divisor function is multiplicative, the

Theorem.

The divisor function (http://planetmath.org/TauFunction) is multiplicative.

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).

Title	divisor function is multiplicative, the
Canonical name	DivisorFunctionIsMultiplicativeThe
Date of creation	2013-03-22 15:03:47
Last modified on	2013-03-22 15:03:47
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	9
Author	yark (2760)
Entry type	Theorem
Classification	msc 11A25