Prime numbers and the Riemann Hypothesis digits are all intertwined. A prime number is defined as a number that can only be divided by itself and one ( 1 ) ( 1, 2, 3, 5, 7 ).
Here's the skinny on prime numbers:
1. Prime numbers, if they are prime numbers, have the digits 1, 3, 7, 9 in column 0 ( farthest right column ) ( 11, 13, 17, 19 ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0). If not try other numbers.
The calculation of the location of the prime numbers on a line or calculating the number of previous primes on a line involve the digits ( ½, 0 & 9 ). 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. I have separately discovered that the location of the prime numbers on the line ( y = ½ ) or any other line for that matter involves the digit 9.
The primary number for the location of a prime on any line from the Riemann Hypothesis is ( .509999 ). For the primes ( 1, 2, 3, ) you multiply them by ( .99 ) since their actual positions are ( 1, 2, 3 ). For the primes (5 to 23 ) you multiply them by ( .509999 ). For the primes 29 & 31 you multiply them by ( .509999 ) and subtract Pi ( 3.141592654 ) to bring them closer to their true position.
For the rest of the primes up to 97 you raise ( .509999 ) to the power of 2 and adjust using either Pi or the natural number e ( 2.718281828 ).
In summary, Riemann anticipated the number ( ½ ) & ( zero ( 0 ) and zero adjustment, but missed my discovery of the number 9 and the necessity of Pi and ( e ).
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).
It would seem from the above, that Riemann anticipated the zeros ( 0 ) and the value of those zeros being ( ½ ) since they were on the line ( y = ½ ). Riemann didn't anticipate the number 9 or the option of adjusting the calculation by adding 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 ) .
Showing posts with label Riemann Hypothesis Solution. Show all posts
Showing posts with label Riemann Hypothesis Solution. Show all posts
Thursday, October 20, 2011
Tuesday, October 18, 2011
The Riemann Hypothesis Resolved Using ½ , 0 and 9
Like a lot of people, I've been fooling around trying to prove Riemann's Hypothesis without much success. Let's suppose for the sake of argument that all the zeros lie on the line ( y = ½ ) . If they do, the solution for the Riemann Hypothesis brings us no closer to showing what the deep connection is between the zeta function and the distribution of the primes. My reading of the literature on the Riemann Hypothesis indicates that the zeros lie on the line ( y = ½ ) and that if the zeroes are manipulated in some fashion you can calculate the location of the primes to a greater and greater accuracy. If you know the location of one prime you know how many primes precede it because you are counting ( 1, 2, 3, 4 etc. ). Riemann's genius or intuition was that the distribution of the primes involve the number zero ( 0 ) and ½. The line is important because if you count something, that something is being placed on a line. If the something is a prime it lies on a line. Riemann's Hypothesis says that all the zeros lie on a line so for the sake of argument you have primes on a line and zeros ( 0 ) on a line. Riemann said that if you manipulate the zeros ( 0 ), you can bring the prime sitting on the line closer and closer to its' true position. What the line ( y = ½ ) means is a mystery, although the number ( ½ ) may prove useful. The Riemann Hypothesis equation brings the zeros ( 0 ) to the line ( y = ½ ) depending on what you are using for ( t ) or the other variables. Let's suppose the primes lie on a straight line which also has zeros ( 0 ) on it.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891). You will see from these calculations that the best multiplier is .5099999 since the multiplication ( answer ) is the closest to 6.
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
I calculate that the best usable number for the location of a prime is ( .5099999 ). Riemann said that the proportion of primes is about ( 1 / ln (x) ) where ( ln(x) is the natural logarithm of x to base ( e ) or ( 2.718281828 ) . Using Riemann's formula the proportion of primes below prime number 11 is ( 1 / ln(11) or ( .417032391 ). The number of primes below ( 11 ) using this formula is ( 11 X .4107032391 = 4.587356306) or in round numbers ( 5 ). My calculation using ( .5099999 ) is ( 6 ) in round numbers which is the exact location of prime number ( 11 ). Riemann worked out that if the zeros really do lie on the critical line, then the primes stray from the ( 1/ln(x) ) distribution exactly as much as a bunch of coin tosses stray from the 50:50 distribution law. I have proved that the non-trivial zeros have to lie on the line, because my calculation for the location of the primes uses Riemann's non-trivial zeros that have to be on the line ( y = ½ ) or a line since you are calculating the location of the primes in a sequence ( or the number of preceding primes ).
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only conjectures the zeros are on the line ( y = ½ ) . The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. I believe that I have proved that the Riemann zeros are all on the line whether it be ( y = ½ ) or some other line, because those zeros have to be used in a calculation while being on a line to prove that the location of the primes are on a straight line and not somewhere else in space.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891). You will see from these calculations that the best multiplier is .5099999 since the multiplication ( answer ) is the closest to 6.
