4.1: Definitions and Properties
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
An arithmetic function is a function whose domain of definition is the set \(\mathbb{N}\) of positive integers.
An arithmetic function \(f\) is called multiplicative if \(f(ab)=f(a)f(b)\) for all \(a,b\in\mathbb{N}\) such that \((a,b)=1\).
An arithmetic function \(f\) is called completely multiplicative if \[f(ab)=f(a)f(b)\] for all positive integers \(a,b\).
The function \(f(a)=1\) where \(k\) is a completely multiplicative function since \[f(ab)=1=f(a)f(b).\] Notice also that a completely multiplicative function is a multiplicative function but not otherwise.
We now prove a theorem about multiplicative functions. We will be interested in studying the properties of multiplicative functions rather than the completely multiplicative ones.
Given a multiplicative function \(f\). Let \(n=\prod_{k=1}^sp_k^{a_k}\) be the prime factorization of \(n\). Then \[f(n)=\prod_{k=1}^sf(p_k^{a_k}).\]
We prove this theorem by induction on the number of primes in the factorization of \(n\). Suppose that \(n=p_1^{a_1}\). Thus the result follow easily. Suppose now that for \[n=\prod_{k=1}^sp_k^{a_k},\] we have \[f(n)=\prod_{k=1}^sf(p_k^{a_k}).\] So we have to prove that if \[n=\prod_{k=1}^{s+1}p_k^{a_k},\] then \[f(n)=\prod_{k=1}^{s+1}f(p_k^{a_k}).\] Notice that for \[n=\prod_{k=1}^{s+1}p_k^{a_k},\] we have \((\prod_{k=1}^{s}p_k^{a_k},p_{s+1}^{a_{s+1}})=1\). Thus we have get \[f(n)=f(\prod_{k=1}^{s+1}p_k^{a_k})=f(\prod_{k=1}^{s}p_k^{a_k})f(p_{s+1}^{a_{s+1}})\] which by the inductive step gives \[f(\prod_{k=1}^{s+1}p_k^{a_k})=f(n)=\prod_{k=1}^{s+1}f(p_k^{a_k}).\]
From the above theorem, we can see that to evaluate a multiplicative function at an integer, it will be enough to know the value of the function at the primes that are in the prime factorization of the number.
We now define summatory functions which represents the sum of the values of a given function at the divisors of a given number.
Let \(f\) be an arithmetic function. Define \[F(n)=\sum_{d\mid n}f(d)\] Then \(F\) is called the summatory function of \(f\).
This function determines the sum of the values of the arithmetic function at the divisors of a given integer.
If \(f(n)\) is an arithmetic function, then \[F(18)=\sum_{d\mid 18}f(d)=f(1)+f(2)+f(3)+f(6)+f(9)+f(18).\]
If \(f\) is a multiplicative function, then the summatory function of \(f\) denoted by \(F(n)=\sum_{d\mid n}f(d)\) is also multiplicative.
We have to prove that \(F(mn)=F(m)F(n)\) whenever \((m,n)=1\). We have \[F(mn)=\sum_{d\mid mn}f(d).\] Notice that by Lemma 6, each divisor of \(mn\) can be written uniquely as a product of relatively prime divisors \(d_1\) of \(m\) and \(d_2\) of \(n\), moreover the product of any two divisors of \(m\) and \(n\) is a divisor of \(mn\). Thus we get \[F(mn)=\sum_{d_1\mid m, d_2\mid n}f(d_1d_2)\] Notice that since \(f\) is multiplicative, we have \[\begin{aligned} F(mn)&=& \sum_{d_1\mid m, d_2\mid n}f(d_1d_2)\\&=&\sum_{d_1\mid m, d_2\mid n}f(d_1)f(d_2)\\&=&\sum_{d_1\mid m}f(d_1)\sum_{d_2\mid n} f(d_2)=F(m)F(n) \end{aligned}\]
Exercises

Determine whether the arithmetic functions \(f(n)=n!\) and \(g(n)=n/2\) are completely multiplicative or not.

Define the arithmetic function \(g(n)\) by the following. g(n)=1 if \(n=1\) and 0 for \(n>1\). Prove that \(g(n)\) is multiplicative.
Contributors
Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution (CC BY) license.