Proof of Goldbach s Cojecture Reza Javaherdashti farzijavaherdashti@gmail.com Abstract After certai subsets of Natural umbers called Rage ad Row are defied, we assume (1) there is a fuctio that ca produce prime umbers ad (2) each eve umber greater tha 2, like A, ca be represeted as the sum of prime umbers. We show this by DC(A). Each Row is similar to each other i properties,(so is each Rage). It is prove that i a arbitrary Row for ay eve umber greater tha 2, DC(A)=2, that is to say, each eve umber greater tha two is the sum of two prime umbers. So Goldbach s cojecture is proved. 1.Historical Backgroud: Of still-usolved problems o prime umbers oe ca metio Goldbach s cojecture. Goldbach (1690-1764) i his letter to Euler i 1742, asked if ay eve umber greater tha 2 could be writte as the sum of two prime umbers. Euler could ot aswer or could he fid ay couter-example. The mai problem with Goldbach s cojecture is that i most of theorems i arithmetic, prime umbers appear as products, however, i Goldbach s cojecture it is the additio of prime umbers that makes all the problem. I 1931, a youg, ot that famous Russia mathematicia amed Schirelma (1905-1938) proved that ay positive iteger could have bee represeted, at most, as the sum of 300,000 prime umbers. The reasoig was costructive ad direct without givig ay practical use to decompose a give iteger ito the sum of prime umbers. Some years after him, the Russia mathematicia Viogradoff by usig ad improvig methods iveted by the Eglish mathematicias, Hardy ad Littlewood, ad their great Idia colleague Ramauja, could decrease umber of the metioed prime umbers from 300,000 to 4. Viogradoff s approach has bee proved to be true for itegers large eough. With exact words, Viogradoff proves that there exists a iteger like N so that for ay iteger >N, it ca be represeted, at most, as the sum of four prime umbers. He gives oway to determie ad measure N. Vigoradoff s method has actually proved 1
that acceptig the ifiite itegers that caot be show as the sum of, at most, four prime umbers, results i cotradictio. 2.Method ad Basic Assumptios: The method we use i our approach, relies o the followig facts (or, iterpretatios): 1.Cotroversial poits with Goldbach s cojecture are: 1.1.It seems as if there must be a kid of formula that ca produce prime umbers. 1.2. After such a prime umber-producig formula is foud, oe should look for its relatioship with eve umbers greater tha 2. 2.Goldbach assumes that sum of prime umbers gives eve umbers greater tha 2; the problem is how to limit the umber of such prime umbers with oly two. To clarify the approach, we assume that: Assumptio #1: There is a fuctio like f (x) that produces prime umbers. Assumptio #2: Each eve umber greater tha 2, ca be take as to be the sum of prime umber where is a Natural umber. Usig the above-metioed assumptios, we defie Row ad Rage as subsets of Natural umbers. The, by usig DC (A), that desigates that how may prime umbers ca produce eve umber A greater tha 2, we prove DC(A)=2. I other words, It is proved that that the miimum umber of prime umbers that results i a eve umber A, which is greater tha 2, is oly two. Thus, Goldbach s cojecture is proved. Our method cosists of three parts: I. Basic cocepts o Row ad Rage. II. Basic defiitios of f(x) ad DC(A). III. Proof. We will NOT use statistical data or tables of prime umbers i our method. 2
