domain: math math phys sdu

text:
Surajit Ghosh

India Kolkata

NUMBERS ARE 3 DIMENSIONAL

Keywords: Zeta function, Gamma function, Riemann hypothesis, MSC2010:11M26, 11A41

Riemann hypothesis stands proved in three different ways of three different level of complexity.To prove Riemann hypothesis from the functional equation concept of Delta function and periodic harmonic conjugate of both Gamma and Delta functions are introduced similar to Gamma and Pi function. Other two proofs are derived using Eulers formula and elementary algebra. Analytically continuing zeta function to an extended domain, poles and zeros of zeta values are redefined. Other prime conjectures like Goldbach conjecture, Twin prime conjecture etc.. are also proved in the light of new understanding of primes. Numbers are proved to be three dimensional as worked out by Hamilton. Logarithm of negative and complex numbers are redefined using extended number system. Factorial of negative and complex numbers are redefined using values of Delta function and periodic harmonic conjugate of both Gamma and Delta functions.

Part 1. The Introduction

Let us begin with a short Introduction

The Riemann zeta function and the Riemann Hypothesis are two interconnected concepts in mathematics, specifically in the field of number theory. Let's start with a brief introduction to the Riemann zeta function.The Riemann zeta function, denoted by ζ(s), is a mathematical function that was first introduced by the German mathematician Bernhard Riemann in 1859. It is defined for complex numbers s with a real part greater than 1, and it can be extended analytically to other complex numbers as well.The Riemann zeta function is defined by the infinite series: ζ(-s) = 1 (-s) + 2 (-s) + 3 (-s) + 4 (-s) + ... This series converges (i.e., it has a finite value) when the real part of s is greater than 1. For example, when s = 2, the series becomes:

ζ(2) = 1 2 + 2 2 + 3 2 + 4 2 + ... This is known as the Basel problem, and it was famously solved by the Swiss mathematician Leonhard Euler, who found that ζ(2) equals π 2 6 . 1.1. Euler the great grandfather of zeta function. In 1737, Leonard Euler published a paper where he derived a tricky formula that pointed to a wonderful connection between the infinite sum of the reciprocals of all natural integers (zeta function in its simplest form) and all prime numbers.

ζ(s) = ∞ n=1 1 n s = p 1 + 1 p s + 1 p 2s + 1 p 3s + 1 p 4s ...
Equivalent to:

ζ(s) = ∞ n=1 1 n s = p 1 1 -P -s
To carry out the multiplication on the right, we need to pick up exactly one term from every sum that is a factor in the product and, since every integer admits a unique prime factorization; the reciprocal of every integer will be obtained in this manner -each exactly once. This was originally the method by which Euler discovered the formula. There is a certain sieving property that we can use to our advantage: where all elements having a factor of 3 or 2 (or both) are removed.

It can be seen that the right side is being sieved. Repeating infinitely for 1 p s where p is prime, we get:

. . . 1 - 1 11 s 1 - 1 7 s 1 - 1 5 s 1 - 1 3 s 1 - 1 2 s ζ(s) = 1
Dividing both sides by everything but the ζ(s) we obtain:

ζ(s) = 1 1 -1 2 s 1 -1 3 s 1 -1 5 s 1 -1 7 s 1 -1 11 s .
. . This can be written more concisely as an infinite product over all primes p:

ζ(s) = p prime 1 1 -p -s
To make this proof rigorous, we need only to observe that when ℜ(s) > 1, the sieved right-hand side approaches 1, which follows immediately from the convergence of the Dirichlet series for ζ(s).

1.2. Riemann the grandfather of zeta function. Riemann might had seen the following relation between zeta function and eta function (also known as alternate zeta function) which converges for all values Re(s) > 0.

ζ(s) = ∞ n=1 1 n s =⇒ ∞ n=1 2 (2n) s = 1 2 s-1 ζ(s)
Now subtracting the latter from the former we get:

1 - 1 2 s-1 ζ(s) = 1 1 s - 1 2 s + 1 3 s -. . . = ∞ n=1 (-1) n-1 1 n s =: η(s) =⇒ ζ(s) = 1 -2 1-s -1 η(s)
Then Riemann might had realised that he could analytically continue zeta function from the above equation for 1 ̸ = Re (s) > 0 after re-normalizing the potential problematic points. In his seminal paper Riemann showed that zeta function have the property of analytic continuation in the whole complex plane except for s=1 where the zeta function has its pole. Zeta function satisfies Riemann's functional equation.

ζ(s) = 2 s π (s-1) sin πs 2 Γ(1 -s)ζ(1 -s)
Riemann Hypothesis is all about non trivial zeros of zeta function. There are trivial zeros which occur at every negative even integer. There are no zeros for s > 1. All other zeros lies at a critical strip 0 < s < 1. Now, let's turn our attention to the Riemann Hypothesis. The Riemann Hypothesis is one of the most famous unsolved problems in mathematics, named after Bernhard Riemann. It states that all non-trivial zeros of the Riemann zeta function lie on a critical line inside the critical strip. This critical line has a real part of 1 2 .To put it simply, if the Riemann Hypothesis is true, it would mean that all the interesting solutions of the equation ζ(s) = 0 lie on this critical line. The Riemann Hypothesis has far-reaching consequences in number theory and has connections to the distribution of prime numbers. Many results in number theory have been proven assuming the truth of the Riemann Hypothesis.Despite its significance, the Riemann Hypothesis remains unproven. Mathematicians have made significant progress in understanding the behavior of the zeta function, but a proof or disproof of the Riemann Hypothesis still eludes us. It stands as one of the most challenging and intriguing problems in mathematics, attracting the attention of researchers for over a century. Showing that there are no zeros with real part 1 -Jacques Hadamard and Charles Jean de la Vallée-Poussin independently prove the prime number theorem which essentially says that if there exists a limit to the ratio of primes up to a given number and that numbers natural logarithm that should be equal to 1. When I started reading about number theory I wondered that if prime number theorem is proved then what is left. The biggest job is done. I questioned myself why zeta function cannot be defined at 1. Calculus has got set of rules for checking convergence of any infinite series, sometime especially when we are encapsulating infinities into unity, those rules may fall short to check the convergence of infinite series. In spite of that Euler was successful proving sum to product form and calculated zeta values for some numbers by hand only. Leopold Kronecker proved and interpreted Euler's formulas is the outcome of passing to the right-sided limit as s → 1 + . I decided I will stick to Great Grandpa Euler's approach in attacking the problem and I will investigate further how zeta function blows up to infinity at 1 and renormalize the infinity introducing some new mathematical tools which may be missing now in the present context.

Part 2. The Lemmas

Let us work out the lemmas related to factorial

In this section we will work out factorial related some new mathematics which will help us to prove Riemann Hypothesis.

2.1. Recalling Gamma and Pi function. Euler in the year 1730 proved that the following indefinite integral gives the factorial of x for all real positive numbers,

x! = Π(x) = ∞ 0 t x e -t dt, x > 1
Euler's Pi function satisfies the following recurrence relation for all positive real numbers.

Π(x + 1) = (x + 1)Π(x)

In 1768, Euler defined Gamma function, Γ(x), and extended the concept of factorials to all real negative numbers, except zero and negative integers. Γ(x), is an extension of the Pi function, with its argument shifted down by 1 unit.

Γ(x) = ∞ 0 t x-1 e -t dt
Euler's Gamma function is related to Pi function as follows: Γ(x + 1) = Π(x) = x! Euler's Gamma function have the following properties:

• Γ(z) = ∞ 0 e -t t z-1 dt for z ∈ C, ℜ(z) > 0;
• Γ is analytic on C\{0, -1, -2, . . .}; • Γ has a simple pole with residue (-1) n n! at z = -n for n = 0, 1, 2, ...; • Γ(z + 1) = zΓ(z) for z ∈ C\{0, -1, -2, . . .}; • Γ(x) = (x -1)! for x ∈ Z > 0.

2.2. Introduction of Delta function. Now let us extend factorials of negative integers by way of shifting the argument of Gamma function further down by 1 unit. Hoping that capital Delta will not be confused with Dirac's small delta function let us define capital Delta function as follows:

Lemma 1.

∆(x) =

∞ 0 t x-2 e -t dt I am not trying to somehow imprint my name in this function as neither of my names initial can be related to Delta. Still I choose Delta because in cyclic sense Epsilon will sit exactly beween Pi and Gamma in Greek Alphabet making a semi circle, but nature has a complete cycle, my following work will complete the circle, so the very next letter after Gamma should the best candidate to represent 1+ something, 1+ something squared pattern or something -1, something squared -1 pattern. Hope people will recognise this biased definition of mine. Let z ∈ C with ℜ(z) > 0. Then using integration by parts,

∆(z + 2) = ∞ 0 t z+1 e -t dt = -t z+1 e -t ∞ 0 + ∞ 0 (z + 1)t z+1-1 e -t dt = lim t→∞ (-t z+1 e -t ) -(0e -0 ) + (z + 1) ∞ 0 t z+1-1 e -t dt = (z + 1) ∞ 0 t z e -t dt = (z + 1)∆(z + 1) = (z + 1) ∞ 0 t z e -t dt = (z + 1) -t z e -t ∞ 0 + ∞ 0 zt z-1 e -t dt = (z + 1) lim t→∞ (-t z e -t ) -(0e -0 ) + z ∞ 0 t z-1 e -t dt = (z + 1)z ∞ 0 t z-1 e -t dt = (z + 1)z∆(z)
Similar to Gamma function Delta function will then have the following properties:

• ∆(z) = ∞ 0 e -t t z-2 dt for z ∈ C, ℜ(z) > 0;
• ∆ is analytic on C\{-1, -2, . . .}; • ∆ has a simple pole with residue (-1) n n! at z = -n for n = 1, 2, ...;

• ∆(z + 1) = z∆(z) for z ∈ C\{-1, -2, . . .}; • ∆(x) = (x -2)! for x ∈ Z ≥ 0.

