Foundations - The Fundamental Theorem of Arithmetic

 

Mathematical Conundrum Series

This post is part of a series of posts that should be read in order:

Mathematical Conundrum Series

• Introduction
• Foundations - The Counting Numbers
• Foundations - The Fundamental Theorem of Arithmetic
• The Fundamental Theorem of Arithmetic - Part 2
• The Inverse Naturals and The Euler Product
• The Conundrum
• The Conundrum - Free Will and Choice

I am continuing my previous post, 'Foundations - The Counting Numbers', where I spoke of some of the shortcuts we are taught kind of unknowingly in mathematics. This post is about another one of those shortcuts that at first glance seems very trivial or even non-sensical. It is about factorization.

We know that the atoms per se of all natural numbers, inverse natural numbers, and rational numbers are the prime numbers, and this is all wrapped up in the Fundamental Theorem of Arithmetic, the Unique Factorization Theorem. 

The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be represented uniquely as a product of prime numbers, up to the order of the factors. This means that every natural number is either a prime number or a composite number made of primes or powers of primes.

$$137,200=2^4\cdot 5^2\cdot 7^3$$
You will notice that in the above factorization there is no representation for the prime number 3. It may seem silly to write the factorization as this:
$$137,200=2^4\cdot 3^0\cdot 5^2\cdot 7^3$$ but it is entirely acceptable. It is one of those shortcuts. Why include $$3^0$$ in this factorization? $$3^0$$ is just 1 and 1 times a number is just the number.

This actually becomes important in later posts I am going to make. For now, I will provide a reference to the Canonical representation of factorization here: Canonical representation of a positive integer, which includes the following: 

In fact, any positive integer can be uniquely represented as an infinite product taken over all the positive prime numbers 

The quoted text basically means this:
$$137,200=2^4\cdot 3^0\cdot 5^2\cdot 7^3\cdot {11}^0\cdot {13}^0\cdot {17}^0\cdot {19}^0\cdot {23}^0\cdot {29}^0\cdot {31}^0\cdot \dots P_n^0$$

There is one caveat to this. We are working in the natural numbers, the positive integers. The Fundamental Theorem of Arithmetic is defined for the natural numbers, the positive integers. All natural numbers, positive integers, are finite. There are countably infinite elements in the set of natural numbers, but every single element is a finite number. This means writing any natural number as its factorization with infinite terms requires that there are only finitely many terms with prime factors with powers that are non-zero. No matter how large the number, writing its infinitely termed factorization will have a point in its progression to the right where every further prime power term is the respective prime raised to the zero power.

So every natural number, positive integer, will have a specific form or signature in its infinitely termed factorization, where everything in the rightward progression eventually becomes the respective prime to the zero power, or 1, 1, 1, 1, 1, 1, 1,... This is for every natural number, positive integer, no matter how large.

The Fundamental Theorem of Arithmetic directly implies the following:

The Fundamental Theorem of Arithmetic is an infinite product of prime geometric power series.

$$\left(\begin{array}{c}\\ 2^n\\ +\\ ⋮\\ 2^5\\ +\\ 2^4\\ +\\ 2^3\\ +\\ 2^2\\ +\\ 2^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ 3^n\\ +\\ ⋮\\ 3^5\\ +\\ 3^4\\ +\\ 3^3\\ +\\ 3^2\\ +\\ 3^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ 5^n\\ +\\ ⋮\\ 5^5\\ +\\ 5^4\\ +\\ 5^3\\ +\\ 5^2\\ +\\ 5^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ 7^n\\ +\\ ⋮\\ 7^5\\ +\\ 7^4\\ +\\ 7^3\\ +\\ 7^2\\ +\\ 7^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ {11}^n\\ +\\ ⋮\\ {11}^5\\ +\\ {11}^4\\ +\\ {11}^3\\ +\\ {11}^2\\ +\\ {11}^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ {13}^n\\ +\\ ⋮\\ {13}^5\\ +\\ {13}^4\\ +\\ {13}^3\\ +\\ {13}^2\\ +\\ {13}^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ {17}^n\\ +\\ ⋮\\ {17}^5\\ +\\ {17}^4\\ +\\ {17}^3\\ +\\ {17}^2\\ +\\ {17}^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ {19}^n\\ +\\ ⋮\\ {19}^5\\ +\\ {19}^4\\ +\\ {19}^3\\ +\\ {19}^2\\ +\\ {19}^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ {23}^n\\ +\\ ⋮\\ {23}^5\\ +\\ {23}^4\\ +\\ {23}^3\\ +\\ {23}^2\\ +\\ {23}^1\\ +\\ 1\\ \end{array}\right)\times \left(\begin{array}{c}\\ {29}^n\\ +\\ ⋮\\ {29}^5\\ +\\ {29}^4\\ +\\ {29}^3\\ +\\ {29}^2\\ +\\ {29}^1\\ +\\ 1\\ \end{array}\right)\times \cdots$$

This is the Fundamental Theorem of Arithmetic.

Now several people will say, "Prove that the above infinite product of prime geometric power series is the Fundamental Theorem of Arithmetic. If that is requested, I will provide it. It will be simple but long. I did it 30 years ago by pencil and paper and 20 years ago by Excel spreadsheet. It goes back to 2300 years ago and the sieve of Eratosthenes and its removal sets / Smallest Common Factor (SCF) sets. Analyzing the SCF sets shows that they are the above construction of an infinite product of prime geometric power series.

Continue to - The Fundamental Theorem of Arithmetic - Part 2