The Reimann Hypothesis Imaginary Math

The Riemann zeta function states all non-trivial zeros have a real part equal to ( ½ ) . When you think about it, this statement is rather profound . In our real world mathematics a zero is either a placeholder in a number ( 301 ) or has a value of zero ( 0 ) . Riemann didn't say so but as soon as these zeros ( 0 ) crossed the line ( y = ½ ) that zero had a value of ( ½ ) . Riemann went on to extend the original equation into the imaginary plane and that meant that zero ( 0 ) had a value of ( ½ + it ) on the line ( y = ½ + it ) where.”i” is imaginary and “t” is real . Riemann's unspoken thought suggests that there is undiscovered mathematics .

When you think about this statement zero can have a value of :

1. zero ( 0 )
2. ½
3. ( ½ + it )

Riemann took the original equation into the complex plane which is represented by “i” . Since zero is a single digit, the undiscovered mathematics seems to indicate that if the digits in a number are added they can be placed on a line that has a single value. Since Riemann took the original equation into the complex plane, this means that zero ( 0 ) can have a single value that is part real and part imaginary . For instance if the digits in the number ( 47 ) are added ( 4 + 7 = 11 ) and ( 1 + 1= 2 ) , then ( 47 ) is on the line ( y = 2 ) . ( 47 ) can also be on the line ( y = 11 ) because ( 4 + 7 = 11 ) . Zero ( 0 ) doesn't have a “real” digit so it can't be on a line that has a “real” meaning ( y = 0 ) . Zero ( 0 ) in the numerical system seems to be equivalent to “i” in the symbolic mathematical system, except that zero can have any value whereas “i” is limited to √ - 1 . The Riemann Hypothesis has something to do with primes . A prime is a number that can only be divided evenly by itself and one ( 1 ) . A prime number has the digits ( 1, 3, 7, 9 ) in column ( 0 ) which is the far right column . The single digit primes are ( 1, 2, 3, 5, 7 ) . Primes with more than one digit are on the lines ( 1, 2, 4, 5, 7, 8 ) . The number of primes are infinite . The ordinary numbers that aren't primes are also infinite .That being said, Riemann developed a formula for the number of primes less than a number . The “number” was defined as the sum of the zeros ( 0 ) in the zeta function. Riemann had taken the original equation into the complex plane which meant that the value of the zero ( 0 ) was now ( ½ + it ) on the line ( y = ½ + it ) . The formula said that the magnitude of the oscillations of the primes around their expected position is controlled by the real parts of the zeta function . If the zeros are on the line ( y = ½ + it ) , their value is ( ½ + it ) which means ( it ) has to have a “real” value. If ( it ) is real you add up the numbers obtaining a real number . Lastly , the error term in the prime number theorem is closely related to the position of the zeros . When you think about the statement “ the “number” was defined as the sum of the zeros ( 0 ) on the zeta function “ it really doesn't matter because the number of primes are always less than any other number. The first prime numbers are ( 1, 2, 3 ) . Their positions are ( 1, 2, 3 ) . The error term in the prime number theorem is closely related to the position of the zeros . If you multiply the primes ( 1, 2, 3 ) by 100% you get the prime's positions ( 1, 2, 3 ) . If you multiply the primes ( 1, 2, 3 ) by 10% you obtain ( .1, .2, .3 ) for the position of the primes . You dropped one ( 1 ) zero and your calculation was out. Therefore the error term in the prime number theorem is closely related to the position of the zeros. The single digit primes are ( 1, 2, 3, 5, 7 ) . Prime number 5 is in the fourth ( 4^th ) position . If you multiply prime 5 by .8 you get 4 . The number ( .8 ) is obtained by adding using the zeta function . Prime number 7 is in the fifth position ( 5^th ) . If you multiply 7 X .714285714 you get 5. This term is obtained by adding the zeta function . If you add zeros to .714285714 or put zeros somewhere within .714285714 ( for instance .7140285714 ) you will soon see that the error term in the prime number theorem is closely related to the position of the zeros .

The Weird Riemann Hypothesis

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 ????

