What is f(0) if f(x) is a rational number on [0,1]?

[victoranderson]

Please see attached.
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?

 

[jbunniii]

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

View attachment 67599

Please see attached.
The question is to determine f(0).

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

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 $x \in [0,1]$.

If f(x)=1, then

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:

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

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

2. Obviously, f(1) = 1

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

so f(0) = 1

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.

 

[victoranderson]

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.

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..

 

[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.