Riemann Hypothesis Solved Using A Quantum Mechanical System ( Revised )

Riemann's Hypothesis is that all the non-trivial zeros lie on the line ( y = ½ ) and these zeros have something to do with prime numbers. Since Riemann's Hypothesis includes the line ( y = ½ ), let's chose a quantum energy system in which the energy levels are 2. Riemann's equation also included complex numbers. In a quantum energy system Riemann's complex numbers could be represented by zero's ( 0 ) since ( 2 + 0 = 2 ). ( 1 + 1= 2 ) also equals 2. Riemann also went on to say that the prime number locations were influenced by the position of the zeros. To extend our ( 1 + 1 = 2 ) analogy, we now have ( 1 + 0 + 1 = 2 ). We can convert these additions to numbers ( 11, 101, etc. ). We can now say that we have a real axis ( y = 2 ) with infinite numbers with zeros between their ones ( 11, 101, 1001, 10001, etc. ) on an infinite real axis line ( y = 2 ). We can also say that each of the zero's in ( 101, 1001, 10001, etc. ) are on the line ( y = 2) since the numbers ( 101, 1001, 10001, etc. are also on this line. Riemann also said that the values of the zeros ( 0's ) on the line are equal to ( ½ ) which is also true when the line ( y = 2 ) is flipped. We could also extend this argument to one which says that the value of the ones ( 1's ) are also ( ½ ) since the reciprocal of one ( 1 ) is ( 1 ) and the fraction ( 1 / 1 ) when added ( 1 / 101 ) has the value ½ on the line ( y = ½ ) . These numbers would also be evenly spaced on the real axis line ( y = 2 ) since each infinite number would total energy level 2 which means the distance between the numbers would be ( 2 + 2 = 4 ) or a spacing of 4. Flipped the distance between the numbers would be one ( 1 ) since ( ½ + ½ = 1 ). Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ) and this is what we've proved up to now since the non-trivial zeros are in the numbers ( 101, 1001, 10001, etc. ). If we flip our infinite real axis line ( y = 2 ), we create a real axis line ( y = ½ ) with flipped infinite real numbers ( 1/11, 1/ 101, 1/1001, etc. ) all the way to infinity. If we add ( ½ + 1/101 ) we obtain ( .50990099 ). Prime number ( 11 ) is the 6th prime ( 1, 2, 3, 5, 7, 11 ). ( 11 X .50990099 = 5.6089108911 ) or 6 rounded to one digit. The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's and can be adjusted to get a more accurate estimate.. As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .50990099 ) X ( .50990099 )) you get ( 25.22969903 )which is very close to 26. If you adjust the Riemann zeros to form the number ( .509999 ) and do the multiplication ( 97 x (.509999 X .509999) you get ( 25.22960106 ) which is further away from 26. You can see from these calculations that if the Riemann zeros are manipulated as Riemann has suggested to obtain the location of the prime ( or calculate the number of primes up to that point ) you can vary the distance to the true position of the prime. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.