A Quantum Solution For The Riemann Hypothesis

The Riemann Hypothesis equation is constructed for our 4 dimensional world in which space is the prime criteria. The thrust of the Riemann Hypothesis is to prove that all the zeros lie on the line ( y = ½ ) and if they do there is some relationship to zeros, primes and energy. Maybe if we took a look at the Riemann hypothesis from a quantum viewpoint we can resolve it.

The quantum world is all about energy-time. The energy in the quantum world shows all the states , possibilities or information at one spot or superlocation at the same time because physical space doesn't exist. It exists in the abstract in as much as we can say something is right here or over there but we can't travel through space because it doesn't exist.

The universe is built on the number 3. What this means is that when we consider a welter of possibilities most decisions come down to:

1. Yes – 1
2. No – 0
3. Maybe – ½ .

In the case of the Riemann hypothesis since quantum energy shows all the states, possibilities or information the zero is always ( yes – 1 ), no ( no – 0 ) or Maybe ( ½ ) in the same superlocation since space doesn't physically exist except in the abstract.

Riemann used the formula ( s = ½ + it ) in his real world formula. The “i” in “it” represented the imaginary space dimension. In electronic formulas the “i” represents time ( √-1 ). In the quantum world we only have energy and not space because space doesn't physically exist except in the abstract. Time in the quantum world has the energy value of 9 since space doesn't exist. Riemann said the zeros have something to do with the oscillation of the primes around their expected ( superlocation ) position. Riemann said that the zeros lie on the line ( y = ½ ) in the real world where space is the primary criteria. In the quantum world the primary criteria is all states, possibilities or information.

Combining all quantum realities, states or information in the quantum world we arrange the quantum energy levels by using the digits ( ½ , 0, 9 ) as a multiplier . ( ½ ) means Maybe. Maybe the prime is in that superlocation. ( 0 ) is Riemann's zero ( 0 ) which is in the same superlocation. ( 9 ) is the quantum value of time in that superlocation.

Here's how it works with primes using quantum time = 9.

1. 1 X .99 = .99 – Location ( 1 )
2. 2 X .99 = 1.98 – Location ( 2 )
3. 3 X .99 = 2.97 - Location ( 3 )

Using the digits ( ½ , 0, 9 ) this is how it works:

1. 5 X ( .509999 ) = 2.549995 – Location (4)
2. 7 X ( .509999 ) = 3.569993 – Location ( 5 )
3. 11 X ( .509999 ) = 5.609989 – Location ( 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 13^th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). 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. ( ½ ) in the quantum world is maybe and is a quantum constant. ( 9 ) in the quantum world is the quantum value of time and is a constant. ( 0 ) in the quantum world is a variable because although it appears as a constant, its' digital position can be changed and hence ( 0 ) is a quantum variable. 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 1000^th prime. If you add Pi ( 3.141592654 ) to 995.2299524 you get ( 998.3715451 ) which is ( 1000 – 998.371545 = 1.628454946 ) short of 1000.