

I don't see any mistake, silly or not. Obviously, there exist an irrational number between any two rationals so if a function is continuous and takes on two different values, then it must take on all values between them and so some irrational values.

Result: if f is continuous and takes on only rational values, then f can only have one value. What kind of function has that property?

5. ## Re: Intermediate Value Theorem

Oh, this makes sense.
Are we to assume that a=/=b, as if it did would we be able to conclude anything about f at all? Or is there a part of the question that states this through definition?

And does that mean that f must be in the form f(x) = a, a is an element of the reals?
Is there a name for this kind of function?

Thanks

6. ## Re: Intermediate Value Theorem

Originally Posted by MarkJacob
Oh, this makes sense.
Are we to assume that a=/=b, as if it did would we be able to conclude anything about f at all? Or is there a part of the question that states this through definition?

And does that mean that f must be in the form f(x) = a, a is an element of the reals?
Is there a name for this kind of function?

Thanks
I think a should only be an element of the rationals. And this kind of function is called a constant function.

7. ## Re: Intermediate Value Theorem

Originally Posted by HallsofIvy
Result: if f is continuous and takes on only rational values, then f can only have one value. What kind of function has that property?
Is it $f(x)=\alpha \cdot {\cal X}_\mathbb{Q}(x)$ where ${\cal X}_\mathbb{Q}(x)=\left\{ \begin{array}{l}1,\, if\;\; x\in \mathbb{Q};\\0, otherwise.\end{array}\right.$ and $\alpha \in \mathbb{R}$.

Or is it possible at all for $f:\mathbb{Q}\to\mathbb{R}$ to be continuous over interval $[a,b]$? There are too many points in an interval at which f is undefined, or? Now I'm confused... totally.
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