Proof. Let $t=mn$ with $m,n$ coprime. Applying the fundamental theorem of arithmetic, we can write $$ m=p_1^{a_1}p_2^{a_2}\cdots p_{r}^{a_r}, \qquad n=q_1^{b_1}q_2^{b_2}\cdots q_s^{b_s}, $$ where each $p_j$ and $q_i$ are prime. Moreover, since $m$ and $n$ are coprime, we conclude that $$ t=p_1^{a_1}p_2^{a_2}\cdots p_{r}^{a_r}q_1^{b_1}q_2^{b_2}\cdots q_s^{b_s}. $$

Now, each divisor of $t$ is of the form $$ t=p_1^{k_1}p_2^{k_2}\cdots p_{r}^{k_r}q_1^{h_1}q_2^{h_2}\cdots q_s^{h_s}. $$ with $0\leq k_j\leq a_j$ and $0\leq h_i\leq b_i$ , and for each such divisor we get a divisor of $m$ and a divisor of $n$ , given respectively by $$ u=p_1^{k_1}p_2^{k_2}\cdots p_{r}^{k_r}, \qquad v=q_1^{h_1}q_2^{h_2}\cdots q_s^{h_s}. $$

Now, each respective divisor of $m$ , $n$ is of the form above, and for each such pair their product is also a divisor of $t$ . Therefore we get a bijection between the set of positive divisors of $t$ and the set of pairs of divisors of $m$ , $n$ respectively. Such bijection implies that the cardinalities of both sets are the same, and thus $$ d(mn)=d(m)d(n). $$