Posted on by Mike Hotchkiss

Another Approach to Proof of the Twin Prime Conjecture

Summary

The Hotchkiss proof of the Twin Prime Conjecture presents a comprehensive argument demonstrating the impossibility of a largest twin prime pair. Beginning with the assumption of the existence of such a pair and its implications for fundamental theorems like the Prime Number Theorem and Euclid’s Theorem, the proof methodically deconstructs the notion of a largest twin prime pair. Utilizing the Hotchkiss Prime Theorem alongside established principles in number theory, the proof establishes the independence and mutual exclusivity of primes in defined sets, highlighting the interaction between these sets and composite values. By demonstrating that any assumed largest twin prime pair leads to a contradiction with the concept of infinitely many primes and the inclusive nature of the defined sets, the proof concludes that there will always be another twin prime pair beyond any assumed largest pair. Overall, the Hotchkiss proof provides a rigorous and compelling argument for the validity of the Twin Prime Conjecture.

Optimized Primes

Rationale

Assume for contradiction that there exists a largest twin prime pair, denoted as (p, q), where p and q are both prime numbers and there are no twin prime pairs beyond this pair.

Existence of a Largest Twin Prime Pair:

  • Let (p, q) be the largest twin prime pair, where p and q are consecutive primes such that q = p + 2. By assumption, there are no twin prime pairs beyond (p, q).

Consequence for the Prime Number Theorem:

  • The prime number theorem states that the number of primes less than or equal to a given number x is approximately x / ln(x). If there were a largest twin prime pair (p, q), it would imply that there is a finite limit to the number of primes, contrary to the prime number theorem, which asserts that the number of primes is unbounded. Therefore, the existence of a largest twin prime pair contradicts the prime number theorem.

Consequence for Euclid’s Theorem:

  • Euclid’s theorem states that there are infinitely many primes. If there were a largest twin prime pair (p, q), it would imply that there is a largest prime, namely q, and thus a finite number of primes. This contradicts Euclid’s theorem, which guarantees the existence of infinitely many primes.

Overall Contradiction:

  • The prime number theorem and Euclid’s theorem are both well-established and widely accepted in mathematics. The assumption of a largest twin prime pair leads to a contradiction with these fundamental theorems. Since the prime number theorem and Euclid’s theorem have been rigorously proven and are integral to number theory, the existence of a largest twin prime pair cannot hold true.
  • Therefore, we conclude that the existence of a largest twin prime pair would invalidate both the prime number theorem and Euclid’s theorem, leading to a contradiction. Consequently, the Twin Prime Conjecture, which asserts the existence of infinitely many twin primes, must be true.

In order to prove this, we will apply the Hotchkiss Prime Theorem.

1. “Hotchkiss Prime Theorem”:

Let A = {6x + 5 | x ∈ ℤ} be the set of all numbers of the form 6x + 5, and B = {6y + 7 | y ∈ ℤ} be the set of all numbers of the form 6y + 7. Let AA, AB, and BB represent the sets of products:

AA = {(6x + 5)(6y + 5) | x, y ∈ ℤ}

AB = {(6x + 5)(6y + 7) | x, y ∈ ℤ}

BB = {(6x + 7)(6y + 7) | x, y ∈ ℤ} Then, any number that is an element of A or B but not an element of AA, AB, or BB is a prime number.

Proof:

Assumption: Assume there exists a number k that is:

  • Composite (not prime).
  • An element of either set A or B (i.e., it’s of the form 6x + 5 or 6y + 7).
  • Not an element of AA, AB, or BB.

Case 1: k is of the form 6x + 5 (k ∈ A)

Since k is composite, it has at least two factors, say a and b, where a > 1 and b > 1. Since k is odd, both a and b must be odd. Considering the possible forms of odd numbers in relation to multiples of 6, we have the following subcases:

  • Subcase 1.1: a = (6x + 1) and b = (6y + 1)k = a * b = (6x + 1)(6y + 1) = 36xy + 6x + 6y + 1, which is an element of AA.
  • Subcase 1.2: a = (6x + 1) and b = (6y + 5)k = a * b = (6x + 1)(6y + 5) = 36xy + 36x + 5, which is an element of AB.
  • Subcase 1.3: a = (6x + 5) and b = (6y + 5)k = a * b = (6x + 5)(6y + 5) = 36xy + 60x + 25, which is an element of AA.
  • Subcase 1.4: a = (6x + 5) and b = (6y + 1)k = a * b = (6x + 5)(6y + 1) = 36xy + 30x + 5, which is an element of AB.

