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Question:

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

Discussion:

Take $$n\in \mathbb{N} \cup \{0\}$$.

By the given equation $$\phi(n\times 0)=\phi(n)+\phi(0)$$.

This means $$\phi(0)=\phi(n)+\phi(0)$$.

Oh! This means $$\phi(n)=0$$. $$n\in \mathbb{N} \cup \{0\}$$ was taken arbitrarily. So…

$$\phi(n)=0$$ for all $$n\in \mathbb{N} \cup \{0\}$$.

There is only one such map.