Divisor Count Function is Multiplicative

Theorem

The divisor count function:

$\ds \tau: \Z_{>0} \to \Z_{>0}: \map {\sigma_0} n = \sum_{d \mathop \divides n} 1$

is multiplicative.

Proof

Let $f_1: \Z_{>0} \to \Z_{>0}$ be the constant function:

$\forall n \in \Z_{>0}: \map {f_1} n = 1$

Thus we have:

$\ds \map {\sigma_0} n = \sum_{d \mathop \divides n} 1 = \sum_{d \mathop \divides n} \map {f_1} d$

But from Unity Function is Completely Multiplicative, $f_1$ is multiplicative.

The result follows from Sum Over Divisors of Multiplicative Function.

$\blacksquare$

Sources

• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): divisor function