The extended Delta function shall have the following recurrence relation.

∆(x + 2) = (x + 2)∆(x + 1) = (x + 2)(x + 1)∆(x) = x! 2.3. Removing Poles of Gamma and Pi function. Newly defined Delta function is related to Euler's Gamma function and Pi function as follows:

Lemma 2.

∆(x + 2) = Γ(x + 1) = Π(x)
To make it simple we will be taking examples on integers but we will not forget the same may be true in case other numbers such as fractions, complex numbers etc. within the extended domain.Whereever necessary we will use that extended domain without recalling. Plugging into x = 2 above

∆(4) = Γ(3) = Π(2) = 2 Plugging into x = 1 above ∆(3) = Γ(2) = Π(1) = 1
Plugging into x = 0 above ∆(2) = Γ(1) = Π(0) = 1 Now these three factorial functions are behaving like triangle of harmonic-trio based on the fact they all are analytic around -1. Gamma is playing the role of the central function whereas Pi and Delta are acting like its harmonic pair. When we look deeper we will see following the definition of Delta function zeta function is harmonised first if not simultaneously (see the proof of Riemann hypothesis) which in turn harmonises these factorial functions.Γ(z) = u + iv analytic means u and v which can be expressed as variants of zeta series must be harmonic conjugates and so are u and -v because iΓ(z) = -v + iu . The same kind of harmonic relation is observed above, therefore harmonicity among Gamma, Pi and Delta functions are established both based on the harmonicity of zeta series or independent of zeta series. The real and imaginary parts of Γ(z) = 1 z must be harmonic away from the origin given the fact Γ(z) can be defined on zero now.Since the real and imaginary parts of Γ(z) = 1 z are harmonic, the same must be true of the respective integrals, which is limit of linear combinations of such functions. Since the circle is complete and it is finite and Gamma, Pi and Delta functions trio are continuous, interchanging the order of integration is not a problem now. We can bring two more harmonic function centered around Delta function into the picture to unify the symmetries among these fantastic five factorial functions as follows. Plugging into x = -1 above we can remove poles of Gamma and Pi function as follows:

∆(1) = Γ(0) = Π(-1) = 0 -∞ t 1-1 e t dt = e t 0 -∞ Refer:2.5 =⇒ ∆(1) = Γ(0) = Π(-1) = (e 0 ) -lim t→-∞ e t = 1 =⇒ ∆(1) = Γ(0) = Π(-1) = 1 = 1.∆(0) = Γ(-1) = Π(-2) =⇒ ∆0 = Γ(-1) = Π(-2) = 1
Plugging into x = -2 above we can remove poles of Gamma and Pi function as follows:

∆(0) = Γ(-1) = Π(-2) = 0 -∞ t 2-2 e t dt = e t 0 -∞ Refer:2.6 =⇒ ∆(0) = Γ(-1) = Π(-2) = (e 0 ) -lim t→-∞ e t = 1 = -2.∆(-1) =⇒ ∆(0) = Γ(-1) = Π(-2) = 1 = -1.∆(-1) = Γ(-2) = Π(-3) =⇒ ∆(-1) = Γ(-2) = Π(-3) = -1
Continuing further we can remove poles of Gamma and Pi function as follows:

Plugging into x = -3 above and equating with result found above

∆(-1) = Γ(-2) = Π(-3) = -2. -1.∆(-2) = -1 =⇒ ∆(-2) = Γ(-3) = Π(-4) = - 1 2
Plugging into x = -4 above and equating with result found above ∆(-2) = Γ(-3) = Π(-4) = -3. -2.∆(-3) = -1 2 =⇒ ∆(-3) = Γ(-4) = Π(-5) = -1 12

Plugging into x = -5 above and equating with result found above ∆(-3) = Γ(-4) = Π(-5) = -4. -3.∆(-4) = -1 12 =⇒ ∆(-4) = Γ(-5) = Π(-6) = -1 48

Plugging into x = -6 above and equating with result found above

∆(-4) = Γ(-5) = Π(-6) = -5. -4.∆(-5) = - 1 48 =⇒ ∆(-5) = Γ(-6) = Π(-7) = - 1 240
Plugging into x = -7 above and equating with result found above

∆(-5) = Γ(-6) = Π(-7) = -6. -5.∆(-6) = - 1 240 =⇒ ∆(-6) = Γ(-7) = Π(-8) = - 1 1440
Plugging into x = -8 above and equating with result found above

∆(-6) = Γ(-7) = Π(-8) = -7. -6.∆(-7) = - 1 1440 =⇒ ∆(-7) = Γ(-8) = Π(-9) = - 1 10080 . . .
And the pattern continues up to negative infinity.

Relation between Delta and Zeta function

. Lemma 3. ∆(x) = ∞ 0 t x-2 e -t dt substituting t = nu =⇒ dt = ndn we get =⇒ ∆(x) = ∞ 0 (nu) x-2 e -nu ndu =⇒ ∆(x) = n x n ∞ 0 u x-2 e -nu du =⇒ ∆(x) n n x = ∞ 0 u x-2 e -nu du =⇒ ∆(x) ∞ n=1 n n x = ∞ n=1 ∞ 0 u x-2 e -nu du =⇒ ∆(x) ∞ n=1 1 n x ∞ n=1 n = ∞ n=1 ∞ 0 u x-2 e -nu du =⇒ ∆(x)ζ(x)ζ(-1) = ∞ 0 u x-2 ∞ n=1 e -nu ndu = ∞ 0 u x-2 1 1 -e -u -1 du =⇒ 1 2 ∆(x)ζ(x) = ∞ 0 u x-2 e -u 1 -e -u du = ∞ 0 u x-2 du e u -1 2.5. Periodic harmonic conjugate of Gamma function.
Lemma 4. We can define periodic Gamma function as harmonic conjugate of capital Gamma function as follows:

Γ p (x) = 0 -∞ t x+1 e t dt
Let z ∈ C with ℜ(z) < 0. Then using integration by parts,

Γ p (z -1) = 0 -∞ t z e t dt = t z e t 0 -∞ - 0 -∞ zt z-1 e t dt = (0e 0 ) -lim t→-∞ (t z e t ) + z ∞ 0 t z-1 e t dt = z ∞ 0 t z-1 e t dt = zΓ(z) = Γ(z + 1) = -zΓ p (z -2)
Similar to Gamma function periodic Gamma function will then have the following properties:

• Γ p (z) = 0 -∞ e t t z+1 dt for z ∈ C, ℜ(z) < 0; • Γ p is analytic on C\{0, +1, +2, . . .};
• Γ p has a simple pole with residue (-1) n n! at z = n for n = 0, 1, 2, ...;

• Γ p (z + 1) = zΓ p (z) for z ∈ C\{0, +1, +2, . . .};
The periodic Gamma function shall have the following recurrence relation.

Γ p (x + 1) = xΓ p (x)
2.6. Periodic harmonic conjugate of Delta function.

Lemma 5. We can define periodic Delta function as harmonic conjugate of capital Delta function as follows:

∆ p (x) = 0 -∞ t x+2 e t dt
Let z ∈ C with ℜ(z) < 0. Then using integration by parts,

∆ p (z -2) = 0 -∞ t z+1 e t dt = t z+1 e t 0 -∞ - 0 -∞ (z + 1)t z+1-1 e t dt = (0e 0 ) -lim t→-∞ (t (z+1) e t ) + (z + 1) ∞ 0 t z e t dt = (z + 1) ∞ 0 t z e t dt = (z + 1)Γ(z + 1) = Γ(z + 2) = -(z + 1)∆ p (z -1) = -z(z + 1)∆ p (z -2)
Similar to Delta function periodic Delta function will then have the following properties:

• ∆ p (z) = 0 -∞ e t t z+1 dt for z ∈ C, ℜ(z) < 0; • ∆ p is analytic on C\{ +1, +2, +3, . . .}; • ∆ p has a simple pole with residue (-1) n n! at z = n for n = 1, 2, 3, ...; • ∆ p (z + 1) = z∆ p (z) for z ∈ C\{ +1, +2, +3, . . .};
The periodic Delta function shall have the following recurrence relation.

∆ p (x + 2) = x∆ p (x + 1)
2.7. Closure of factorial operation. Now these fantastic five factorial functions shall have the following inter functional relationship.

