:tocdepth: 2

.. _positive_divisors:

Positive Divisors
============================================================================

.. contents:: :local:

.. highlight:: python

The outline of the this notebook is as follows:

Today we will spend time looking at the functions :math:`\tau` which
counts the number of positive the divisors of :math:`n` and
:math:`\sigma` which is the sum of positive divisors of :math:`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 :math:`f` defined
on the set of positive integers. That is
:math:`f : \mathbb{Z}_{>0}\to\mathbb{C}`. An arithmetic function
:math:`f` is said to be multiplicative if it is not identically zero and
if

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

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

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

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

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

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   <ol>

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   <li>

Is :math:`\tau` an arithmetic function?

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   <ul>

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   <li>

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

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   </li>

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   </ul>

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   </li>

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   <li>

For any two positive integers :math:`a` and :math:`b` with
:math:`\gcd(a,b) = 1`, is :math:`\tau(a\,b) = \tau(a)\,\tau(b)`?

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   <ul>

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   <li>

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

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   </li>

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   </ul>

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   </li>

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   </ol>

How will you define :math:`\tau` in SageMath?