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
I calculate that the best usable number for the location of a prime is ( .5099999 ). Riemann said that the proportion of primes is about ( 1 / ln (x) ) where ( ln(x) is the natural logarithm of x to base ( e ) or ( 2.718281828 ) . Using Riemann's formula the proportion of primes below prime number 11 is ( 1 / ln(11) or ( .417032391 ). The number of primes below ( 11 ) using this formula is ( 11 X .4107032391 = 4.587356306) or in round numbers ( 5 ). My calculation using ( .5099999 ) is ( 6 ) in round numbers which is the exact location of prime number ( 11 ). Riemann worked out that if the zeros really do lie on the critical line, then the primes stray from the ( 1/ln(x) ) distribution exactly as much as a bunch of coin tosses stray from the 50:50 distribution law. I have proved that the non-trivial zeros have to lie on the line, because my calculation for the location of the primes uses Riemann's non-trivial zeros that have to be on the line ( y = ½ ) or a line since you are calculating the location of the primes in a sequence ( or the number of preceding primes ).
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only conjectures the zeros are on the line ( y = ½ ) . The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. I believe that I have proved that the Riemann zeros are all on the line whether it be ( y = ½ ) or some other line, because those zeros have to be used in a calculation while being on a line to prove that the location of the primes are on a straight line and not somewhere else in space.
Sunday, October 16, 2011
The Position Of Primes Using The Riemann Hypothesis Numbers & The One That Was Missed
Like a lot of people, I've been fooling around trying to prove Riemann's Hypothesis without much success. Let's suppose for the sake of argument that all the zeros lie on the line ( y = ½ ) . If they do, the solution for the Riemann Hypothesis brings us no closer to showing what the deep connection is between the zeta function and the distribution of the primes. My reading of the literature on the Riemann Hypothesis indicates that the zeros lie on the line ( y = ½ ) and that if the zeroes are manipulated in some fashion you can calculate the location of the primes to a greater and greater accuracy. If you know the location of one prime you know how many primes precede it because you are counting ( 1, 2, 3, 4 etc. ). Riemann's genius or intuition was that the distribution of the primes involve the number zero ( 0 ) and ½. The line is important because if you count something, that something is being placed on a line. If the something is a prime it lies on a line. Riemann's Hypothesis says that all the zeros lie on a line so for the sake of argument you have primes on a line and zeros ( 0 ) on a line. Riemann said that if you manipulate the zeros ( 0 ), you can bring the prime sitting on the line closer and closer to its' true position. What the line ( y = ½ ) means is a mystery, although the number ( ½ ) may prove useful. The Riemann Hypothesis equation brings the zeros ( 0 ) to the line ( y = ½ ) depending on what you are using for ( t ) or the other variables. Let's suppose the primes lie on a straight line which also has zeros ( 0 ) on it.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891).
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only proves the zeros either cross the line or are on the line ( y = ½ ). The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. The most fascinating part is that the tweaking involves Pi and the natural number ( e ) but that is another story.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891).
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only proves the zeros either cross the line or are on the line ( y = ½ ). The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. The most fascinating part is that the tweaking involves Pi and the natural number ( e ) but that is another story.
Saturday, October 08, 2011
Riemann Hypothesis & Strings
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. The calculations done up to the present time have shown that the zeros are on the line ( y = ½ ) and there are yet more calculations to go. Let us suppose that each of Riemann's zeros are part of separate strings of numbers and each time the Riemann equation crosses the line ( y = ½ ) it does so where a zero is located. To make it simple, suppose that the zeros are in the middle of the number and the number 1 is on both ends ( 101, 1001, 10001 etc. ). If we add up the digits in these numbers we find the total is always 2 ( 1 + 0 + 1 = 2 ), ( 1 + 0 + 0 + 1 = 2 ), ( 1 + 0 + 0 + 0 +1 = 2 ). To take it one step further we could say these numbers are on the line ( y = 2 ) since all the numbers total 2. Riemann's Hypothesis is that the non- trivial zeros lie on the line ( y = ½ ). If we flip our analogy, the numbers on the line ( y = ½ ) become ( 1 / 101, 1/1001, 1/ 10001, etc. ). Riemann's equation still crosses the line ( y = ½ ) where the zeros are located. Riemann's equation isn't exact in terms of the positions of the zeros. It has been calculated that the first position of the zeros is around 14 and the second calculated position around 21. Our analogy is exact because we can create these numbers to infinity and the principle is the same. Riemann said that the zeros have a real value of ½ or .5 . If we add ( ½ + 1/101 ) we obtain ( .50990099 ). Prime number ( 11 ) is the 6th prime ( 1, 2, 3, 5, 7, 11 ). ( 11 X .50990099 = 5.6089108911 ) or 6 rounded to one digit. The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's and can be adjusted to get a more accurate estimate. As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .50990099 ) X ( .50990099 )) you get ( 25.22969903 )which is very close to 26. If you adjust the Riemann zeros to form the number ( .509999 ) and do the multiplication ( 97 x (.509999 X .509999) ) you get ( 25.22960106 ) which is further away from 26. You can see from these calculations that if the Riemann zeros are manipulated as Riemann has suggested to obtain the location of the prime ( or calculate the number of primes up to that point ) you can vary the distance to the true position of the prime. As a matter of interest Riemann's equation first cuts the line ( y = ½ ) at a position just over 14. The 14th prime is 41. ( 41 X ( ½ + 1/ 101) X ( ½ + 1 / 101 ) is ( 10.664099959 ) which is just off 14 by the value of Pi ( 3.141592554 ). The 21st prime is 71. It's position is calculated at 18 using the same method ( 71 X ( ½ + 1/ 101 ) X ( ½ + 1 / 101 )) which is very close to 21. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.