∆(x + 2) = Γ(x + 1) = Π(x) = Γ p (x -1) = ∆ p (x -3)
Lemma 6. We can use fantastic five factorial functions for closure of factorials. For negative even numbers the formulas are as follows. For odd numbers we just need to multiply or divide the result by the corresponding succeeding or preceding odd number.

-1

∆(-3) = -1 Γ(-4) = -1 Π(-5) = -1 -Γ p (-6) = -1 -∆ p (-7) = -2! = 2 -1 ∆(-5) = -1 Γ(-6) = -1 Π(-7) = -1 -Γ p (-8) = -1 -∆ p (-9) = -4! = 12 -1 ∆(-7) = -1 Γ(-8) = -1 Π(-9) = -1 -Γ p (-10) = -1 -∆ p (-11) = -6! = 720 -1 ∆(-9) = -1 Γ(-10) = -1 Π(-11) = -1 -Γ p (-12) = -1 -∆ p (-13) = -8! = 40320 -1 ∆(-11) = -1 Γ(-12) = -1 Π(-13) = -1 -Γ p (-14) = -1 -∆ p (-15) = -10! = 3628800 -1 ∆(-13) = -1 Γ(-14) = -1 Π(-15) = -1 -Γ p (-16) = -1 -∆ p (-17) = -12! = 479001600 -1 ∆(-15) = -1 Γ(-16) = -1 Π(-17) = -1 -Γ p (-18) = -1 -∆ p (-19) = -14! = 87178291200 . . .
And the pattern continues up to negative infinity.

For x < 0, x = -2, -4, -6... we can generalise the pattern as follows.

-1

∆(x -1) = -1 Γ(x -2) = -1 Π(x -3) = -1 -Γ p (x -4) = -1 -∆ p (x -5) = x!
For x < 0, x = -3, -5, -7... we can generalise the pattern as follows.

-1 ∆(x-1)

x + 1 = -1 Γ(x-2) x + 1 = -1 Π(x-3) x + 1 = -1 -Γp(x-4) x + 1 = -1 -∆p(x-5) x + 1 = x!
We can evaluate factorial of all complex argument z! = (x + iy)! = ∆(x + iy + 2) = Γ(x + iy + 1)

Example 1.

∆(2 + i) = Γ(1 + i) = i! = iΓ(i) ≈ 0.498 -0.155i ∆(1 + i) = Γ(i) = (i -1)! = (i -1).i! ≈ -0.343 + 0.653i ∆(2 -i) = Γ(1 -i) = -i! = -iΓ(-i) ≈ 0.498 + 0.155i ∆(1 -i) = Γ(-i) = (-i -1)! = (-i -1). -i! ≈ -0.343 -0.653i

Let us work out the lemmas related to zeta function

In this section we will work out zeta function related some new mathematics which will help us to prove Riemann Hypothesis.

3.1. Harmonic zeta function and alternate functional equation. Multiplying both side of Riemann's functional equation by (1 -s) we get

(1 -s)ζ(s) = 2 s π (s-1) sin πs 2 (1 -s)Γ(1 -s)ζ(1 -s) Putting (1 -s)Γ(1 -s) = Γ(2 -s) we get: (1 -s)ζ(s) = 2 s π (s-1) sin πs 2 Γ(2 -s)ζ(1 -s) s → 1we get: ∵ lim s→1 (s -1)ζ(s) = 1 ∴ (1 -s)ζ(s) = -1 2 s π (s-1) sin πs 2 Γ(2 -s)ζ(1 -s) = -1
Similarly multiplying both numerator and denominator right hand side of Riemann's functional equation by (1 -s)(2 -s) before applying any limit we get :

ζ(s) = 2 s π (s-1) sin πs 2 (1 -s)(2 -s)Γ(1 -s)ζ(1 -s) (1 -s)(2 -s) Putting (1 -s)(2 -s)Γ(1 -s) = Γ(3 -s) we get: ζ(s) = 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s) (1 -s)(2 -s)
Multiplying both side of the above equation by (1 -s) we get

(1 -s)ζ(s) = 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s) (2 -s) s → 1we get: ∵ lim s→1 (s -1)ζ(s) = 1 ∴ (1 -s)ζ(s) = -1 -1 = 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s) (2 -s)
Multiplying both side of the above equation further by (2 -s) we get:

(s -2) = 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
Multiplying both side of the above equation by ζ(s -1) we get

(s -2)ζ(s -1) = 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)ζ(s -1) s → 2 we get: ∵ lim s→2 (s -2)ζ(s -1) = 1 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)ζ(s -1) = 1
Which can also be written as:

2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s) = 1 ζ(s -1)
The above equation is consistent at s = 1 following analytical continuation of Riemann's functional

equation at ζ(1 -s) = ζ(s -1) = ζ(0) = -1
2 . Following harmonic conjugate theorem the above kind of consistency make the expression a harmonic conjugate function around the pole of zeta function.The real and imaginary parts of ζ(z) = 1 z around a unit disc centered at 1 must be harmonic away from the origin given the fact ζ(1) is analytic now. The above analytic function will not have any zeros as proven by Jacques Hadamard and Charles Jean de la Vallée-Poussin. Therefore the above function will be an entire function. But overall we see zeta function have got lots of zeros. That is possible only another harmonic away from the origin. Due to the fact that harmonic conjugate appears in pairs, the domain of unit disk area gets bifurcated. Since the real and imaginary parts of ζ(z) = 1 z are harmonic, the same must be true of the respective functional equations. Since the circle is complete and it is finite changing the sign of the equation multiplying by -1 is not a problem and also we can set 1 ζ(s-1) = 1 due to its harmonicity and we can rewrite the above equation as follows:

-2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s) = -1
Both the above boxed forms are equivalent to Riemann's original functional equation therefore Riemann's original functional equation can be analytically continued further which will lead us to zeros of zeta function as follows:

Lemma 7.

ζ(s) = -2 s π (s-1) sin πs 2 ∆(4 -s)ζ(1 -s)
This can be rewritten in terms of Gamma function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
This again can be rewritten in terms of Pi function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 Π(2 -s)ζ(1 -s)
To manually define zeta function such a way that it takes value 1 or mathematically ∃!s ∈ N; ζ(s-1) = 1 , Euler's induction approach was applied and it was observed that zeta function have the potential unit value as demonstrated in the results section. Justification of the definition we set for ζ(1) = 1 and consistency of the above forms of functional equation have been cross checked and it was found that the proposition complies with all the theorems used in complex analysis. Justification of the definition we set for ζ(-1) = 1 2 and consistency of the above forms of functional equation have been cross checked in the results section. ζ(-1) = 1 2 must be the second solution to ζ(-1) apart from the known Ramanujan's proof ζ(-1) = -1 12 . One has to accept that following the zeta functions analytic and its harmonic conjugal behaviour zeta values can be multivalued if seen as a continuam or alternatively it can be seen as a multi zeta function.

Integral representation of

ζ(1) from Bose integral. Lemma 8. -ζ(1) = ∞ 0 dx e x -1 = ∞ 0 e -x 1 -e -x dx = ∞ 0 ∞ n=1 e -x .(e -x ) n-1 dx = ∞ n=1 ∞ 0 e -nx dx substituting nx = u we get =⇒ ∞ n=1 ∞ 0 e -u n du = ∞ n=1 1 n ∞ 0 e -u du = - ∞ n=1 1 n substituting x = ln(2) in -ζ(1) = ∞ 0 dx e x -1
and changing the limit we get

ζ(1) = 1 0 dx e ln (2) -1 = x 2 -1 1 0 = 1 3.3. Integral representation of ζ(-1) from Ramanujan summation. Lemma 9. ζ(-1) = i 3 ∞ 0 f (6it) -f (-6it) e 2πt -1 dt = i 3 ∞ 0 12it e 2πt -1 dt = 12i 4 ∞ 0 t e 2πt -1 dt Substituting u = 2πt we get ζ(-1) = 12 ∞ 0 u 2π e u -1 . du 2π = 12 4π 2 ∞ 0 udu e u -1 = 12 4π 2 . π 2 6 = 1 2
4. Let us work out the lemmas related to logarithm

In this section we will work out logarithm related some new mathematics which will help us to interpret Riemann Hypothesis.