Case 2: k is of the form 6y + 7 (k ∈ B)

This case follows a similar logic to Case 1. We analyze the possible forms of factors a and b(both must be odd) and arrive at similar contradictions:

  • Subcase 2.1: a = (6x + 1) and b = (6y + 1)k = a * b = (6x + 1)(6y + 1) = 36xy + 6x + 6y + 1, which is an element of BB.
  • Subcase 2.2: a = (6x + 1) and b = (6y + 7)k = a * b = (6x + 1)(6y + 7) = 36xy + 42x + 7, which is an element of AB.
  • Subcase 2.3: a = (6x + 7) and b = (6y + 7)k = a * b = (6x + 7)(6y + 7) = 36xy + 84x + 49, which is an element of BB.
  • Subcase 2.4: a = (6x + 7) and b = (6y + 1)k = a * b = (6x + 7)(6y + 1) = 36xy + 42y + 7, which is an element of AB.

Contradiction: In all subcases, we’ve shown that if k is a composite number of the form 6x+ 5 or 6y + 7, it must be an element of AA, AB, or BB. This contradicts our initial assumption that k is not an element of those sets.

Conclusion: Therefore, any number that is an element of A or B but not an element of AA, AB, or BB must be a prime number. This completes the proof.

2. Theorem: There are infinitely many prime numbers in sets A and B.

Proof:

Euclid’s Theorem:

Euclid’s theorem states that there are infinitely many prime numbers. This means there is no largest prime number, and the set of prime numbers is infinite.

Characterization of Sets A and B:

Sets A and B are defined as follows:

  • Set A: {6k + 5 | k ∈ Z} (All integers of the form 6k + 5)
  • Set B: {6k + 7 | k ∈ Z} (All integers of the form 6k + 7)

Infinite Primes in Sets A and B:

Consider the primes in sets A and B. These primes are of the form 6k±1 for some integer k. 

Since there are infinitely many prime numbers, and every prime greater than 3 can be expressed in the form 6k±1, there are infinitely many prime numbers in sets A and B.

Conclusion: Because there are infinitely many primes in sets A and B, the theorem is proven to be true.

This proof establishes the connection between the infinitude of prime numbers and the presence of infinitely many prime numbers in sets A and B. It clarifies the theorem and provides a logical argument supported by Euclid’s theorem and the characterization of sets A and B.

3. Theorem: Identification of Twin Prime Pairs by Hotchkiss Sets A, B, AA, AB, and BB

Definitions:

  • Set A: {6k + 5 | k ∈ Z} (All integers of the form 6k + 5)
  • Set B: {6k + 7 | k ∈ Z} (All integers of the form 6k + 7)
  • Set AA: {a × a | a ∈ A} (All products of two elements in A)
  • Set AB: {a × b | a ∈ A, b ∈ B} (All products of one element in A and one in B)
  • Set BB: {b × b | b ∈ B} (All products of two elements in B)

Theorem: Two prime numbers, p and q, form a twin prime pair if and only if:

  • Both p and q are in sets A and B respectively (or vice versa), and
  • Neither p nor q are in sets AA, AB, or BB, and
  • p and q differ by 2 (i.e., p = q ± 2).

Proof:

Part 1: If two prime numbers p and q form a twin prime pair, then they meet the conditions of the theorem.

  • Condition 1: If p and q form a twin prime pair, then they must differ by 2. If p is in set A (p = 6k + 5), then q must be in set B (q = 6k + 7) or vice versa.
  • Condition 2: Since p and q are prime numbers, they cannot be factored into two smaller integers. Therefore, neither p nor q can be formed by the product of two elements from sets A and B. Thus, they are not in sets AA, AB, or BB.
  • Condition 3: This is a direct consequence of the definition of twin primes.

Part 2: If two prime numbers p and q meet the conditions of the theorem, then they form a twin prime pair.

  • Condition 1: Since p and q are in sets A and B respectively (or vice versa), they are both prime numbers.
  • Condition 2: Since neither p nor q is in AA, AB, or BB, they cannot be factored into two smaller integers.
  • Condition 3: Since p and q differ by 2, they fulfill the definition of a twin prime pair.

Conclusion: We have shown that two prime numbers p and q form a twin prime pair if and only if they meet the conditions of the theorem. Therefore, the sets A, B, AA, AB, and BB, along with the requirement that the prime numbers differ by 2, can effectively identify twin prime pairs.

4. Theorem: All pairs of twin primes greater than (3,5) are contained in sets A and B.

Proof:

Characterization of Twin Primes:

Twin primes are pairs of prime numbers that differ by 2. For any twin prime pair (p,q), we have q=p+2.

