Prove $f(a + b) = f(a) + f(b)$ in an ordered field

Question

Let $F$ be an ordered field with identities $0$ and $1'$, and define $f: \mathbb{N} \to F$ by: $f(1) = 1'$, $f(x + 1) = f(x) + 1'$ (the addition for the right hand-side is addition in the field). So basically, $f(x) = 1' + 1' + ... + 1'$, $x$ times (again, the addition here being addition in the field).

Show $f(a + b) = f(a) + f(b)$ (addition for the right hand-side is addition in the field).

I attempted to use induction, because I didn't know of any other method to prove this. I did induction on $b$. So first I need to prove $f(a + 1) = f(a) + f(1)$, but this is true from the definition so we're done here. For the induction step, I need to show $f(a + (b + 1)) = f(a) + f(b + 1)$. Or equivalently, I need to prove $f(a + (b + 1)) = f(a) + (f(b) + 1')$. How do I prove this step?

Or, is there an alternative method of proof which would work here that's more elegant?

Answer 1

I think @Hurkyl means:

For b = 1, clearly, f(a + 1) = f(a) + 1 for all a. Assume that, for all a, f(a + b)=f(a) + f(b). Then, for all a, f(a + (b + 1)) = f((a + b) + 1) = f(a + b) + 1. Thus we can conclude by induction that f(a + b) = f(a) + f(b) for all a and b.

Answer 2

Transpose your proof. Rather than fixing $a$ and iterating over $b$, you should instead iterate over $b$, and prove it for all $a$ at each step.