Strings Solve The Riemann Hypothesis

Let's look at the Riemann Hypothesis from the aspect of a string. Suppose we have 9 strings numbered from one to nine ( 1 – 9 ). The only rule for any number on any particular string is that the total of the digits total the value of that string. For instance, string 2, would have digits totaling 2 ( 2, 11, 101, 1001, etc. ) since the digits in all the numbers total 2 ( 1 + 0 + 1 = 2 ) etc.. The question remaining is what do we do about string zero ( 0 ). String zero ( 0 ) in strings has the same function as √ -1 in particle math using numbers, formulas, and symbols. This is so because string zero ( 0 ) cannot be realistically graphed so it means anything as a real string. If string ( 0 ) is imaginary it is on the imaginary plane at right angles to string 2 or any other string for that matter. Each string number ( 2, 11, 101, 101 ) has a tail ( 2 ) on string 2 because the digits total 2 ( 101 = ( 1 + 0 + 1 = 2 )). The next question is concerning the distance between each number that is on string 2. If we add the number 11 ( 2 ) and 101 ( 2 ) number we get 4 ( 2 + 2 = 4 ) so we have a distance of 4 between each number similar to the distance in Calculus. At each location of 4 units we have an imaginary String 0 running vertically from String 2. The Riemann Hypothesis says that all the non-trivial zeros ( 0 ) are on the line ( y = ½ ) running off a line represented by the equation 1/2 + it. Our non-trivial zeros ( 0 ) become real when they cross string 2 at the “4” locations. If we can add zero to the addition of the digits without changing the digit total our zeros ( 0 ) are real. The Riemann Hypothesis, on the other hand, using particle math of numbers, formulas and symbols has to prove the zeros ( 0 ) have to cross ( y = ½ ) . Riemann's formula also uses fractions and imaginary numbers which he adds. Strings can also use fractions and imaginary String 0 which is equivalent to Riemann's method.

String 2 Whole Numbers:

2
101
1001
10001
100001
etc.

String 2 Fractions:

½
( 1/101)
(1 / 1001)
( 1 / 10001)
( 1/ 100001)
etc.

Take the fractions ( 1 / 101 ) etc. and raise them by the power of 2, since the numbers are on string 2. Fraction ( 1 / 101 ) ^ 2 = .2583624924.

Sum the powers which peak around .2583635005.

Prime number ( 97 ) which is the last of the two digit primes is the 26th prime. If we multiply ( 97 X .2583635005 ) we get ( 25.0612595501).

The Riemann Hypothesis says that the location of the prime numbers is dependent on the position of the zeros in the calculation.

If we take the square root of ( 25.0612595501) we get ( .5082946985 ).

Experimentation will show that the square root containing the most accurate zero (s) for the calculation is ( .509999999 ).
Prime number ( 97 ) which is the last of the two digit primes is the 26th prime. If we take the new square root ( .509999999 ) and raise it to the power of 2 ( .509999999 ^2 ) we get ( .260099999). If we multiply ( 97 X .260099999 ) we get ( 25.229699995) which is closer to 26 than using ( .5082946985) in the calculation.

Prime number 29 is the 11th prime. If we multiply ( 29 X (( .50999999 )^1) we get ( 14.78999997 ) which is approximately off 11 by the value of Pi ( 3.141592654 ). Therefore subtract the value of Pi.

Prime number 43 is the 15th prime. If we multiply ( 43 X (( .50999999 )^1) we get ( 11.18429998 ) which is approximately off 15 by the value of Pi ( 3.141592654 ). Therefore add the value of Pi.

Prime number 149 is the 36th prime. If we multiply ( 149 X (( .50999999 )^2) we get ( 38.75489992) which is approximately off 36 by the value of e^1 ( 2.718281828 ). Therefore subtract ( 2.718281828 ).

In summary:

1. The Riemann Hypothesis says that the location of the prime numbers is dependent on the position of the zeros in the calculation. The basic number which involves zero(s) in the calculation is the number ( .50999999 ). The prime number 191 is the 44th prime number The basic number which involves zero(s) in the calculation for the position ( 44th ) of the prime number 191 is the number ( .50099999 ).

2. You may have to add or subtract Pi ( 3.141592654 ) from the calculated position to obtain the most accurate position.

3. You may have to add or subtract the natural number ( e ) or ( 2.718281828 ) from the calculated position to obtain the most accurate position.