# Homework Help: F(x) is a rational number

1. Mar 13, 2014

### victoranderson

The question is to determine f(0).

In my opinion the question wants me to find f(x)..

If f(x)=1, then

1. f:[0,1] -> R is continuous (do I need to prove?)

2. Obviously, f(1) = 1

3. f(x) is a rational number for all xε[0,1]

so f(0) = 1

Am I correct?

2. Mar 13, 2014

### jbunniii

Editing because I think I misunderstood your question in my first reply.

Yes, that's reasonable. Technically you only need to find $f(0)$, but enough information is given that you can find $f(x)$ for all $x \in [0,1]$.

I take this to mean you want to set $f(x) = 1$ for all $x \in [0,1]$ and verify that it meets the conditions. And it does meet the conditions:

Assuming you know that a constant function is continuous, then you don't need to prove it again.

Yes.

However, all you have done is to show that the constant function $f(x) = 1$ satisfies the conditions. Can you prove that it is the ONLY function that meets the conditions? How do I know there isn't some other function with $f(0) = 0$ which works?

Indeed, you can answer this problem without even stating the values of $f(x)$ for $0 < x < 1$. The key facts you need are that $f$ is continuous and that $f$ takes on only rational values. To do this, suppose that $f(0) \neq 1$ and look for a contradiction. Hint: there's a fundamental theorem about continuous functions that makes this a one-liner.

Last edited: Mar 13, 2014
3. Mar 13, 2014

### victoranderson

Thanks a lot.
I think I am done in this question but I really want to know more...

If I prove f(0)=1 by contradiction
which theorem can we use?
I think extreme values theorem , IVT are unrelated..

4. Mar 13, 2014

### HallsofIvy

What you need to prove is that if f is continuous and f takes only rational number values for all x in [0, 1], then f is a constant function. Use IVT. If f(a) is NOT equal to f(b) for some a and b in [0,1]) then f takes on all values between f(a) and f(b). That is an interval and any interval contains irrational numbers.