# Math212

Applied And Computational Mathematics

## Sunday, December 24, 2017

In General:

It is obvious that the term Prime are strictly related to numbers that can not have any other factors except for the number one and itself.
For Instance:
1. The number $$14$$ is not Prime since it has more than two factors we discussed, i.e. the numbers $$(1,2,7,14)$$ are divisors for the number $$14$$.
2. On the other hand, the Prime number $$11$$ has just two divisors which are $$(1,14)$$ only.

But, according to this definition, the number $$1$$ are an eligible candidate for this definition. So, what other definition mathematicians have related to ?

The general definition above is not sufficient enough for determining whether the number $$1$$ is prime or not!

However, the prime-factorization theorem have the answer for this ambiguity. It is also called: "The Fundamental Theorem of Arithmetic", and that theorem states:
$$\forall M \in \mathbb{N}$$ $$M$$ can be written as a unique product of primes

Discussion:
1.  The Number $$24$$ has the unique product of primes representation which is: $$24 = 2\times 2\times 2\times 3$$.
2. The Prime Number $$17$$ has the unique prime product representation which is the number $$17$$ itself.
So, the last theorem will be violated if the number $$1$$ considered to be prime, and here is a counter example:

Let $$C \in \mathbb{N}$$ be a composite number which has the following unique n-primes product $$p_{1}\times p_{2}\times p_{3}\times ... \times p_{n}$$, i.e. $$C= p_{1}\times p_{2}\times p_{3}\times ... \times p_{n}$$. Now, Assume that $$1$$ is prime number, then the composite number $$C$$ has infinite number of primes product representations as follow:
$$C= p_{1}\times p_{2}\times p_{3}\times ... \times p_{n}$$
$$C= 1\times p_{1}\times p_{2}\times p_{3}\times ... \times p_{n}$$
$$C= 1\times 1\times p_{1}\times p_{2}\times p_{3}\times ... \times p_{n}$$
$$C= 1\times 1\times 1\times p_{1}\times p_{2}\times p_{3}\times ... \times p_{n}$$
and so on ...
Then, No unique prime product factorization will be existed and the upper theorem will be violated.
Hence, $$1$$ is not prime number.

Numerically consider the number $$30$$ which has the following unique prime product: $$2\times 3\times 5$$. But, if the number $$1$$ is prime then there will be no such uniqueness in prime product representation for each of the positive whole numbers.
And here is how it looks like if $$1$$ considered to be prime:
$$30= 2\times 3\times 5$$
$$30= 1\times 2\times 3\times 5$$
$$30= 1\times 1\times 2\times 3\times 5$$
$$30= 1\times 1\times 1\times 2\times 3\times 5$$ and so on...
Thus, $$1$$ is not prime number.