# General Function/Number Proof

1. Nov 6, 2011

### icestone111

1. The problem statement, all variables and given/known data
Let f: Z → Z be a function such that f(a + b) = f(a) + f(b) for all a,b ε Z. Prove that there exists an integer n such that f(a) = an for all a ε Z.

3. The attempt at a solution
I'm a little bit confused here if this is just supposed to be really simple, or if there's more to it that I'm just completely missing.

What I am thinking right now is that, couldn't n be any integer and you can just define f as a function f(x) = xn and therefore f(a + b) = (a + b)n = an + bn = f(a) + f(b) which satisfies the original statement?

Thanks!

2. Nov 6, 2011

### SammyS

Staff Emeritus
So, you're saying that not only is this true for a particular integer, n, but it's true no matter what integer is used for n?

That does appear to be true.

3. Nov 6, 2011

### icestone111

Thanks for your reply, so I do not have to analyze this question any further?

4. Nov 7, 2011

### HallsofIvy

Staff Emeritus
SammyS may be misinterpreting your problem. As stated, with the function from Z to Z, this is true: if f(x+y)= f(x)+ f(y), there exist n such that f(x)= nx.

I recommend you do this in "sections". First prove that for k any positive integer, f(k)= kf(1) by induction on k. Then show, by looking at f(k+ 0), that f(0)= 0. Finally, show, by looking at f(k+(-k)), that f(-k)= -f(k)= -kf(1). Of course, "n" is just f(1).

5. Nov 7, 2011

### Deveno

i think the point of this problem is to show that f is completely determined by f(1).

as HallsofIvy pointed out, it's best to prove this first for the natural numbers using induction, and then show that f(0) must be 0, and then that f(-k) must be -(f(k)).

alternatively, any integer m is (m)(1). you could handle all 3 cases (positive integers, 0 and negative integers) by using induction on |m|, although this is a bit trickier, and would involve proving the cases f(m+1) and f(m-1) separately, from the induction hypothesis (you'd still have to show that f(-1) = -f(1)). it pretty much works out to be the same amount of work.

6. Nov 7, 2011

### icestone111

Thanks guys, this was a bunch of help. So there was more than expected, and it makes sense now!