Number theory proof for why $\tau$, the number of divisors of $n$ is multiplicative

Question

Could someone please help me with a proof as to why $\tau$ is multiplicative?

Answer 1

Given a number $a$ with prime decomposition $p_1^{x_1}p_2^{x_2}\cdots p_m^{x_m}$ (with $x_i \in \Bbb N$ and $p_i \neq p_j$ if $i \neq j$), the number of divisors of $a$ is $$ \tau(a) = (x_1 + 1)(x_2+1)\cdots (x_m + 1). $$ Given another number $b$ with prime decomposition $q_1^{y_1}q_2^{y_2}\cdots q_m^{y_m}$ such that $\gcd(a, b) = 1$, this means that for any $i\leq m, j\leq n$ we have $p_i \neq q_j$. Then what is the prime decomposition of $ab$? What is $\tau(ab)$, using the formula above? Is this equal to $\tau(a)\tau(b)$?

Answer 2

Let $D(n)$ be the set of divisors of $n$.

If $\gcd(a, b) = 1$, then consider $f\colon D(a) \times D(b) \to D(ab)$ given by $(x,y) \mapsto xy$. Now prove:

• $f$ is indeed a function, that is, $xy$ is a divisor of $ab$.

• $f$ is surjective, that is, every divisor of $ab$ is of the form $xy$.

• $f$ is injective, that is, every divisor of $ab$ is of the form $xy$ in unique way.