Constant to Power of Number of Distinct Prime Divisors is Multiplicative Function

From ProofWiki
Jump to: navigation, search

Theorem

Let $c \in \R$ be a constant.

Let $f: \N \to \R$ denotes the mapping defined as:

$\forall n \in \N: f \left({n}\right) = c^k$

where $k$ is number of distinct primes that divide $n$.


Then $f$ is multiplicative.


Proof

Let $r, s \in \Z$ such that $r \perp s$.

Let $r$ be composed of $p$ distinct primes:

$r_1, r_2, \ldots r_p$

Thus:

$f \left({r}\right) = c^p$

Let $s$ be composed of $q$ distinct primes:

$s_1, s_2, \ldots s_q$

Thus:

$f \left({s}\right) = c^q$

As $r \perp s$, all the $r_k$ and $s_k$ are distinct.

Thus $r s$ is composed of:

the $p$ distinct primes $r_1, r_2, \ldots r_p$

and:

the $q$ distinct primes $s_1, s_2, \ldots s_q$

which is a total of $p + q$ distinct primes.

Thus:

\(\displaystyle f \left({r s}\right)\) \(=\) \(\displaystyle c^{p + q}\) Definition of $f$
\(\displaystyle \) \(=\) \(\displaystyle c^p c^q\) Exponent Combination Laws: Product of Powers
\(\displaystyle \) \(=\) \(\displaystyle f \left({r}\right) f \left({s}\right)\) from above

Hence the result.

$\blacksquare$


Also see


Sources