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). Let $\mathbb{N}' = \left \{ f(x) \mid x \in \mathbb{N} \right \}$. Show that $f$ is a bijection from $\mathbb{N}$ to $\mathbb{N}'$.
To prove that $f$ is injective, we have to prove that $f(x) = f(y) \implies x = y$. But, we don't have any $f(x)$ or $f(y)$ to work with in order to get $x = y$, just $f(1)$ and $f(x + 1) = f(x) + 1'$ which doesn't seem to help. Is there another way to prove that this function is injective?
Assume $f(x)=f(y)$. Without loss of generality we can assume there is $k\in\mathbb{N}$ such that $y=x+k$. Then
$$f(x)=f(y)=f(x+k)=f(x)+k \times 1'$$
Therefore $k\times 1'=0$. Since it is an ordered field its characteristic is zero. Therefore $k=0$ and $x=y$.