Form of Twin Primes:

Twin primes are of the form 6k±1 (except for 3 and 5). This means, for some integer k, p and q can be expressed as 6k−1 and 6k+1, respectively.

Prime Number Representation:

All prime numbers greater than 3 can be expressed as either 6k+1 or 6k+56, where k is a non-negative integer. This corresponds to sets A and B, respectively.

Twin Primes as A or B:

Since twin primes are of the form 6k±1, they must belong to either set A or set B.

  • For 6k−1, this corresponds to set A.
  • For 6k+1, this corresponds to set B.

Exclusion from AA, AB, BB:

By the definition of twin primes, p and q cannot be products of elements from sets A and B. Therefore, they cannot be in sets AA, AB, or BB.

Conclusion:

All pairs of twin primes, being of the form 6k±1 and not being products of elements from sets A and B, are indeed contained within sets A and B.

Thus, we have proven that all pairs of twin primes are contained within sets A and B, as per the given theorem.

5. Theorem: Independence and Mutual Exclusivity of Primes in Sets A and B

Let A = {6x + 5 | x ∈ ℤ} and B = {6y + 7 | y ∈ ℤ} be defined as sets containing numbers of the form 6x + 5 and 6y + 7, respectively. Let AA, AB, and BB represent the composite sets formed by the products within sets A and B as follows:

AA = {(6x + 5)(6x’ + 5) | x, x’ ∈ ℤ}

AB = {(6x + 5)(6y + 7) | x, y ∈ ℤ}

BB = {(6y + 7)(6y’ + 7) | y, y’ ∈ ℤ}

Then, the following theorem holds:

  • Independence of Primes in Sets A and B: The primes in sets A and B are independent variables, meaning that the occurrence of a prime in one set does not affect the likelihood of finding a prime in the other set.
  • Mutual Exclusivity of Primes in Sets A and B: The primes in sets A and B are mutually exclusive sets, indicating that a number cannot simultaneously belong to both sets A and B.
  • Interactions in Composite Sets AA, AB, and BB: The composite values within sets AA, AB, and BB represent interactions between elements of sets A and B. These composite values constitute interactions that invalidate the possibility of a prime number existing within sets A or B due to their composite nature.

Proof:

Independence of Primes in Sets A and B:

  • The primes in sets A and B are of the form 6x + 5 and 6y + 7, respectively, where x and y are integers.
  • The occurrence of a prime in set A does not affect the form of primes in set B, and vice versa. Therefore, the primes in sets A and B are independent variables.

Mutual Exclusivity of Primes in Sets A and B:

  • By definition, a number of the form 6x + 5 cannot simultaneously be of the form 6y + 7, and vice versa.
  • Therefore, a prime number belonging to set A cannot belong to set B, and vice versa. This establishes mutual exclusivity.

Interactions in Composite Sets AA, AB, and BB:

  • The composite values within sets AA, AB, and BB represent the products of elements from sets A and B.
  • These composite values result from interactions between elements of sets A and B.
  • Since these interactions produce composite values, any number in sets AA, AB, or BB cannot be a prime number.
  • Thus, the presence of composite values within sets AA, AB, and BB invalidates the possibility of primes existing within sets A or B.

Conclusion: The theorem demonstrates the independence and mutual exclusivity of primes in sets A and B, while also highlighting the interactions within composite sets AA, AB, and BB that preclude the existence of primes within sets A or B.

6. Theorem: Let’s assume that there exists a largest twin prime pair (p, q) such that p and q are both elements of sets A and B, respectively, and there are no twin prime pairs beyond this largest pair.

Proof:

Let (p, q) be the largest twin prime pair, where p=6k−1 and q=6k+1 for some integer k.

By assumption, there are no twin prime pairs beyond (p, q).

Consider the sets A and B:

  • Set A contains all numbers of the form 6x+5, which includes primes greater than 3.
  • Set B contains all numbers of the form 6y+7, which also includes primes greater than 3.

Since sets A and B include all primes greater than 3, and there are infinitely many primes according to Euclid’s Theorem, there must be infinitely many values of k such that the pairs (6k – 1, 6k + 1) form twin primes.

Therefore, if we assume there is a largest twin prime pair (p, q), there would always be another twin prime pair (6k−1,6k+1) for some k beyond this largest pair.

This contradicts our initial assumption that there exists a largest twin prime pair.

Conclusion: The assumption that there exists a largest twin prime pair leads to a contradiction with the concept of infinitely many primes and the comprehensive inclusion of primes greater than 3 in sets A and B. Therefore, there will always be another twin prime pair beyond any assumed largest pair.

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