4.1. The unit circle, the unit sphere revisited. The unit circle is a circle centered at the origin (0, 0) in a coordinate plane, with a radius of 1 unit. In trigonometry and complex analysis, the unit circle is commonly used to study the relationships between angles and points on the circle.In trigonometry, the unit circle is used to define the trigonometric functions (sine, cosine, tangent, etc.) for any angle. Each point on the unit circle corresponds to a unique angle, and the coordinates of that point represent the values of the trigonometric functions for that angle.If we closely observe the trigonometric unit circle we will see a pattern. After every 30 degree angle, value of the trigonometric ratios are same in absolute sense, I mean Pythagoras formula on sine and cosine results 1 . So it takes one sixth Pi rotation to complete a cycle of three. There are two semi circles, four quadrants inside the unit circle. Therefore we need to go through six cycles of three to complete a semi cycle and twelve cycles of three to complete a cycle. In complex analysis, the unit circle is also significant. When complex numbers are represented in polar form the unit circle plays a crucial role. The angle theta corresponds to the argument or phase of the complex number, while the point on the unit circle represents the direction of the complex number in the complex plane.Euler's famous formula for complex numbers e iθ = icosθ + isinθ connects the exponential function with trigonometric functions and bridges between trigonometry,algebra,analytical geometry. Concept of trigonometric unit circle is not the only unit circle we know. Our knowledge of mathematics is not restricted to trigonometry alone. We have got arithmetic, algebra. Four basic operation of arithmetic i.e, addition, subtraction, multiplication and division also exhibits the same cycle of 3.Two anti operation pair out of those four operations on any numbers can always be zero through the results additive identity, or an unity through the results multiplicative identity. When things are linear the cycles are obvious, we start suspecting cycles when things starts geting non-linear in higher dimensions. We have got non-linear algebraic functions such as polynomial, exponential, logarithmic functions etc. which should also exhibit the same cyclic pattern as these functions also obeys two fundamental laws of algebra i.e. the law of additive inverse and the law of multiplicative inverse. Let us have little more clarity on cycles. Logarithms of negative numbers are not defined in the real numbers in the same way that the square roots of negative numbers are not defined in real numbers. The logarithm of negative numbers is undefined. The logarithms of negative numbers are defined in complex numbers from Euler's identity e iπ = -1. Still there is a caveat as we note that the complex logarithm is a multi-valued function. This means that for any given complex number z, there are infinitely many complex logarithms that satisfy e w = z. Each of these logarithms differs by an integer multiple of 2iπ.To define a single-valued logarithm, called the principal logarithm, we typically choose a branch cut in the complex plane. The branch cut is a curve that removes a portion of the complex plane to make the logarithm function continuous. Common choices for the branch cut include the negative real axis or the branch cut extending from the origin to negative infinity.To complete the cycle and to define complex logarithm in an unique way we need to define some real value for imaginary number i. This should be done assigning an unique lowest possible value to i which can reach atleast theoratically all real numbers along the number line.

x y

0 • 30 • 60 • 90 • 120 • 150 • 180 • 210 • 240 • 270 • 300 • 330 • 360 • π 6 π 3 π 2 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π e ln ( √ 3
2 ) , e ln ( 12 )

e ln ( 1 2 ) , e ln ( √ 3
2 )

e ln (-

√ 3
2 ) , e ln ( 12 )

e ln (-1 2 ) , e ln ( √ 3

2 )

e ln (-

√ 3 2 ) , e ln (- 1
2 )

e ln (-1 2 ) , e ln (-

√ 3 2 ) e ln ( √ 3
2 ) , e ln (-1

2 )

e ln ( 1 2 ) , e ln (-

√ 3 2 ) (e iπ = -1, 0) (e 0 = 1, 0) e i 3π 2 = -i = e ?
? e i π 2 = i = e ? ? z = r(cos x + i sin x) is the trigonometric form of complex numbers. Using Euler's formula e ix = cos x + i sin x we can write z = re ix . Putting x = π in Euler's formula we get , e iπ = -1.Putting x = π 2 we get e iπ 2 = i. So the general equation of the points lying on unit circle |z| = |e ix | = 1. But that's not all because i is not defined yet. From the trigonometric unit circle we can predict existence of some absolute real value for i as the coordinates of i numerically obeys arithmetic/ algebraic fundamental law of additive inverse and the law of multiplicative inverse. -

If x = π 3 in trigonometric form then z = cos( π 3 ) + i.sin( π 3 ) = 1 2 ( √ 3 + i).So |z| = r = ( √ 3 2 ) 2 + ( 1 2 ) 2 = 1 2 . √ 4 = 1 2 .2 = 1.
3 -2 -1 1 2 3 -2 2 Re(z) Im(z)
The d-unit circle together with the general unit circle completes the algebraic cycle via |z| = 1 2 e ix = 1. Therefore taking the imaginary number i out of the unit circle as a constant scale factor we can visulise the imaginary axis maintaining the rotational symmetries arising out of trigonometric cycle of 3. The imaginary cycle gets reflected along the real axis and the dynamics keep on cycling alternately. Five lattice points 1, i, -1, -i, 0 works as mid value between i[ 1+ i -1 -i = 0 ] and [ 1+ i -1 -i = 0 ]. After all the concept of place hoder zeros in decimal number system is just replication of completing the complex unit circle.When Unit circle in complex plane is stereo-graphically projected to unit sphere the points within the area of unit circle gets mapped to southern hemisphere, the points on the unit circle gets mapped to equatorial plane, the points outside the unit circle gets mapped to northern hemisphere. d-unit circle can also be easily projected to Riemann sphere. Projection of d-unit circle to d-unit sphere will have three parallel disc (like three dimensions hidden in one single dimension of numbers) for three (equivalent unit values in three different sense) magnitude of 1 2 , 1, 2 in the southern hemisphere, on the equator, in the northern equatorial sphere respectively as shown in the following diagram. The polar sphere will be projecting points outside the d-unit circle.

South pole=0

Three parallel surfaces in a single cone (the one way absolute view) will look like as follows.

Explanation 1. One may attempt to show that |z| = 1 2 e ix = 1 will mean 1= 2. This may not be right interpretation. Correct way to interpret is given here under.

We know: e ix = r(cos θ + i sin θ). Taking derivative both side we get ie ix = (cos θ + i sin θ) dr dx + r(-sin θ + i cos θ) dθ dx .

Now Substituting r(cos θ + i sin θ) for e ix and equating real and imaginary parts in this formula gives dr dx = 0 and dθ dx = 1. Thus, r is a constant, and θ is x + C for some constant C. Now if we assign r = 1 2 and ix = ln 2 then re ix = 1 2 .e ln 2 = 1 The initial value x=1 then gives i = ln 2. That means in 4D the imaginary number i turns into a complete real number ln 2 in logarithmic way. Now let see how quaternion helps in justifying the definition for imaginary number i. For simplification let us use a single alphabet for expressing quaternion. Let us recall the power addition identity, which is, e (a+b) = e a .e b However this only applies when 'a' and 'b' commute, so it applies when a or b is a scalar for instance. The more general case where 'a' and 'b' don't necessarily commute is given by: e q = e q 1 .e q 2 where: q = q 1 + q 2 + q 1 × q 2 + 1 3 (q 1 × (q 1 × q 2 ) + q 2 × (q 2 × q 1 )) + ...

Where: × = vector cross product. This shows that when a and b become close to becoming parallel then q 1 × q 2 approaches zero and q approaches q 1 + q 2 so the rotation algebra approaches vector algebra. As we have seen all the three unit discs appear parallel to each other our life gets easier and we can do complex exponentiation and logarithm as we do it for real numbers. We have seen in Euler's formula e iπ = -1 complex exponentials turn into real numbers. Applying the same trick we can turn quaternion exponentials into real numbers following the cycle of e iπ + 1 = 0 . This time we need to complete the cycle of [ 1+ i -1 -i = 0 ] by way of showing e q = -1 which will prove the formula |z| = 1 2 e ix = 1 for generalised complex numbers. Let us consider the following infinite series. e q 1 .e q 2 . q 3 .e q 4 .e q 5 .e q 6 . q 7 ... =e q 1 +q 2 +q 3 +q 4 +q 5 +q 6 +q 7 ... =e 2. ln (2) * =e ln (2) .e ln (2) = -e -ln (2) .e ln (2) = -1 2 .2

= -1 * q 1 + q 2 + q 3 + q 4 + q 5 + q 6 + q 7 ...

= i( 0 s=-∞ ζ(s)) + j( 0 s=-∞ ζ(s)) + k( 0 s=-∞ ζ(s)) + ∞ s=-∞ ζ(s) + ∞ s=-∞ ζ(s) + ∞ s=-∞ ζ(s) + ∞ s=-∞ ζ(s) = i(ζ(-1) + ζ(0)) + j(ζ(-1) + ζ(0)) + k(ζ(-1) + ζ(0)) + η(1) + η(1) * * = i.0 + j.0 + k.0 + 2.η(1) = 2.

ln (2)

There is a cycle of three in four infinite series of infinite zeta values which emerges as two infinite alternate zeta series following 1 2 .e ln 2 = 1. This proves the formula |z| = 1 2 e ix = 1.Thus we see ix = ln(cos θ + i sin θ) is a multivalued function not only because of infinite rotation around the unit circle but also due to different real solutions to i in higher dimensional number system of quaternion discovered by Hamilton involving higher dimensional algebra like Clifford algebra, or at least involving 3 dimensional vector algebra connected to four dimensional zeros through algebraic cycles proposed by Hodge, Tate, Weil and others existence of which are not proven algebraically, numerically they actually exists, using zeta results I have outlined the minimal proof, someone has to work on the rigorous proof. Now coming back to the explanation, when we completed the circle of [ 1+ i -1 -i = 0 ] we actually completed two semi circle [ 1 -1 = 0 ] following e iπ + 1 = 0. Thus we cannot interpret 1=2 or 2=1, we have to interpret either 2=2 or 1=1.

4.2.

Closure of logarithmic operation under quaternion.