Numbers basically break down into the following three areas:

1. Rational
2. Imaginary
3. Irrational
   Rational Numbers have a beginning and an end. Examples of rational numbers are ( 1, 2, 3, 4 ) . Imaginary numbers are numbers such as the square root of ( - 1 ) ( √ -1 ) . Irrational numbers have a beginning but no end, and seem to go on forever. Examples of Irrational numbers are Pi, “e” or Euler's number . It suddenly occurred to me that when it was realized that Riemann's zeta function said that the magnitude of the oscillations of primes around their expected position is controlled by the real parts of the zeros they were really talking about irrational numbers except that these irrational numbers were governed by zeros in the digits of the number . If you put all the prime numbers on a straight line and numbered them, you'd know their location on that straight line . Of course this is impossible . If Riemann's zeta function said that the magnitude of the oscillations of primes around their expected position is controlled by the real parts of the zeros there is an implication that an irrational number controlled by the placement of zeros ( 0 ) in the digits when multiplied by the prime number would indicate the true position of the prime on that line and by extension all the primes that preceded it . If you multiply a number by an irrational number that has a beginning and no end you will soon see that you get different answers. Riemann also said that all his non-trivial zeros had a value of ( ½ ) which seems odd when you think about it . How can a zero have the value of ( ½ ) ???? Exercising our imagination, which seems to be as temporarily irrational as is our theme, it seems that every time a zero is on Riemann's line ( y = ½ ), it has the value of ½ . If you are also looking for a major brain freeze, you can also say that if all the primes are placed on the line ( y = ½ ) from 1 to infinity they also have the value of ( ½ ) . Weird as it may seem, we are busy finding non-trivial zeros on the line ( y = ½ ) that have a value of ( ½ ) because they are on the line ( y = ½ ) . Consequently, we now have primes and zeros on the line ( y = ½ ) which all have the value of ( y = ½ ) because they are on the line ( y = ½ ) . A prime is a number that can only be divided by itself and one ( 1 ). The position of the primes from ( 1 to 31 ) can be calculated by multiplying the primes ( 1 to 31 ) by ( .509999999 ^ 1). From ( 37 to 97 ) the primes can be multiplied by ( .509999999 ^ 2 ) to keep the calculated distance close to the actual distance. This is how it works in principle by manipulating zeros ( 0 ) and the addition or subtraction of two irrational numbers Pi and “e” .

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 ). 

The Riemann Hypothesis Solved Using Calculus Limits

Mathematics basically breaks down into the following three concepts:

1. Numbers.
2. Symbols.
3. Lengths
   The 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. Calculus was the next major step. Calculus is the study of limits which involve fractions. Limits are concerned with how close you can get to something I call a brick wall ( limit ) without actually touching it. How close is actually the calculation of length before you hit the wall or limit. Here you are dealing with fractions. Calculus, using symbols, is the study of the change in the length of a formula containing symbols that change in length. If you have symbols in a formula, what happens to the formula when that symbol or symbols change in length ???? To make life simpler, the initial formula is simplified to lessen the potential confusion. In any language, as in mathematics, there are many ways to express an idea .
   Riemann expressed his idea using the following formulas:

Riemann said that the “s” in the above equation was ( s = ½ + it ). The “i” in it represented an imaginary number and the “t” represented a real number. He also said that all his non – trivial zeros lie on the line ( y = ½ ) . Innumerable calculations up to the present time have proven that Riemann's conjecture is true but there also may be an exception yet to be discovered. The Riemann Hypothesis also implies that if the zeros consistently cross the line ( y = ½ ) then these zeros have something to do with the distribution of primes. There is an implicit idea here that every crossing of zeros on the line ( y = ½ ) is related to the position of a relative prime beginning at the beginning of the line ( y = ½ ) and 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 in terms of distance of an unknown prime from the beginning of ( y = ½ ) would be accurate. The only variable in the equation is in the one ( s = ½ + it ) or ( it ). 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 unknown prime to its' actual position on the line ( y = ½ ) ???? The Riemann function backs up this concept by saying 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. In particular the amount of error is closely related to the position of the zeros . This means that the unknown value of ( it ) has adjustable zeros as well as a real number which is used to fine tune the position or location of a particular prime on the line ( y = ½ ) . Primes are numbers that can only be divided evenly by themselves and one ( 1 ). The single digit prime numbers are 1, 2, 3, 5, 7 . If you draw a straight line and put some or all the primes on it, you will soon see that there isn't a standard distance between the primes. You can also count the primes from the beginning of the line but there isn't any way of accurately saying how many primes precede a prime without counting all the primes from the beginning up to that particular prime. Let's assume that the non-trivial zeros lie on the line ( y = ½ ) . For the sake of convenience we can also put all the primes from one ( 1 ) to infinity on the line ( y = ½ ) and indicate their position on the line ( y = ½ ) from the beginning of the line ( y = ½ ) . The single digit primes are five ( 5 ) in number consisting of 1, 2, 3,. 5, 7. If you arbitrarily multiply the primes by ( ½ ) you will soon see that your answer has usually very little relationship to the primes true position on the line ( y = ½ ) . In other words, by multiplying by ( ½ ) you will soon see that if all the numbers are written down from zero ( 0 ) to infinity that ( ½ ) of them aren't primes. The Riemann Hypothesis also implies that if the zeros consistently cross the line ( y = ½ ) then these zeros have something to do with the distribution of primes. If the zeros consistently cross the line ( y = ½ ) than these zeros ( 0 ) can be used in the calculation . We now have the value of the line ( ½ ) plus the zeros to use in some way to see if we can calculate the position of the primes on the line ( y = ½ ) from the beginning. Multiplying by ( ½ ) or ( .5 ) or ( .50 ) doesn't change your original answer. Using ( .05 ) also doesn't improve things to any great extent. Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes. To accomplish this manipulation, place some digits after the zero ( 0 ) in ( .50 ) such as forming ( .509999 ) . Therefore:

1. 5 X ( .509999 ) = 2.549995 (4)
2. 7 X ( .509999 ) = 3.569993 ( 5 )

3. 11 X ( .509999 ) = 5.609989 ( 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 ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). 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. Riemann's intuition also told him that the zeros can be manipulated ( position changed or zeros added ) which is also true. This manipulation of zeros ( 0 ) is really using the principles of limits in calculus to see how close you can find a factor when multiplied by a prime that will give you the position of that prime on the line ( y = ½ ). 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. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).

The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I've also proved that the manipulation of the zeros as implied by Riemann can bring a prime closer to its' position using the principles of Calculus.

Another Way To Look At The Riemann Hypothesis

Let us assume that all the primes to infinity lie on a horizontal line . While it is acknowledged that the primes lie on this horizontal line to infinity and are in order, we don't know precisely where the primes lie on this horizontal line in terms of their distance from each other. Most of the time we picture a number on a line as being at a specific point on the line, even though we may not draw that point on the line when we write that particular number down . Let us assume that wherever the infinite primes are on the horizontal line at varying distances from each other, we have a vertical line ( y = ½ ) off the horizontal line which forms an upside down “T” ( ! ! ! ! etc. ) . The infinite primes are at the intersection of a vertical line ( y = ½ ) and an infinite line holding all the prime numbers . Riemann said all his zeros are on the line ( y = ½ ) and have a value of ( ½ ) when they are on this line ( y = ½ ) . If this is true, then we can use Riemann's zeros ( 0 ) in the calculation of the location of the prime on the infinite horizontal line containing all the primes because the infinite horizontal line containing the primes intersect the ( y = ½ ) line. At the present time we are calculating to prove that all of Riemann's non-trivial zeros line on the line ( y = ½ ) and so far so good . Riemann also said his zeros can be added or subtracted or placed in different positions in order to adjust the position of the primes from the beginning of the horizontal line . This adjustment as an integer also establishes the number of primes preceding that prime number .The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ). The prime numbers from position 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see from the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.

If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.

Here’s how the system works for numbers in general.

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 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.

It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. The power zeros can also be adjusted. For instance (( .5099999999 ) ^2 = (..260099999 )). If we change the digit 6 to zero ( 0 ) creating ( .200099999 ) and multiplying it by the prime number 347 we get ( 69.43469965 ) Prime number 347 is the 69th prime. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range if the calculated answer is close but too far out.

In Summary:

1. ( .509999999 ) may be adjusted using Riemann zeros.
2. ( .509999999 ) may be raised to a power to increase the zeros.
3. ( .509999999 ) raised to a power may have its' power zeros adjusted.
4. Tweaking can be done by adding or subtracting Pi or “e”.

