The 'The Sieve of Eratosthenes - Part 2

Previous post: The Sieve of Eratosthenes - Backing Up

To continue examining the data sets formed by the 'Sieve of Eratosthenes', let's look at some definitions for the sieve sets and remainder sets. Refer to the previous post for an explanation of the process and for examples of the sets, The Sieve of Eratosthenes - Backing Up.

Reviewing the 'Sieve of Eratosthenes'

Let's define the process a little better and define the resultant sets of numbers from each iteration of the process.

1. The starting point is the entire set of all natural numbers.

2. On the first iteration we sieve out all the numbers that are divisible by the prime number 2.

This produces the first sieve set soe2, and the first remainder set soe2_rem.

The soe2 set contains all numbers that have a smallest common prime factor of 2. 

The soe2_rem set contains all numbers not having any factor of the prime number 2. 

The soe2_rem set only has numbers with factors of prime numbers 3 and greater. 

3. On the second iteration we are only working with the soe2_rem set.

From the soe2_rem set we sieve out all numbers that are divisible by the prime number 3.

This produces the second sieve set soe3, and the second remainder set soe3_rem.

The soe3 set contains all numbers that have a smallest common prime factor of 3. 

The soe3_rem set contains all numbers not having any factors of the prime numbers 2, 3.  

The soe3_rem set only has numbers with factors of prime numbers 5 and greater. 

4. On the third iteration we are only working with the soe3_rem set.

From the soe3_rem set we sieve out all numbers that are divisible by the prime number 5.

This produces the third sieve set soe5, and the third remainder set soe5_rem.

The soe5 set contains all numbers that have a smallest common prime factor of 5. 

The soe5_rem set contains all numbers not having any factors of the prime numbers 2, 3, 5.  

The soe5_rem set only has numbers with factors of prime numbers 7 and greater. 

5. On the fourth iteration we are only working with the soe5_rem set.

From the soe5_rem set we sieve out all numbers that are divisible by the prime number 7.

This produces the fourth sieve set soe7, and the fourth remainder set soe7_rem.

The soe7 set contains all numbers that have a smallest common prime factor of 7. 

The soe7_rem set contains all numbers not having any factors of the prime numbers 2, 3, 5, 7.  

The soe7_rem set only has numbers with factors of prime numbers 11 and greater. 

We continue this iterative process for each prime number.

Sieve of Eratosthenes set definitions:

For each P (prime),

Each soe set can be defined as: 

The subset of all natural numbers that have a smallest common prime factor of P.

Each soe_rem set can be defined as: 

The subset of all natural numbers that only contains the number 1 and the natural numbers with prime factors greater than the prime number P.

Factorization Model for the 'Sieve of Eratosthenes' sets and remainder sets:

To get to our conclusion that the infinite series of all natural numbers has an equality with an infinite product where each term of the infinite product is a prime geometric series we need to look at factorization and the model for representing any natural number (positive integer) as an infinite product. For this we can refer to Wiki, Canonical representation of a positive integer.

Any positive integer can be uniquely represented as an infinite product taken over all the positive prime numbers, as $${\displaystyle n=2^{n_1}{3}^{n_2}{5}^{n_3}{7}^{n_4}\cdots =\prod _{i=1}^{\mathrm{\infty }}{p}_i^{n_i},}$$ where a finite number of the n_i are positive integers, and the others are zero.

Imagine if you can, going to a web page where you could create any natural number. The web page would have a form with the prime numbers and a drop down for each exponent value. The page would have infinite width to accommodate all the prime numbers and the dropdowns for each exponent selection would include every infinite positive integer and also 0. In fact let's create an abbreviated version of what that would look like, except instead of a dropdown for the choice of an exponent for each prime number, let's just show the selections themselves. This is of course an abbreviated version for demonstration purposes.

You have to imagine a column for every prime number and a row for continuing every integer exponent for each prime number.

20	30	50	70	110	130	170	190	230	290
21	31	51	71	111	131	171	191	231	291
22	32	52	72	112	132	172	192	232	292
23	33	53	73	113	133	173	193	233	293
24	34	54	74	114	134	174	194	234	294
25	35	55	75	115	135	175	195	235	295
26	36	56	76	116	136	176	196	236	296
27	37	57	77	117	137	177	197	237	297
28	38	58	78	118	138	178	198	238	298
29	39	59	79	119	139	179	199	239	299

With this you can actually select the prime number with its exponent value. You must select a prime number with its exponent value for each and every prime number, (make one selection from each column), then push a button to create the number and you have your number.

Factorization of the soe2 and soe2_rem sets:

Now let's change it up a bit and see what would happen if I restricted your choice of exponents for the prime number 2.

If I restricted the choice of exponents for the prime number 2, so that you can only select 2⁰, what numbers could you create?

20	30	50	70	110	130	170	190	230	290
	31	51	71	111	131	171	191	231	291
	32	52	72	112	132	172	192	232	292
	33	53	73	113	133	173	193	233	293
	34	54	74	114	134	174	194	234	294
	35	55	75	115	135	175	195	235	295
	36	56	76	116	136	176	196	236	296
	37	57	77	117	137	177	197	237	297
	38	58	78	118	138	178	198	238	298
	39	59	79	119	139	179	199	239	299

The only numbers you could create are the odds, the soe2_rem set.
The subset of all natural numbers that only contains numbers with prime factors greater than the prime number 2, with the exception that the number 1 is always in the remainder set.
( 1, 3, 5, 7, 9, 11,... )

If I then changed that up and restricted the choice of exponents for the prime number 2, so that you can only select integer powers of 2 and you could never select 2⁰, what numbers could you create?

