Completely Multiplicative Function is Multiplicative

Theorem

Let $f: \Z \to \Z$ be a function on the integers $\Z$.

Let $f$ be completely multiplicative.

The validity of the material on this page is questionable. In particular: Complete multiplicativity is defined for fields, but $\Z$ is not a field. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Then $f$ is multiplicative.

The validity of the material on this page is questionable. In particular: Multiplicativity is defined for $f : \N \to \N$, undefined for $f : \Z \to \Z$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Proof

By definition of complete multiplicativity:

$\forall m, n \in \Z: \map f {m n} = \map f m \map f n$

Hence by True Statement is implied by Every Statement:

$\forall m, n \in \Z: m \perp n \implies \map f {m n} = \map f m \map f n$

So $f$ is multiplicative.

$\blacksquare$