## Monday, October 31, 2011

### The Riemann Hypothesis Resolved ( Hopefully finally)

Riemann said that all his non – trivial zeros lie on the line ( y = ½ ) . These zeros are at the location that the primes form a right angle line to ( y = ½ ) or in other words, are vertical to the line ( y = ½ ) . Prime numbers are defined as numbers that can only be divided by themselves and ( 1 ). An example of one digit primes are ( 1, 2, 3, 5, 7 ). You can see from these numbers that primes aren't linear ( 1, 2, 3, 4 ) but form harmonics. If all the primes were played musically, you would get a tune. The physical distance between the primes would be similar to the value of the rests in a musical piece. Calculations have been done on the location that some of Riemann's zeros lie on the line ( y = ½ ) using Riemann's formula ( ½ + bi ). The first few zeroes were calculated around  14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. These numbers are not close to the position of the beginning one digit primes ( 1, 2, 3, 5, 7 ). You will see that the distance between ( 1, 2, 3, ) is ( 1 ) and the distance between 3 & 5, 5 & 7, is 2. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes which is equivalent to altering the value of the rests in a musical piece.

1. 1 X .99 = .99 ( 1 )

2. 2 X .99 = 1.98 ( 2 )

3. 3 X .99 = 2.97 ( 3 )

If we add the digit ( ½ ) or ( .5 ) in Riemann's line ( y = ½ ) and Riemann's zero ( 0 ) to .99, we form the number ( .509999 ).

1. 5 X ( .509999 ) = 2.549995 (4)

2. 7 X ( .509999 ) = 3.569993 ( 5 )

3. 11 X ( .509999 ) = 5.609989 ( 6 )

This basic principle calculates the position of the primes fairly accurately up to prime 31.

Using the same multipliers on the next prime ( 37 ) we get ( 37 X .509999 = 18.869963. Prime ( 37 ) is actually the 13th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). If we take ( .509999 ) and multiply it by itself ( ( .509999 X .509999 = .26009898 ).

1. ( 37 X .509999 X ,509999 = 9.62366226 ( 13 )

This calculation shows that the calculated position is short of the true position by about ( 13 – 9.62366226 = 3.37633774 ). The difference is approximately equal to Pi ( 3.141592654 ). In some calculations the difference is about the natural number ( e ) ( 2.71828`828 ).

In summary, Riemann's intuition told him that the zeros had a real value of ( ½ ) which was true since the digit ( ½ ) is used in the calculation. Riemann's intuition also told him that the zeros can be manipulated ( position changed or zeros added ) which is also true. What Riemann missed was that the calculation involved the number 9 and that Pi ( 3.141592654 ) and the natural number ( e ) ( 2.718281828 ) might have to be added or subtracted from the final answer. Riemann also missed that powers would also have to be used.

In general for calculating the location of any prime you:

1. Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).

The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I have also found the missing pieces to Riemann's Hypothesis which was correct as far as it went.