After having read Sabbagh’s book a General Solution to the Riemann
Hypothesis popped up in my tiny little brain.

Here it is.

ζ(0) = negative one half = non trivial zero.

Proof:

1. Superimpose two unit circles.

2. Designate the vertical diameters as 2i and the horizontal diameters
as 2x.

3. Starting at zero degrees and rotating counterclockwise, the
points of axes along the circumference then assume the values (1, 0); (0, i);
(-1, 0) and (0,-i).

4. Pin the common point of the two unit circles at (1; 0) such that
this point is restricted from moving.

5. Reduce the diameter of one of the unit circles by half. This
creates an inner circle of diameter i vertically and diameter x horizontally,
which is pegged/pinned to the larger unit circle at point (1; 0):

6. This pegged inner circle is identical in every respect to the
larger parent unit circle except it has been scaled to half the size of the
larger unit circle.

7. The origin of the parent unit circle assumes various equivalent
values Cartesian: zero, plus or minus 1

Fermatian: zero, limit, derivative, maxima, minima, plus and minus
1, one half, i

Newtonian: zero, shrinking the 2i secant to zero= i = limit =
derivative = maxima = minima:

Inner Circle shift: The (-1; 0) point of the reduced circle now
occupies the origin of the larger
parent circle.

8. The origin of the reduced/inner circle is identical to the origin
of the larger parent circle. They differ only in location.

9. Thus since i = 0 then the entirety of the i diameter of the
reduced /inner circle is, and must be, zero.

10. The same result may be obtained by shrinking the 2i diameter/secant
of the parent circle to the length of the i = zero diameter/secant of the inner
circle.

11. Thus the i diameter comprises the critical line and since its
entirety is composed of zeroes then these zeroes are, obligatorily, non-trivial
zeroes.

12. Given step 7, i is equivalent to the tangent =
m = slope.

13. Vis a vis the parent circle, this slope is
manifest as the hypotenuse (tangent line) of a right triangle of height i and
base .5x or one half x.

14. This results in the formation of a negative .5 slope or negative
one half slope or a positive 2 precessional rate where slope = m = negative one
half.

15. Therefore m = i = 0 = nontrivial zero = negative one half.

16. Therefore critical line equals

i = 0 = non trivial zero= negative one half = Zeta(0).

17. Thus is it proven that Zeta(0) =negative one half=non trivial
zero.

QED

What people usually DON'T ask is the following: pick a random
integer between 1 and N. What is the probability that the integer you picked is
prime? In the polynomial case, they say: pick a random polynomial of degree d,
with coefficients restricted. What is the probability that the polynomial I
pick is irreducible?

It is known by the Prime Number Theorem that the fraction of
integers between 1 and N that are prime is roughly 1 / log(N). Since the
denominator grows toward infinity as N does, the probability that the integer
is prime is very close to 0 when N is very large.

Some pointers not addressed in Sabbagh's book:

(1) Prime numbers are distributed throughout the infinite set of
integers, where the number of primes below N is roughly ~ N/ln (N). The set of
integers is well ordered by definition, i.e.: 1,2,3,4,5,6, . . . .

(2) IFF it were possible to similarly order the set (or some subset)
of compact polynomials with integer coefficients, would the set of Prime
polynomials in the larger set obey something like a logarithmic or exponential
distribution? For instance: Where the of NON-prime polynomials
might approach ln (N/N), and the density of Prime-polynomials might be ~ (N -
ln (N/N)), or some powers of these?, i.e., the density of factorable
polynomials decreases toward zero as N goes to infinity.

Has anyone ever regularized a polynomial version of the Riemann Zeta
function for polynomials? I am not surprised no one hasn't since in some sense
the algorithmic universe would have to be vast, or nature would be very simple.
But I just don't know. I haven't been up to speed on Math Papers for the past
years.

I'd have liked this exposition on the Riemann Hypothesis if Sabbagh had jnot concentrared so much on Louis de Branges efforts at proving it. Scary to think that so much was placed on the Riemann Hypothesis (all his life actually; this book was written in 2004; it's 2019 and no one has come nowhere close to solving it - forget Atiyah's attempt because it's bogus). It's probably impossible to understand the allure and the importance of the Riemann Hypothesis without some mathematical background, but if you have a decent first year college mathematical background, it's doable.

Here's a more detailed view, also written by Sabbagh, on the de Branges attempt.

**Bottom-line**: A fun book. As another very
concise but hugely influential example there's 'P=NP'. That's pretty easy to
get on a t-shirt. And like the Riemann Hypothesis we're fairly sure what the
outcome is ('P!=NP' in this case) but a proof could be very helpful. In the
end, Math is too hard for mathematicians. They still cannot solve the
Navier-Stokes equations. Engineers like myself routinely "solve" them
by experimentation and clever approximations, the Reynolds-averaged
Navier-Stokes equations that is. What do I care about proofs...? The Riemann
Hypothesis really means that the primes are pretty much randomly distributed
(with the probability of N being prime going like 1/log(N)), i.e., they're not
concentrated and deconcentrated in certain waves. The proof of the Riemann
Hypothesis could really be some clever analogy of the proof of the fact that
there are infinitely many primes. If there were a finite set of primes, their
products plus one would be another prime that would be a contradiction. A
Riemann Hypothesis proof would be seismic. I do think, though, that the media
would mainly ignore a valid proof. Very very few know who Riemann was outside
of the math/physics community, let alone his hypothesis. If a proof of the
Riemann Hypothesis is done through "traditional" ways I would expect
it will be very hard to understand, and that one will also learn a lot of
subtle complex analysis on the way. If the proof is easy because the approach
is radically new, then I would expect that the approach turns out to be more
interesting than the Riemann Hypothesis itself.

NB: ζ = Zeta Function.