	30	50	70	110	130	170	190	230	290
21	31	51	71	111	131	171	191	231	291
22	32	52	72	112	132	172	192	232	292
23	33	53	73	113	133	173	193	233	293
24	34	54	74	114	134	174	194	234	294
25	35	55	75	115	135	175	195	235	295
26	36	56	76	116	136	176	196	236	296
27	37	57	77	117	137	177	197	237	297
28	38	58	78	118	138	178	198	238	298
29	39	59	79	119	139	179	199	239	299

Now you could not create any odd numbers, the only numbers you could create are the evens, the soe2 set.
The subset of all natural numbers that have a smallest common prime factor of 2.
( 2, 4, 6, 8, 10, 12,... )

But what is this telling us?

It is telling us this, the evens are composed exactly of each power of 2 times the odds.

$${\displaystyle 2^1\times (1,3,5,7,9,11,...)}$$ $${\displaystyle 2^2\times (1,3,5,7,9,11,...)}$$ $${\displaystyle 2^3\times (1,3,5,7,9,11,...)}$$ $${\displaystyle ⋮}$$ $${\displaystyle 2^n\times (1,3,5,7,9,11,...)}$$

This directly implies that the evens (the soe2 set) is equal to the 2 geometric series times the odds (the soe2_rem set). We should literally be able to write the infinite series of all evens as equal to the product of the 2 geometric series times the infinite series of the odds and this should preserve every term of the infinite series of evens:

$$2+4+6+8+10+12+14+16+18+20+\cdots =\left(\begin{array}{c}\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )$$

This allows us to write the following equation:
$$\begin{array}{}\text{(1)}& x=1+2+3+4+5+6+7+8+9+10+\cdots \end{array}$$

We can reorder the infinite series on the right hand side, to be the soe2 infinite subseries (the evens) + the soe2_rem infinite subseries (the odds).
$$x=(2+4+6+8+10+\cdots )+(1+3+5+7+9+\cdots )$$

Since we know from our factorization model that the soe2 infinite subseries (the evens) is just the prime number 2 geometric series times the soe2_rem infinite subseries (the odds), we can write the right hand side of our equation as:
$$\begin{array}{}\text{(2)}& x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )\end{array}$$

Notice we have added the term 2⁰, to the first term of the product on the right hand side. This preserves all of the terms that are the odds from the infinite series expression on the right hand side of equation $\text{(1)}$.

And all of the terms that are the integer powers of 2, in the first term of the product on the right hand side produces all of the evens from the infinite series expression on the right hand side of equation $\text{(1)}$.

If we substitute for the variable x in equation $\text{(2)}$ with the value of x from equation $\text{(1)}$, we get the following:

$$\begin{array}{}\text{(3)}& 1+2+3+4+5+6+7+8+9+10+11+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )\end{array}$$

Since the second term of the product on the right hand side of equation $\text{(3)}$ is the odds (the soe2_rem set as an infinite series), we can examine the factorization of that expression.

The soe2_rem set (odds) factorization model.
The subset of all natural numbers that only contains numbers with prime factors greater than the prime number 2, with the exception that the number 1 is always in the remainder set.

Factorization of the soe2_rem set (the odds) into the soe3 and soe3_rem sets:

Once again you have to imagine a column for every prime number greater than 2 and a row for continuing every integer exponent for each prime number. 

The below table can create all of the natural numbers representing the soe2_rem set (odds).

30	50	70	110	130	170	190	230	290
31	51	71	111	131	171	191	231	291
32	52	72	112	132	172	192	232	292
33	53	73	113	133	173	193	233	293
34	54	74	114	134	174	194	234	294
35	55	75	115	135	175	195	235	295
36	56	76	116	136	176	196	236	296
37	57	77	117	137	177	197	237	297
38	58	78	118	138	178	198	238	298
39	59	79	119	139	179	199	239	299

Once again you must select a prime number with its exponent value for each and every prime number, (make one selection from each column), then push a button to create the number and you have your number.

Let's change it up a bit and see what would happen if I restricted your choice of exponents for the prime number 3.

If I restricted the choice of exponents for the prime number 3, so that you can only select 3⁰, what numbers could you create?

30	50	70	110	130	170	190	230	290
	51	71	111	131	171	191	231	291
	52	72	112	132	172	192	232	292
	53	73	113	133	173	193	233	293
	54	74	114	134	174	194	234	294
	55	75	115	135	175	195	235	295
	56	76	116	136	176	196	236	296
	57	77	117	137	177	197	237	297
	58	78	118	138	178	198	238	298
	59	79	119	139	179	199	239	299

The only numbers you could create are numbers in the soe3_rem set.
The subset of all natural numbers that only contains numbers with prime factors greater than the prime number 3, with the exception that the number 1 is always in the remainder set.
( 1, 5, 7, 11, 13, 17,... )

If I then changed that up and restricted the choice of exponents for the prime number 3, so that you can only select integer powers of 3 and you could never select 3⁰, what numbers could you create?

	50	70	110	130	170	190	230	290
31	51	71	111	131	171	191	231	291
32	52	72	112	132	172	192	232	292
33	53	73	113	133	173	193	233	293
34	54	74	114	134	174	194	234	294
35	55	75	115	135	175	195	235	295
36	56	76	116	136	176	196	236	296
37	57	77	117	137	177	197	237	297
38	58	78	118	138	178	198	238	298
39	59	79	119	139	179	199	239	299

Now you could not create any numbers of the soe3_rem set, the only numbers you could create are numbers of the soe3 set.
The subset of all natural numbers that have a smallest common prime factor of 3.
( 3, 9, 15, 21, 27, 33, 39,... )

But what is this telling us?

