For Every Zero Crossing Riemann's Line ( y = ½ )

Sometimes in life, you stumble across something that holds more meaning than first meets the eye. Meeting your future husband or wife for the first time and not realizing it is a good example. Mathematically, Dirac's equation is probably the most famous something. Dirac had the ability to visualize an equation in terms of geometry . Dirac came up with an equation that was more insightful than he first realized which you can look up in the Internet . Riemann's something was when it was realized that his zeta function said that the magnitude of the oscillations of primes around their expected position is controlled by the real parts of the zeros of the zeta function . You and I can read the words but what is the meaning without repeating the sentence as an answer ????  

Mathematics basically breaks down into the following three areas:

1. Numbers.
2. Symbols.
3. Lengths
   Numbers can represent anything. Symbols such as x or y are usually equal to something which is up to you to discover. These symbols may be in a formula expressing relationships which are familiar to anyone that has ever studied algebra. These relationships in algebra are generally a static value as only one numerical answer is needed. Lengths are usually the distance from the beginning of something. An example of this concept is the measurement of a board before you cut it with a saw. The Riemann Hypothesis also hints at the concept of length as it relates to the number of primes. A prime is a number that can only be divided evenly by itself and one ( 1 ) . The five ( 5 ) single digit prime numbers are ( 1. 2, 3, 5, 7 ) . Riemann came up with the formula ( s = ½ + it ) that related to another formula. Without filling in the background, the Riemann Hypothesis implies that if the zeros consistently cross the line ( y = ½ ) in relation to the formula ( s = ½ + it ) then these zeros have something to do with the distribution of primes. There is an implicit idea here that every crossing of a zero on the line ( y = ½ ) is related to the position of a corresponding prime from the beginning of the line ( y = ½ ) to the end of the line ( y= ½ ) at infinity. Theoretically, if the values of Riemann's formula were chosen correctly, the calculation of the zeros crossing the line would give you the distance of a known prime from the beginning of ( y = ½ ) . This means that if the location of the zero crossing the line ( y = ½ ) were accurate that location would correspond with the position of a prime on the line ( y = ½ ) . Riemann's formula is really a disguised calculus problem using zeros ( 0 ) to adjust the placement of a prime on the line ( y = ½ ) . In other words, if you adjust the zero in a mathematical expression how close can you get the relevant known prime to its' actual position on the line ( y = ½ ) ???? This position calculation would give you the number of primes from the beginning of the line ( y = ½ ) . My previous blogs on this topic will give you an idea as to how the calculation is done. Calculus, using symbols, is the study of the change in the length of a formula containing symbols that change in length. One of these symbols is ( c ) which represents a constant ( whole number ) which never changes. The accepted length of a constant ( c ) in calculus ( d (c)) is zero ( 0 ) but that isn't correct because Riemann's intuition told him that the zeros which are also a constant had a real value of ( ½ ) on his line ( y = ½ ) consisting of primes. The actual length of any ordinary constant is the sum of its' digits. 47 would have a length of 11 or 2 ( 4 + 7 = 11 ) or ( 1 + 1 = 2 ). 

