How many mappings $\phi:\Bbb{N}\cup\{0\}\to\Bbb{N}\cup\{0\}$ exist such that $\phi(ab)=\phi(a)+\phi(b)$?

Question

How many mappings $\phi:\Bbb{N}\cup\{0\}\to\Bbb{N}\cup\{0\}$ exist such that $\phi(ab)=\phi(a)+\phi(b)$?

My book says that the answer is finite. However, I am getting infinite as the answer. Let the prime numbers be mapped to any natural numbers of their choice of their choice, and let $0$ and $1$ map to $0$. Then numbers can be expressed in their prime factorization form, and be mapped to the corresponding sum.

For example, let us arbitrarily decide that $\phi(2)=3$ and $\phi(3)=7$. Then $\phi(2^43^2)=12+14=28$.

What is wring with this method?

For any number $n$, $\phi(n\times0) = \phi(0) = \phi(n)+\phi(0)$, so that $\phi(n)=0$ for all $n$. — zarathustra, Dec 12 '14 at 17:45

score 2 · Answer 1 · answered Dec 12 '14 at 20:12

2

Note that for any number $n$, we have $\phi(n)+\phi(0)=\phi(n\times 0) = \phi(0)$, i.e., $\phi(n)=0$. Therefore there is only one such function, which is identically $0$.

answered Dec 12 '14 at 20:12

zarathustra

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How many mappings $\phi:\Bbb{N}\cup\{0\}\to\Bbb{N}\cup\{0\}$ exist such that $\phi(ab)=\phi(a)+\phi(b)$?

1 Answers1

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