Primes & The Riemann Hypothesis

Here’s how Primes and The Riemann Hypothesis are connected. If you write down all the primes from 1 to 97 you will find there are 26 of them. All numbers are basically a connected string so you can use String Mathematics on them. Riemann said that all the zeros of his zeta function lay on the line ( y = ½ ). It turns out that the fraction ½ in the equation ( y = ½ ) can be used to estimate the location of any prime in a prime list. The single digit string primes are ( 1, 2, 3, 5, 7 ). If you multiply the single digit primes ( 1, 2, 3, 5, 7 ) by ½ you get the approximate location of these primes in a list of primes. The two digit string primes start at 11 and go to 97. Here it gets trickier because not all numbers ending in 1, 3, 7, 9 in column zero ( the far right column ) are primes. Two digit string primes have two digits. In estimating the location of two digit primes sometimes you multiply the two digit prime by ½ and sometimes by ( ½ X ½ ). If you use string mathematics to solve the Riemann Hypothesis the limit of the sum is (0.258363501). If you multiply the limit of the sum ((0.258363501) X 97) you get 25 in round numbers and 97 is the 26th prime. The square root of the limit of the sum (0.258363501) is ( .508 ) to three digits. If you use ( .508 ) in your calculations of the location of the primes ( 1 to 97 ) you will find your answer is generally closer to the real position in the list. ( 7919 ) is the 1000th prime. Using string mathematics you will see the number of string digits in ( 7919 ) is 4. In this particular case you multiply (½) 3 times to get an answer close to 1000 ( 7919 X .5 X .5 X .5 = 989.875 ). Multiplying ( .508 ) 3 X gives ( 1038 ) but using String Mathematics add a zero between 5 and 8 giving ( .5008 ). ( 7919 X (.5008 X .5008 X .5008 ) = 994.63 ) which is about 5 short of 1000 which is more accurate than multiplying ( 7919 X .5 X .5 X .5 ) = 989.875) The calculation of the location of any prime probably involves the adjustment of zeros (0) in (.508) and then deciding how many times you should multiply the revised figure ( for example (.5008 ) before multiplying it with the suspected prime number ending in 1, 3, 7, 9.

The general approach is possibly the following:

(.508 ) multiplied twice for primes 2 to 3 digits in width since there is an overlap in 3 digit prime numbers.

(.5008 ) multiplied three times for primes 3 to 4 digits in width ( see 7919 example ) since there is an overlap in 4 digit prime numbers.

( .50008 ) multiplied 4 times for primes 4 - 5 digits in width since there is an overlap in 5 digit prime numbers.

( .5 ) multiplied a number of times equal to one ( 1 ) less than the number of digits will give you a rough position ( ex. 7919 has 4 digits, so multiply ( .5 ) 3 times since ( 4 - 1 = 3 ).

Here’s the explanation:

The location of the real part of the not obvious or unseen zeros ( 0 ) is exactly ½ in the equation ( y = ½ ), since the number of primes in this region are denser, so the formula gives a relatively precise location. As the primes spread out, multiplying with ( .5 ) and the not obvious zeros ( .50000 etc. ) becomes increasingly inaccurate, so the obvious zeros ( 0 ) have to be added between (.5 ) and ( .008 ) to form numbers that are less close to ½. ( .508, .5008, .50008, etc. ). If the primes did not spread themselves in comparison to the other numbers you wouldn’t have to spread the obvious zeros ( 0 ) between (.5) and (.008). The not obvious zeros ( 0 ) lie on the vertical complex plane because they aren’t required for the real calculation. Therefore it can be said that the positive zeros ( 0 ) lie on the complex plane on one side of the real line ( y = ½ ) and the negative zeros ( - 0 ) lie on the other side of ( y = ½ ) on the complex plane vertical to ( y = ½ )

Using String Mathematics, the Riemann Hypothesis can be resolved like this:

The universe consists of particles / objects, strings and frames. Each object or particle has a string attached to it which we may call a tail, trail, or highway as well as anything else. Most strings do not have a location unless you state a formula such as ( y = 2 ) or ( y = ½ ). A string is a total of the object’s digits. For instance 97 has a string total of ( 9 + 7 = 16 ) or ( 9 + 7 = 16, 1 + 6 = 7 ). Therefore, as an example, 97 has two strings which are 16 and 7. The formula ( y = 2 ) has a string value of 2 and a particle / object value of 2. Here’s the solution to Riemann’s Hypothesis using string mathematics. Riemann said that all the zeros of his zeta function lie on the line ( y = ½ ) as the result of summing. In the formula ( y = 2 ), the string holds an infinity of numbers providing their sum doesn’t exceed 2. These numbers can be created by adding zeros to a number ( 101, 1001, etc. ). The string can also hold the number 2 and 11 since ( 1 + 1 = 2 ).

Here’s a condensed version of the Riemann Hypothesis proof based on string mathematics and the conversion of ( y = 2 ) to ( y = ½ ) :

1. Write down the number 2 which represents both the string length of number 2 and number 2 itself.

2. Write down the number 11 since ( 1 + 1 = 2 ) and the string length of number 2 is still intact.

3. Put a series of zeros between the two 1’s creating numbers 101, 1001, 10001, ----- 100000001. ( the string length of number 2 is still intact ).

4. Write the fractions ½, 1/11, 1/101, 1/1001, 1/1001, ------ 1/100000001.

5. Raise each fraction to the power of 2 ( this is the value of the string / tail and not the number 2 ).

6. Sum the fractions to the power of string / tail value 2.

7. Depending on the power of your calculator / spreadsheet / perseverance, the limit or convergence will be around (0.258363501).

So what??? When you were doing your calculation involving the zeros between the "1" digits ( 101, 1001, etc. ) you were actually using the string / tail values in the string which was attached to the mathematical value of 2 which was on the line ( y = 2 ). When you wrote the fractions, you were converting the line ( y = 2 ) to the line ( y = ½ ). The summation of the fractions ( ½, 1/11, 1/ 101, etc. ) produced the limit of (0.258363501). The zeros never left the string / tail when the line was converted and therefore the Riemann Hypothesis is proved using string mathematics. The same principle applies to any other equation converted to its’ reciprocal. The zeros can be placed anywhere to make a number in the string .