The universe is based on the number 3. If you ever have listened to Sesame Street you will remember that the Count always started the program by saying " This program is being brought to you by the number ????. The Count when welcoming you into this universe when you were born could have said " This universe is brought to you by the number 3". If you believe in a higher power simply because this universe is too complex to be random then you now know the higher power pushed her pen towards the number 3 . If you don’t believe me how do you account for the natural number "e" (2.718281828) and Pi (3.141592654) which is close to 3 ??? Mathematics is no different. At the moment we have 2 known main categories of Mathematics. What’s the 3rd???

They are:

1. Particle / object

2. Geometry

Particle / object mathematics is the mathematics that we use all the time. We add, subtract, multiply and divide these numbers. We use equations with knowns, unknowns and variables commonly known as algebra and calculus. Geometry has to do with diagrams and from the diagrams you calculate things such as angles and distances. This is trigonometry and geometry. There is a third category called strings. Strings are everywhere in the universe. Sometimes strings are called waves or dark energy. Quantum mechanics talks of quantum particles and wave like behavior which is strings by another name. We have something called String Theory. Each mathematical particle / object has an associated string attached with it. Most of the time we only calculate things using the particle / object and ignore the string. Mathematics can also use the string part and ignore the particle / object part.

This is how it works:

1. Total the number of digits in a number. For instance ( 39 ) has two digits whose total is 12 ( 3 + 9 = 12 ). This means that the particle / object (39) has a string or tail equal to 12.

2. The string contains numbers, whose digits, when added can’t exceed the value of the string which in the example (39) is (12). Higher numbers can be formed by adding zeros ( 0 ) between the numbers. For instance for particle / object (39) which has a string / tail ( 12 ) also contains the number ( 10902 ) because ( 1 + 0 + 9 + 0 + 2 = 12 ).

3. A string can hold more than one number of the same value and contains an infinite value of numbers because of the ability to place a zero ( 0 ) in the number without changing the string total ( 12 ).

4. Any mathematical operation that can be used by particles / objects can be used with strings.

The Mathematical Categories are:

1. Particles / Objects

2. Geometry

3. Strings

## Sunday, November 30, 2008

## Saturday, November 29, 2008

### 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.

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.

## 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.

## Saturday, November 22, 2008

### 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 .

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 .

## 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 .

## Tuesday, November 18, 2008

### Particles, Strings & Dark Stuff

This universe has particles, strings and dark stuff. At a quantum level, we talk about particles. At an eyeball level we talk about objects which is really the same as particles. Objects are such things as you and me and buildings. Things we can see and touch. Objects can also be abstract concepts like events that just happen. Math has its’ particles which are variables, unknowns, and constants. Computer programming has its’ objects. Just look at Java. Strings are a little harder to grasp. If you look at mathematics, any formula is really a string that holds some objects ( variables, unknowns and constants ) together which can be manipulated into something that provides an answer. Language consists of objects which we call letters. You arrange these letters in some combination and you have a word. The word has a meaning. You take this string of letters ( objects ) and make another string called a sentence or paragraph. This sentence, paragraph, novel, etc. or string has a meaning. The scientists have a string theory which is basically that all matter comes with strings. Ever hear of someone telling you that this gift (object) comes with strings attached??? Same principle. Black Holes, for instance, is just a bunch of compressed or intertwined strings. If that concept is correct, and if Black Holes do crazy things to gravity, time and space than it stands to reason that gravity, time and space are strings. A Black Hole generally sucks everything into it, which means, by extension, that gravity, time, and space exist as strings surrounding us even though we aren’t experiencing their existence in that form. The universe is composed of Dark Matter which we can’t see. This Dark Matter is 90% of the universe and generally keeps things in place. If things are kept in place in the universe by Strings it would seem to me that a string is dark matter and provides a path and tail for a particle or object since we can see where the object has been. A path or highway is another name for a string. Chemical bonds are a form of string acting in the same principle as gravity holding chemicals in place. If an object is an abstract event that won’t move we talk in terms of inertia which is gravity by another name. The chemical bond is formed at the string of each chemical or a compound. Essentially two strings of two different chemicals joining together. At a quantum level we have a nucleus held together by quantum strings. If we tear the string ( nuclear fission ) we create a quick release of energy which is really a series of strings gone berserk. If we develop a method for compressing a string in a nucleus without bending it, we have nuclear fusion which produces a controlled form of energy . On it goes but this essay will give you a quick overview of the concept. Just look around for some examples!!!