It is telling us this, the numbers of the soe3 set are composed exactly of each power of 3 times the numbers of the soe3_rem set.

$${\displaystyle 3^1\times (1,5,7,11,13,17,...)}$$ $${\displaystyle 3^2\times (1,5,7,11,13,17,...)}$$ $${\displaystyle 3^3\times (1,5,7,11,13,17,...)}$$ $${\displaystyle ⋮}$$ $${\displaystyle 3^n\times (1,5,7,11,13,17,...)}$$

This directly implies that the soe3 set is equal to the 3 geometric series times the soe3_rem set. We should literally be able to write the infinite series of all numbers of the soe3 set as equal to the product of the 3 geometric series times the infinite series of the soe3_rem set and this should preserve every term of the infinite series of the soe3 set:

$$3+9+15+21+27+33+39+45+51+\cdots =\left(\begin{array}{c}\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\\ \end{array}\right)\times (1+5+7+11+13+17+19+\cdots )$$

Since we are examining the factorization of the soe2_rem set (the odds) into the soe3 and soe3_rem set, the above equation only accounts for the soe3 set and is missing the soe3_rem set. Working with the entire soe2_rem set (odds) as a series, we can write:

Re-order the terms of the soe2_rem series (odds) into the soe3 subseries plus the soe3_rem subseries: $$1+3+5+7+9+11+13+15+17+\cdots =(3+9+15+21+27+33+39+\cdots )+(1+5+7+11+13+17+19+\cdots )$$ Re-write the right hand side of the equation into a product, the 3 geometric series times the soe3_rem subseries. $$1+3+5+7+9+11+13+15+17+\cdots =\left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times (1+5+7+11+13+17+19+\cdots )$$ You will notice in the product on the right hand side of the above equation, we have included 3⁰ as the first term of the 3 geometric series. This maintains all of the terms of the soe3_rem subseries and then the integer powers of the 3 geometric series maintains all of the terms of the soe3 subseries, the entirety of which is equal to the soe2_rem (odds) series on the left hand side of the equation.

We can now return to equation $\text{(3)}$, where the second term of the product on the right hand side is the soe2_rem (odds) subseries, and re-write that subseries into the right hand side of the equation we just derived above.

Equation 3: $$1+2+3+4+5+6+7+8+9+10+11+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )$$

Re-writing the second term of the product on the right hand side (soe2_rem subseries) into a product.

$$\begin{array}{}\text{(4)}& 1+2+3+4+5+6+7+8+9+10+11+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\\ \end{array}\right)\times (1+5+7+11+13+17+19+\cdots )\end{array}$$

Where is this all going:

If we continue to recursively examine the factorization of each 'Sieve of Eratosthenes' set and remainder set, we will see that each soe set and soe_rem set have a similar relationship as follows:

Each soe set factor wise, is simply that prime number's geometric series times the soe_rem set (remainder set), expressed as a series.

So we can add to our 'Sieve of Eratosthenes' definitions.

Sieve of Eratosthenes set definitions:

For each P (prime),

Each soe set can be defined as: 

The subset of all natural numbers that have a smallest common prime factor of P.

Each soe_rem set can be defined as: 

The subset of all natural numbers that only contains the number 1 and the natural numbers with prime factors greater than the prime number P.

Each soe set can be expressed as a product:
Where the first term is the prime number P's geometric series with positive integer exponents and the second term is the infinite series of all of the elements of the soe_rem set.

Or speaking purely in the context of series, the infinite series of all natural numbers:

For each P (prime),

Each soe subseries can be defined as: 

The subseries of all natural numbers that have a smallest common prime factor of P.

Each soe_rem subseries can be defined as: 

The subseries of all natural numbers that only contains one and the natural numbers with prime factors greater than the prime number P.

Each soe subseries can be expressed as a product:
Where the first term of the product is the prime number P's geometric series with positive integer exponents, and the second term of the product is the infinite subseries of all natural numbers that only contains one and the natural numbers with prime factors greater than the prime number P. (This is the property that we are concerned with for the 'Sieve of Eratosthenes'.)

With the definition of this new property of the 'Sieve of Eratosthenes', we can use the following logic to pursue the equality we are looking for, namely 'the infinite series of all natural numbers has an equality as an infinite product of prime geometric series'.

The 'Sieve of Eratosthenes' performed on the infinite series of all natural numbers:

$$\begin{array}{}\text{(1a)}& x=1+2+3+4+5+6+7+8+9+10+\cdots \end{array}$$

Sieving pass for the prime number 2.
We can reorder the infinite series on the right hand side, to be the soe2 infinite subseries (the evens) + the soe2_rem infinite subseries (the odds).
$$x=\underset{\text{soe2}}{\underbrace{(2+4+6+8+10+\cdots )}}+\underset{\text{soe2\_rem}}{\underbrace{(1+3+5+7+9+\cdots )}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \underset{\text{soe2\_rem}}{\underbrace{(1+3+5+7+9+\cdots )}}$$

Sieving pass for the prime number 3.
We can re-order the terms of the last term of the product on the right hand side, sieving by the next prime number, 3.
This gives us an soe3 subseries and an soe3_rem subseries.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \underset{\text{soe3}}{\underbrace{((3+9+15+21+27+\cdots )}}+\underset{\text{soe3\_rem}}{\underbrace{(1+5+7+11+13+\cdots ))}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \underset{\text{soe3\_rem}}{\underbrace{(1+5+7+11+13+\cdots )}}$$

Sieving pass for the prime number 5.
We can re-order the terms of the last term of the product on the right hand side, sieving by the next prime number, 5.
This gives us an soe5 subseries and an soe5_rem subseries.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \underset{\text{soe5}}{\underbrace{((5+25+35+55+65+\cdots )}}+\underset{\text{soe5\_rem}}{\underbrace{(1+7+11+13+17+\cdots ))}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \underset{\text{soe5\_rem}}{\underbrace{(1+7+11+13+17+\cdots )}}$$