Wednesday, October 05, 2011
Riemann Hypothesis Solved Using A Quantum Mechanical System ( Revised )
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 ). We can also say that each of the zero's in ( 101, 1001, 10001, etc. ) are on the line ( y = 2) since the numbers ( 101, 1001, 10001, etc. are also on this line. Riemann also said that the values of the zeros ( 0's ) on the line are equal to ( ½ ) which is also true when the line ( y = 2 ) is flipped. We could also extend this argument to one which says that the value of the ones ( 1's ) are also ( ½ ) since the reciprocal of one ( 1 ) is ( 1 ) and the fraction ( 1 / 1 ) when added ( 1 / 101 ) has the value ½ on the line ( y = ½ ) . 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. Flipped the distance between the numbers would be one ( 1 ) since ( ½ + ½ = 1 ). Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ) and this is what we've proved up to now since the non-trivial zeros are in the numbers ( 101, 1001, 10001, etc. ). 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 + 1/101 ) we obtain ( .50990099 ). Prime number ( 11 ) is the 6th prime ( 1, 2, 3, 5, 7, 11 ). ( 11 X .50990099 = 5.6089108911 ) or 6 rounded to one digit. The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's and can be adjusted to get a more accurate estimate.. As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .50990099 ) X ( .50990099 )) you get ( 25.22969903 )which is very close to 26. If you adjust the Riemann zeros to form the number ( .509999 ) and do the multiplication ( 97 x (.509999 X .509999) you get ( 25.22960106 ) which is further away from 26. You can see from these calculations that if the Riemann zeros are manipulated as Riemann has suggested to obtain the location of the prime ( or calculate the number of primes up to that point ) you can vary the distance to the true position of the prime. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.
Sunday, October 17, 2010
Riemann Hypothesis Resolved Using Prime Numbers
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. It is known that there are infinitely many zeros on the line 1/2 + it as t ranges over the real numbers. A prime number is any number that is evenly divided by itself and 1. Therefore, a prime number is a specialized real number which exists on the x axis. It is also known that any prime number can create a real number that isn’t a prime. A real number is a number that is sequentially placed on the x - axis and therefore has a specific location that can be calculated through subtraction. A prime number, however, cannot be specifically located on the x - axis because the distance between the prime numbers aren’t consistent. Riemann extended his equation into complex numbers which oscillate on a vertical plane at right angles to the x - axis or in other words across the line ( y = ½ ). The non-trivial zeros do not become real until the oscillating complex zero crosses ( y = ½ ) and becomes real. The complex zero is really on an imaginary string before it lies along the line ( y = ½ ). At this point the real zero can be used to calculate the location of the prime number on the x - axis. It is also known that there are an infinite number of prime numbers and by extension an infinite number of real numbers because prime numbers can be used to construct real numbers. So far the calculated zeros lie on the critical line, but the prime’s location strays because the calculation process behaves just like a 50:50 coin toss. The 50:50 distribution law says that coins will divide themselves proportionately over the long term, but on the specific terms they generally will not distribute themselves evenly. There are always exceptions. The most important thing that Riemann said was 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 and the error term in the prime number theorem is closely related to the position of the zeros.
Here are some facts about prime numbers.
1. Prime numbers, if they are prime numbers, have the numbers 1, 3, 7, 9 in column 0 ( farthest right column ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0).
