Value of Multiplicative Function is Product of Values of Prime Power Factors

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f: \N \to \C$ be a multiplicative function.

Let $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ be the prime decomposition of $n$.


Then:

$\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$


Proof

We have:

$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$

We also have:

$\forall i, j \in \closedint 1 n: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$

So:

$\map f {p_i^{k_i} p_j^{k_j} } = \map f {p_i^{k_i} } \, \map f {p_j^{k_j} }$


It is a simple inductive process to show that $\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$.

$\blacksquare$