# Continuous Functions in Real Analysis

1. Mar 31, 2010

### magnoliamkt

1. The problem statement, all variables and given/known data

Let f, g be continuous from R to R (the reals), and suppose that f(r) = g(r) for all rational numbers r. Is it true that f(x) = g(x) for all x $$\in$$ R?

2. Relevant equations

3. The attempt at a solution

Basically, this seems trivial, but is probably tricky after all. I know that for f(x) to equal g(x) would mean that f(q) = g(q) where q is irrational as well as f(r) = g(r) as stated. I cannot think of example functions that are uniformly continuous on the Real line where this would fail, but yet, I also cannot think of a way to empirically prove that this is always true. Any help or a good starting point beyond this would be greatly appreciated. Note - this question follows the section of my text on "Combinations of Continuous Functions" but since this doesn't actually seem to combine f and g, beyond possibly the fact that f(x) = g(x) $$\Rightarrow$$ f(x) - g(x) = a continuous function h(x) as f, g continuous, I don't know of any other useful info in the text through this section.

2. Mar 31, 2010

### Dick

It's not that tricky. Use that if x is irrational then there is a sequence of rationals r_n such that limit r_n=x as n->infinity. Have you proved that?

3. Mar 31, 2010

### magnoliamkt

I could see how it might be useful, but no we haven't proved it yet, so I don't know if I could validly use it.

Last edited: Mar 31, 2010
4. Mar 31, 2010

### Dick

It's easy enough to prove. For any integer n consider all of the rationals k/n where k is any integer. Pick r_n to be a rational k/n which is closest to x. How big can |x-r_n| be? The name for this sort of property is saying that the rationals are dense in the reals.

5. Mar 31, 2010

### Jamin2112

I'm a little confused. What d'ya mean?

6. Mar 31, 2010

### magnoliamkt

Let me make sure I'm following you: Choosing a sequence r_n of rational numbers k/n close to an irrational number x $$\Rightarrow$$ $$\left|x-r_n \right|$$< $$\epsilon$$ (epsilon) maybe? And then... lim $$\ r_n$$ = x. So we have that because lim of both f(r) and g(r) equal to an irrational number x (or q in my original post), implying that f(q)=g(q)?

7. Mar 31, 2010

### Dick

That's a little confusing. The point is that if you can find rationals r_n->q (the irrational), then limit f(r_n)=f(q) and limit g(r_n)=g(q), since f and g are continuous. What's r in your post?

8. Mar 31, 2010

### magnoliamkt

r was a rational number (as per the original question statement). I think I do understand this now though - sorry for the confusing post (I'm new to using this forum and the math language). Thanks for your help! :)