Sieving pass for the prime number 7.
We can re-order the terms of the last term of the product on the right hand side, sieving by the next prime number, 7.
This gives us an soe7 subseries and an soe7_rem subseries.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \underset{\text{soe7}}{\underbrace{((7+49+77+91+119+\cdots )}}+\underset{\text{soe7\_rem}}{\underbrace{(1+11+13+17+19\cdots ))}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ 7^2\\ +\\ 7^3\\ +\\ 7^4\\ +\\ 7^5\\ +\\ ⋮\\ 7^n\end{array}\right)\times \underset{\text{soe7\_rem}}{\underbrace{(1+11+13+17+19\cdots ))}}$$

This process can be continued for every prime number and the result is:

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ 7^2\\ +\\ 7^3\\ +\\ 7^4\\ +\\ 7^5\\ +\\ ⋮\\ 7^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ {11}^2\\ +\\ {11}^3\\ +\\ {11}^4\\ +\\ {11}^5\\ +\\ ⋮\\ {11}^n\end{array}\right)\times \left(\begin{array}{c}\\ {13}^0\\ +\\ {13}^1\\ +\\ {13}^2\\ +\\ {13}^3\\ +\\ {13}^4\\ +\\ {13}^5\\ +\\ ⋮\\ {13}^n\end{array}\right)\times \left(\begin{array}{c}\\ {17}^0\\ +\\ {17}^1\\ +\\ {17}^2\\ +\\ {17}^3\\ +\\ {17}^4\\ +\\ {17}^5\\ +\\ ⋮\\ {17}^n\end{array}\right)\times \left(\begin{array}{c}\\ {19}^0\\ +\\ {19}^1\\ +\\ {19}^2\\ +\\ {19}^3\\ +\\ {19}^4\\ +\\ {19}^5\\ +\\ ⋮\\ {19}^n\end{array}\right)\times \cdots$$

Substituting for the value of x in equation $\text{(1a)}$ we get:

The infinite series of all natural numbers has an equality as an infinite product of prime geometric series:

$$1+2+3+4+5+6+7+8+9+10+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ 7^2\\ +\\ 7^3\\ +\\ 7^4\\ +\\ 7^5\\ +\\ ⋮\\ 7^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ {11}^2\\ +\\ {11}^3\\ +\\ {11}^4\\ +\\ {11}^5\\ +\\ ⋮\\ {11}^n\end{array}\right)\times \left(\begin{array}{c}\\ {13}^0\\ +\\ {13}^1\\ +\\ {13}^2\\ +\\ {13}^3\\ +\\ {13}^4\\ +\\ {13}^5\\ +\\ ⋮\\ {13}^n\end{array}\right)\times \left(\begin{array}{c}\\ {17}^0\\ +\\ {17}^1\\ +\\ {17}^2\\ +\\ {17}^3\\ +\\ {17}^4\\ +\\ {17}^5\\ +\\ ⋮\\ {17}^n\end{array}\right)\times \left(\begin{array}{c}\\ {19}^0\\ +\\ {19}^1\\ +\\ {19}^2\\ +\\ {19}^3\\ +\\ {19}^4\\ +\\ {19}^5\\ +\\ ⋮\\ {19}^n\end{array}\right)\times \cdots$$

Summarizing the 'Sieve of Eratosthenes' Process

This 'Sieve of Eratosthenes' Process can be applied to the natural numbers, the inverse natural numbers, and the rational numbers.

In the inverse natural numbers, the sieving process is the same, except we are working with 0 and negative integer exponent values. We even see the 'Sieve of Eratosthenes' sets and remainder sets employed in Euler's Theorem 7, Euler's infinite product over the primes. 

In the rational numbers, the sieving process is the same, except that we need to sieve in both the numerator and denominator for each iteration, and the prime geometric series then includes exponents 0, positive integers, and negative integers. The property where each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries is also true in the rationals.

Example: The first iteration of sieving all rationals, we would sieve all rational numbers with a factor of 2 in the numerator or the denominator and our remainder set would then have only rational numbers with prime factors of 3 or greater in the numerator or denominator. From that first remainder set then we would sieve out all rational numbers with a factor of 3 in the numerator or denominator and the new remainder set would then have only rational numbers with prime factors of 5 or greater in the numerator or denominator. I can provide an additional post to step through this is needed.

In the inverse natural numbers the prime geometric series is convergent and we have an algebraic solution to prove that the series of inverse natural numbers has an equality as an infinite product of prime geometric series, à la Euler's infinite product over the primes. This however does not mean that the natural numbers and rational numbers do not have the same type of equality, it is just that the prime geometric series in the naturals and the rationals are not convergent and this prevents them from having an algebraic solution similar to Euler's Theorem 7.

The Sieve of Eratosthenes - Backing Up

I have realized much of the information I have provided requires a greater level of context and simplification. This is the beginning of the effort to provide better context with a much simpler foundation. My idea or proposal that the infinite series of all natural numbers has an equality as an infinite product of prime geometric series, just as Euler's product is an infinite product of inverse prime geometric series, requires a simpler foundation and mathematical logic. That simpler foundation and mathematical logic is the 'Sieve of Eratosthenes'.

I will use the explanation for the the 'Sieve of Eratosthenes' from Wikipedia, https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes.

Overview

A prime number is a natural number that has exactly two distinct natural number divisors: the number 1 and itself.

To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method:

1. Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n).
2. Initially, let p equal 2, the smallest prime number.
3. Enumerate the multiples of p by counting in increments of p from 2p to n, and mark them in the list (these will be 2p, 3p, 4p, ...; the p itself should not be marked).
4. Find the smallest number in the list greater than p that is not marked. If there was no such number, stop. Otherwise, let p now equal this new number (which is the next prime), and repeat from step 3.
5. When the algorithm terminates, the numbers remaining not marked in the list are all the primes below n.