Riemann Hypothesis Hopefully, Finally, Resolved

Let us assume that all the primes to infinity lie on a horizontal line . While it is acknowledged that the primes lie on this horizontal line to infinity and are in order, we don't know precisely where the primes lie on this horizontal line in terms of their distance from each other. Let us assume that wherever the infinite primes are on the horizontal line at varying distances from each other, we have a vertical line ( y = ½ ) off the horizontal line which forms an upside down “T” ( ! ! ! ! etc. ) . The infinite primes are at the intersection of a vertical line ( y = ½ ) and an infinite line holding all the prime numbers . Riemann said all his zeros are on the line ( y = ½ ) and have a value of ( ½ ) when they are on this line ( y = ½ ) . At the present time we are calculating to prove that all of Riemann's non-trivial zeros line on the line ( y = ½ ) and so far so good . If we can establish that the calculation of the position of these primes use ( ½ ) and Riemann's zero's ( 0 ) , then we have established that Riemann's zeros are all on the line ( y = ½ ) because ( y = ½ ) intersects with the horizontal line holding the infinite primes . Riemann also said his zeros can be added or subtracted or placed in different positions in order to adjust the position of the primes from the beginning of the horizontal line . This adjustment as an integer also establishes the number of primes preceding that prime number .The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ). The prime numbers from position 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see from the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.

If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.

Here’s how the system works for numbers in general.

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 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.

It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. The power zeros can also be adjusted. For instance (( .5099999999 ) ^2 = (..260099999 )). If we change the digit 6 to zero ( 0 ) creating ( .200099999 ) and multiplying it by the prime number 347 we get ( 69.43469965 ) Prime number 347 is the 69th prime. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range if the calculated answer is close but too far out.

In Summary:

1. ( .509999999 ) may be adjusted using Riemann zeros.

2. ( .509999999 ) may be raised to a power to increase the zeros.

3. ( .509999999 ) raised to a power may have its' power zeros adjusted.

4. Tweaking can be done by adding or subtracting Pi or “e”.

Primes, Fractions & The Riemann Hypothesis

The Riemann Hypothesis involves an understanding of fractions . The largest fraction you can have is ( ½ ) or ( .5 ) . This may seem counter-intuitive, but to prove it simply subtract any ( 1 / number ) fraction from ( ½ ) and you will end up with a positive fraction . The Riemann Hypothesis uses the formula ( 1 + ½ ^s + 1/3 ^s + ---- ) to infinity . In this formula you have ( s = ½ + it ) . You may be familiar with calculus , but if you aren't calculus is concerned with how close you can get a fraction to a number without actually touching ( reaching ) the number. This getting close as a numerical figure is called a limit . You can think of it as how close you can get your nose to a brick wall without touching / hurting your nose . If you add fractions to infinity, you will find that the distance from ( ½ ) becomes smaller and smaller . If you add fractions to infinity that have been raised to a power, you will find that the distance from ( ½ ) is closer to ( ½ ) then by simply adding unraised fractions . Since the distance from ( ½ ) is getting closer, you can think of all the added fractions as getting closer to the line ( y = ½ ) . You may not know this, but if you raise a fraction to the power of zero ( 0 ), you get one ( 1 ) ( ½ ^0 = 1 ) . Zero ( 0 ) functions more as a placeholder than a value . Of course, you can argue that if you don't have anything you have something which is zero ( 0 ) . If the raised fraction to a power is getting closer and closer to ( ½ ), then the power to which it is being raised is getting closer and closer to zero ( 0 ) since ( ½ ) raised to the power of zero ( 0 ) is one ( 1 ) which means in the extreme ( ½ ) doesn't change because it becomes one or itself ( think philosophically here ) . One of the problems in mathematics is that you can't graph zero . You can, of course, draw a line and say this is line zero ( 0 ) meaning something you are considering starts here . If the power to which a fraction ( ½ ) is getting closer and closer to zero ( 0 ), then that zero ( 0 ) is getting closer and closer to ( ½ ) which is on the line ( y = ½ ) . Therefore it can be argued that zero ( 0 ) has the value of ( ½ ) on the line ( y = ½ ) . This is what Riemann said. Calculations have been done and continue to be done to prove that Riemann's zeros ( 0 ) lie on the line ( y = ½ ) and so far, so good. Riemann also said that his zeros ( 0 ) had something to do with primes . A prime is a number that can only be evenly divided by itself and one ( 1 ) . He also went on to say that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. If we can calculate the location of a prime, we know how many primes precede it because the primes would be in order . To state the obvious , the position of any prime in a series is less than the prime itself. For instance, the first one ( 1 ) digit primes are 1, 2, 3, 5, 7 . The primes ( 1, 2, 3 ) are in the positions ( 1, 2, 3 ) . The primes 5 and 7 are in positions 4 and 5 . The primes ( 1, 2, 3 ) can be multiplied by one ( 1 ) ( 1 X 1, 2 X 1, 3 X 1 ) to get their true position ( 1, 2, 3 ) . If we multiply prime 5 by .9099 we get ( 5 X ..9099 = 3.63996) which is mathematically close to 4 . If we multiply prime 7 by .70999 we get ( 7 X .70999 = 4.96993 which is mathematically close to 5 . These zeros ( 0 ) are Riemann zeros .

