Goldbach's Conjecture — Towards the Inconsistency of Arithmetic

1 Goldbach's Conjecture — Towards the Inconsistency of Arithmetic Ralf Wüsthofen Abstract. This paper proves that ZFC and Peano arithmetic (PA) are inconsistent, the latter result being a corollary of the former. We introduce a metamathematical extension of ZFC that allows us to use statements in the proof that express logical consequence. We then show, by explicitly stating a contradiction, that the extended theory is inconsistent and that this immediately leads to the existence of a contradiction in ZFC. The contradiction is triggered by the conjunction of two properties of an infinite set, by means of which we express a strengthened form of the strong Goldbach conjecture. We use elementary number theory, with the constructive role of prime numbers within the natural numbers being an essential point. Notations. Let denote the natural numbers starting from 1, let a denote the natural numbers starting from a > 1 and let 3 denote the prime numbers starting from 3. Let ZFCμ denote the metamathematical extension of ZFC that allows us to make statements of the form "we have a proof of P" for some statement P expressible in ZFC used in this paper, which means that this paper contains a proof in ZFC that P holds. Analogous to the syntactic entailment ⊢ P, i.e. "there exists a proof of P", we symbolize "we have a proof of P" by ⊢w P. Let SSGB denote the following strengthened form of the strong Goldbach conjecture: Every even number greater than 6 is the sum of two distinct odd primes. Theorem. ZFC is inconsistent. [1], [2] Proof. In the following we use ZFCμ and produce a formula φ along with both a proof of φ from ZFCμ and a proof of φ from ZFCμ. We then show that this immediately implies that there exists a contradiction in ZFC. We define the set Sg := { (pk, mk, qk) | k, m ; p, q 3, p < q; m = (p + q) / 2 }. Sg has the following two properties. 2 First, the whole range of 3 can be expressed by the triple components of Sg (”covering”). We prove this by dividing it into the following three cases. (i) x 3 is prime. Then, x = pk with p 3, k = 1. (ii) x 3 is composite and not a power of 2. Then, x = pk with p 3, k ≠ 1. (iii) x 3 is a power of 2. Then, x = (p + q)k / 2 with p = 3, q = 5, k = (a power of 2). So we have (C) x 3 Ǝ (pk, mk, qk) Sg x = pk x = mk. Second, all pairs (p, q) of distinct odd primes are used in the definition of the set Sg (“maximality”). So we have (M) p, q 3, p < q k (pk, mk, qk) Sg, where m = (p + q) / 2. SSGB is equivalent to saying that every integer greater than or equal to 4 is the arithmetic mean of two distinct odd primes. So, under the assumption SSGB there is an n 4 that is different from all m defined in Sg, whereas under the assumption SSGB there is no such n. The following steps are independent of the choice of n if there is more than one. For example, the minimal such n works. The property (C) implies that for the above n, every nk', k' , equals a component of some Sg triple. The property (M) excludes the possibility that n is the arithmetic mean of a pair of distinct odd primes not used in Sg. So, (M) excludes the possibility that the question of whether SSGB holds or not depends on whether (M) holds or not. The basic idea is now the following. 3 Since the properties (C) and (M) hold, under the assumption SSGB the set Sg can be written as the union of the following triples, which would otherwise be impossible. (a) Sg triples of the form (pk = nk', mk, qk) with k = k' if n is prime (b) Sg triples of the form (pk = nk', mk, qk) with k ≠ k' if n is composite and not a power of 2 (c) Sg triples of the form (3k, 4k = nk', 5k) if n is a power of 2 (d) all other Sg triples of the form (pk = nk', mk, qk), (pk, mk = nk', qk) or (pk, mk, qk = nk') (e) Sg triples of the form (pk ≠ nk', mk ≠ nk', qk ≠ nk'). Under the assumption SSGB, the set Sg can be written as the union of triples of the above types (a) to (e), where n is to be replaced by any y 3. Therefore, the Sg triples look the same regardless of whether n exists or not, i.e. regardless of whether SSGB holds or not. This is contradicted by the fact that under the assumption SSGB the numbers m defined in Sg take all integer values ≥ 4 whereas under the assumption SSGB they don’t. To formalize this idea, we proceed as follows. We split Sg into two complementary subsets in the following way. For any y 