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

$$m=p_1^{a_1}{p}_2^{a_2}\mathrm{\cdots }{p}_r^{a_r},n=q_1^{b_1}{q}_2^{b_2}\mathrm{\cdots }{q}_s^{b_s},$$

where each $p_j$ and $q_i$ are prime. Moreover, since $m$ and $n$ are coprime, we conclude that

$$t=p_1^{a_1}{p}_2^{a_2}\mathrm{\cdots }{p}_r^{a_r}{q}_1^{b_1}{q}_2^{b_2}\mathrm{\cdots }{q}_s^{b_s}.$$

Now, each divisor of $t$ is of the form

$$t=p_1^{k_1}{p}_2^{k_2}\mathrm{\cdots }{p}_r^{k_r}{q}_1^{h_1}{q}_2^{h_2}\mathrm{\cdots }{q}_s^{h_s}.$$

with $0\le k_j\le a_j$ and $0\le h_i\le b_i$, and for each such divisor we get a divisor of $m$ and a divisor of $n$, given respectively by

$$u=p_1^{k_1}{p}_2^{k_2}\mathrm{\cdots }{p}_r^{k_r},v=q_1^{h_1}{q}_2^{h_2}\mathrm{\cdots }{q}_s^{h_s}.$$

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