## 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.