The main idea here is that every value given to p will be prime, because if it were composite it would be marked as a multiple of some other, smaller prime. Note that some of the numbers may be marked more than once (e.g., 15 will be marked both for 3 and 5).

SKopp at German Wikipedia, CC BY-SA 3.0, via Wikimedia Commons

For the data that we are interested in from the 'Sieve of Eratosthenes', we want the sets of numbers that are the result of each iteration, and only the numbers that have not been previously marked. We also want to know what the remainder set is after each iteration. The list of integers that we begin with also starts with the number 1 and therefore the number 1 will always be in each remainder set.

Example: For the prime number 3 iteration, the original sieve marks out 3,6,9,12,15,18,21,24,27,... However 6,12,18,24 have already been marked out by the previous prime number 2 iteration. We only want the set of numbers from each iteration that have not been previously marked out. Also, we want to include the prime number in each iteration as a member of that iteration's sieve set. The easiest method to accomplish this is to only mark out multiples of the next prime number in the previous iteration's remainder set.

Here is a table with the 'Sieve of Eratosthenes' sets we are looking for marked out.

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20
21	22	23	24	25	26	27	28	29	30	31	32	33	34	35	36	37	38	39	40
41	42	43	44	45	46	47	48	49	50	51	52	53	54	55	56	57	58	59	60
61	62	63	64	65	66	67	68	69	70	71	72	73	74	75	76	77	78	79	80
81	82	83	84	85	86	87	88	89	90	91	92	93	94	95	96	97	98	99	100
101	102	103	104	105	106	107	108	109	110	111	112	113	114	115	116	117	118	119	120
121	122	123	124	125	126	127	128	129	130	131	132	133	134	135	136	137	138	139	140
141	142	143	144	145	146	147	148	149	150	151	152	153	154	155	156	157	158	159	160
161	162	163	164	165	166	167	168	169	170	171	172	173	174	175	176	177	178	179	180
181	182	183	184	185	186	187	188	189	190	191	192	193	194	195	196	197	198	199	200
201	202	203	204	205	206	207	208	209	210	211	212	213	214	215	216	217	218	219	220

Legend:

Prime	Set Name	SOE Sieve Set	Remainder Name	SOE Sieve Remainder Set
2	soe2	2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40, 42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78, 80,82,84,86,88,90,92,94,96,98,100,102,104,106,108,110,112, 114,116,118,120,122,124,126,128,130,132,134,136,138,140, 142,144,146,148,150,152,154,156,158,160,162,164,166,168, 170,172,174,176,178,180,182,184,186,188,190,192,194,196, 198,200,202,204,206,208,210,...	soe2_rem	1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39, 41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77, 79,81,83,85,87,89,91,93,95,97,99,101,103,105,107,109,111, 113,115,117,119,121,123,125,127,129,131,133,135,137,139, 141,143,145,147,149,151,153,155,157,159,161,163,165,167, 169,171,173,175,177,179,181,183,185,187,189,191,193,195, 197,199,201,203,205,207,209,...
3	soe3	3,9,15,21,27,33,39,45,51,57,63,69,75,81,87,93,99,105,111, 117,123,129,135,141,147,153,159,165,171,177,183,189,195, 201,207,...	soe3_rem	1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55,59, 61,65,67,71,73,77,79,83,85,89,91,95,97,101,103,107,109,113, 115,119,121,125,127,131,133,137,139,143,145,149,151,155,157, 161,163,167,169,173,175,179,181,185,187,191,193,197,199,203, 205,209,...
5	soe5	5,25,35,55,65,85,95,115,125,145,155,175,185,205,...	soe5_rem	1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73, 77,79,83,89,91,97,101,103,107,109,113,119,121,127,131,133, 137,139,143,149,151,157,161,163,167,169,173,179,181,187,191, 193,197,199,203,209,...
7	soe7	7,49,77,91,119,133,161,203,...	soe7_rem	1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83, 89,97,101,103,107,109,113,121,127,131,137,139,143,149,151, 157,163,167,169,173,179,181,187,191,193,197,199,209,...

Interestingly, the soe sets and remainder sets are all periodic and symmetric. 

This is easy to notice starting in the soe3_rem remainder set.

soe3_rem: {1,5} for the primorial length of 6.

{1,5} +6n, where n = 0→∞

*** This is where we get the statement that every prime number greater than 3, must be of the form 6n ±1

soe5 set: {5,25} for the primorial length of 30. 

{5,25} +30n, where n = 0→∞

soe5_rem set: {1,7,11,13,17,19,23,29} for the primorial length of 30.

{1,7,11,13,17,19,23,29} +30n, where n = 0→∞

*** The soe5_rem set is the foundation of Gary Croft's: 'The Prime Spiral Sieve' https://www.primesdemystified.com

soe7 set: {7,49,77,91,119,133,161,203} for the primorial length of 210. 

{7,49,77,91,119,133,161,203} +210n, where n = 0→∞

soe7_rem set: {too big to fit here, see below} for the primorial length of 210.

{1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,121,127,131,137,139,143,149,151,157,163,167,169,173,179,181,187,191,193,197,199,209} +210n, where n = 0→∞

*** The soe7_rem set is the foundation of a recent paper entitled “The Periodic Table of Primes" [4]. Where the authors argue that primes can be predicted using a periodic table-like structure." 

https://phys.org/news/2024-04-breakthrough-prime-theory-primes.html

https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4742238

These 'Sieve of Eratosthenes' sets and remainder sets are also the basis for James McCanney's work, 'New Definition of Prime Numbers with Sppn Tables and Proofs by Induction'.

https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4800997#:~:text=James%20McCanney,-jmccanneyscience%2C%20LLC&text=The%20prime%20numbers%20are%20calculated,with%20unique%20ancestors%20and%20descendants.

