Completely Multiplicative Function of Quotient
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Theorem
Let $f: \R \to \R$ be a completely multiplicative function.
Then:
- $\forall x, y \in \R, y \ne 0: \map f {\dfrac x y} = \dfrac {\map f x} {\map f y}$
whenever $\map f y \ne 0$.
Proof
Let $z = \dfrac x y$.
Then:
\(\ds z\) | \(=\) | \(\ds \dfrac x y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y z\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f x\) | \(=\) | \(\ds \map f y \map f z\) | Definition of Completely Multiplicative Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\map f x} {\map f y}\) | \(=\) | \(\ds \map f z\) | dividing by $\map f y$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\map f x} {\map f y}\) | \(=\) | \(\ds \map f {\frac x y}\) | Definition of $z$ |
$\blacksquare$