## Sunday, November 23, 2008

### Riemann Hypothesis Resolved

The Riemann Hypothesis is all about location since Riemann said all the zeros of his function lie on the line ( y = ½ ). Let’s assume they all do lie along the line ( y = ½ ). Up to the present time using Riemann’s equation of trail it has been established that the first 100 billion zeros do lie on the line ( y = ½ ) calculating each value at a time, but there is a possibility one or several don’t lie on the line. In any event no one has developed a simple solution to prove that the trail of all zeros ( 0 ) end up on the line ( y = ½ ) no matter what values are used in the calculation. Perhaps in some way, as yet not understood, Riemann’s equation is flawed even though it has been proved right. The question arises because imaginary numbers are used and although we can work them mathematically to get real solutions, perhaps our understanding of them isn’t really complete and that incompleteness is showing itself in the Riemann formula. There is also a belief that the Riemann Zeta Function has a relationship to the distribution of prime numbers. Prime numbers are numbers that can only be divided evenly by themselves and 1. An example of primes are 1, 2, 3, 5, 7, 11, however, they aren’t evenly distributed and no one formula up to the present time has been discovered that will calculate the position of any prime.

For instance the following primes have the following locations ( 1 to 6 ):

1. 1

2. 2

3. 3

4. 5

5. 7

6. 11

The problem is that it is almost impossible to list all the primes because the primes aren’t readily discernible and for a second part extend to infinity. Therefore the problem arises as to just where does any arbitrarily selected prime fit into a list or conversely how many primes precede this prime. The resolution of this conundrum is within the Riemann Hypothesis. Riemann didn’t realize it when he formulated his hypothesis because it hadn’t been invented but he is really talking about an aspect of string theory when he said:

All the zeros of the Riemann zeta function lie on the line y = ½.

Let’s use string mathematics. Every object has a string or tail associated with it. In string mathematics the tail of any number ( object ) is the sum of its’ digits. The location of each tail of a number is generally unknown, but in the Riemann Hypothesis case, Riemann said the tail lies on the line y = ½ as the result of summing. Before summing, the string or tail existed on the line y = 2 as that was the string tail value of any mathematical object that was 2. For instance 97’s digits total 16 ( 9 + 7 = 16 ) and ( 1 + 6 = 7 ). This means 97 has both a string tail of ( 16 ) and ( 7 ). The string tail of all numbers can be reduced to a figure from 1 to 9. which is its’ tail or string in one digit. The digit tail zero ( 0 ) for the number zero ( 0 ) is imaginary in the sense that you can’t do anything with zero ( 0 ) except use it as a place holder in a number ( ex. 609 ) or treat it as the square root of ( - 1 ). In string theory the equation ( y = 2 ) means that any number ( y ) also has a string equal to 2 as well as a number value 2. Since zeros do not increase the string / tail value of 2, 2 can be any number that contains zeros (0) in the string part. If you accept this premise then any zero remains on the string y = 2 by definition. If a calculation is performed on ( y = 2 ) the string tail is automatically recalculated to reflect the new line. This means if the line ( y = 2 ) is changed to ( y = ½ ), the string / tail values become ( 1 ) over a number containing zeros ( 0 ) so that the total of the denominator is 2 ( y = ½ ).Riemann said that the zeros go into the line y = ½ or their location is on the line y = ½. What Riemann was actually saying even though he didn’t know it was that any number containing zeros whose digits totaled two representing a string had its’ zeros on the line y = ½ which was 2 as a fraction. Here is a solution to Riemann’s Hypothesis based on string theory and limits. If we add 1 + ½ + 1/3 + etc. we find that the fraction gets larger and larger and never reaches a limit from here to infinity. If we add 1 + (½)^2 + (1/3)^2 + etc. we might soon see that the addition soon converges to a value or limit.

Here’s a condensed version of the Riemann Hypothesis proof based on string theory and limits:

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).

8. Write down the primes from 1 to 97.

9. Number the primes.

10. If you multiply 97 X (0.258363501) the location of the prime will be approximately at 25 and 97 is at location 26.

11. If you multiply 97 X (½) ^ 2 you get approximately 24.

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 because they were the zeros in the numbers 101, 1001 etc.. The Riemann Hypothesis is proved.