Multiplicative Functions

Introduction

Hello students 👋 In this lesson, you will learn about multiplicative functions, one of the most important ideas in arithmetic functions and number theory. These functions help us describe patterns in divisors, prime numbers, and factorization. They show up often in Midterm 1 topics because they connect simple counting ideas to deeper structures in integers.

Learning goals

By the end of this lesson, you should be able to:

explain what a multiplicative function is,
recognize when an arithmetic function has the multiplicative property,
use prime factorization to compute values of multiplicative functions,
connect multiplicative functions to divisor sums and other number theory tools,
apply examples to solve problems accurately.

A big idea in number theory is that many complicated problems become easier when we break an integer into its prime factors. Multiplicative functions make that strategy powerful ✨

What is a multiplicative function?

An arithmetic function is a function whose input is a positive integer and whose output is usually a number, often another integer or real number. Examples include $\tau(n)$, the number of divisors of $n$, and $\sigma(n)$, the sum of the divisors of $n$.

A function $f$ is called multiplicative if:

$f(1)=1$,
and whenever $m$ and $n$ are relatively prime, meaning $\gcd(m,n)=1$, we have

$$f(mn)=f(m)f(n).$$

The word “multiplicative” means the function respects multiplication, but only when the two inputs share no common prime factors. That condition matters a lot.

If the rule works for all positive integers $m$ and $n$, even when they are not relatively prime, then $f$ is called completely multiplicative. In that case,

$$f(mn)=f(m)f(n)$$

for all $m,n\ge 1$.

Most of the functions in a standard arithmetic functions unit are multiplicative, but not completely multiplicative. Knowing the difference is essential.

Simple examples

The identity function $f(n)=n$ is completely multiplicative, because

$$f(mn)=mn=f(m)f(n).$$

The function $f(n)=n^k$, where $k$ is a fixed positive integer, is completely multiplicative because

$$f(mn)=(mn)^k=m^k n^k=f(m)f(n).$$

The Möbius function $\mu(n)$ is multiplicative, but not completely multiplicative.

The divisor-counting function $\tau(n)$ is multiplicative.

The divisor-sum function $\sigma(n)$ is multiplicative.

These examples matter because they appear in many problems involving factorization and divisor sums.

Why relatively prime numbers matter

The condition $\gcd(m,n)=1$ means that $m$ and $n$ have no shared prime factors. This is the key reason multiplicative functions are useful: prime factors can be handled separately.

Suppose

$$n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$$

is the prime factorization of $n$, where the $p_i$ are distinct primes. Since the prime-power pieces $p_i^{a_i}$ are pairwise relatively prime, a multiplicative function satisfies

$$f(n)=f\left(p_1^{a_1}\right)f\left(p_2^{a_2}\right)\cdots f\left(p_r^{a_r}\right).$$

This is one of the most useful formulas in number theory. It turns a value at a complicated integer into a product of values at prime powers.

Real-world style example

Imagine students is organizing a set of $60$ items into boxes. The number $60$ factors as

$$60=2^2\cdot 3\cdot 5.$$

If a multiplicative function measures something about each prime-power block separately, then the whole value can be found by combining those smaller parts. This is much faster than studying $60$ all at once.

Computing multiplicative functions from prime powers

To use a multiplicative function well, you usually need its values on prime powers. Once those are known, the value on any positive integer can be found from prime factorization.

Example: the divisor-counting function $\tau(n)$

$$n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r},$$

then the number of divisors is

$$\tau(n)=(a_1+1)(a_2+1)\cdots(a_r+1).$$

Why? For each prime $p_i$, a divisor can include that prime with exponent from $0$ to $a_i$. That gives $a_i+1$ choices for each prime, and the choices multiply because the prime powers are independent.

Example

Find $\tau(60)$.

Since

$$60=2^2\cdot 3^1\cdot 5^1,$$

we get

$$\tau(60)=(2+1)(1+1)(1+1)=3\cdot 2\cdot 2=12.$$

So $60$ has $12$ positive divisors.

Example: the divisor-sum function $\sigma(n)$

For a prime power $p^a$, the divisors are

$$1,p,p^2,\dots,p^a,$$

so the sum of divisors is

$$\sigma\left(p^a\right)=1+p+p^2+\cdots+p^a.$$

This is a geometric sum, which can also be written as

$$\sigma\left(p^a\right)=\frac{p^{a+1}-1}{p-1}.$$

Because $\sigma$ is multiplicative, if

$$n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r},$$

then

$$\sigma(n)=\sigma\left(p_1^{a_1}\right)\sigma\left(p_2^{a_2}\right)\cdots\sigma\left(p_r^{a_r}\right).$$

Example

Find $\sigma(12)$.