Part 1): Basic cocepts o Row ad Rage 1.1.Defiitio of Row: Row-that we show as r(x i x f )- is a term used for represetig ay subset of Natural umbers, N, that has all of the properties below: I. r(x i x f ) N, that is, each Row is a subset of Natural umbers. If [r(x i x f )] shows the umber of elemets of the set r(x i x f ), the : [r(x i x f )] = d d N The above meas that i a Row, umber of elemets is limited. II. (property of havig a smallest ad a greatest elemet i a Row): Each r(x i x f ) has just oe smallest ad just oe greatest elemet, that is, each r(x i x f ), at most, has oe smallest elemet like x i i r(x i x f ) such that for all x r(x i.x f ), x>x i. I the same way, each r(x i x f ) has oe greatest elemet III. IV. like x f such that for all x r(x i x f ), x < x f. So x f is the greatest elemet of r(x i x f ) ad there is o elemet larger tha it. (property of a ordered Row): I each Row r(x i x f ) all of its elemets are orderable, from left to right, ad from the smallest to the largest elemet. (property of costat differece betwee elemets of a Row) For all x r(x i x f ), if x x i ad x x f, the x-1 < x < x+1. This meas that each x r(x i x f ) is less tha or larger tha a umber immediately before or after by a costat differece, which is oe. 1.1.1. Examples: A = {1,2,3,4} A is a Row so A = r(14) as x i = 1 ad x f = 4 ad the differece betwee each immediate successive elemet is uity. B = {25,26,27} B is a Row so B = r(2527) as x i = 25 ad x f = 27 C = {4,3,2,1} C is NOT a Row; III is ot held. D = {5,9,10,11,14} D is NOT a Row; IV is ot held. E = {49,51,53,55} E is NOT a Row; IV is ot held. 3
Covetio.1: 1- From ow o, istead of r(x i x f ), the symbol r(if) is used. 2- Accordig to III, x< x i or x> x f is ot defied i r(if). 3- Ay r(if), schematically ca be show as follows: x i x 1 x 2 x x f r(if) = { x i, x 1, x 2,., x, x f } x < x f 1.2.Defiitio of Rage: Rage-that we show as R(x I x F )-is a term used for represetig ay subset of Natural umbers that has all of the properties below: [1]. R(x I x F ) N [2]. r(if) R(x I x F ) [3]. [R(x I x F )] = D D N, D>d [4]. Each R(x I x F ) has, at most, oe smallest elemet show as x i, ad oe greatest elemet x F,ie; for each X R(x I x F ): X < x F, X > x I [5]. For each X R(x I x F ) givig X x I ad X x F : X-1< X< X+1 [6]. All the elemets of a Rage R(x I x F ) ca be ordered, from left to right ad from the smallest to the largest elemet. 1.1.2.Examples: F = {1,2,3,,98,99,100} F is a Rage so F = R(1100),r(110) F, so [2] is held. G = {5,6,7,,22,23} G is rage so G = R(523) H = {30,31,32,35,36,37,39,41,42,45} H is NOT a Rage, [5] is ot held. I = {31,32,50,33,50,1,2,4,10,2000} I, is NOT a Rage,; -[2] is ot held as oly {1,2} ad {31,32} have the smallest ad the largest elemets, so IV is ot held for these subsets. 4
-[5] is ot held. -[6] is ot held. Covetio.2: 1-From ow o, R(x I x F ) is show as R(IF). 2-By cosiderig [2] ad [4], it appears that i some cases X I = x i or X F = x f,for example, i F, X I = 1 ad also i A, x i = 1 (A F). 3-By usig Row ad Rage cocepts, N (atural umbers set) ca be subdivided ito subsets that have certai properties as R(IF) N ad r(if) R(IF). There are ifiite umber of R(IF) sets but defiite ad limited umber of r(if) subsets, for istace, there are iifiite umber of sets like R(1100), R(101200), R(201300) etc. but i each of these sets there are limited ad defiite umber of r(if);as a example, i R(1100),there are te subsets (=Rows) such as r(110), r(1120), r(2130),, r(91100). So, ay theorem that is proved or ay coclusio that is made for arbitrary r(if) ad R(IF), ca the be geeralised for all similar Rows ad Rages so that what is prove, will be applicable to whole Natural umbers. Covetio.3: For each x which is a elemet of a Row r(if), we use the followig symbolism: x r(if) or (x)r(if) ad for each X which is a elemet of R(IF), we use the followig symbolism: X R(IF) or (X)R(IF) Ad so forth. Covetio.4: I a Row r(ij), its smallest elemet-that is x i ca be show as either (x i )r(ij) or A( i ) ad its greatest elemet-x f -ca be show as (x f )r(ij) ad so forth. Result of Covetio.4: I a Row r(pq), A(p) is the smallest elemet. I a Row r(st), A(s) is the smallest elemet ad so forth. Theorem.1 Take two Rows r(ij) ad r(kl) i a rage R(IF). Prove that for all x i ad x f elemets of these Rows, we have: (x f )r(ij) ± p = (x i )r(kl) 5
