total number of mapping How many maps $\phi\, $ are there from $ N\, \cup\, {{0}}$  to $ N\, \cup\, {{0}}$, such that the property $ \phi(ab) \, = \, \phi(a)\,+\,\phi(b)\,$ is satisfied for all $a\, b \,\in N\, \cup\, {{0}}$? I came up with only one, mapping all numbers to zero.
 A: If $\phi$ is really supposed to be a map from $\mathbb{N}\cup\{0\}$ to $\mathbb{N}\cup\{0\}$, then indeed that's the only example. HINT: for any $a$, we have $\phi(0)=\phi(0a)=\phi(0)+\phi(a)$ . . .
Meanwhile, suppose we're only looking at maps $\mathbb{N}\rightarrow\mathbb{N}$ (so $0$ is not included). Then there are many: for example, the evenness of a number $n$, $even(n)$, defined as $even(n)=k$ iff $2^k$ divides $n$ but $2^{k+1}$ does not.
A: $\phi(0) = \phi(0b) = \phi(0) + \phi(b)$ so $\phi(b) = 0$.
EDIT:
However, if you only want $\phi$ defined on positive integers, i.e. a homomorphism of semigroups $(\mathbb N, \cdot) \to (\mathbb N \cup \{0\}, +)$,
then you have infinitely many: $\phi$ is determined by its values on the primes:
$$\phi\left(\prod_j p_j^{d_j}\right) = \sum_j d_j \phi(p_j)$$
A: Hints:
For any natural number n,
ϕ(0*n) = ϕ(0) + ϕ(n)
A: Let's indeed assume the map is from $\mathcal{N}\to\mathcal{N}\cup \{0\}$. Seeing that $\phi(0)=0$ and $\phi(1)=0$ is easy. Note also that $\phi(p^{k}) = k.\phi(p)$ for all p and k, which helps us to see that if we take arbitrary number $n=p_{1}^{\alpha_1}p_{2}^{\alpha_2}\ldots p_{l}^{\alpha_l}$ (this is the unique prime factorization of n), then:
$$\phi(n) = \phi(p_{1}^{\alpha_1}p_{2}^{\alpha_2}\ldots p_{l}^{\alpha_l}) = \alpha_{1}\phi(p_{1}) + \alpha_{2}\phi(p_{2}) +\ldots +\alpha_{l}\phi(p_{l}) $$
This means that the value of the map are completely determined by the values assigned to the prime numbers. I think that any assignment of such values leads to a map that works!
A: Assuming that the question is "How many functions $f:\mathbb{N}\rightarrow\mathbb{N}$ have the property $f(ab)=f(a)+f(b)$, for all $a,b\in\mathbb{N}$?", where I am letting $0\in\mathbb{N}$, as is standard in logic, the resolution is as follows:
Suppose $f$ is such a function. Then $f(0)=f(0\cdot n)=f(0)+f(n)$ for all $n\in\mathbb{N}$. In particular, $n=0$ gives $f(0)=f(0)+f(0)$. So $f(0)=0$. Applying our identity once more yields $$0=f(0)=f(0)+f(n)=0+f(n)=f(n)$$ for all $n\in\mathbb{N}$.
On the other hand, if we use the postive integers, $\mathbb{Z}^+$ as our domain, then we have the following:
$f(1)=f(1)+f(1)$. Therefore, $f(1)=0$. For any $a,b\in\mathbb{N}$, $f(a^b)=bf(a)$ can easily be seen by induction. Therefore, if $n=p_1^{a_1}\cdots p_r^{a_r}$ is the prime factorization of a natural number (where $p_k$ is the $k$th prime and the exponents are non-negative), then $$f(n)=a_1f(p_1)+\ldots+a_rf(p_r).$$ On the other hand, if $b_1,b_2,b_3,\ldots$ is any sequence of non-negative integers, define $f(n)=a_1b_1+\ldots+a_rb_r$. Then $f(p_r)=0b_1+\ldots+0b_{r-1}+1b_r=b_r$ and if $m=p_1^{c_1}\cdots p_l^{c_l}$, then $$f(mn)=f\left(p_1^{a_1+c_1}\cdots p_{e}^{a_e+c_e}\right)=(a_1+c_1)f(p_1)+\ldots+(a_e+c_e)f(p_e)=f(n)+f(m),$$ where $e$ is the greater of $r,l$ and $a_i,c_i$ are zero for $i$ greater than $r,l$ respectively. So we have a different such function for each sequence of positive integers.
