Can we find a surjective function from $N$ to Q/{0} such that $f(xy)=f(x)f(y)$?

Question

I tried a proof by contradiction, but I don't know how to proceed it using the theorems

Answer 1

Let's try to construct it by thinking what happens to the first few natural numbers:

• Putting $y=1$: $f(x\times 1)=f(x)\times f(1)$, but $f(x\times 1)=f(x)$, so $f(x)=f(x)\times f(1)$, which means $f(1)=1$.
• What about $y=2$? $f(x\times 2)=f(x)\times f(2)$ but we have no more information, so the choice of $f(2)$ is free so far. However, this determines the value of $f(2^n)$: $f(4)=f(2)\times f(2)$ which is fixed, then (edit) $f(8)=f(4)\times f(2)$ which are both fixed... By induction we define the values of $f(2^n)$.
• $y=3$. $f(x\times 3)=f(x)\times f(3)$, and the choice of $f(3)$ is again free. Just as before, this defines the values for all the powers of 3, $f(3^n)$. However, we also have defined the values of numbers whose only prime factors are $2$ and $3$. For instance, $f(6)=f(2)\times f(3)$ is fixed, and then $f(12)=f(2)\times f(6)$ and $f(18)=f(3)\times f(6)$ are fixed...
• Generalising, we only have the freedom to define the values of $f(p)$ where $p$ is a prime. $f(1)=1$ and $f(n)$ for $n$ a composite number depends on the values of $f(p_1), f(p_2),\dots$, where $p_1, p_2, \dots$ are the prime factors of $n$.

This is enough. There are infinitely many prime numbers, therefore the set $\mathbb{P}$ of all primes is countably infinite. The set $\mathbb{Q}\setminus\{0\}$ is also countable, and therefore we can find a bijection between $\mathbb{P}$ and $\mathbb{Q}\setminus\{0\}$. Now define $f(1)=1$, and define $f(p)$ for $p$ prime using the bijection. This makes $f$ surjective (it hits each point of $\mathbb{Q}\setminus\{0\}$ at least once), and well-defined since the values $f(n)$ for $n$ composite are fully specified.