where p is a costat. Proof: Assume that p = 0, the (x f )r(ij) = (x i )r(kl). Now take R(IF) as: (1) (r(ij) R(IF) & (r(kl) R(IF)) R(IF) = r(ij) U r(kl), where U is uio sig, results i: (2) (x R(IF)) (x r(ij)) & (x r(kl)) as (x i )r(kl) r(kl) ad (x f )r(ij) r(ij), the both (x i )r(kl) ad (x f )r(ij) must be elemets of R(IF) but i this case: 1.the differece betwee (x i )r(kl) ad (x f )r(ij) will ot be equal to uity ( [5] ad IV are ot held). 2.R(IF) will ot be ordered ( [6] is ot held) So, R(IF) will ot be a Rage but this is i cotradictio with our assumptio. Therefore, p 0. If each elemet of r(ij) is smaller tha each elemet of r(kl) the, (x f )r(ij) + p = (x i )r(kl) ad so forth. Defiitio.1: If (x f )r(ij) + p = (x i )r(kl), r(ij) ad r(kl) are called Successive Rows. Remider: Rage s are arbitrary, radom sample sets from set of atural umbers. Row s are arbitrary, radom sample sets from rages. By itroducig the cocepts of rage ad row, it is followed that if somethig is prove for these radom cuts from atural umbers set, it has bee prove for WHOLE atural umbers. However, defiitios of rage ad row clearly show that NOT ANY radom set from atural umbers set ca be picked up as rage ad/or row (see examples 1.1.1. ad 1.1.2.). Part.2): Basic defiitios of f(x) ad DC(A) Defiitio 2: We defie a fuctio f(x) so that for ay x = a, f(a) be a prime umber. Defiitio 3: The degree of complexity fuctio that we show it as DC is the umber of prime umbers to be added to each other with a + sig betwee each to yield a eve umber greater tha two. DC will itself be a Natural umber. 2.2.1.Example: -For 8 = 2 + 2 + 2 + 2 DC (8) = 4 6
-For 8 = 5 + 3 DC (8) = 2 -For 216 = 213 + 3 DC (216) = 2 Coclusio from defiitio 3: The least value for DC(A), where A is a eve umber greater tha 2, is 2; ie, DC (A) 2. Defiitio 4: I ay Row r(ij) of the Rage R(IF), umber of eve umbers (2K), odd umbers (2K + 1) ad prime umbers (f(x)) are show, respectively, as Γ(2K), Γ(2K + 1), ad Γf(x). So i the Rage R(1100) ad Row r(110), Γ(2K) = 5 that is to say, (2,4,6,8,10), Γ(2K + 1) = 5 ie (1,3,5,7,9) ad Γf(x) = 4 (2,3,5,7). Part.3): Proof Assumptios: 1) Assume i Row r(ij) of Rage R(IF), there exists at least oe prime umber like f(x). 2) Assume i ay Row r(ij), the umber of eve ad odd umbers are equal to each other (i ay Row there is as may odd umbers as there is eve umbers). So i ay Row r(ij) there exists odd ad eve umbers alteratively (after each odd umber there is a eve umber ad vice versa). 3) Assume i a Row r(ij), umber of odd umbers be more tha prime umbers i the same Row, i other words, ay prime umber greater tha 2 is a odd umber BUT ay odd umber is NOT a prime umber. From assumptios 1) ad 3), oe cocludes: (3) 1 Γf(x) Γ(2K + 1) For eve umbers greater tha 2, like A, degree of complexity fuctio ca be writte as DC (A). Accordig to the coclusio from defiitio 3, DC (A) 2, ie, at least two prime umbers must be added to each other to yield A. As A is a eve umber greater tha 2, the umber of eve umbers that are required to be added to each other to yield A, will be less tha the umber of existig eve umber i the Row r(ij), so i r(ij) the umber of prime umbers to be added up to yield A is: (4) DC (A) Γ(2K) By addig Γ(2K + 1) to right-had sides of iequalities (3) ad (4) it yields: (5) Γf(x) Γ(2K + 1) + Γ(2K + 1) 7
(6) DC (A) Γ(2K) + Γ(2K + 1) Accordig to assumptio 2) Γ(2K) = Γ(2K + 1); by applyig this to (5), it yields: (7) Γf(x) Γ(2K) + Γ(2K + 1) by addig each side of (6) ad (7) to each other, we take: (8) DC (A) + Γf(x) 2[Γ(2K) + Γ(2K + 1)] or (9) DC (A) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) Combiig the coclusio from defiitio 3 with iequality (9) yields: (10) 2 DC (A) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) The above relatio ca be decomposed ito the followig three iequalities: (11-1) DC (A) > 2 (11-2) DC (A) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) (11-3) DC (A) = 2 We will prove that (11-1) ad (11-2) will be resultig i cotradictios so that they will ot be held. Therefore, the oly remaiig relatio will be (11-3) that states that the umber of prime umbers to be added up to yield a eve umber greater tha 2, is two. Assume (11-1) holds, ie, DC > 2. This meas: (12) DC (A) = a 1 + a 2 + = i= a i 3 Where m shows total umber of umbers-eve, odd ad prime umbers-existig i a give Row r(ij); a i (i = 1,2,3,,m) shows the prime umbers to yield A. Relatio (12) may be re-writte as (13): i m (13) DC (A) = 2 + = i= a i i m Where is a arbitrary umber less tha m. Oe should otice that i (12) a 1 ad a 2 are two eve umbers where i (13) some prime umbers like umber like 2. Equatio (13) yields: i = m a i i= are added to a (14) DC (A) 2 = m a i 8