3. A prime number is defined as being only evenly divisible by itself and one ( 1 ).
Here’s the proof:
A prime number is defined as any number that can be only divided evenly by itself and one. Furthermore a prime number, if it is a prime number, only has the digits, 1, 3, 7, 9 in column 0 which is the farthest right column. I have also discovered that if the sum of the digits of any number ending in 1, 3, 7, 9 total a multiple of 3, except for prime number 3 ( for example total 6, 9, 12, etc. ) then it isn’t a prime number. The single digit prime numbers are 1, 2, 3, 5, 7, if we ignore the convention of no longer considering 1 as a prime number. A real number which includes prime numbers are located somewhere on the x - axis. If we multiply each single digit prime number ( 1, 2, 3, 5, 7 ) by ½ we get 1 at approximately ½, 2 at approximately 1, 3 at approximately 1.5, 5 at approximately 2.5 and 7 at approximately 3.5. 1 is actually at 1, 2 at 2, 3 at 3, 5 at 4 and 7 at 5. 97 is a prime and if we multiply 97 by ½, we get 48.5. The prime 97 isn’t anywhere close to being the 48th prime. The Riemann Hypothesis says 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. Our initial calculation indicated that the prime number 97 was oscillating around position 48.5 on the x axis which isn’t correct. The Riemann Hypothesis says 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. What to do???? We know that the prime number 97 lies on the x axis as it is a specialized real number. The Riemann Hypothesis says 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. Therefore we make up a multiplier. The first digit is ½ or .5. The second digit is one of the Riemann Hypothesis zeros so we now have ( .50 ) . ( .50 ) X 97 is no better off than multiplying 97 X ( .5 ). We know that the prime numbers have 1, 3, 7, or 9 in column 0. Therefore arbitrarily add the 9 digit to ( .50 ) forming ( .509 ). If we multiply prime 97 X ( .509 ) the answer is worse. If, however, we take ( .509 ^ 2 ) we get ( .259081 ) and 97 X ( .259081 ) = 25.130857. Prime number 97 is in actuality the 26th prime. The Riemann Hypothesis says 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. Using the Riemann Hypothesis zeros between .5 and 9 ( .5----9 ) and raising it to a power we can oscillate the prime 97 around its’ position. Thus the error can be controlled by adjusting the Riemann Hypothesis zeros.
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 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust error accordingly using Riemann Hypothesis zeros.
As a matter of interest, if we are looking for the largest prime number in existence, the simplest way of doing it is to add digits to the left of any number ending in 1, 3, 7, 9 in column zero ( 0 ). Find the total of all the digits added together and divide by 3. If the result is an integer with no remainder, it is not a prime number. This truth can be verified by seeing if the number is only evenly divisible by itself and 1. It is also interesting that any number ending in 1, 3, 7, 9, if it isn’t a prime, is usually divisible by some number ending in 1, 3, 7, 9 in column 0.
If all the zeros used for calculating the position of primes on strings are real, then the primes themselves are real. Primes can be combined to create real numbers as well as fractions, so all the zeros for those numbers are real. Therefore the Riemann Hypothesis is true.
Actual Location Primes Multiplier 2 To Power 2 Power 2 Calculated Location
1 1 0.509999999 1 0.509999999 0.509999999
2 2 0.509999999 1 0.509999999 1.019999999
3 3 0.509999999 1 0.509999999 1.529999998
4 5 0.509999999 1 0.509999999 2.549999997
5 7 0.509999999 1 0.509999999 3.569999996
6 11 0.509999999 1 0.509999999 5.609999994
7 13 0.509999999 1 0.509999999 6.629999993
8 17 0.509999999 1 0.509999999 8.669999991
9 19 0.509999999 1 0.509999999 9.68999999
10 23 0.509999999 1 0.509999999 11.72999999
11 29 0.509999999 1 0.509999999 14.78999997
12 31 0.509999999 1 0.509999999 15.80999998
13 37 0.509999999 2 0.260099999 9.623699981
14 41 0.509999999 2 0.260099999 10.66409998
15 43 0.509999999 2 0.260099999 11.18429998
16 47 0.509999999 2 0.260099999 12.22469998
17 53 0.509999999 2 0.260099999 13.78529997
18 59 0.509999999 2 0.260099999 15.34589997
19 61 0.509999999 2 0.260099999 15.86609997
20 67 0.509999999 2 0.260099999 17.42669997
21 71 0.509999999 2 0.260099999 18.46709996
22 73 0.509999999 2 0.260099999 18.98729996
23 79 0.509999999 2 0.260099999 20.54789996
24 83 0.509999999 2 0.260099999 21.58829996
25 89 0.509999999 2 0.260099999 23.14889995
26 97 0.509999999 2 0.260099999 25.22969995
The position of the primes was calculated by multiplying the primes ( 1 to 31 ) by ( .509999999 ^ 1). From ( 37 to 97 ) the primes were multiplied by ( .509999999 ^ 2 ) to keep the calculated distance close to the actual distance.