Lemma 10. i = ln (2) comes as a solution to indeterminacy of negative logarithm and a filler for the caveat in the definition of principal logarithm in terms of complex logarithm. We need quaternion to do the job. If we visualise principal logarithm as logarithm a set of quaternion instead of product of two pairs of i then we can arrive zero at par with the definition of logarithm and solve the issue of indeterminacy of the principal value i.e.

ln (1) = 0 = ln (-1. -1) = ln (-1. -1. -1. -1) = ln (i 2 .j 2 .k 2 .i.j.k) = 3(ln i + ln j + ln k)
. Any guess what angle can make vector-sum of three equal vectors equal to zero? As shown in Riemann hypothesis proof, its 120 degree in 3D or 60 degree in 4D. This way numbers are very complexly 3 dimensional hidden in other hidden dimensions of quaternion although we do not feel its requirement in our everyday use of numbers. Following the above definition we can generalise the definitions further as follows.

ln (-1) = -1 -1 = - 1 1 ln (-2) = -2 -1 = - 1 2 ln (-3) = -3 -1 = - 1 3
And the pattern continues up to negative infinity.

But in complex sense we can do the closure of logarithmic operation under quaternion as follows:

Example 2. Find natural logarithm of -1 using first quaternion solution of i ln(-1) = ln(i 2 ) = 2ln(i) = -0.733025841(approx) Part 3. The Proofs

Let us prove Riemann Hypothesis

In this section we will prove Riemann Hypothesis. 5.1. An elegant proof using Euler's product form. Euler's Product form of zeta Function in Euler's exponential form of complex numbers is as follows:

ζ(s) = ∞ n=1 1 n s = p 1 + re iθ + r 2 e i2θ +
r 3 e i3θ ... Now any such factor 1 + re iθ + r 2 e i2θ + r 3 e i3θ ... will be zero if

re iθ + r 2 e i2θ + r 3 e i3θ ... = -1 = e iπ
Comparing both side of the equation and equating left side to right side on the unit circle we can say: *

θ + 2θ + 3θ + 4θ... = π r + r 2 + r 3 + r 4 .... = 1
We can solve θ and r as follows:

θ + 2θ + 3θ + 4θ... = π θ(1 + 2 + 3 + 4...) = π θ.ζ(-1) = π θ. -1 12 = π θ = -12π r + r 2 + r 3 + r 4 .... = 1 r(1 + r + r 2 + r 3 + r 4 ....) = 1 r 1 1 -r = 1 r = 1 -r r = 1 2
We can determine the real part of the non trivial zeros of zeta function as follows:

r cos θ = 1 2 cos(-12π) = 1 2
Therefore Principal value of ζ( 1 2 ) will be zero, hence Riemann Hypothesis is proved.

Explanation 2. *We can try back the trigonometric form then the algebraic form of complex numbers do the summation algebraically and then come back to exponential form as follows:

re iθ + r 2 e i2θ + r 3 e i3θ ... = (r cos θ + ir sin θ) + (r 2 cos 2θ + ir 2 sin 2θ) + (r 3 cos 3θ + ir 3 sin 3θ).... = (x 1 + iy 1 ) + (x 2 + iy 2 ) + (x 3 + iy 3 ) + (x 4 + iy 4 ) + (x 5 + iy 5 ).... = (x 1 + x 2 + x 3 + x 4 + x 5 + ...) + i(y 1 + y 2 + y 3 + y 4 + y 5 + ...)

= R cos Θ + iR sin Θ = (r + r 2 + r 3 + r 4 ....)e i(θ+2θ+3θ+4θ...)

Explanation 3. One may attempt to show that (re iθ + r 2 e i2θ + r 3 e i3θ ...) = -1 actually results re iθ 1-re iθ which implies in absurdity of 0 = -1. Correct way to evaluate re iθ 1-re iθ is to apply the complex conjugate of denominator before reaching any conclusion. re iθ (1+re iθ )

(1-re iθ )(1+re iθ ) then shall result to re iθ = -1 which points towards the unit circle. In the present proof we need to go deeper into the d-unit circle and come up with the interpretation which can explain the Riemann Hypothesis.

Explanation 4. One may attempt to show inequality of the reverse calculation 1 2 1 + 1 2 2 + 1 2 3 ... = 1 ̸ = -1. re iπ = -1 need to be interpreted as the exponent which then satisfies 1 -1 = 1 or 2.2 -1 = 1 on the unit or d-unit circle. There is nothing called t-unit circle satisfying 3.3 -1 = 1.

Explanation 5. Essentially proving log 2 ( 1 2 ) = -1 in a complex turned simple way is equivalent of saying log(1) = 0 in real way. Primes other than 2 satisfy log p ( 1 2 ) = e iθ also in a pure complex way.

5.

2. An elementary proof using alternate product form. Euler's alternate Product form of zeta Function in Euler's exponential form of complex numbers is as follows:

∞ n=1 1 n s = p 1 1 -1 re iθ = p re iθ re iθ -1
Multiplying both numerator and denominator by re iθ + 1we get:

∞ n=1 1 n s = p re iθ (re iθ + 1) (re iθ -1)(re iθ + 1)
Now any such factor re iθ (re iθ +1) (r 2 e i2θ -1) will be zero if re iθ (re iθ + 1) is zero:

re iθ (re iθ + 1) = 0 re iθ (re iθ -e iπ ) = 0 r 2 e i2θ -re i(π-θ) * = 0 r 2 e i2θ = re i(π-θ)
We can solve θ and r as follows:

2θ = (π -θ) 3θ = π θ = π 3 r 2 = r r 2 r = r r r = 1
We can determine the real part of the non trivial zeros of zeta function as follows:

r cos θ = 1. cos( π 3 ) = 1 2
Therefore Principal value of ζ( 1 2 ) will be zero, and Riemann Hypothesis is proved. -θ) is arrived as follows:

Explanation 6. * e i(
e iθ = e iθ 1 = e iθ 1 -1 = e iθ -1 1 = e iθ i 2 1 = e iθ i 2 = e -iθ

5.

3. An exhaustive proof using alternate functional equation. Multiplying both side of Riemann's functional equation by (1 -s) we get

(1 -s)ζ(s) = 2 s π (s-1) sin πs 2 (1 -s)Γ(1 -s)ζ(1 -s) Putting (1 -s)Γ(1 -s) = Γ(2 -s) we get: ζ(1 -s) = (1 -s)ζ(s) 2 s π (s-1) sin πs 2 Γ(2 -s) s → 1we get: ∵ lim s→1 (s -1)ζ(s) = 1 ∴ (1 -s)ζ(s) = -1 and Γ(2 -1) = Γ(1) = 1 ζ(0) = -1 2 1 π 0 sin π 2 = - 1 2
Examining the functional equation we shall observe that the pole of zeta function at Re(s) = 1 is attributable to the pole of Gamma function. In the critical strip 0 < s < 1 Delta function holds equally good if not better for factorial function. As zeta function has got the holomorphic property the act of stretching or squeezing preserves the holomorphic character. Using this property we can remove the pole of zeta function. Introducing Delta function for factorial we can remove the poles of Gamma and Pi function and rewrite the functional equation in terms of its harmonic conjugate function as follows (see above):

ζ(s) = -2 s π (s-1) sin πs 2 ∆(4 -s)ζ(1 -s)
This can be rewritten in terms of Gamma function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
This again can be rewritten in terms of Pi function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 Π(2 -s)ζ(1 -s)
Now Putting s = 1we get:

ζ(1) = -2 1 π (1-1) sin π 2 Γ(3 -1)ζ(0) = 1
zeta function is now defined on entire C , and as such it becomes an entire function. In complex analysis, Liouville's theorem states that every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f (z)| ≤ M for all z in C is constant. Being an entire function zeta function is constant as none of the values of zeta function do not exceed M = ζ(2) = π 2 6 .Maximum modulus principle further requires that non constant holomorphic functions attain maximum modulus on the boundary of the unit circle. Being a constant function zeta function duly complies with maximum modulus principle as it reaches maximum modulus π 2 6 outside the unit circle i.e. on the boundary of the double unit circle. Gauss's mean value theorem requires that in case a function is bounded in some neighbourhood, then its mean value shall occur at the centre of the unit circle drawn on the neighbourhood. |ζ(0)| = 1 2 is the mean modulus of entire zeta function. Inverse of maximum modulus principle implies points on half unit circle give the minimum modulus or zeros of zeta function. Minimum modulus principle requires holomorphic functions having all non zero values shall attain minimum modulus on the boundary of the unit circle. Having lots of zero values holomorphic zeta function do not attain minimum modulus on the boundary of the unit circle rather points on half unit circle gives the minimum modulus or zeros of zeta function. Everything put together it implies that points on the half unit circle will mostly be the zeros of the zeta function which all have ± 1 2 as real part as Riemann rightly hypothesized.

Putting s = 1 2 in ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s) ζ 1 2 = -2 π (1-1 2 ) sin π 2.2 Γ 5 2 ζ 1 2 ζ 1 + 3 √ 2.π.π 4. √ 2 = 0 ζ 1 2 1 + 3π 4 = 0 ζ 1 2 = 0
Therefore principal value of ζ( 12 ) is zero and Riemann Hypothesis holds good.

