Abstract
Two prime numbers{3,5}.{3,5}→{3+3= 6,3+5=8,5+5=10} →{6,8,10}.{{10}→(5+5=10 = 3 + 7) → 7}Increased by 7 →{3, 5, 7}.{3,5,7}→{ 3+3= 6, 3+5=8, 5+5=10, 5+7=12, 7+7=14}→{ 6,8,10,12,14}.{{14}→(7+7=14 = 3 + 11) → 11} Increased by 11 →{3, 5, 7, 11}.Note changes: {3,5,7}→{ 6,8,10,12,14}→ {3, 5, 7,11}.The same logic would be:{3,5,7,11}→{ 6,8,10,12,14,16}→ {3, 5, 7,11,13}.The same logic would be:{3,5,7,11,13}→{ 6,8,10,12,14,16,18,20}→ {3, 5, 7,11,13,17}.If you expand infinitely in the above specified mode: {3, 5, 7, 11, 13, 17 ,...Get: { 6,8,10,12,14,16,18,20,22,...The above is: the continuity of prime numbers can lead to even continuity.Get: Goldbach conjecture holds.If it is mandatory: Authenticity stops at an even number 2n.{{3, 5, 7, 11 ,...,p1}→{ 6,8,10,12,...,2n}.{3, 5, 7, 11, ...,p1} ↛(2n+2).∀p+∀p≠2n+2 } It can be proved that: It violates the "Bertrand Chebyshev" theorem.∴ {3, 5, 7, 11, 13, ...→{ 6,8,10,12,14,...Get: Goldbach conjecture holds.