# Analysis: Prove f(u+v)=f(u)+f(v) implies f(x)=f(1)x

1. Oct 3, 2011

### pcvt

1. The problem statement, all variables and given/known data
Suppose that a function f R->R has the property that f(u+v) = f(u)+f(v). Prove that f(x)=f(1)x for all rational x. Then, show that if f(x) is continuous that f(x)=f(1)x for all real x.

3. The attempt at a solution
I've proved that f(x)=f(1)x for all natural x by breaking up x into f(1)+f(x-1) and using induction, but I can't figure out how to do it for all rationals.

2. Oct 3, 2011

### micromass

What is

$f(1/n)+...+f(1/n)$

??? (if you take n sums)

3. Oct 3, 2011

### Dick

f(1)=f(1/2+1/2)=f(1/2)+f(1/2), right? That means f(1/2)=f(1)*(1/2). Can you show f(1)=f(1/3+1/3+1/3)=f(1/3)+f(1/3)+f(1/3)? How about using induction to show f(1/n)=f(1)*(1/n)??

4. Oct 4, 2011

### pcvt

Ok, I see that I can show it for f(1/n). How can I combine the two though, since I don't have a rule to say f(m/n)=f(1)*(m/n)? Also, how can I include negative numbers?

Thanks!

5. Oct 4, 2011

### HallsofIvy

I would be inclined to look at the basic "sets of numbers" and use their defining properties. For example, the "natural numbers" or "counting numbers" are defined by "induction"- 1 is a natural number and, if n is a natural number, then n+ 1 is also. Obviously, for any number, x, f(x)= f(x)(1). Use induction to show that for any positive integer, n, then f(n)= f(x)n and then take x= 1. (You will use the general "x" below.)

The "whole numbers" are just the positive integers together with 0. And 0 is the "additive identity", for any positive integer, n, n+ 0= n. Look at f(n+ 0)= f(n)+ f(0).

The "integers" includes the negative integers: we now have additive inverse. Any negative integer is of the form -n for some positive integer n. f(n+ (-n))= f(n)+ f(-n) and, of course, f(n+(-n))= f(0).

Finally, look at numbers of the form 1/n, n non-zero. Obviously, n(1/n)= 1 so f(n(1/n))= f(1). But you have already shown that f(nx)= f(x)n. So take x= 1/n.

6. Oct 5, 2011

### pcvt

So helpful, got it from there. Thanks!