Existence of maps on $\mathbb{N} \cup \{0\}$ satisfying $\phi(ab)=\phi(a)+\phi(b)$

Question

How many maps $\phi : \mathbb{N} \cup \{0\} \to \mathbb{N} \cup \{0\} $ are there with the property that $\phi(ab)=\phi(a)+\phi(b)$, for all $a,b \in \mathbb{N} \cup \{0\} $?

My Attempt is $$\phi(0)+\phi(m)=\phi(0) \implies \phi(m)=0\quad \text{ for all } m \in \mathbb{N} \cup \{0\}$$

Hence there is only one such map.

Is it correct?

I would also mention that conversely $\phi(m) \equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map). — Daniel Schepler, Dec 21 '18 at 18:19

score 1 · Answer 1 · answered Dec 21 '18 at 18:09

1

You've got the correct conclusion, but it could use a tiny bit more justification. I'd express it as $$\varphi(0)+\varphi(m)=\varphi(0m)=\varphi(0),$$ just to make it perfectly clear.

answered Dec 21 '18 at 18:09

Cameron Buie

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Existence of maps on $\mathbb{N} \cup \{0\}$ satisfying $\phi(ab)=\phi(a)+\phi(b)$

1 Answers1