m a i represets the umber of prime umbers i a Row r(ij) which is less tha total umber of existig prime umbers of r(ij), that is to say: (15) m a i Γf(x) replacig (14) ito (15) yields: (16) DC (A) 2 = Γf(x) Assumptio 3) of Part.3):Proof about odd ad prime umbers gives: (17) m a i Γ(2K + 1) replacig (14) ito (17) yields: (18) DC (A) 2 = Γ(2K + 1) Or (19) DC (A) Γ(2K + 1) + 2 From (16), (20) is resulted: (20) DC (A) Γf(x) + 2 By addig sides of (19) ad (20) to each other: (21) 2DC (A) Γ(2K) + Γf(x) + 4 as DC (A) < 2DC (A), oe may coclude (22). To let the iequality hold with more force, we add Γ(2K) to right-had side of (21) too; (22) 2DC (A) Γ(2K) + Γ(2K + 1) Γf(x) + 4 The right-had side of (22) may be writte as (23): (23) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) = [Γ(2K) + Γ(2K + 1) + Γf(x) + 4] + [Γ(2K) + Γ(2K + 1) - 2Γf(x) - 4] (24) Γ(2K) + Γ(2K + 1) + Γf(x) + 4 < 2[Γ(2K) + Γ(2K + 1)] - Γf(x) To combie (22) ad (24) yields: (25) DC (A) < Γ(2K) + Γ(2K + 1) + Γf(x) + 4 < 2[Γ(2K) + Γ(2K + 1)] - Γf(x) The relatio (25) results i: (26) DC (A) < 2[Γ(2K) + Γ(2K + 1)] - Γf(x) Iequality (26) resembles (11-2). So we will cosider (26) more precisely: Oe sees that DC (A) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) shows the umber of prime umbers i a give Row r(ij) to be added to each other to result i a umber like A, which is a eve umber greater tha2. Therefore, 2[Γ(2K) + Γ(2K + 1)] - Γf(x) must 9
be less tha the umber of existig odd umbers i r(ij) (Assumptio 3) of Part.3):Proof). O the other had, accordig to assumptio 2) Part.3):Proof at least half of the total m umbers i a Row r(ij) i other words, -are existig odd umbers of r(ij). As it 2 was stated i (12), if m be total umber of existig umbers i a Row r(ij), the: m (27) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) 2 as Γ(2K) + Γ(2K + 1) = m (Assumptio 2) of Part.3):Proof)): m (28) 2m - Γf(x) 2 (29) 1.5m Γf(x) as m < 1.5m the: (30) m < 1.5m < Γf(x) or: (31) m < Γf(x) Iequality (31) is ot held as it states that i a Row r(ij), the umber of prime umbers of the Row is larger tha total umbers of the Row, which is IMPOSSIBLE. I the same way, oe may prove that if (26) is writte as: DC (A) = 2[Γ(2K) + Γ(2K + 1)] - Γf(x), agai we will come up with the cotradictio stated as i (31). The above discussio shows that (11-2) is NOT held. Oe may show (25) as a combiatio of the followig: (32-1) DC (A) > 2 (32-2) DC (A) < Γ(2K) + Γ(2K + 1) Γf(x) + 4 (32-3) DC (A) < 2[Γ(2K) + Γ(2K + 1)] - Γf(x) We call (32-1) through (32-3), as p, q ad r, respectively so that we traslate (25) to the logical expressio (33)as a cojuctio: (33) p ( q & r) where & represets cojuctio sig. Accordig to what we have gaied so far:: 1.r is false, ie, relatios (27) to (31) prove that : DC (A) 2[Γ(2K) + Γ(2K + 1)] - Γf(x), is NOT held. 2.as r is false, the (q & r) is false. So p must be false also, that is to say: 10
DC (A) > 2 is NOT held. Coclusio 01. (11-1) isot held so DC (A) > 2 isot true. 02. (11-2) isot held so DC (A) 2[Γ(2K) + Γ(2K + 1)] - Γf(x) isot true. 03. (11-3) holds as the oly possibility so DC (A) = 2 is true. I fact Coclusio 03 states that a eve umber greater tha 2 ca be writte as the sum of two prime umbers. So Goldbach s cojecture is true. Summary: By defiig certai subsets i Natural umbers set, after all possibilities are cosidered, we prove that Goldbach s cojecture is held i oe arbitrary. The proof ca be geeralised to all the subsets so that the cojecture is prove for all Natural umbers. 11