## Tuesday, November 11, 2008

### Prime Secrets

1. Primes are defined as any number that can only be divided evenly by itself and 1.

2. Primes end in 1, 3, 7, 9 in column "0" ( far right column ).

3. Not all numbers ending in 1, 3, 7, 9 are prime numbers.

4. Numbers that aren’t prime numbers but end in 1, 3, 7, 9 are evenly divisible by some number ending in 1, 3, 7, 9 in column "0" ( far right column ).

5. Except for the prime number 3, any number ending in 1, 3, 7, 9, say (39), when their digits are totaled ( 3 + 9 = 12 ) are not a prime if the total of their digits can be evenly divided by 3 ( 3 + 9 = 12) ( 12 / 3 = 4 ) ( 39 / 13 = 3 ).

6. The number of primes preceding a prime number can be approximated by first taking a prime number, say (7919), and counting all its’ digits. Prime number ( 7919 ) has 4 digits ( 7,9,1,9 = 4 digits ). The other method is to count all of the prime’s digits and subtract (1). Prime number ( 7919 ) has 4 digits and subtracting ( 1 ) leaves 3 digits. Prime number ( 7919 ) has 4 digits ( 7, 9, 1, 9 = 4 digits ), and subtracting (1) leaves ( 4 - 1 = 3 ) digits. Next take the total number of its’ digits or the total number of its’ digits minus 1 and multiple the prime ((7919) X (½ ^ 4) = 494.94) or ((7919) X (½ ^ 3) = 989.88). Since it’s unlikely that 7919 only has approximately 494 primes behind it, the likelier answer is around 989. Add 2, 4, 6, 8, 10 to 989 to get a closer approximate number. In this case add 10. ( 989.88 + 10 = 999.88 ). 7919 is the 1000th prime according to the tables. The approximation requires some obvious juggling but if you apply points 1 to 5 to the approximation you will come close to a probable answer.

2. Primes end in 1, 3, 7, 9 in column "0" ( far right column ).

3. Not all numbers ending in 1, 3, 7, 9 are prime numbers.

4. Numbers that aren’t prime numbers but end in 1, 3, 7, 9 are evenly divisible by some number ending in 1, 3, 7, 9 in column "0" ( far right column ).

5. Except for the prime number 3, any number ending in 1, 3, 7, 9, say (39), when their digits are totaled ( 3 + 9 = 12 ) are not a prime if the total of their digits can be evenly divided by 3 ( 3 + 9 = 12) ( 12 / 3 = 4 ) ( 39 / 13 = 3 ).

6. The number of primes preceding a prime number can be approximated by first taking a prime number, say (7919), and counting all its’ digits. Prime number ( 7919 ) has 4 digits ( 7,9,1,9 = 4 digits ). The other method is to count all of the prime’s digits and subtract (1). Prime number ( 7919 ) has 4 digits and subtracting ( 1 ) leaves 3 digits. Prime number ( 7919 ) has 4 digits ( 7, 9, 1, 9 = 4 digits ), and subtracting (1) leaves ( 4 - 1 = 3 ) digits. Next take the total number of its’ digits or the total number of its’ digits minus 1 and multiple the prime ((7919) X (½ ^ 4) = 494.94) or ((7919) X (½ ^ 3) = 989.88). Since it’s unlikely that 7919 only has approximately 494 primes behind it, the likelier answer is around 989. Add 2, 4, 6, 8, 10 to 989 to get a closer approximate number. In this case add 10. ( 989.88 + 10 = 999.88 ). 7919 is the 1000th prime according to the tables. The approximation requires some obvious juggling but if you apply points 1 to 5 to the approximation you will come close to a probable answer.

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