Actual Location Primes String 1 String 2 Multiplier 2 To Power 2 Power 2 Calculated Location
27 101 2 2 0.509999999 2 0.2601 26.27009995
28 103 4 4 0.509999999 2 0.2601 26.79029995
29 107 8 8 0.509999999 2 0.2601 27.83069994
30 109 10 1 0.509999999 2 0.2601 28.35089994
31 113 5 5 0.509999999 2 0.2601 29.39129994
32 127 10 1 0.509999999 2 0.2601 33.03269993
33 131 5 5 0.509999999 2 0.2601 34.07309993
34 137 11 2 0.509999999 2 0.2601 35.63369993
35 139 13 4 0.509999999 2 0.2601 36.15389993
36 149 14 5 0.509999999 2 0.2601 38.75489992
37 151 7 7 0.509999999 2 0.2601 39.27509992
38 157 13 4 0.509999999 2 0.2601 40.83569992
39 163 10 1 0.509999999 2 0.2601 42.39629992
40 167 14 5 0.509999999 2 0.2601 43.43669991
41 173 11 2 0.509999999 2 0.2601 44.99729991
42 179 17 8 0.509999999 2 0.2601 46.55789991
43 181 10 1 0.509999999 2 0.2601 47.07809991
44 191 11 2 0.500999999 2 0.251001 47.9411909
45 193 13 4 0.500999999 2 0.251001 48.4431929
46 197 17 8 0.500999999 2 0.251001 49.4471969
47 199 19 1 0.500999999 2 0.251001 49.9491989
The position of the primes was calculated by multiplying the primes ( 101 to 181 ) by (.5099999999^2) From (191 to 199 ) the primes were multiplied by ( .5009999999 ^ 2 ) to keep the calculated distance close to the actual distance.
The most important thing that Riemann said was 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 and the error term in the prime number theorem is closely related to the position of the zeros. If you multiplied the primes from ( 191 to 199 ) by ( .509999999^2 ) the answer would be further from their real locations and hence the error would be greater.
It is also interesting that by either adding or subtracting Pi or ( e^1 ) depending on what is required you can get the Riemann Zero calculation closer to the actual location of the primes.
Actual Location Calculated Location Add / Subtract Total
1 0.509999999 0 0.509999999
2 1.019999999 0 1.019999999
3 1.529999998 0 1.529999998
4 2.549999997 0 2.549999997
5 3.569999996 0 3.569999996
6 5.609999994 0 5.609999994
7 6.629999993 0 6.629999993
8 8.669999991 0 8.669999991
9 9.68999999 0 9.68999999
10 11.72999999 0 11.72999999
11 14.78999997 -3.141592654 11.64840732
12 15.80999998 -3.141592654 12.66840733
13 9.623699981 3.141592654 12.76529263
14 10.66409998 3.141592654 13.80569263
15 11.18429998 3.141592654 14.32589263
16 12.22469998 3.141592654 15.36629263
17 13.78529997 3.141592654 16.92689263
18 15.34589997 3.141592654 18.48749262
19 15.86609997 3.141592654 19.00769262
20 17.42669997 3.141592654 20.56829262
21 18.46709996 3.141592654 21.60869262
22 18.98729996 3.141592654 22.12889262
23 20.54789996 3.141592654 23.68949261
24 21.58829996 3.141592654 24.72989261
25 23.14889995 0 23.14889995
26 25.22969995 0 25.22969995
27 26.27009995 0 26.27009995
28 26.79029995 0 26.79029995
29 27.83069994 0 27.83069994
30 28.35089994 0 28.35089994
31 29.39129994 0 29.39129994
32 33.03269993 0 33.03269993
33 34.07309993 0 34.07309993
34 35.63369993 0 35.63369993
35 36.15389993 0 36.15389993
36 38.75489992 -2.718281828 36.03661809
37 39.27509992 -2.718281828 36.55681809
38 40.83569992 -2.718281828 38.11741809
39 42.39629992 -3.141592654 39.25470726
40 43.43669991 -3.141592654 40.29510726
41 44.99729991 -3.141592654 41.85570726
42 46.55789991 -3.141592654 43.41630725
43 47.07809991 -3.141592654 43.93650725
44 47.9411909 -3.141592654 44.79959825
45 48.4431929 -3.141592654 45.30160025
46 49.4471969 -3.141592654 46.30560425
47 49.9491989 -2.718281828 47.23091707
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 since ( .509999999 ) and (.599999990) both hold the same digits but the zero ( 0 ) position has been shifted thus putting the position of the primes further out of alignment. 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.
Here are some facts about prime numbers.
1. Prime numbers, if they are prime numbers, have the numbers 1, 3, 7, 9 in column 0 ( farthest right column ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0).
