The Riemann Hypothesis involves an understanding of fractions . The largest fraction you can have is ( ½ ) or ( .5 ) . This may seem counter-intuitive, but to prove it simply subtract any ( 1 / number ) fraction from ( ½ ) and you will end up with a positive fraction . The Riemann Hypothesis uses the formula ( 1 + ½ ^s + 1/3 ^s + ---- ) to infinity . In this formula you have ( s = ½ + it ) . You may be familiar with calculus , but if you aren't calculus is concerned with how close you can get a fraction to a number without actually touching ( reaching ) the number. This getting close as a numerical figure is called a limit . You can think of it as how close you can get your nose to a brick wall without touching / hurting your nose . If you add fractions to infinity, you will find that the distance from ( ½ ) becomes smaller and smaller . If you add fractions to infinity that have been raised to a power, you will find that the distance from ( ½ ) is closer to ( ½ ) then by simply adding unraised fractions . Since the distance from ( ½ ) is getting closer, you can think of all the added fractions as getting closer to the line ( y = ½ ) . You may not know this, but if you raise a fraction to the power of zero ( 0 ), you get one ( 1 ) ( ½ ^0 = 1 ) . Zero ( 0 ) functions more as a placeholder than a value . Of course, you can argue that if you don't have anything you have something which is zero ( 0 ) . If the raised fraction to a power is getting closer and closer to ( ½ ), then the power to which it is being raised is getting closer and closer to zero ( 0 ) since ( ½ ) raised to the power of zero ( 0 ) is one ( 1 ) which means in the extreme ( ½ ) doesn't change because it becomes one or itself ( think philosophically here ) . One of the problems in mathematics is that you can't graph zero . You can, of course, draw a line and say this is line zero ( 0 ) meaning something you are considering starts here . If the power to which a fraction ( ½ ) is getting closer and closer to zero ( 0 ), then that zero ( 0 ) is getting closer and closer to ( ½ ) which is on the line ( y = ½ ) . Therefore it can be argued that zero ( 0 ) has the value of ( ½ ) on the line ( y = ½ ) . This is what Riemann said. Calculations have been done and continue to be done to prove that Riemann's zeros ( 0 ) lie on the line ( y = ½ ) and so far, so good. Riemann also said that his zeros ( 0 ) had something to do with primes . A prime is a number that can only be evenly divided by itself and one ( 1 ) . He also went on to say that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. If we can calculate the location of a prime, we know how many primes precede it because the primes would be in order . To state the obvious , the position of any prime in a series is less than the prime itself. For instance, the first one ( 1 ) digit primes are 1, 2, 3, 5, 7 . The primes ( 1, 2, 3 ) are in the positions ( 1, 2, 3 ) . The primes 5 and 7 are in positions 4 and 5 . The primes ( 1, 2, 3 ) can be multiplied by one ( 1 ) ( 1 X 1, 2 X 1, 3 X 1 ) to get their true position ( 1, 2, 3 ) . If we multiply prime 5 by .9099 we get ( 5 X ..9099 = 3.63996) which is mathematically close to 4 . If we multiply prime 7 by .70999 we get ( 7 X .70999 = 4.96993 which is mathematically close to 5 . These zeros ( 0 ) are Riemann zeros .
Obviously, you can diddle the multiplier any way you want to get the answer, but since we are dealing with the Riemann Hypothesis , Riemann zeros and the number ½ on the line ( y = ½ ) it generally works like this:
The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ). The prime numbers from position 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see from the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.
If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.
Here’s how the system works for numbers in general.
Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.
It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. The power zeros can also be adjusted. For instance (( .5099999999 ) ^2 = (..260099999 )). If we change the digit 6 to zero ( 0 ) creating ( .200099999 ) and multiplying it by the prime number 347 we get ( 69.43469965 ) Prime number 347 is the 69th prime. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range if the calculated answer is close but too far out.
1. ( .509999999 ) may be adjusted using Riemann zeros.
2. ( .509999999 ) may be raised to a power to increase the zeros.
3. ( .509999999 ) raised to a power may have its' power zeros adjusted.
4. Tweaking can be done by adding or subtracting Pi or “e”.