Product of Multiplicative Functions is Multiplicative

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Theorem

Let $f: \N \to \C$ and $g: \N \to \C$ be multiplicative functions.

Then their pointwise product:

$f \times g: \Z \to \Z: \forall s \in S: \map {\paren {f \times g} } s := \map f s \times \map g s$

is also multiplicative.


Proof

Let $f$ and $g$ be multiplicative.

Let $m \perp n$.

Then:

\(\ds \map {f \times g} {m \times n}\) \(=\) \(\ds \map f {m \times n} \times \map g {m \times n}\) Definition of Pointwise Multiplication of Integer-Valued Functions
\(\ds \) \(=\) \(\ds \map f m \times \map f n \times \map g m \times \map g n\) Definition of Multiplicative Arithmetic Function
\(\ds \) \(=\) \(\ds \map f m \times \map g m \times \map f n \times \map g n\) Integer Multiplication is Commutative
\(\ds \) \(=\) \(\ds \paren {\map {f \times g} m} \times \paren {\map {f \times g} n}\) Definition of Pointwise Multiplication of Integer-Valued Functions

Hence the result by definition of multiplicative function.

$\blacksquare$


Sources