Divisor Sum Function is Multiplicative

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

is multiplicative.

Proof
Let $I_{\Z_{>0}}: \Z_{>0} \to \Z_{>0}$ be the identity function:


 * $\forall n \in \Z_{>0}: I_{\Z_{>0}} \left({n}\right) = n$.

Thus we have:


 * $\displaystyle \sigma \left({n}\right) = \sum_{d \backslash n} d = \sum_{d \backslash n} I_{\Z_{>0}} \left({d}\right)$.

But from Identity Function is Completely Multiplicative, $I_{\Z_{>0}}$ is multiplicative.

The result follows from Sum Over Divisors of Multiplicative Function.