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.