Obviously, you can diddle the multiplier any way you want to get the answer, but since we are dealing with the Riemann Hypothesis , Riemann zeros and the number ½ on the line ( y = ½ ) it generally works like this:

The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ). The prime numbers from position 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see from the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.

If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.

Here’s how the system works for numbers in general.

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 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.
It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. The power zeros can also be adjusted. For instance (( .5099999999 ) ^2 = (..260099999 )). If we change the digit 6 to zero ( 0 ) creating ( .200099999 ) and multiplying it by the prime number 347 we get ( 69.43469965 ) Prime number 347 is the 69th prime. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range if the calculated answer is close but too far out.

In Summary:

1. ( .509999999 ) may be adjusted using Riemann zeros.

2. ( .509999999 ) may be raised to a power to increase the zeros.

3. ( .509999999 ) raised to a power may have its' power zeros adjusted.

4. Tweaking can be done by adding or subtracting Pi or “e”.

Fractions & The Riemann Hypothesis

The Riemann Hypothesis involves an understanding of fractions . The largest fraction you can have is ( ½ ) or ( .5 ) . This may seem counter-intuitive, but to prove it simply subtract any fraction from ( ½ ) and you will end up with a positive fraction . The Riemann Hypothesis uses the formula ( 1 + ½ ^s + 1/3 ^s + ---- ) to infinity . In this formula you have ( s = ½ + it ) . You may be familiar with calculus , but if you aren't calculus is concerned with how close you can get to a fraction without actually touching ( reaching ) it . This getting close as a numerical figure is called a limit . You can think of it as how close you can get your nose to a brick wall without touching / hurting your nose . If you add fractions to infinity, you will find that the distance from ( ½ ) becomes smaller and smaller . If you add fractions to infinity that have been raised to a power, you will find that the distance from ( ½ ) is closer to ( ½ ) then by simply adding unraised fractions . Since the distance from ( ½ ) is getting closer, you can think of all the added fractions as getting closer to the line ( y = ½ ) . You may not know this, but if you raise a fraction to the power of zero ( 0 ), you get one ( 1 ) ( ½ ^0 = 1 ) . Zero ( 0 ) functions more as a placeholder than a value . Of course, you can argue that if you don't have anything you have something which is zero ( 0 ) . If the raised fraction to a power is getting closer and closer to ( ½ ), then the power to which it is being raised is getting closer and closer to zero ( 0 ) since ( ½ ) raised to the power of zero ( 0 ) is one ( 1 ) which means in the extreme ( ½ ) doesn't change because it becomes one or itself ( think philosophically here ) . One of the problems in mathematics is that you can't graph zero . You can, of course, draw a line and say this is line zero ( 0 ) meaning something you are considering starts here . If the power to which a fraction ( ½ ) is getting closer and closer to zero ( 0 ), then that zero ( 0 ) is getting closer and closer to ( ½ ) which is on the line ( y = ½ ) . Therefore it can be argued that zero ( 0 ) has the value of ( ½ ) on the line ( y = ½ ) . This is what Riemann said . The Riemann Hypothesis says that all the non-trivial zeros ( 0 ) are on the line ( y = ½ ) ( along with all the fractions added to ( ½ )).

Riemann's Hypothesis – Calculating Zeros On ( y = ½ )

Sometimes in life you stumble upon something that is basically true and it's up to other people to prove it. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. Riemann provided a formula to prove his belief which was ( s = ½ + it ). The current method of calculating the position of the zeros is by plugging numbers into the formula and calculating. I don't have any quarrel with this method except that there isn't any guarantee that you haven't missed a zero which is supposed to be on the line ( y = ½ ) and is actually off the line ( y = ½ ) . The rub in the calculation is that if you miss a “t” or an “i”, it might be the number that proves Riemann's Hypothesis is incorrect. Personally, I don't think it matters because Riemann accidentally stumbled onto something that is correct so missed numbers in “t” or “i” are academic. I don't know what numbers they are using but the available literature says the first few zeroes were calculated around  14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. Even if you proved that all the zeros lay on the line ( y = ½ ), that proof still doesn't tell you how all these zeros relate to the position of the primes on the line ( y = ½ ) in useable terms.

