Completely Multiplicative Function is Multiplicative
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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.
$\blacksquare$