I would have had to consider two extreme cases: (1) that the RH is
completely false just as FLT is completely false when NaturalNumbers =
p-adics. If taking that extreme route would have had me find out what
was flawed in my claimed two proofs below.

(2) the second extreme case is to say that something lies on the 1/2
Real line but not the NaturalNumbers of the illdefined notion of
finite-integers but rather instead the p-adics. Only the line is not a
straight line. And my second *alleged proof below* using a spiral sort
of touches or hints of a curved line.

TWO PROOFS OF THE RIEMANN HYPOTHESIS

PROOFS: Two proofs of the Riemann Hypothesis follows as A
and B.

Proof (A) is a geometrical proof. It was proved that the Riemann
Hypothesis is equivalent to the following-- the Moebius function mu of
x, m(x), and adding-up the values of m(x) for all n less than or equal
to N giving M(N). Then M(N) grows no faster than a constant multiple k
of (N^1/2)(N^E) as N goes to infinity (E is arbitrary but greater than
0). Figure1, by setting-up a logarithmic spiral in a rectangle of
whirling squares where the squares are the sequences:
1,1,2,3,5,8,13,21,34,55,89, . . . 2,2,4,6,10,16,26, . . .
3,3,6,9,15,24,39, . . . then every number appears in at least one of
these sequences because every number will start a sequence. Since all
numbers are represented uniquely by prime factors (the unique prime
factorization theorem or called the fundamental theorem of arithmetic)
and The Prime Number Theorem: the distribution of prime numbers is
governed by a logarithmic function, where (An/n)/(1/Ln of n) tends to
1
as n increases, where An denotes the number of primes below the
positive integer n, and where An/n is called the density of the primes
in the first n positive integers. The density of the primes, An/n, is
approximated by 1/(Ln of n), and as n increases, the approximation
gets
better. The distribution of prime numbers is governed by a
logarithmic function where these two math concepts-- one of prime
numbers, and the other, logarithms seem unconnected at first
appearance, but in reality they are totally connected. Geometrically,
the logarithmic spiral exhausts every positive integer, see figure 1.
The area of the rectangles containing the logarithmic spiral is always
greater, since the spiral is always inside the rectangles. Thus the
Moebius function k (N^1/2)(N^E) is satisfied since the area of the
logarithmic spiral is less than the rectangle whose area represents
the
number N, and whose sides represent its factors. The area of a
logarithmic spiral is represented by A=(r)(e^(Hj)) , and so depending
on where the point of origin for the spiral is taken rsubO determines
k, and depending on the value of H, H determines the E value for N,
when H=0 then the curve is a circle. The logarithmic spiral inside
rectangles of whirling squares implies that for any number N then
N^1/2
is the limit of the factors for N, for example, given the number 28,
then 28^1/2 = 5.2915. . and so looking for the factors of 28, it is
useless to try beyond 5 because the factors repeat, 4x7 then repeats
as
7x4. But if the Moebius function was false then there must exist a
number M such that M^1/2 is not the limit of the factors for M and the
spiral is outside of the square, which is impossible, hence the
Moebius
function is true. Therefore the Riemann Hypothesis is proved. Q.E.D.

My second proof (B) of the Riemann Hypothesis uses a reductio
ad absurdum argument. Euler proved that a formula encoding the
multiplication of primes was equal to the zeta function. Euler's
formula in complex variable form is as follows:
(1/(1-(1/(2^c))))x(1/(1-(1/(3^c))))x(1/(1-(1/(5^c))))x(1/(1-(1/(7^c))))x
(1/(1-(1/(11^c))))x . . . , where c is a complex variable, c=u+iv. The
Riemann zeta function is as follows. Re(c) =
1+(1/(2^c))+(1/(3^c))+(1/(4^c))+. . . , where c is a complex variable,
c=u+iv. Euler's formula involves multiplication of terms and the
Riemann zeta function involves addition of terms of a sequence. Taking
Re(c) > 0, suppose the Riemann Hypothesis is false then there is a 0
such that Re(c)=0 and c not equal 1/2 +iy, which implies there is
another 0 which is not on the 1/2 real line. Which means another real
number other than 1/2 works as an exponent resulting in a zero for the
Riemann zeta function, and a zero in the Euler formula. Thus, Riemann
zeta function subtract Euler formula must equal zero. This implies
for
any other real number exponent, either rational or irrational numbers,
such as for example the rational exponents: 1/3,1/4,1/5, . . . (Note:
any other exponent y/x , where y and x are Real numbers and where the
Real number of A^(y/x) such that y not equal 1, immediately transforms
to a number A^y(1/x), so that exponents with a 1 in the numerator
entail all of the Real exponents). To make clear of the above, for
example, 2^2/3 is 4^1/3. So then back to the proof. Then for exponent
1/3 there has to exist a number M not equal 0 where (M+M+M)^1/M =
(MXMXM)^1/M = M. Then for exponent 1/4 there has to exist a number M
not equal 0 where (M+M+M+M)^1/M = (MXMXMXM)^1/M = M, and so on.
Including the infinite number of cases where the x denominator is
irrational are impossible. Only the real number 1/2 works since 2 does
not equal 0, and (2+2)^1/2 = (2X2)^1/2 = 2, and so (2+2)^1/2
- (2X2)^1/2 = 0. In all of Reals and the Complex numbers, 2 is the
only number N which has the encoding ((N+N)^1/N) = ((NxN)^1/N) = N.
Unlike 0, the number 2, its sum equals its product and where the sum
and product is a new number 4. If RH were false, then another number
other than 2 would satisfy a generalized encoding formula ((N+N)^1/N)
= ((NxN)^1/N) = N. False, hence the proof. QED (Quantum
Electrodynamics)

(snip my old post of 1996) And below is a re-post of Riemann
Hypothesis that was on my website (now moving to a new location and
will be up and running in the next several months).

COMMENTS about RH: If I had stuck with doing and concentrated on math
from 1996 to 2004, one of the main topics on my agenda would have been
to reconcile RH with the idea that the NaturalNumbers were the
p-adics. And that precarious situation would have forced me patch up
or mend the notion that the NaturalNumbers lie on the 1/2 Real line.

I would have had to consider two extreme cases: (1) that the RH is
completely false just as FLT is completely false when NaturalNumbers =
p-adics. If taking that extreme route would have had me find out what
was flawed in my claimed two proofs below.

(2) the second extreme case is to say that something lies on the 1/2
Real line but not the NaturalNumbers of the illdefined notion of
finite-integers but rather instead the p-adics. Only the line is not a
straight line. And my second *alleged proof below* using a spiral sort
of touches or hints of a curved line.

As far as I know from 1993-1996 in conversation with Karl Heuer about
my RH proofs the number 2 as N in the encoding of ((N+N)^1/N) =
((NxN)^1/N) = N remains the same whether we think of 2 as a p-adic or
as a finite integer. So that encoding formula is not corrupted.