Location Of Primes

If you look at a table of Prime Numbers, you will notice that they aren’t evenly spaced which raises the question of how many primes precede any arbitrarily selected number. On the other hand, if you look at a table of prime numbers they seem to follow a pattern. You will also notice that the prime numbers get rarer, as you look at higher and higher numbers. Most formulas that have been developed to find the location of any prime number seem to be a hit and miss affair because the prime numbers aren’t evenly spaced. George Riemann, who formulated the Riemann Hypothesis, found that his formula produced the occasional zero as an answer which he hypothesized lay on the line ( y = ½ ). So far, calculating on an individual basis, it has been found that over a billion zeros using Riemann’s formula lie on this line but since Riemann hypothesized all zeros the search goes on. So far, no one has developed a proof that definitely establishes all zeros lie on the line ( y = ½ ). Here’s a way of resolving the conundrum concerning Riemann’s zeros (0’s) using the concept of strings. If we add the digits of any number (N) like ( 7919 ), for example, we get a total ( 7 + 9 + 1 + 9 = 26 ) which represents its’ string ( 26 ) or line ( y = 26 ). If we continue the addition of the digits ( 26 ) we get ( 2 + 6 = 8 ) or a line ( y = 8 ) for number ( 7919 ). Riemann had the idea that when his equation produced a zero (0) in the calculation, that zero for non- trivial numbers was on the line ( y = ½ ). Working in reverse, we can take any arbitrary line ( y = k ), with ( k = ( 0 to 9 )), because all numbers ( N ) sum to a number from ( 0 to 9 ) as one digit. Any line ( y = k ) can contain an infinite number of numbers, since zero ( 0 ) does not change the value of ( y = k ). For instance, if ( y = 2 ), it can hold the numbers ( 2, 11, 101, 110, etc. ) since the total digits do not exceed (2). It can readily be seen that all zeros lie on the line ( y = k ) no matter what the value of k whether it be one digit or multiple digits and thus Riemann was correct when he said all zeros lie on the line ( y = ½ ), since ( y = ½ ) is the reciprocal of ( y = 2 ) and contains the numbers ( ½, 1/11, 1/101, 1/110, etc.. The same conclusion can be applied to any arbitrary line ( y = k ) or ( y = 1/k ) for values of k from ( zero (0) to infinity ) of any number (N) which has an infinite number of digits (D) If Riemann had changed his formula to produce zeros ( 0’s ) for other non-trivial values for other lines ( y = k ) or ( y = 1/k ) he would have ended up hypothesizing that all zeros ( 0’s ) also lie on these lines. There is also the suggestion that the line ( y = ½ ) has a relationship to a formula for locating the position of any prime / number of primes preceding that particular prime. If ( y = 2 ), then the line ( y = 2 ) contains the numbers (2, 11, 101, 1001, 10001, 100001, etc.) since the sum of the digits does not exceed 2. If ( y = ½ ), then the line ( y = ½ ) contains the numbers ( ½, 1/11, 1/101, 1/1001, 1/10001, 1/100001, etc. ) since the sum of the digits does not exceed ½ and ½ is the reciprocal of 2 or ( y = 2 ) . Raise each number ( ½, 1/11, 1/101, 1/1001, 1/10001, 1/100001, etc. ) to the power of 2. Sum the numbers which have a limit of (0.258363501). The square root of (0.258363501) is (0.508294698). Similarly, If ( y = ½ ), then the line ( y = ½ ) contains the numbers ( ½, 1/11, 1/110, 1/1100, 1/11000, 1/110000, etc. ) since the sum of the digits does not exceed ½ and ½ is the reciprocal of 2 or ( y = 2 ) . Raise each number ( ½, 1/11, 1/110, 1/1100, 1/11000, 1/110000, etc. ) to the power of 2. Sum the numbers which have a limit of (0.258347942). The square root of (0.258347942) is (0.508279394). The important number is (.508) to 3 figures. The calculation of the location of any prime probably involves the adding of zeros (0’s) to the middle of (.508) to equal the number of digits in the probable prime after 3 digits. For instance ( 7919 has 4 digits so therefore extend ( .508 ) to ( .5008 )) by adding 1 zero to the middle and then raising (.5008) to the power equal to one less than the digit in the probable prime ending in ( 1, 3, 7, 9 ).

The general approach is the following:

Extend ( .508) to equal the number of digits in the probable prime starting after 3 digits and then raise ( .508 extended ) to a power equal to the number of digits in the probable prime minus 1. For instance ( 7919 ) contains 4 digits so extend ( .508 ) to 4 digits by adding zero (0) to the middle ( .5008 ). (7919) has 4 digits, so raise ( .5008 ) to the power of 3 ( 4 digits - 1 = 3 ). ( .5008 ^ 3 = .12560096 ) Multiply ( 7919 X .12560096 ) which equals ( 994.6340063 ). 7919 is the 1000th prime which means this method is approximately 5 short of 1000. The same method can be used using ( ½ or .5 ) but the answer is less accurate (7919 X (.5^3) = 989.875) which is ( 10.125 ) short of 1000 ( 7919 is the 1000th prime).

For numbers, 3 digits or under, such as ( 97 ) multiply ( 97 X (.508^2)) which equals approximately 25. As a matter of interest ( 97 ) is the 26th prime.

For 1 digit numbers multiply by ( .508 ).

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.