The general set operation to determine all prime numbers in a selected range

PRIME NUMBER SERIES 4

Let $\mathbb{P}$ be the infinite set of all prime numbers.

Proposition:

$\mathbb{P}$ = (ℕ \ {0,1}) \ (ℕ \ {0,1})²

Proof:

ℕ = {0, 1, 2, …}

ℕ \ {0,1} = {2, 3, 4, …}

(ℕ \ {0,1})² = (ℕ \ {0,1}) x (ℕ \ {0,1}) = {2,3,4,…} x {2,3,4,…} =

(ℕ \ {0,1})² = {{4,6,8,…},{6,9,12,…},{8,12,16,…},…}

Set ℕ consists of the numbers 0 and 1 as well as all composite numbers and all not composite numbers, i.e. prime numbers.

Definition: A composite number is the product of two natural numbers, both > 1.

Mind: It does not mean two prime factors. That would be wrong. But: In the case a composite number is built of more than two prime factors it is always possible to bring it into the form of two natural numbers.

ℕ = {{0,1}, {a*b ʌ a, b ϵ ℕ ʌ a, b > 1}, $\mathbb{P}$ }

Because all prime numbers p ϵ ℕ ʌ p > 1, {0,1} can not be a partial set of $\mathbb{P}$ .

The difference set

ℕ \ {0,1} = {2, 3, 4, …}

covers all natural numbers >1. This difference set consists of the integral divisible numbers and set $\mathbb{P}$ .

With

(ℕ \ {0,1}) x (ℕ \ {0,1}) = {2, 3, 4, …} x {2, 3, 4, …} =

= {{4, 6, 8, …}, {6, 9, 12, …}, {8,12,16, …}, {10, 15, 20, …}, {12,18,24}, {…}, …}

are covered all possible products of the natural numbers >1. For this reason all integral divisible natural numbers are given.

This proves: The difference set

$\mathbb{P}$ = (ℕ \ {0,1}) \ (ℕ \ {0,1})²

can only consist of the infinite set of all prime numbers.

Example: It shall be determined every prime number 1000<p<1020 in case existing:

The range of numbers to be examined is set $\overset{1019}{\underset{1001}{\mathbb{N}}}$ with

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ = {n | n ϵ ℕ ʌ 1000<n<1020} = {1001,1002,1003,1004,1005,1006,1007,1008,1009,1010,1011,1012,1013,1014,

1015,1016,1017,1018,1019}

The set Z of composite numbers in set N is determined by

Z = {z = a*b ʌ a, b ϵ ℕ ʌ a,b > 1}

2 ≤ a ≤ √1019

2 ≤ a ≤ 31

There has not to be used every a for the determination, but it is enough to use every prime factor in the calculated range 2 ≤ a ≤ 31.

With a = 2 and 500 < b < 510 the resulting product set is

Z₂ = {1002, 1004, 1006, 1008, 1010, 1012, 1014, 1016, 1018}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ = {1001, 1003, 1005, 1007, 1009, 1011, 1013, 1015, 1017, 1019}

With a = 3 and 333 < b < 340 the resulting product set is

Z₃ = {1002, 1005, 1008, 1011, 1014, 1017}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ = {1001, 1003, 1007, 1009, 1013, 1015, 1019}

With a = 5 and 200 < b < 204 the resulting product set is

Z₅ = {1005, 1010, 1015}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ = {1001, 1003, 1007, 1009, 1013, 1019}

With a = 7 and 142 < b < 146 the resulting product set is

Z₇ = {1001, 1008, 1015}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ = {1003, 1007, 1009, 1013, 1019}

With a = 11 and 90 < b < 93 the resulting product set is

Z₁₁ = {1001, 1012}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁= {1003, 1007, 1009, 1013, 1019}

With a = 13 and 76 < b < 79 the resulting product set is

Z₁₃ = {1001, 1014}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁\ Z₁₃= {1003, 1007, 1009, 1013, 1019}

With a = 17 and 58 < b < 60 the resulting product set is

Z₁₇ = {1003}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁\ Z₁₃\ Z₁₇= {1007, 1009, 1013, 1019}

With a = 19 and 52 < b < 54 the resulting product set is

Z₁₉ = {1007}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁\ Z₁₃\ Z₁₇\ Z₁₉= {1009, 1013, 1019}

With a = 23 and 43 < b < 45 the resulting product set is

Z₂₃ = {1012}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁\ Z₁₃\ Z₁₇\ Z₁₉\ Z₂₃= {1009, 1013, 1019}

With a = 29 and 34 < b < 36 the resulting product set is

Z₂₉ = {1015}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁\ Z₁₃\ Z₁₇\ Z₁₉\ Z₂₃\ Z₂₉= {1009, 1013, 1019}

With a = 31 and 32 < b < 33 the resulting product set is

Z₃₁ = {}

$\overset{1019}{\underset{1001}{\mathbb{N}}}$ \ Z₂ \ Z₃ \ Z₅ \ Z₇ \ Z₁₁\ Z₁₃\ Z₁₇\ Z₁₉\ Z₂₃\ Z₂₉\ Z₃₁= {1009, 1013, 1019}

$\mathbb{P}$ = {1009, 1013, 1019}

Result: There are the prime numbers 1009, 1013, 1019 in the range 1000<p<1020.

Munich, 27 October 2019

Gottfried Färberböck

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THE PRIME NUMBER SERIES

The general set operation to determine all prime numbers in a selected range