These 'Sieve of Eratosthenes' sets and remainder sets are also the basis for 'Liu Fengsui's Prime Formula' and subsequent papers.

https://www.primepuzzles.net/problems/prob_037.htm

The 'Sieve of Eratosthenes' sets and remainder sets have a lot of interesting properties. However we are interested in a different property of these sets and that will be the topic of my next post.

Continuation: The 'The Sieve of Eratosthenes - Part 2

The Conundrum - Free Will and Choice

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

In my previous post, I presented 'The Conundrum'. In this post, I will look deeper at 'The Conundrum' and ask the reader, is it real or is it not?

In my previous post, I looked at Euler's infinite product over the primes derived from the harmonic series (the inverse natural numbers). Euler's infinite product over the primes is in fact an infinite product of inverse prime geometric power series. I consider this mathematical object the generating object of all inverse natural numbers. It kind represents the reverse of factorization. Instead of taking the inverse natural numbers and determining what prime factors a specific inverse natural number is composed of, we are instead generating/creating all inverse natural numbers from their generating object that supplies all of those prime factors, and this is basically just simple algebra, multiplicative distribution of the infinite product of inverse prime geometric power series.

But then I closed my prior post with this.

The Conundrum

What is this?

The above infinite paths, which can also be represented in the infinite multiplicative sequences model, never terminate to 1,1,1,1,1,... in their rightward progression. There is always another term in the rightward progression that is not a selected value of 1. The question is, do they really exist?

Remember that all of this information comes from a math paper that I was asked to write, and it took about six months to complete the paper, and I posted the paper in the forum and thread of the original requester asking me to write it. I received very few responses (actually 2 members responded, neither of which was the original requester). One of the responses from an individual that seemed to kind of grasp what I was trying to point out was this:

Things like the harmonic series are certainly fun. I would tend to think your infinitely factored set is related to the harmonic series set having infinite members (even though no member itself is infinite). Infinities are headache inducing and divergent infinite series means you can't even just check for approaching a limit but it actually is fun reading your work.

It has been 30 months since I received that response. I never replied back, as I was baffled. I couldn't understand how the infinitely factored (multiplicatively distributive) paths could not exist, how anyone could not see from both the visual trace path model and the infinite multiplicative sequences model that these infinitely factored inverse numbers existed. I also did not think the part of the response about infinities being divergent and difficult was even relevant to the question. The question is whether the infinitely factored numbers exist or not, not that they may be difficult to deal with if they do exist. That difficulty has nothing to do with whether they are extant or not. I had hoped someone would actually look at the logic and methods, and comment on the validity of the logic and methods. If the logic and methods are wrong, then why are they wrong?

I have come to the conclusion that not being able to see the infinitely factored numbers may come down to what I started this series of posts about to begin with, and that is the shortcuts and assumptions built into our mathematical education whose foundation is the counting numbers 1, 2, 3, 4, 5,...

More Than the Sum of Its Parts?

Euler's infinite product over the primes was derived from the harmonic series, the infinite summation series of all inverse natural numbers (all inverse counting numbers). How could the derived infinite product result have more numbers in it than Euler initially started with? It is counterintuitive. But that is indeed the assumption drilled into everyone from our mathematical education based on these counting numbers. The counting numbers are the foundation of everything we have today in mathematics and probably in society as a whole, all science, engineering, physics, etc. This foundation is drilled into us from when we are three years old and dominates our world view, with all of its assumptions and shortcuts. The counting numbers are basically the first principles in the hierarchy of mathematical education from day one. The question is, should they be the first principles that everything else is built on?

To me, the answer to my previous question is No. I think the infinite product of prime geometric power series should be the base, the foundation, the fundamental object, the first principles, and what arises from it becomes the second step.

If we examine the definition of the counting numbers or inverse counting numbers (naturals or inverse naturals), every element of each set has a finite value, meaning a finite set of prime factors. That is their definition. But if we take the infinite product of prime geometric power series as the foundation, it is plain as daylight that there are infinite multiplicative distributive paths that never have that infinite progression to the right where all values progress with an infinite signature of all ones, 1,1,1,1,1,... The only way to get rid of the infinite paths that always have another term to the right that is not a selected value of 1 is to impose a rule, an artificial constraint from the outside on the infinite product of prime geometric power series. We would have to create a rule that said, "No distinct/unique infinite multiplicative distributive path can have a signature where the infinite progression on the righthand side never becomes 1,1,1,1,1,... Now we can create that rule and inject it into our mathematical logic, but to me that is nonsense.

If we look at the infinite product of inverse prime geometric power series and use the infinite multiplicative sequences model as a proxy for the multiplicative distribution and we add that artificial new rule we just created, what would that mean?

Let's look at that.

In the infinite multiplicative sequences model, each term corresponds to a term in the infinite product of inverse prime geometric power series. There are infinite terms in the infinite product of inverse prime geometric power series, so we must have infinite terms in our infinite multiplicative sequences model. The values that are available to choose for each term of a multiplicative sequence are the corresponding values available from the corresponding term of the infinite product of inverse prime geometric power series; those values are either 1 (one) or an integer power of the corresponding prime number.

But that new rule we have just created limits us. It tells us that every distinct/unique combination of sequences we create must, at some point in its infinite rightward progression, somewhere there in that progression, we no longer have a choice. For a number to be finitely factored, every distinct/unique sequence you come up with, somewhere in that sequence, the choice of selecting either the number 1 (one) or a corresponding prime power for the next term, there is no choice, from that point rightward, you can only select the value 1 (one) for all of the rest of the infinite terms to the right. In effect, at some point you have no choice, your choice has already been determined for you, choice is no longer possible.