3. A prime number is defined as being only evenly divisible by itself and one ( 1 ).
Here’s the proof:
A prime number is defined as any number that can be only divided evenly by itself and one. Furthermore a prime number, if it is a prime number, only has the digits, 1, 3, 7, 9 in column 0 which is the farthest right column. I have also discovered that if the sum of the digits of any number ending in 1, 3, 7, 9 total a multiple of 3, except for prime number 3 ( for example total 6, 9, 12, etc. ) then it isn’t a prime number. The single digit prime numbers are 1, 2, 3, 5, 7, if we ignore the convention of no longer considering 1 as a prime number. A real number which includes prime numbers are located somewhere on the x - axis. If we multiply each single digit prime number ( 1, 2, 3, 5, 7 ) by ½ we get 1 at approximately ½, 2 at approximately 1, 3 at approximately 1.5, 5 at approximately 2.5 and 7 at approximately 3.5. 1 is actually at 1, 2 at 2, 3 at 3, 5 at 4 and 7 at 5. 97 is a prime and if we multiply 97 by ½, we get 48.5. The prime 97 isn’t anywhere close to being the 48th prime. The Riemann Hypothesis says 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. Our initial calculation indicated that the prime number 97 was oscillating around position 48.5 on the x axis which isn’t correct. The Riemann Hypothesis says 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. What to do???? We know that the prime number 97 lies on the x axis as it is a specialized real number. The Riemann Hypothesis says 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. Therefore we make up a multiplier. The first digit is ½ or .5. The second digit is one of the Riemann Hypothesis zeros so we now have ( .50 ) . ( .50 ) X 97 is no better off than multiplying 97 X ( .5 ). We know that the prime numbers have 1, 3, 7, or 9 in column 0. Therefore arbitrarily add the 9 digit to ( .50 ) forming ( .509 ). If we multiply prime 97 X ( .509 ) the answer is worse. If, however, we take ( .509 ^ 2 ) we get ( .259081 ) and 97 X ( .259081 ) = 25.130857. Prime number 97 is in actuality the 26th prime. The Riemann Hypothesis says 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. Using the Riemann Hypothesis zeros between .5 and 9 ( .5----9 ) and raising it to a power we can oscillate the prime 97 around its’ position. Thus the error can be controlled by adjusting the Riemann Hypothesis zeros.
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 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust error accordingly using Riemann Hypothesis zeros.
As a matter of interest, if we are looking for the largest prime number in existence, the simplest way of doing it is to add digits to the left of any number ending in 1, 3, 7, 9 in column zero ( 0 ). Find the total of all the digits added together and divide by 3. If the result is an integer with no remainder, it is not a prime number. This truth can be verified by seeing if the number is only evenly divisible by itself and 1. It is also interesting that any number ending in 1, 3, 7, 9, if it isn’t a prime, is usually divisible by some number ending in 1, 3, 7, 9 in column 0.
If all the zeros used for calculating the position of primes on strings are real, then the primes themselves are real. Primes can be combined to create real numbers as well as fractions, so all the zeros for those numbers are real. Therefore the Riemann Hypothesis is true.
Actual Location Primes Multiplier 2 To Power 2 Power 2 Calculated Location
1 1 0.509999999 1 0.509999999 0.509999999
2 2 0.509999999 1 0.509999999 1.019999999
3 3 0.509999999 1 0.509999999 1.529999998
4 5 0.509999999 1 0.509999999 2.549999997
5 7 0.509999999 1 0.509999999 3.569999996
6 11 0.509999999 1 0.509999999 5.609999994
7 13 0.509999999 1 0.509999999 6.629999993
8 17 0.509999999 1 0.509999999 8.669999991
9 19 0.509999999 1 0.509999999 9.68999999
10 23 0.509999999 1 0.509999999 11.72999999
11 29 0.509999999 1 0.509999999 14.78999997
12 31 0.509999999 1 0.509999999 15.80999998
13 37 0.509999999 2 0.260099999 9.623699981
14 41 0.509999999 2 0.260099999 10.66409998
15 43 0.509999999 2 0.260099999 11.18429998
16 47 0.509999999 2 0.260099999 12.22469998
17 53 0.509999999 2 0.260099999 13.78529997
18 59 0.509999999 2 0.260099999 15.34589997
19 61 0.509999999 2 0.260099999 15.86609997
20 67 0.509999999 2 0.260099999 17.42669997
21 71 0.509999999 2 0.260099999 18.46709996
22 73 0.509999999 2 0.260099999 18.98729996
23 79 0.509999999 2 0.260099999 20.54789996
24 83 0.509999999 2 0.260099999 21.58829996
25 89 0.509999999 2 0.260099999 23.14889995
26 97 0.509999999 2 0.260099999 25.22969995
The position of the primes was calculated by multiplying the primes ( 1 to 31 ) by ( .509999999 ^ 1). From ( 37 to 97 ) the primes were multiplied by ( .509999999 ^ 2 ) to keep the calculated distance close to the actual distance.