Part 4. The Results

The Results

6.1. Infinite product of positive zeta values converges.

ζ(1) = 1 + 1 2 1 + 1 3 1 + 1 4 1 ... = 1 + 1 2 1 + 1 2 2 + 1 2 3 ... 1 + 1 1 + 1 3 2 + 1 3 3 ... ... ζ(2) = 1 + 1 2 2 + 1 3 2 + 1 4 2 ... = 1 + 1 2 2 + 1 2 4 + 1 2 6 ... 1 + 1 2 + 1 3 4 + 1 3 6 ... ... ζ(3) = 1 + 1 2 3 + 1 3 3 + 1 4 3 ... = 1 + 1 2 3 + 1 2 6 + 1 2 9 ... 1 + 1 3 + 1 3 6 + 1 3 9 ... ... ζ(1)ζ(2)ζ(3)... =
From the side of infinite sum of negative exponents of all natural integers:

= 1 + 1 2 1 + 1 3 1 + 1 4 1 ... 1 + 1 2 2 + 1 3 2 + 1 4 2 ... 1 + 1 2 3 + 1 3 3 + 1 4 3 ... ... = 1 + 1 2 1 + 1 2 2 + 1 2 3 ... + 1 3 1 + 1 3 2 + 1 3 3 ... + 1 4 1 + 4 2 + 1 4 3 ... ... = 1 + 1 + 1 2 1 + 1 3 1 + 1 4 1 + 1 5 1 + 1 6 1 + 1 7 1 + 1 8 1 + 1 9 1 ... = 1 + ζ(1)
From the side of infinite product of sum of negative exponents of all primes:

1 + 1 2 1 + 1 2 2 + 1 2 3 ... 1 + 1 3 1 + 1 3 2 + 1 3 3 ... 1 + 1 5 1 + 1 + 1 5 3 ... ... 1 + 1 2 2 + 1 2 4 + 1 2 6 ... 1 + 1 3 2 + 1 3 4 + 1 3 6 ... 1 + 1 5 2 + 1 + 1 5 6 ... ... 1 + 1 2 3 + 1 2 6 + 1 2 9 ... 1 + 1 3 3 + 1 3 6 + 1 3 9 ... 1 + 1 5 3 + 1 + 1 5 9 ... ... = 1 + 1 1 + 1 3 1 + 1 3 2 + 1 3 3 ... 1 + 1 5 1 + 1 5 2 + 1 5 3 ... ... 1 + 1 2 2 + 1 2 4 + 1 2 6 ... 1 + 1 3 2 + 1 3 4 + 1 3 6 ... 1 + 1 5 2 + 1 + 1 5 6 ... ... 1 + 1 2 3 + 1 2 6 + 1 2 9 ... 1 + 1 3 3 + 1 3 6 + 1 3 9 ... 1 + 1 5 3 + 1 + 1 5 9 ... ...
continued to next page.... continued from last page....

Simultaneously halving and doubling each factor and writing it as sum of two forms

= 2 1 2 1 + 1 3 1 -1 3 + 1 + 1 3 1 + 1 3 2 + 1 3 3 ... 1 2 1 + 1 5 1 -1 5 + 1 + 1 5 1 + 1 5 2 + 1 5 3 ... ... 1 2 1 + 1 4 1 -1 4 + 1 + 1 2 2 + 1 2 4 + 1 2 6 ... 1 2 1 + 1 9 1 -1 9 + 1 + 1 3 2 + 1 3 4 + 1 3 6 ... ... 1 2 1 + 1 8 1 -1 8 + 1 + 1 2 3 + 1 2 6 + 1 2 9 ... 1 2 1 + 1 27 1 -1 27 + 1 + 1 3 3 + 1 3 6 + 1 3 9 ... ... = 2 1 2 1 + 1 2 + 1 + 1 3 1 + 1 3 2 + 1 3 3 ... 1 2 1 + 1 4 + 1 + 1 5 1 + 1 5 2 + 1 5 3 ... ... 1 2 1 + 1 3 + 1 + 1 2 2 + 1 2 4 + 1 2 6 ... 1 2 1 + 1 8 + 1 + 1 3 2 + 1 3 4 + 1 3 6 ... ... 1 2 1 + 1 7 + 1 + 1 2 3 + 1 2 6 + 1 2 9 ... 1 2 1 + 1 26 + 1 + 1 3 3 + 1 3 6 + 1 3 9 ... ... = 2 1 + 1 2 1 2 + 1 3 1 + 1 3 2 + 1 3 3 ... 1 + 1 2 1 4 + 1 5 1 + 1 5 2 + 1 5 3 ... ... 1 + 1 2 1 3 + 1 2 2 + 1 2 4 + 1 2 6 ... 1 + 1 2 1 8 + 1 3 2 + 1 3 4 + 1 3 6 ... ... 1 + 1 2 1 7 + 1 2 3 + 1 2 6 + 1 2 9 ... 1 + 1 2 1 26 + 1 3 3 + 1 3 6 + 1 3 9 ... ... = 2 1 + 1 2 1 2 1 + 1 3 1 + 1 4 1 ... + 1 2 1 + 1 3 1 + 1 4 1 ... = 2 1 + 1 2 2ζ(1) -2 = 2(1 -1 + ζ(1)) = 2ζ(1)
Now comparing two identities we get:

1 + ζ(1) = 2ζ(1)) ζ(1) = 1
Hence Infinite product of positive zeta values converges to 2.

6.2. Infinite product of negative zeta values converges.

ζ(-1) = 1 + 2 1 + 3 1 + 4 1 + 5 1 ... = 1 + 2 + 2 2 + 2 3 ... 1 + 3 + 3 2 + 3 3 ... ... ζ(-2) = 1 + 2 2 + 3 2 + 4 2 + 5 2 ... = 1 + 2 2 + 2 4 + 2 6 ... 1 + 3 2 + 3 4 + 3 6 ... ... ζ(-3) = 1 + 2 3 + 3 3 + 4 3 + 5 3 ... = 1 + 2 3 + 2 6 + 2 9 ... 1 + 3 3 + 3 6 + 3 9 ... ...
From the side of infinite sum of negative exponents of all natural integers:

ζ(-1)ζ(-2)ζ(-3)... = 1 + 2 1 + 3 1 + 4 1 + 5 1 ... 1 + 2 2 + 3 2 + 4 2 + 5 2 ... 1 + 2 3 + 3 3 + 4 3 + 5 3 ... ... = 1 + 2 + 2 2 + 2 3 ... + 3 + 3 2 + 3 3 ... + 4 + 4 2 + 4 3 ... ... = 1 + 1 + 2 + 2 2 + 2 3 ... -1 + 1 + 3 + 3 2 + 3 3 ... -1 + 1 + 4 + 4 2 + 4 3 ... -1 ... = 1 + - 1 1 -1 + - 1 2 -1 + - 1 3 -1 + - 1 4 -1 ... = 1 - 1 + 1 2 + 1 3 + 1 4 ... + 1 + 1 + 1 + 1... = 1 -ζ(1) + ζ(0) = 1 -1 - 1 2 = 1 2
From the side of infinite product of sum of negative exponents of all primes:

1 + 2 + 2 2 + 2 3 ... 1 + 3 + 3 2 + 3 3 ... 1 + 5 + 5 2 + 5 3 ... ... 1 + 2 2 + 2 4 + 2 6 ... 1 + 3 2 + 3 4 + 3 6 ... 1 + 5 2 + 5 4 + 5 6 ... ... 1 + 2 3 + 2 6 + 2 9 ... 1 + 3 3 + 3 6 + 3 9 ... 1 + 5 3 + 5 6 + 5 9 ... ... = 1 + 2 1 + 3 1 + 4 1 + 5 1 ... =⇒ ζ(-1) = 1 2 33 of 95 Therefore ζ(-1) = 1 2 must be the second solution apart from ζ(-1) = -1 12 .
Using Delta function instead of Gamma function we can rewrite the functional equation applicable as follows:

ζ(s) = -2 s π (s-1) sin πs 2 ∆(4 -s)ζ(1 -s)
This can be rewritten in terms of Gamma function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
This again can be rewritten in terms of Pi function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 Π(2 -s)ζ(1 -s) Putting s = -1we get: ζ(-1) = -2 -1 π (-1-1) sin -π 2 Γ(3 -s)ζ(2) = 1 2
To proof Ramanujan's Way

σ = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9..... 2σ = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9... + 1 + 1 + 1 + 1 + 1 + 1 + 1... *
Subtracting the bottom from the top one we get:

-σ = 0 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1... + 1 + 1 + 1 + 1 + 1 + 1 + 1... σ = -(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1........) σ = 1 2
*The second part is calculated subtracting bottom from the top before doubling.

