Mathematics
basically breaks down into the following three concepts:

- Numbers.
- Symbols.
- LengthsThe symbols such as x or y are usually equal to something which is up to you to discover. These symbols may be in a formula expressing relationships which are familiar to anyone that has ever studied algebra. These relationships in algebra are generally a static value as only one numerical answer is needed. Calculus was the next major step. Calculus is the study of limits which involve fractions. Limits are concerned with how close you can get to something I call a brick wall ( limit ) without actually touching it. How close is actually the calculation of length before you hit the wall or limit. Here you are dealing with fractions. Calculus, using symbols, is the study of the change in the length of a formula containing symbols that change in length. If you have symbols in a formula, what happens to the formula when that symbol or symbols change in length ???? To make life simpler, the initial formula is simplified to lessen the potential confusion. In any language, as in mathematics, there are many ways to express an idea .Riemann expressed his idea using the following formulas:

Riemann
said that the “s” in the above equation was ( s = ½ + it ). The
“i” in it represented an imaginary number and the “t”
represented a real number. He also said that all his non – trivial
zeros lie on the line ( y = ½ ) . Innumerable calculations up to the
present time have proven that Riemann's conjecture is true but there
also may be an exception yet to be discovered. The Riemann
Hypothesis also implies that if the zeros consistently cross the line
( y = ½ ) then these zeros have something to do with the
distribution of primes. There is an implicit idea here that every
crossing of zeros on the line ( y = ½ ) is related to the position
of a relative prime beginning at the beginning of the line ( y = ½ )
and the end of the line ( y= ½ ) at infinity. Theoretically, if the
values of Riemann's formula were chosen correctly the

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

- 5 X ( .509999 ) = 2.549995 (4)
- 7 X ( .509999 ) = 3.569993 ( 5 )

3.
11 X ( .509999 ) = 5.609989 ( 6 )

This
basic principle calculates the position of the primes fairly
accurately up to prime 31.

Using
the same multipliers on the next prime ( 37 ) we get ( 37 X .509999 =
18.869963. Prime ( 37 ) is actually the 13

^{th}prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). If we take ( .509999 ) and multiply it by itself ( ( .509999 X .509999 = .26009898 ).- ( 37 X .509999 X ,509999 = 9.62366226 ( 13 )

This
calculation shows that the calculated position is short of the true
position by about ( 13 – 9.62366226 = 3.37633774 ). The difference
is approximately equal to Pi ( 3.141592654 ). In some calculations
the difference is about the natural number ( e ) ( 2.71828`828 ).

In
summary, Riemann's intuition told him that the zeros had a real value
of ( ½ ) which was true since the digit ( ½ ) is used in the
calculation. Riemann's intuition also told him that the zeros can be
manipulated ( position changed or zeros added ) which is also true.
This manipulation of zeros ( 0 ) is really using the principles of
limits in calculus to see how close you can find a factor when
multiplied by a prime that will give you the position of that prime
on the line ( y = ½ ). What Riemann missed was that the calculation
involved the number 9 and that Pi ( 3.141592654 ) and the natural
number ( e ) ( 2.718281828 ) might have to be added or subtracted
from the final answer. Riemann also missed that powers would also
have to be used.

In general for
calculating the location of any prime you:

- Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 which equals 995.2299524. 7919 is the 1000
^{th}prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).

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