Actual Location Primes String 1 String 2 Multiplier 2 To Power 2 Power 2 Calculated Location
27 101 2 2 0.509999999 2 0.2601 26.27009995
28 103 4 4 0.509999999 2 0.2601 26.79029995
29 107 8 8 0.509999999 2 0.2601 27.83069994
30 109 10 1 0.509999999 2 0.2601 28.35089994
31 113 5 5 0.509999999 2 0.2601 29.39129994
32 127 10 1 0.509999999 2 0.2601 33.03269993
33 131 5 5 0.509999999 2 0.2601 34.07309993
34 137 11 2 0.509999999 2 0.2601 35.63369993
35 139 13 4 0.509999999 2 0.2601 36.15389993
36 149 14 5 0.509999999 2 0.2601 38.75489992
37 151 7 7 0.509999999 2 0.2601 39.27509992
38 157 13 4 0.509999999 2 0.2601 40.83569992
39 163 10 1 0.509999999 2 0.2601 42.39629992
40 167 14 5 0.509999999 2 0.2601 43.43669991
41 173 11 2 0.509999999 2 0.2601 44.99729991
42 179 17 8 0.509999999 2 0.2601 46.55789991
43 181 10 1 0.509999999 2 0.2601 47.07809991
44 191 11 2 0.500999999 2 0.251001 47.9411909
45 193 13 4 0.500999999 2 0.251001 48.4431929
46 197 17 8 0.500999999 2 0.251001 49.4471969
47 199 19 1 0.500999999 2 0.251001 49.9491989
The position of the primes was calculated by multiplying the primes ( 101 to 181 ) by (.5099999999^2) From (191 to 199 ) the primes were multiplied by ( .5009999999 ^ 2 ) to keep the calculated distance close to the actual distance.
The most important thing that Riemann said was 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 and the error term in the prime number theorem is closely related to the position of the zeros. If you multiplied the primes from ( 191 to 199 ) by ( .509999999^2 ) the answer would be further from their real locations and hence the error would be greater.
It is also interesting that by either adding or subtracting Pi or ( e^1 ) depending on what is required you can get the Riemann Zero calculation closer to the actual location of the primes.
Actual Location Calculated Location Add / Subtract Total
1 0.509999999 0 0.509999999
2 1.019999999 0 1.019999999
3 1.529999998 0 1.529999998
4 2.549999997 0 2.549999997
5 3.569999996 0 3.569999996
6 5.609999994 0 5.609999994
7 6.629999993 0 6.629999993
8 8.669999991 0 8.669999991
9 9.68999999 0 9.68999999
10 11.72999999 0 11.72999999
11 14.78999997 -3.141592654 11.64840732
12 15.80999998 -3.141592654 12.66840733
13 9.623699981 3.141592654 12.76529263
14 10.66409998 3.141592654 13.80569263
15 11.18429998 3.141592654 14.32589263
16 12.22469998 3.141592654 15.36629263
17 13.78529997 3.141592654 16.92689263
18 15.34589997 3.141592654 18.48749262
19 15.86609997 3.141592654 19.00769262
20 17.42669997 3.141592654 20.56829262
21 18.46709996 3.141592654 21.60869262
22 18.98729996 3.141592654 22.12889262
23 20.54789996 3.141592654 23.68949261
24 21.58829996 3.141592654 24.72989261
25 23.14889995 0 23.14889995
26 25.22969995 0 25.22969995
27 26.27009995 0 26.27009995
28 26.79029995 0 26.79029995
29 27.83069994 0 27.83069994
30 28.35089994 0 28.35089994
31 29.39129994 0 29.39129994
32 33.03269993 0 33.03269993
33 34.07309993 0 34.07309993
34 35.63369993 0 35.63369993
35 36.15389993 0 36.15389993
36 38.75489992 -2.718281828 36.03661809
37 39.27509992 -2.718281828 36.55681809
38 40.83569992 -2.718281828 38.11741809
39 42.39629992 -3.141592654 39.25470726
40 43.43669991 -3.141592654 40.29510726
41 44.99729991 -3.141592654 41.85570726
42 46.55789991 -3.141592654 43.41630725
43 47.07809991 -3.141592654 43.93650725
44 47.9411909 -3.141592654 44.79959825
45 48.4431929 -3.141592654 45.30160025
46 49.4471969 -3.141592654 46.30560425
47 49.9491989 -2.718281828 47.23091707
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 since ( .509999999 ) and (.599999990) both hold the same digits but the zero ( 0 ) position has been shifted thus putting the position of the primes further out of alignment. 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.
Sunday, November 23, 2008
Riemann Hypothesis Resolved
The Riemann Hypothesis is all about location since Riemann said all the zeros of his function lie on the line ( y = ½ ). Let’s assume they all do lie along the line ( y = ½ ). Up to the present time using Riemann’s equation of trail it has been established that the first 100 billion zeros do lie on the line ( y = ½ ) calculating each value at a time, but there is a possibility one or several don’t lie on the line. In any event no one has developed a simple solution to prove that the trail of all zeros ( 0 ) end up on the line ( y = ½ ) no matter what values are used in the calculation. Perhaps in some way, as yet not understood, Riemann’s equation is flawed even though it has been proved right. The question arises because imaginary numbers are used and although we can work them mathematically to get real solutions, perhaps our understanding of them isn’t really complete and that incompleteness is showing itself in the Riemann formula. There is also a belief that the Riemann Zeta Function has a relationship to the distribution of prime numbers. Prime numbers are numbers that can only be divided evenly by themselves and 1. An example of primes are 1, 2, 3, 5, 7, 11, however, they aren’t evenly distributed and no one formula up to the present time has been discovered that will calculate the position of any prime.