First factorize:

$$12=2^2\cdot 3^1.$$

Then

$$\sigma(12)=\sigma(2^2)\sigma(3^1).$$

Now compute each part:

$$\sigma(2^2)=1+2+4=7,$$

$$\sigma(3^1)=1+3=4.$$

$$\sigma(12)=7\cdot 4=28.$$

That matches the actual divisors of $12$: $1,2,3,4,6,12$, whose sum is $28.

How to prove a function is multiplicative

To prove that a function $f$ is multiplicative, you usually must show two things:

$f(1)=1$,
if $\gcd(m,n)=1$, then $f(mn)=f(m)f(n)$.

A common strategy is to use prime factorization and the fact that coprime numbers have disjoint sets of prime factors.

Example idea

Suppose a function is defined by a formula involving divisors of $n$. If the divisors of $mn$ can be described in terms of divisors of $m$ and $n$ when $\gcd(m,n)=1$, then the function may be multiplicative. This is exactly how many divisor-related functions are proved multiplicative.

A classic theorem in number theory says that if $f$ and $g$ are multiplicative, then their Dirichlet convolution $f*g$ is also multiplicative. In many classes, this is a major reason multiplicative functions are important: they are stable under useful operations.

Common examples and non-examples

It helps to know what counts as multiplicative and what does not.

Multiplicative examples

$f(n)=n^k$ for fixed $k$ is completely multiplicative.
$\tau(n)$ is multiplicative.
$\sigma(n)$ is multiplicative.
The Möbius function $\mu(n)$ is multiplicative.
The Euler totient function $\varphi(n)$ is multiplicative.

For example, $\varphi(n)$ counts the integers from $1$ to $n$ that are relatively prime to $n$. If

$$n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r},$$

then

$$\varphi(n)=n\prod_{i=1}^r\left(1-\frac{1}{p_i}\right).$$

This formula is another sign of how useful multiplicative functions are: they turn a counting problem into a product over prime factors.

Non-example

The function $f(n)=n+1$ is not multiplicative. For instance,

$$f(2)f(3)=(2+1)(3+1)=12,$$

but

$$f(6)=6+1=7.$$

Since $2$ and $3$ are relatively prime and $f(6)\ne f(2)f(3)$, the function is not multiplicative.

Why multiplicative functions matter in arithmetic functions

Multiplicative functions are central to the study of arithmetic functions because they let you move from general integers to prime powers. That is a huge simplification.

In Midterm 1 topics, they often connect to:

divisor sums like $\sigma(n)$,
divisor counts like $\tau(n)$,
totients like $\varphi(n)$,
Möbius inversion and related identities,
prime factorization methods.

This means multiplicative functions are not just a definition to memorize. They are a tool for organizing many number theory ideas into a consistent pattern.

Practical reasoning tip

When you see a function on $n$, ask:

Can I write $n$ as a product of prime powers?
Is the function known to be multiplicative?
Can I compute the function on each prime power separately?

If the answer is yes, the problem often becomes much easier.

Conclusion

Multiplicative functions are a major part of arithmetic functions in number theory because they let us break integers into prime-power pieces and rebuild answers through multiplication. The key definition is simple: $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $\gcd(m,n)=1$. From that idea, many important formulas follow, including those for $\tau(n)$ and $\sigma(n)$. students, if you understand how multiplicative functions work on prime powers, you have a powerful method for solving many Midterm 1 problems 📘

Study Notes

An arithmetic function takes a positive integer input.
A function $f$ is multiplicative if $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $\gcd(m,n)=1$.
A function is completely multiplicative if $f(mn)=f(m)f(n)$ for all positive integers $m$ and $n$.
If $n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$, then a multiplicative function satisfies

$$f(n)=f\left(p_1^{a_1}\right)f\left(p_2^{a_2}\right)\cdots f\left(p_r^{a_r}\right).$$

The divisor-counting function satisfies

$$\tau(n)=(a_1+1)(a_2+1)\cdots(a_r+1).$$

The divisor-sum function satisfies

$$\sigma\left(p^a\right)=1+p+p^2+\cdots+p^a=\frac{p^{a+1}-1}{p-1}.$$

Multiplicative functions are useful because they reduce problems about $n$ to problems about prime powers.
Important examples include $\tau(n)$, $\sigma(n)$, $\varphi(n)$, and $\mu(n)$.
A quick test for non-multiplicativity is to find relatively prime numbers $m$ and $n$ such that $f(mn)\ne f(m)f(n)$.
Multiplicative functions are a core part of Midterm 1 and arithmetic functions because they connect prime factorization, divisor sums, and counting arguments.