Divisor Count Function is Multiplicative

Theorem
The tau function:
 * $\displaystyle \tau: \Z_{>0} \to \Z_{>0}: \tau \left({n}\right) = \sum_{d \backslash n} 1$

is multiplicative.

Proof
Let $f_1: \Z_{>0} \to \Z_{>0}$ be the constant function:
 * $\forall n \in \Z_{>0}: f_1 \left({n}\right) = 1$

Thus we have:
 * $\displaystyle \tau \left({n}\right) = \sum_{d \backslash n} 1 = \sum_{d \backslash n} f_1 \left({d}\right)$

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

The result follows from Sum Over Divisors of Multiplicative Function.