6.3. Infinite sum of Positive Zeta values converges.

ζ(1) = 1 + 1 2 1 + 1 3 1 + 1 4 1 ... ζ(2) = 1 + 1 2 2 + 1 3 2 + 1 4 2 ... ζ(3) = 1 + 1 2 3 + 1 3 3 + 1 4 3 ... . . . ζ(1) + ζ(2) + ζ(3)... = 1 + 1 2 1 + 1 3 1 + 1 4 1 ... + 1 + 1 + 1 + 1 + ... = ζ(1) + ζ(0) = 1 - 1 2 = 1 2 6.4. Infinite sum of Negative Zeta values converges. ζ(-1) = 1 + 2 1 + 3 1 + 4 1 + 5 1 ... ζ(-2) = 1 + 2 2 + 3 2 + 4 2 + 5 2 ... ζ(-3) = 1 + 2 3 + 3 3 + 4 3 + 5 3 ... . . . ζ(-1) + ζ(-2) + ζ(-3)... = 1 + 2 1 + 3 1 + 4 1 + 5 1 ... + 1 + 1 + 1 + 1 + ... = ζ(-1) + ζ(0) = 1 2 - 1 2 = 0 6.5. Infinite product of All Zeta values converges. ζ(-1)ζ(-2)ζ(-3)...ζ(1)ζ(2)ζ(3)... = 2.ζ(-1).ζ(1) = 2.1. 1 2 = 1 6.6. Infinite sum of All Zeta values converges. ζ(-1) + ζ(-2) + ζ(-3)...ζ(1) + ζ(2) + ζ(3)... = 0 + 1 2 = 1 2 6.7.
Counter to Nicole Oresme's logic of divergence. Nicole Oresme in around 1350 proved divergence of harmonic series by comparing the harmonic series with another divergent series. He replaced each denominator with the next-largest power of two.

+ 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 ... > 1 + 1 2 + 1 4 + 1 4 + 1 8 + 1 8 + 1 8 + 1 8 ... + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 ... > 1 + 1 2 + 1 4 + 1 4 + 1 8 + 1 8 + 1 8 + 1 8 + ... + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 ... > 1 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 ...
He then concluded that the harmonic series must diverge as the above series diverges. It was too quick to conclude as we can go ahead and show:

R.H.S = 1 + 1 2 1 + 1 + 1 + 1 + 1 + 1 + 1 + ... = 1 + 1 2 . -1 2 = 1 - 1 4
If we consider ζ(1) = 1 then also it passes the comparison test. Therefore We need to come out of the belief that harmonic series diverges.

= 1 + + 1 2 + 1 2 + 1 2 1 + 1 + 1... = 1 + + 1 2 . -1 2 = 1 + - 1 4 = 1 + -1 -2 + 3 -4 + ... = 1 + - 1 + 2 + 3... -2 1 + 2 + 4... = 1 + - 1 2 -2 1 + 1 + 1... = 1 + - 1 2 -2 -1 2 = 1 + - 1 2 + 1 = 1 + - 3 2 = 1 = 1 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 1 + 1 + 1... = 1 + 5 2 + 1 2 . -1 2 = 1 + 5 2 - 1 4 = 1 + 5 2 -1 -2 + 3 -4 + ... = 1 + 5 2 - 1 + 2 + 3... -2 1 + 2 + 4... = 1 + 5 2 - 1 2 -2 1 1 -2 = 1 + 5 2 - 1 2 + 2 = 1 + 5 2 - 1 + 4 2 = 1 + 5 2 - 5 2 = 1
According to residue theorem we can have a Laurent expansion of an analytic function at the pole f (s) = ∞ n=-∞ a n (s -s 0 ) n of f in a punctured disk around s 0 , and therefrom we can have Res (f (s); s 0 ) = a -1 , i.e. the residue is the coefficient of (s -s 0 ) -1 in that expansion. For the pole ζ(1), we know the start of the Laurent series (since it is a simple pole, there is only one term with a negative exponent), namely ζ(s) = 1 s-1 + γ + . . . so we have Res (ζ(s); 1) = 1. At the pole zeta function have zero radius of convergence ( also known as infinite radius of convergence ). Interpreting zeta function at the pole to be divergent is extreme arbitrary, contradictory and void of rationality. The pole neither falls outside the radius of convergence resulting ζ(1) = ∞ nor inside the radius of convergence resulting ζ(1) = 1 to be absolutely true, its just on the zero radius of convergence suggesting both values to be equally good. Since none of the above value is more natural than the others, both of them can be incorporated into a multivalued zeta function (We should not say, it's not a function at all which gives unique results, we should keep it in mind that ultimately it's two different zeta function seen as a continuum) which is again totally consistent with harmonic conjugate theorem and allows us to interpret

⇒ 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 ... = 1 6.8. Integral representation of ζ(1). -ζ(1) = ∞ 0 dx e x -1 = ∞ 0 e -x 1 -e -x dx = ∞ 0 ∞ n=1 e -x .(e -x ) n-1 dx = ∞ n=1 ∞ 0 e -nx dx substituting nx = u we get =⇒ ∞ n=1 ∞ 0 e -u n du = ∞ n=1 1 n ∞ 0 e -u du = - ∞ n=1 1 n substituting x = ln(2) in -ζ(1) = ∞ 0 dx e x -1
and changing the limit we get

ζ(1) = 1 0 dx e ln (2) -1 = x 2 -1 1 0 = 1 6.9. Integral representation of ζ(-1). ζ(-1) = i 3 ∞ 0 f (6it) -f (-6it) e 2πt -1 dt = i 3 ∞ 0 12it e 2πt -1 dt = 12i 4 ∞ 0 t e 2πt -1 dt Substituting u = 2πt we get ζ(-1) = 12 ∞ 0 u 2π e u -1 . du 2π = 12 4π 2 ∞ 0 udu e u -1 = 12 4π 2 .
π 2 6 = 1 2 6.10. Positive Odd values from harmonic conjugate. Earlier we found that following harmonic conjugate theorem Riemann's functional equation which is an extension of real valued zeta function can also be represented as its harmonic conjugate function which mimics the extended function.

ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
. What shall be the harmonic conjugate of this functional equation? We can shift the argument of zeta function following the fact that the product of Extentended Gamma and Extended Zeta function should also be analytic we can figure out the harmonic conjugate as follows.

ζ(s) = 2 s π (s-1) sin πs 2 Γ(3 -s)ζ(3 -s)
We can get the closed form of odd positive zeta values as follows:

When s=3 ζ(3) = 2 3 π (3-1) sin 3π 2 Γ(3 -3)ζ(3 -3) = 4π 2 But When x=-2, ζ(-2) = 2 -2-1 π (-2) 1 Γ(-2) cos( -2π
2 )

ζ(1 + 2) = ζ(3) 4π 2 Therefore we can say, ζ(-2) = 4π 2 4π 2 = 1 Similarly When s=5 ζ(5) = 2 5 π (5-1) sin 5π 2 Γ(3 -5)ζ(3 -5) = -(2π) 4 But When x=-4, ζ(-4) = 2 -4-1 π (-4) 1 Γ(-4) cos( -4π 2 ) ζ(1 + 4) = 3ζ(5) 8π 4
Therefore we can say, ζ(-4) = -3(2π) 4 8π 4 = -6

Similarly When s=7

ζ(7) = 2 7 π (7-1) sin 7π 2 Γ(3 -7)ζ(3 -7) = -2 5 π 6 But When x=-6, ζ(-6) = 2 -6-1 π (-6) 1 Γ(-6) cos( -6π 2 ) ζ(1 + 6) = 15ζ(7) 8π 6
Therefore we can say, ζ(-6) = -15.2 5 π 6 8π 6 = -60

Similarly When s=9 ζ(9) = 2 9 π (9-1) sin 9π 2 Γ(3 -9)ζ(3 -9) = 2 7 π 8 But When x=-8, ζ(-8) = 2 -8-1 π (-8) 1 Γ(-8) cos( -8π 2 ) ζ(1 + 8) = 315ζ(9) 16π 8
Therefore we can say, ζ(-8) = 315.2 7 π 8 16π 8 = 2520 . . . And the pattern continues for all negative even numbers up to negative infinity. 6.11. Negative even values redefining trivial zeros. We can apply Euler's reflection formula for Gamma function

Γ(1 -s)Γ(s) = π sin(πs) , s ̸ ∈ Z in Riemann's functional equation ζ(s) = 2 s π (s-1) sin πs 2 Γ(1 -s)ζ(1 -s)
to get another representation of zeta function as follows:

ζ(s) = 2 s π (s-1) sin πs 2 π Γ(s) sin(πs) ζ(1 -s) =⇒ ζ(s) = 2 s π (s) sin πs 2 1 Γ(s)2 sin( πs 2 ) cos( πs 2 ) ζ(1 -s) =⇒ ζ(s) = 2 s-1 π (s) 1 Γ(s) cos( πs 2 ) ζ(1 -s) When x=-2, ζ(-2) = 2 -2-1 π (-2) 1 Γ(-2) cos( -2π
2 )

ζ(1 + 2) = ζ(3) 4π 2 = 4.π 2 4π 2 = 1 When x=-4, ζ(-4) = 2 -4-1 π (-4) 1 Γ(-4) cos( -4π 2 ) ζ(1 + 4) = 3ζ(5) 8π 4 = 3. -(2π) 4 8π 4 = -6 When x=-6, ζ(-6) = 2 -6-1 π (-6) 1 Γ(-6) cos( -6π 2 ) ζ(1 + 6) = 15ζ(7) 8π 6 = 15. -2 5 π 6 8π 6 = -60 When x=-8, ζ(-8) = 2 -8-1 π (-8) 1 Γ(-8) cos( -8π 2 ) ζ(1 + 8) = 315ζ(9) 16π 8 = 315.2 7 π 8 16π 8 = 2520 . . .
And the pattern continues for all negative even numbers up to negative infinity.

