Tuesday, January 13, 2009
Riemann Hypothesis Resolved Using Strings
If you look at a table of Prime Numbers, you will notice that they aren’t evenly spaced which raises the question of how many primes precede any arbitrarily selected number. On the other hand, if you look at a table of prime numbers they seem to follow a pattern. You will also notice that the prime numbers get rarer, as you look at higher and higher numbers. Most formulas that have been developed to find the location of any prime number seem to be a hit and miss affair because the prime numbers aren’t evenly spaced. George Riemann, who formulated the Riemann Hypothesis, found that his formula produced the occasional zero as an answer which he hypothesized lay on the line ( y = ½ ). So far, calculating on an individual basis, it has been found that over a billion zeros using Riemann’s formula lie on this line but since Riemann hypothesized all zeros the search goes on. So far, no one has developed a proof that definitely establishes all zeros lie on the line ( y = ½ ). Here’s a way, of resolving the conundrum concerning Riemann’s zeros (0’s) using the concept of strings. If we add the digits of any number (N) like ( 7919 ), for example, we get a total ( 7 + 9 + 1 + 9 = 26 ) which represents its’ string ( 26 ) or line ( y = 26 ). If we continue the addition of the digits ( 26 ) we get ( 2 + 6 = 8 ) or a line ( y = 8 ) for number ( 7919 ). Riemann had the idea that when his equation produced a zero (0) in the calculation, that zero for non- trivial numbers was on the line ( y = ½ ). Working in reverse, we can take any arbitrary line ( y = k ), with ( k = ( 0 to 9 )), because all numbers ( N ) sum to a number from ( 0 to 9 ) as one digit. Any line ( y = k ) can have an infinite number of numbers, since zero ( 0 ) does not change the value of ( y = k ). For instance, if ( y = 2 ), it can hold the numbers ( 11, 101, 110, etc. ) since the total digits do not exceed (2). It can readily be seen that all zeros line on the line ( y = k ) no matter what the value of k whether it be one digit or multiple digits and thus Riemann was correct when he said all zeros lie on the line ( y = ½ ), since ( y = ½ ) is the reciprocal of ( y = 2 ) and contains the numbers ( 1/11, 1/101, 1/110, etc.. The same conclusion can be applied to any arbitrary line ( y = k ) or ( y = 1/k ) for values of k from ( zero (0) to infinity ) of any number (N) which has an infinite number of digits (D) If Riemann had changed his formula to produce zeros ( 0’s ) for other non-trivial values for lines ( y = k ) or ( y = 1/k ) he would have ended up hypothesizing that all zeros ( 0’s ) also lie on these lines.