# divisor function is multiplicative, the

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

 $t=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{r}^{a_{r}}q_{1}^{b_{1}}q_{2}^{b_{2}}% \cdots q_{s}^{b_{s}}.$
 $t=p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{r}^{k_{r}}q_{1}^{h_{1}}q_{2}^{h_{2}}% \cdots q_{s}^{h_{s}}.$

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

 $d(mn)=d(m)d(n).$
Title divisor function is multiplicative, the DivisorFunctionIsMultiplicativeThe 2013-03-22 15:03:47 2013-03-22 15:03:47 yark (2760) yark (2760) 9 yark (2760) Theorem msc 11A25