# Intermediate Value Theorem

• Apr 18th 2012, 06:43 PM
MarkJacob
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???

Mark
• Apr 18th 2012, 07:15 PM
Prove It
Re: Intermediate Value Theorem
Quote:

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???

Mark

How could it possibly be continuous when it only takes on rational values? There are always irrationals in between any two rational numbers...
• Apr 18th 2012, 07:22 PM
MarkJacob
Re: Intermediate Value Theorem
Silly mistake.
• Apr 19th 2012, 06:08 AM
HallsofIvy
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?
• Apr 19th 2012, 06:19 AM
MarkJacob
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
• Apr 19th 2012, 06:28 AM
Prove It
Re: Intermediate Value Theorem
Quote:

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.
• Apr 19th 2012, 06:37 AM
MathoMan
Re: Intermediate Value Theorem
Quote:

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.