Power Function is Completely Multiplicative

Theorem

Let $K$ be a field.

Let $z \in K$.

Let $f_z: K \to K$ be the mapping defined as:

$\forall x \in K: f_z \left({x}\right) = x^z$

Then $f_z$ is completely multiplicative.

Integers

Let $c \in \Z$ be an integer.

Let $f_c: \Z \to \Z$ be the mapping defined as:

$\forall n \in \Z: \map {f_c} n = n^c$

Then $f_c$ is completely multiplicative.

Proof

Let $r, s \in K$.

Then:

\(\ds f_z \left({r s}\right)\)

\(=\)

\(\ds \left({r s}\right)^z\)

\(\ds \)

\(=\)

\(\ds \left({r}\right)^z \left({s}\right)^z\)

Power of Product in Abelian Group

\(\ds \)

\(=\)

\(\ds f_z \left({r}\right) f_z \left({s}\right)\)

$\blacksquare$