To me, this almost comes down to a philosophical argument. Do we have free will? Do we have choice? That is why I included 'Free Will and Choice' in the title of this post. Does Free Will and Choice exist?

To me, the answer is Yes. To me, I am unable to impose that artificial constraint on the infinite product of prime geometric power series. This is also why I think it is the 'infinite product of prime geometric power series' mathematical object that should be foundational, fundamental. If we make the 'infinite product of prime geometric power series' mathematical object foundational and we do not put any external artificial constraints on it, then what we see emerge from it are two sets of numbers. One where every member of the set has finite factors, and a second set where every member of the set has infinite factors.

Further observations of what I call this foundational mathematical object lead me to think that of all the numbers created by this foundational mathematical object, 1/2 are finite and 1/2 are infinite, and you cannot have one of the sets (finitely valued members) without the other set (infinitely valued members). In fact, both emerge bilaterally, dually, and simultaneously. The mathematical object responsible for creating members of both sets is a prime geometric power series term in the infinite product. If you remove, for instance, the inverse two power series from the infinite product, not only do all finitely factored members of the set of inverse natural numbers disappear from the result (all finitely factored numbers with a factor of 2 in them), but also all members of the parallel infinitely factored set also disappear (infinitely factored numbers with a factor of 2 in them). The two sets are bonded together, one cannot exist without the other existing.

So if these infinitely factored numbers do exist and I cannot see how they do not, then where are they in the literature? Well, basically, they are not in the literature. The closest thing in the literature is the formulation of the surreal numbers with the dyadics, birthdays, and day w, or the recent formulation of non-standard analysis and its notion of infinitesimals. But as I have mentioned before, this information (the parallel sets of infinitely factored numbers) could easily be derived 2300 years ago simply from Euclid's Elements or the Sieve of Eratosthenes, which were both available somewhere around the third century BCE.

I will not go into further information in this post on expanding this observation (the parallel infinitely factored sets arising from the infinite product of prime geometric power series), as proceeding further requires the acceptance that these infinitely factored numbers arise naturally just as the finitely factored numbers arise naturally from the infinite product of prime geometric power series, and until that Conundrum is decided upon, going further would be futile because all further expansion of the concept requires the acceptance of these infinitely factored numbers.

I will, however, briefly mention where my math paper goes that all this information comes from.

Accepting that the infinitely factored numbers exist, my paper continues and expands the same observations to the rational numbers and then to the real numbers via the introduction of a real valued variable into the exponents of every prime power series, similar to Euler's zeta function, then it continues to expand to two additional levels. In all, my paper proposes that there are six levels of numbers that are created by the infinite product of prime geometric power series.

1. The inverse natural numbers
2. The natural numbers
3. The rational numbers
4. The real numbers via an Euler like zeta function
5. A first level where only infinitely factored numbers reside
6. A second, even greater level where only infinitely factored numbers reside

In the first four levels, all resultant numbers created by the infinite product of prime geometric power series produce 1/2 of the numbers as being finite and 1/2 of the numbers as being infinite.

 

The next two levels I have no name for, levels 5 and 6, have no finite numbers at all. All members of these two levels are infinite.

Thus was born 'The Conundrum'.

Will this effort of mine lead anywhere or will it collapse? 

Stay tuned...

The Conundrum

 

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 will provide a little bit of background information here once more. The main point of these posts so far is that my own personal experience in mathematical education has left me thinking that I was misled (to some degree) with assumptions and shortcuts that were built into that education. This post will delve into more of the specifics of that feeling I have.

About 5-7 years ago, I came upon information that did not make sense to me regarding the foundations of arithmetic. I knew for years that the fundamental theorem of arithmetic directly implied the infinite product of prime geometric power series in the natural numbers, but I wanted to also see if the same structure existed in the inverse natural numbers, and I found that information in Leonhard Euler's 'Theorem 7', also known as Euler’s Product Formula, from his 1744 published work 'Variae observationes circa series infinitas' Several Remarks on Infinite Series. I already covered this information in the prior post, 'The Inverse Naturals and The Euler Product'. However, I got the idea to do something further with the Euler Product. I wanted to reverse the process that Euler used, to perform something like a validation of Euler's Theorem 7. Instead of starting with the harmonic series and deriving Euler's infinite product over the primes, I wanted to reverse that process by starting with Euler's result, the infinite product, and recover the harmonic series elements.

This gave birth to The Conundrum.

I gathered a lot of this information and compiled it into a pdf document and posted it to a forum I participate in back in 2019. I received a few responses from participants, but most, I think, did not understand what I was talking about. Then in the fall of 2021, I made another attempt to mention this information that I had gathered in a more scientific and mathematical oriented thread where I thought the audience had more mathematical knowledge. Most of the participants' mathematical knowledge far exceeded my own, including several whose careers are in advanced mathematics and research.  Here is the response I received.

The problem is with your presentation of the results. In mathematics we have definitions and theorems and proofs. Can you condense your results in a theorem, clearly stated, and a proof, clearly written? Then perhaps it will be readable.

The responder was correct. Taking a good look at what I had compiled into a document was very disjointed and would be hard for anyone to discern, especially to someone actually in the field of mathematics. So I responded.

Thank you. I will try.

I have never written a math paper before. I understand and can read math papers (if they are not too advanced). I had written proofs before in that old American 1-12 education decades ago. But the information I had gathered was not very amenable to formal theorems and proofs. Still, I thought I could make the information more discernible and logical by using a math paper structure with logic and observations. I have used some of that information in these blog posts already and I will continue to present more of that information and logic.