3, we write Sg = Sg+(y) ∪ Sg-(y), with Sg+(y) := { (pk, mk, qk) Sg | Ǝ k' pk = yk' mk = yk' qk = yk' } Sg-(y) := { (pk, mk, qk) Sg | k' pk ≠ yk' mk ≠ yk' qk ≠ yk' }. We define S1 := { (pk, mk, qk) Sg | SSGB } and S2 := { (pk, mk, qk) Sg | SSGB }. I.e., S1 = Sg if SSGB is true, and S1 = { } if SSGB is false and S2 = Sg if SSGB is true, and S2 = { } if SSGB is false. Then, since under both assumptions SSGB and SSGB the properties (C) and (M) hold, we obtain 4 (1.1) ⊢w ( y 3 SSGB => S1 = Sg+(y) ∪ Sg-(y) ) (1.2) ⊢w ( SSGB => S2 = Sg+(n) ∪ Sg-(n) ). So, since Sg+(n) ∪ Sg-(n) is independent of n, (1.1') ⊢w ( y 3 SSGB => S1 = Sg+(y) ∪ Sg-(y) ) (1.2') ⊢w ( y 3 SSGB => S2 = Sg+(y) ∪ Sg-(y) ). Now, we make use of the following principle. If two sets of (possibly infinitely many) z-tuples are equal, then the sets of their corresponding i-th components are equal; 1 ≤ i ≤ z. For this we define M1 := { m | (p, m, q) S1 } and M2 := { m | (p, m, q) S2 }. Then, applying the principle above to the middle component of the triples (p, m, q), the fact that we have a proof for each of the implications in ( (1.1') (1.2') ) implies by transitivity (2.1) ⊢w ( y 3 SSGB => M1 = { m | (p, m, q) Sg+(y) ∪ Sg-(y) } ) (2.2) ⊢w ( y 3 SSGB => M2 = { m | (p, m, q) Sg+(y) ∪ Sg-(y) } ). 5 We define M := { m | (p, m, q) Sg }. Then, since for every y 3 Sg+(y) ∪ Sg-(y) equals Sg by definition, for every y 3 { m | (p, m, q) Sg+(y) ∪ Sg-(y) } equals M by definition. If SSGB is true, M is equal to 4, and if SSGB is false, M is equal to some non-empty proper subset U of 4. It follows that there is exactly one set X { 4, U } that { m | (p, m, q) Sg+(y) ∪ Sg-(y) } is equal to. Since for every y 3 { m | (p, m, q) Sg+(y) ∪ Sg-(y) } = X regardless of whether SSGB or SSGB holds, in ( (2.1) (2.2) ) we can replace { m | (p, m, q) Sg+(y) ∪ Sg-(y) } by X. Thus, we obtain (3) Ǝ! X { 4, U } ( ⊢w ( SSGB => M1 = X ) ⊢w ( SSGB => M2 = X ) ). Since the statements ( SSGB => M1 = X ) and ( SSGB => M2 = X ) depend on X and since X is the unique element of { 4, U } such that these statements hold, we will make use of the following rule. Let P(A) and Q(A) be statements that depend on a set A. Let A be the unique element of { B1, B2, ..., Bz } such that P(A) and Q(A) hold. Then, ( Ǝ! A { B1, B2, ..., Bz } ( ⊢w P(A) ⊢w Q(A) ) ) => ( ( ⊢w P(B1) ⊢w Q(B1) ) ( ⊢w P(B2) ⊢w Q(B2) ) ... ( ⊢w P(Bz) ⊢w Q(Bz) ) ). We apply the above rule with P(A) = ( SSGB => M1 = A ) Q(A) = ( SSGB => M2 = A ) z = 2 B1 = 4 B2 = U. 6 Then, since the left-hand side of the rule is true, we obtain (3.1) ( ⊢w ( SSGB => M1 = 4 ) ⊢w ( SSGB => M2 = 4 ) ) (3.2) ( ⊢w ( SSGB => M1 = U ) ⊢w ( SSGB => M2 = U ) ). This implies (4.1) ⊢w ( SSGB => M2 = 4 ) (4.2) ⊢w ( SSGB => M1 = U ). Now, we will establish a contradiction to ( (4.1) (4.2) ). We have a proof that ( SSGB => M = 4 ) and we have a proof that ( SSGB => M = U ≠ 4 ). Therefore, since SSGB => M1 = M and SSGB => M2 = M by definition, we get (5.1) ⊢w ( SSGB => M1 = 4 ) (5.2) ⊢w ( SSGB => M2 = U ≠ 4 ). 7 Because of ( (5.1) (5.2) ) and because ⊢w ( SSGB => M2 = { } ≠ 4 ) and ⊢w ( SSGB => M1 = { } ≠ U ), we have a proof that ( M2 = 4 ) is false and we have a proof that ( M1 = U ) is false. Therefore, ( (4.1) (4.2) ) yields (6.1) ⊢w ( SSGB => FALSE ) (6.2) ⊢w ( SSGB => FALSE ). And this yields (7.1) ⊢w SSGB (7.2) ⊢w SSGB. On the other hand, since this paper contains neither a proof of SSGB nor of SSGB, the negation of ( (7.1) (7.2) ) also holds. This proves that ZFCμ is inconsistent. By the principle of explosion, it follows that for any statement Q expressible in ZFC used in this paper, both ⊢w Q and ⊢w Q are true. By the definition of ⊢w P, this implies that there is a proof in ZFC that Q holds and that there is a proof in ZFC that Q holds. This means that ZFC is inconsistent. □ 8 Corollary. Peano arithmetic (PA) is inconsistent. Proof. The term Sg from the above inconsistency proof is not a standard part of PA, but it can easily be defined within PA. This also applies to all other sets used in that proof, since they are all based on Sg or on . Therefore, the corollary is proved in the same way by using the analogous extension PAμ of PA. □ References [1] Warning Signs of a Possible Collapse of Contemporary Mathematics, by Edward Nelson (2006). https://web.math.princeton.edu/~nelson/papers/warn.pdf [2] The Consistency of Arithmetic, by Timothy Y. Chow (2018). https://arxiv.org/pdf/1807.05641