12. Positive Divisors

The outline of the this notebook is as follows:

Today we will spend time looking at the functions \tau which counts the number of positive the divisors of n and \sigma which is the sum of positive divisors of n.  Our goal is, once again, to use our minds, paper and pencil, and SageMath to combine to not just conjecture, but start to give logical reasons for our conjectures.

Definition: An arithmetic function is real or complex f defined on the set of positive integers. That is f : \mathbb{Z}_{>0}\to\mathbb{C}. An arithmetic function f is said to be multiplicative if it is not identically zero and if

f(a\,b) = f(a)\,f(b) \quad \text{ whenever } \gcd(a,\,b) = 1.

Let a and b be positive integers and f : \mathbb{Z}_{>0}\to\mathbb{C} is an arithmetic function. If \gcd(a,\,b)=1 and f(a\,b) = f(a)\,f(b) then f is said to be an multiplicative.

f is said to be completely multiplicative if f(a\,b) = f(a)\,f(b) for all a,b \in \mathbb{N}.

Knowing the definition of \tau and what it means for an arithmetic function to be multiplicative, can we proof that \tau and \sigma are both multiplicative function?

Let us first consider \tau. To prove that \tau is a multiplicative function, we must ask ourselves the following questions.

  1. Is \tau an arithmetic function?

    • Yes \tau is an arithmetic function by definition. In particular, for all n \in \mathbb{N}, \tau(n) \in \mathbb{N} and since \mathbb{N} \subset \mathbb{C}, it follows that \tau is an arithmetic function.

  2. For any two positive integers a and b with \gcd(a,b) = 1, is \tau(a\,b) = \tau(a)\,\tau(b)?

    • We want to count the number of divisors of the product a\,b given that \gcd(a,b)=1. From \gcd(a,b) = 1, we know that a does not divide b and vice-versa. Therefore, for a given divisor of a say a_{1}, a set divisors of the product a\,b will be a_{1} multiplied by all the divisors of b. That is, the set \{a_{1} k \, | \, k \text{ divides } b\} with cardinality \tau(b). Repeating this for all the divisors of a is exactly the product \tau(a) \, \tau(b). This completes the proof.

How will you define \tau in SageMath?