It took me six months to find a proper math paper template and I created a very lengthy html file to which I added a math JavaScript library and a few custom JavaScript scripts for custom linking and citations.  I posted the completed document to the forum as a response to the actual post where I was asked to write a math paper. The next morning someone responded very enthusiastically, which was encouraging, but the responder went kind of overboard about their own interests, talking about loop time in my posted document. The responder wanted more time to read my paper in depth before responding further.

I pre-read what you wrote. You've put a lot of work into it. I want to be more specific on this, but I need a little more time for that. I think it will be possible next week.

However, I sincerely congratulate you on some of your insights. This is something that very few human beings will understand. I understand you, but I need to read it a few more times.

Nevertheless, you see very well what I just said a few days ago (loop logic, linear versus loop time, and maybe even a time-light relationship? It can be seen here!), that nobody sees. It is beautiful. But bear in mind that I am a bit crazy. I am in love with time etc.

I will look at what you wrote and I promise to analyze it in depth before speaking. It may take a while, but I will keep it in mind.

Three and one half weeks later, nothing.

So I tried to mention my post and paper in another thread with a much greater audience, a thread where the original first responder who requested the math paper was also commenting. I had some success in this second thread. The original requester of the paper did not respond, but another forum member with an advanced mathematics background did. This was their response.

It would be difficult to connect your work with things like the different types of particles and forces. People will tend to want to produce the "loops" with math that fits better with the math already being used for particles/forces. Sometimes there can actually even be different correct ways to do something but some ways may fit better with what you want to do next.

What is it with these responses about particles, forces, and loops? My paper was about basic simple high school algebra, with observations and logic about something being amiss with our foundations of arithmetic. So I responded to this forum poster with what I was trying to get at in my paper, which I will reveal here shortly in 'The Conundrum', but that seemed to elicit a response from the responder that was a little dismissive, which I understand, as it is a little difficult to look at simple things you may have learned as a child, but that also has to do with the point of my posts here and the assumptions and shortcuts ingrained in all of us by our early mathematical education. Here is the response.

Things like the harmonic series are certainly fun. I would tend to think your infinitely factored set is related to the harmonic series set having infinite members (even though no member itself is infinite). Infinities are headache inducing and divergent infinite series means you can't even just check for approaching a limit but it actually is fun reading your work.  

So I will  now attempt to explain 'The Conundrum' and what this is all about.

The Conundrum, relates back to all my earlier posts. If you haven't already read the previous posts, this will not make a lot of sense. Please read the prior posts, of which there are five. Each takes at most five minutes.

1. Introduction 
2. Foundations - The Counting Numbers
3. Foundations - The Fundamental Theorem of Arithmetic
4. Foundations - The Fundamental Theorem of Arithmetic - Part 2
5. The Inverse Naturals and The Euler Product

In 'The Conundrum', the assumptions and shortcuts I have mentioned become overwhelmingly apparent. 

In the previous posts, I have shown that the natural numbers have a generating object, the infinite product of prime geometric power series. I have also shown that this very same structure exists in the inverse natural numbers via Leonhard Euler's 'Theorem 7', also known as Euler’s Product Formula, where Euler produces via simple algebra the inverse infinite product of prime geometric power series that produce all inverse natural numbers.

I will use Leonhard Euler's inverse infinite product of prime geometric power series that produce all inverse natural numbers for my demonstration of 'The Conundrum'.

Here it is:

In the 'inverse infinite product of prime geometric power series,' we can easily see that the infinite product produces all members of the inverse natural numbers. Using either the visual trace model or the infinite multiplicative sequences model we can see that the 'inverse infinite product of prime geometric power series' produces every single countably infinite inverse natural number. 

In the visual trace model, we can see that every inverse natural number generated has a specific form or signature. That signature is such that every distinct infinite trace path always, eventually becomes a flat line where every value selected to the infinite right progression of the trace path all becomes ones, 1,1,1,1,1,1,....

In the infinite multiplicative sequences model, we see the same thing, and we should see the exact same thing, as each term of the infinite multiplicative sequences model corresponds to a value selected from the corresponding prime power series term of the infinite product. Just as in the trace model, in the infinite multiplicative sequences model, we see that for every natural number, eventually the progression to the right always becomes 1,1,1,1,1,1,...

This is the very definition of an inverse natural number. The set of all inverse natural numbers is countably infinite (meaning there are countably infinite members), but each individual member is finite. If any member of the inverse natural numbers had countably infinite factors, then it would never assume the signature in the rightward progression of 1,1,1,1,1,... If an inverse natural number never acquired that rightward progression signature of 1,1,1,1,1,..., in other words, there was always another term in the rightward  progression that was not a 1, then it would be an infinite number, not a finite number, and by definition it would not be an inverse natural number.

Here is a review of Euler's infinite product, which he derived from the harmonic series (infinite sum of all inverse natural numbers).

The infinite  product of inverse prime geometric power series derived from the harmonic series.

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

Here is the visual trace path model, demonstrating that all inverse natural numbers have an infinite progression to the right (eventually) of 1,1,1,1,1,...

Here is the infinite multiplicative sequences model, also demonstrating that all inverse natural numbers have an infinite progression to the right (eventually) of 1,1,1,1,1,...

$$\begin{array}{cl}\frac{1}{n}& \text{sequence}\\ 1& 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{2}& 2^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{3}& 1,3^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{4}& 2^{-2},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{5}& 1,1,5^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{6}& 2^{-1},3^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{7}& 1,1,1,7^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{8}& 2^{-3},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{9}& 1,3^{-2},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{10}& 2^{-1},1,5^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{11}& 1,1,1,1,{11}^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{12}& 2^{-2},3^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{13}& 1,1,1,1,1,{13}^{-1},1,1,1,1,1,1,1,1,1,1,1,1,...\end{array}$$

The Conundrum

What is this?

Continue to - The Conundrum - Free Will and Choice