For instance the following primes have the following locations ( 1 to 6 ):
1. 1
2. 2
3. 3
4. 5
5. 7
6. 11
The problem is that it is almost impossible to list all the primes because the primes aren’t readily discernible and for a second part extend to infinity. Therefore the problem arises as to just where does any arbitrarily selected prime fit into a list or conversely how many primes precede this prime. The resolution of this conundrum is within the Riemann Hypothesis. Riemann didn’t realize it when he formulated his hypothesis because it hadn’t been invented but he is really talking about an aspect of string theory when he said:
All the zeros of the Riemann zeta function lie on the line y = ½.
Let’s use string mathematics. Every object has a string or tail associated with it. In string mathematics the tail of any number ( object ) is the sum of its’ digits. The location of each tail of a number is generally unknown, but in the Riemann Hypothesis case, Riemann said the tail lies on the line y = ½ as the result of summing. Before summing, the string or tail existed on the line y = 2 as that was the string tail value of any mathematical object that was 2. For instance 97’s digits total 16 ( 9 + 7 = 16 ) and ( 1 + 6 = 7 ). This means 97 has both a string tail of ( 16 ) and ( 7 ). The string tail of all numbers can be reduced to a figure from 1 to 9. which is its’ tail or string in one digit. The digit tail zero ( 0 ) for the number zero ( 0 ) is imaginary in the sense that you can’t do anything with zero ( 0 ) except use it as a place holder in a number ( ex. 609 ) or treat it as the square root of ( - 1 ). In string theory the equation ( y = 2 ) means that any number ( y ) also has a string equal to 2 as well as a number value 2. Since zeros do not increase the string / tail value of 2, 2 can be any number that contains zeros (0) in the string part. If you accept this premise then any zero remains on the string y = 2 by definition. If a calculation is performed on ( y = 2 ) the string tail is automatically recalculated to reflect the new line. This means if the line ( y = 2 ) is changed to ( y = ½ ), the string / tail values become ( 1 ) over a number containing zeros ( 0 ) so that the total of the denominator is 2 ( y = ½ ).Riemann said that the zeros go into the line y = ½ or their location is on the line y = ½. What Riemann was actually saying even though he didn’t know it was that any number containing zeros whose digits totaled two representing a string had its’ zeros on the line y = ½ which was 2 as a fraction. Here is a solution to Riemann’s Hypothesis based on string theory and limits. If we add 1 + ½ + 1/3 + etc. we find that the fraction gets larger and larger and never reaches a limit from here to infinity. If we add 1 + (½)^2 + (1/3)^2 + etc. we might soon see that the addition soon converges to a value or limit.
Here’s a condensed version of the Riemann Hypothesis proof based on string theory and limits:
1. Write down the number 2 which represents both the string length of number 2 and number 2 itself.
2. Write down the number 11 since ( 1 + 1 = 2 ) and the string length of number 2 is still intact.
3. Put a series of zeros between the two 1’s creating numbers 101, 1001, 10001, ----- 100000001. ( the string length of number 2 is still intact ).
4. Write the fractions ½, 1/11, 1/101, 1/1001, 1/1001, ------ 1/100000001.
5. Raise each fraction to the power of 2 ( this is the value of the string / tail and not the number 2 ).
6. Sum the fractions to the power of string / tail value 2.
7. Depending on the power of your calculator / spreadsheet / perseverance, the limit or convergence will be around (0.258363501).
8. Write down the primes from 1 to 97.
9. Number the primes.
10. If you multiply 97 X (0.258363501) the location of the prime will be approximately at 25 and 97 is at location 26.
11. If you multiply 97 X (½) ^ 2 you get approximately 24.
So what??? When you were doing your calculation involving the zeros between the "1" digits ( 101, 1001, etc. ) you were actually using the string / tail values in the string which was attached to the mathematical value of 2 which was on the line ( y = 2 ). When you wrote the fractions, you were converting the line ( y = 2 ) to the line ( y = ½ ). The summation of the fractions ( ½, 1/11, 1/ 101, etc. ) produced the limit of (0.258363501). The zeros never left the string / tail when the line was converted because they were the zeros in the numbers 101, 1001 etc.. The Riemann Hypothesis is proved.
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.
Friday, November 21, 2008
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 .
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 .
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