Here's my proof for what it is worth:

Riemann said that all his non – trivial zeros lie on the line ( y = ½ ) . These zeros are at the location that the primes form a right angle line to ( y = ½ ) or in other words, are vertical to the line ( y = ½ ) . Prime numbers are defined as numbers that can only be divided by themselves and ( 1 ). An example of one digit primes are ( 1, 2, 3, 5, 7 ). You can see from these numbers that primes aren't linear ( 1, 2, 3, 4 ) but form harmonics. If all the primes were played musically, you would get a tune. The physical distance between the primes would be similar to the value of the rests in a musical piece. Calculations have been done on the location that some of Riemann's zeros lie on the line ( y = ½ ) using Riemann's formula ( ½ + it ). The first few zeroes were calculated around  14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. These numbers are not close to the position of the beginning one digit primes ( 1, 2, 3, 5, 7 ). You will see that the distance between ( 1, 2, 3, ) is ( 1 ) and the distance between 3 & 5, 5 & 7, is 2. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes which is equivalent to altering the value of the rests in a musical piece.

1. 1 X .99 = .99 ( 1 )

2. 2 X .99 = 1.98 ( 2 )

3. 3 X .99 = 2.97 ( 3 )

If we add the digit ( ½ ) or ( .5 ) in Riemann's line ( y = ½ ) and Riemann's zero ( 0 ) to .99, we form the number ( .509999 ).

1. 5 X ( .509999 ) = 2.549995 (4)
2. 7 X ( .509999 ) = 3.569993 ( 5 )
3. 11 X ( .509999 ) = 5.609989 ( 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 13th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). 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. 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 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).

The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I have also found the missing pieces to Riemann's Hypothesis which was correct as far as it went.

The Riemann Hypothesis Resolved ( Hopefully finally)

Riemann said that all his non – trivial zeros lie on the line ( y = ½ ) . These zeros are at the location that the primes form a right angle line to ( y = ½ ) or in other words, are vertical to the line ( y = ½ ) . Prime numbers are defined as numbers that can only be divided by themselves and ( 1 ). An example of one digit primes are ( 1, 2, 3, 5, 7 ). You can see from these numbers that primes aren't linear ( 1, 2, 3, 4 ) but form harmonics. If all the primes were played musically, you would get a tune. The physical distance between the primes would be similar to the value of the rests in a musical piece. Calculations have been done on the location that some of Riemann's zeros lie on the line ( y = ½ ) using Riemann's formula ( ½ + bi ). The first few zeroes were calculated around  14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. These numbers are not close to the position of the beginning one digit primes ( 1, 2, 3, 5, 7 ). You will see that the distance between ( 1, 2, 3, ) is ( 1 ) and the distance between 3 & 5, 5 & 7, is 2. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes which is equivalent to altering the value of the rests in a musical piece.

1. 1 X .99 = .99 ( 1 )

2. 2 X .99 = 1.98 ( 2 )

3. 3 X .99 = 2.97 ( 3 )

If we add the digit ( ½ ) or ( .5 ) in Riemann's line ( y = ½ ) and Riemann's zero ( 0 ) to .99, we form the number ( .509999 ).

1. 5 X ( .509999 ) = 2.549995 (4)

2. 7 X ( .509999 ) = 3.569993 ( 5 )

3. 11 X ( .509999 ) = 5.609989 ( 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 13th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). 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. 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 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).

The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I have also found the missing pieces to Riemann's Hypothesis which was correct as far as it went.

