# A Challenge Problem

1. Dec 2, 2009

### altcmdesc

This isn't a homework question or anything, but I came across this challenge problem posted on a Harvard Math 25a webpage and I'm wondering what the solution to it is since no solution is posted on the page.

Suppose that $$f\colon \mathbb{Q} \to \mathbb{Q}$$ satisfies $$f'(x) = f(x)$$ for all $$x \in \mathbb{Q}$$. Must $$f$$ be the zero function?

2. Dec 2, 2009

### Staff: Mentor

Could you please provide a link to where this came from? For it to stay in the general technical forums, we would need to verify that it is not from an assignment.

3. Dec 2, 2009

### altcmdesc

Sure:

http://www.math.harvard.edu/~tomc/math25a/challenge.pdf [Broken]

It's from a class held Spring 2005.

There are other challenge problems posted on the course webpage which are interesting as well.

Last edited by a moderator: May 4, 2017
4. Dec 2, 2009

### Staff: Mentor

Verified, thank you. Good luck!

Last edited by a moderator: May 4, 2017
5. Dec 3, 2009

### Gib Z

I would say it must be, but hopefully I'm not invoking too strong theorems.

Solving the differential equation provided, the family class is $$f(x) = ae^x$$ where a is some real constant. However for any rational argument, the exponential function has an irrational (in fact, transcendental?) value. So the only function from Q to Q will be the case where a=0, ie f(x) =0.

EDIT: The first problem is also relatively straightforward, seeing as its given that the series are absolutely convergent, so we can rearrange terms, those results follow in the same way the finite cases do.

6. Dec 3, 2009

### altcmdesc

That's the conclusion I reached as well, but I figured that it was too simple of an answer to a challenge problem..