# General Function/Number Proof

## Homework Statement

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.

## 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!

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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.

## 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!
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.

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

HallsofIvy
Homework Helper
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).

Deveno