Riemann Hypothesis Solved Using A Quantum Mechanical System

Riemann's Hypothesis is that all the non-trivial zeros lie on the line ( y = ½ ) and these zeros have something to do with prime numbers. Since Riemann's Hypothesis includes the line ( y = ½ ), let's chose a quantum energy system in which the energy levels are 2. Riemann's equation also included complex numbers. In a quantum energy system Riemann's complex numbers could be represented by zero's ( 0 ) since ( 2 + 0 = 2 ). ( 1 + 1= 2 ) also equals 2. Riemann also went on to say that the prime number locations were influenced by the position of the zeros. To extend our ( 1 + 1 = 2 ) analogy, we now have ( 1 + 0 + 1 = 2 ). We can convert these additions to numbers ( 11, 101, etc. ). We can now say that we have a real axis ( y = 2 ) with infinite numbers with zeros between their ones ( 11, 101, 1001, 10001, etc. ) on an infinite real axis line ( y = 2 ). These numbers would also be evenly spaced on the real axis line ( y = 2 ) since each infinite number would total energy level 2 which means the distance between the numbers would be ( 2 + 2 = 4 ) or a spacing of 4. Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ). If we flip our infinite real axis line ( y = 2 ), we create a real axis line ( y = ½ ) with flipped infinite real numbers ( 1/11, 1/ 101, 1/1001, etc. ) all the way to infinity. If we add ( ½ + 1/11+ 1/ 101 + 1001 + etc. ) we eventually start to get numbers resembling ( .509999. .500999, etc. ) The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's. If you raise these numbers to the power of 2, you get somewhere over ¼ . As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .5099999 ) X ( .5099999 ) you get very close to 26. The calculation of the prime numbers are a little bit more complicated than this illustration but this is the basics.

One Small Step For The Riemann Hypothesis

The Riemann Hypothesis says that all the non-trivial zeros ( 0 ) are on the line ( y = ½ ) and that this hypothesis has something to do with prime numbers. Another way of expressing it, is by saying 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. In particular, the error term in the prime number theorem is closely related to the position of the zeros.

In summary:

1. We have a line ( y = ½ ) .
2. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ )
3. The error term in the prime number theorem is related to the position of the zeros.

The Prime Number Theorem is concerned with the number of primes preceding a number. Due to the nature of mathematics when you multiply, it is probably better to have the number as a prime number for considerations of accuracy. In other words, if you can calculate the location of a prime number in the prime number series you automatically know how many prime numbers precede it.

We have the line ( y = ½ ) . Therefore let the first digit of the multiplier be ½ or .5. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ ) . Therefore let the middle digits be zeros ( 0 ). We now have the number .50. To finish our multiplier add the digit 9 to the end forming the number ( .5099999999 ).

The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ).

The prime numbers from 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see in the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.

If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.

Here’s how the system works for numbers in general.

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 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.
It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range. The present method of proving the Riemann Hypothesis consists of calculating whether or not zeros cross or touch the line ( y = ½ ) . While I appreciate the effort, it seems to me that I have proved the Riemann Hypothesis because my zeros can be infinitely added between the digits ½ and 9 ( .5----9 ). It can be seen that the calculation of the position of a prime and the number of primes preceding that prime all depend on the Riemann Hypothesis' real zeros. Thus the Riemann Hypothesis has been proved by illustrating its’ real relationship to the position of the primes and the number of primes preceding it.

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.

Riemann Hypothesis Resolved Using Strings

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 have 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 ( 11, 101, 110, etc. ) since the total digits do not exceed (2). It can readily be seen that all zeros line 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 ( 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 lines ( y = k ) or ( y = 1/k ) he would have ended up hypothesizing that all zeros ( 0’s ) also lie on these lines.

Riemann Hypothesis Resolved Geometrically

The easiest way to solve the Riemann Hypothesis is by using geometry. Fractions in geometry are actually one line drawn over another line. For instance, ½ , is line ( y = 1 ) drawn over line ( y = 2 ). The distance ( s ) between the line ( y = 1 ) and line ( y = 2 ) is the power to which the fraction ( ½ ) is raised (( ½) ^ (s)). The addition sign ( + ) is the distance between the lines ( y = 1 ) and the line ( y = 2 ) representing ½ and the lines ( y = 1 ) and the line ( y = 3 ) representing 1/3rd etc.. If you measure from Line (1) over Line ( 2 ) representing ½ then the distance is consistent from that line to the other lines and for all intents and purposes the zeros ( 0 ) which are the starting measuring points, are on the geometric line ( y = ½ ). Using individual particle arithmetic its’ been established that the first 100 billion zeros do lie on the line ( y = ½ ).

Reimann accidentally touched on a new way of looking at mathematics and fractions when he said:

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

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.

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 .

String Mathematics

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 .