6.12. Trivial zeros of harmonic conjugate. Like the original functional equation its harmonic conjugate shall also have trivial zeros at every negative even integer due to sine function.

6.13. Negative odd values from harmonic conjugate. Earlier we found that following harmonic conjugate theorem Riemann's functional equation which is an extension of real valued zeta function can also be represented as its harmonic conjugate function which mimics the extended function.

ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
We can get the harmonic conjugates of negative zeta values as follows:

When s=-1 ζ(-1) = -2 -1 π (-1-1) sin -1π 2 Γ(3 + 1)ζ(1 + 1) = 1 2 When s=-3 ζ(-3) = -2 -3 π (-3-1) sin -3π 2 Γ(3 + 3)ζ(1 + 3) = -1 6 When s=-5 ζ(-5) = -2 -5 π (-5-1) sin -5π 2 Γ(3 + 5)ζ(1 + 5) = 1 6 When s=-7 ζ(-7) = -2 -7 π (-7-1) sin -7π 2 Γ(3 + 7)ζ(1 + 7) = -3 10 . . .
And the pattern continues for all negative odd numbers up to negative infinity.

6.14. Negative even values from harmonic conjugate. We can apply Euler's reflection formula for Gamma function

Γ(2 -s)Γ(s -1) = π sin(πs -π) , s ̸ ∈ Z in Riemann's functional equation ζ(s) = -2 s π (s-1) sin πs 2 Γ(3 -s)ζ(1 -s)
to get another representation of zeta function as follows:

ζ(s) = -2 s π (s-1) sin πs 2 π(2 -s) Γ(s -1) sin(πs -π) ζ(1 -s) =⇒ ζ(s) = -2 s π (s-1) sin πs 2 π(2 -s) Γ(s -1) sin(πs) ζ(1 -s) =⇒ ζ(s) = -2 s π (s) sin πs 2 2 -s Γ(s -1)2 sin( πs 2 ) cos( πs 2 ) ζ(1 -s) =⇒ ζ(s) = -2 s-1 π (s) 2 -s Γ(s -1) cos( πs 2 ) ζ(1 -s) When x=-2, ζ(-2) = 2 -2-1 π (-2) 2 + 2 Γ(-3) cos( -2π 2 ) ζ(1 + 2) = ζ(3) π 2 = 4.π 2 π 2 = 4 When x=-4, ζ(-4) = 2 -4-1 π (-4) 2 + 4 Γ(-5) cos( -4π 2 ) ζ(1 + 4) = 9ζ(5) 2π 4 = 3. -(2π) 4 2π 4 = -48 When x=-6, ζ(-6) = 2 -6-1 π (-6) 2 + 6 Γ(-7) cos( -6π
2 )

ζ(1 + 6) = 45ζ(7) π 6 = 15. -2 5 π 6 π 6 = -480 When x=-8, ζ(-8) = 2 -8-1 π (-8) 2 + 8 Γ(-9) cos( -8π
2 )

ζ(1 + 8) = 45ζ(7) π 8 = 315.2 7 π 8 π 8 = 40320 . . .
And the pattern continues for all negative even numbers up to negative infinity.

6.15. Zeta results confirms PNT. In number theory, the prime number theorem (PNT) describes the asymptotic distribution of the prime numbers among the positive integers. It formalizes the intuitive idea that primes become less common as they become larger by precisely quantifying the rate at which this occurs. The theorem was proved independently by Jacques Hadamard and Charles Jean de la Vallée Poussin in 1896 using ideas introduced by Bernhard Riemann (in particular, the Riemann zeta function). The first such distribution found is π(N ) ∼ N log N , where π(N ) is the prime-counting function and log N is the natural logarithm of N. This means that for large enough N, the probability that a random integer not greater than N is prime is very close to 1 log N . The prime number theorem then states that N log N is a good approximation to π(N ) (where log here means the natural logarithm), in the sense that the limit of the quotient of the two functions π(N ) and N log N as N increases without bound is 1:

lim N →∞ π(N ) N log(N )
= 1 known as the asymptotic law of distribution of prime numbers. Using asymptotic notation this result can be restated as π(N ) ∼ N log N

Wherever logarithm is there we can take it guaranteed e = lim n→∞ 1 + 1 n n is working in the background. Now we have got one more formula for euler's number e in the form of:

e = lim n→∞ 2 + 2 n n . lim n→∞ 2 -1 + 2 -1 n n
For this reason prime number theorem works as nicely as primes appear through zeta zeros on critical half line in analytic continuation of zeta function.

6.16. Numbers rotates via Euler's formula. In d-unit circle we have seen |z| = 1 2 e ix = 1 is another form of unit circle. From i 2 = -1 we know that i shall have at least two roots or values, one we have already defined, another we need to find out. We have seen that at π 3 Zeta function attains zero. Let us use Euler's formula to define another possible value of i as Euler's formula deals with unity which comes from the product of exponential and its inverse i.e. logarithm. Now its proven :

z = 1 2 e ix = 1 = 1 2 e ln 2
we can say :

e ix = e ln2
taking logarithm both side :

ix = ln(2)
setting x=1 :

ln(2) = e ln(ln(2)) = e ln(i) = i ≈ e -1 e ≈ 2 -1 2 ≈ e -2 or ln(2) 1 ln(ln(2)) = i 1 ln(i) = e ≈ - 1 ln(i) ≈ 2 + i
we get two more identity like e iπ + 1 = 0:

1 e + ln(i) = 0 = e + 1 ln(i)
again we know i 2 = -1, taking log both side ln (-1) = 2 ln i = 2ln(ln(2))

Lets assume:

e i π 3 = z
taking natural log both side :

iπ 3 = ln(z) Lets set:ln(z) = i + 1 3 iπ = 1 + 3i i(π -3) = 1 i = 1 π -3 π = 3 + 1 i
we get two more identity like e iπ + 1 = 0:

ln(i) -2 = 0 = 1 ln(i) - 1 2 taking log both side of i 2 = -1 ln (-1) = 2 ln i = 2ln 1 π -3 Constant 1.
e iπ = e ln(2).π = 8.824977827 = e 2.17758609 ...(approx) Constant 2.

e iπ = e π π-3 = 4, 324, 402, 934 = e 22.18753992 ...(approx) 6.17. Fundamental formula of numbers. Also I got a nice relationship between sum of numbers and the product of primes which can be regarded as the second fundamental formula of arithmetic.

Theorem 1.

2.

∞ N =1 N = ∞ i=1 P i =⇒ ∞ N =1 N = ∞ i=2 P i
We know :

ζ(-1) = ζ(1) + ζ(0) =⇒ 1 + 1 2 + 1 3 + 1 4 ... + 1 + 1 + 1 + 1 + ... = 1 2 =⇒ 1 + 1 + 1 + 1 2 + 1 + 1 3 + 1 + 1 4 + ... = 1 2 =⇒ 2 1 + 3 2 + 4 3 + 5 4 + 6 5 ... = 1 2 =⇒ 1 + 2 + 3 + 4 + 5 + 6 + 7... * 2.3.5.7.11... * * = 1 2 =⇒ 2. ∞ N =1 N = ∞ i=1 P i
*Series in reverse order.

** LCM steps :

LCM = ∞ i=1 P 1 i .P 2 i .P 3 i ...P 1 i .P 2 i .P 3 i ... LCM = ∞ i=1 P (1+2+3+...)+(1+2+3+...) i LCM = ∞ i=1 P 1 2 + 1 2 i LCM = 2.
3.5.7.11... 6.18. Goldbach's prime theorem.

Theorem 2. Every even integer greater than 2 can be expressed as the sum of two primes.

Proof 1. Let N be a arbitrarily large number. Sum of all the natural numbers up to N shall be N (1+N ) 2 which includes sum of all the primes up to N too. Double of such sum shall be N (1 + N ) which shall include double of sum of all the primes up to N too. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). We can visualise N ln(N ) as a prime number itself and we can allow the prime gaps to change equivalently and complete the number sequence. Now if we take logarithm of N (1 + N ) with respect to the base of N ln(N ) the result shall give us the lower bound of powers that can comfortably be applied on that prime to reach double of the sum of all the natural numbers up to N i.e. N (1 + N ). In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of Goldbach partitions (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set of equations

P 2 ≤ log N ln(N ) N (1 + N ), P 1 + P 2 = E, P log N ln(N ) N (1+N ) 1 ≤ N (1 + N )
. If our resultant exponent is greater than 2 (ideally it should be greater than or equal to 2 as we have ensured all primes are summed up 2 times) then that will be greater than the lower bound of prime gaps 2 which eventually ensures continuity of the pattern up to infinity. Clearly due to infinitude of primes, the result log N ln(N )

N (1 + N ) = log N ln(N ) N + log N ln(N )
(1 + N ) shall be always greater than 2. As 2 Goldbach partition is always lesser than the exponent as calculated above, all the even numbers greater than 2 can be expressed as sum of two primes P 1 + P 2 . Hence Goldbach conjecture stands proved and it can be called as Goldbach theorem. The weaker version of Goldbach conjecture (ternary Goldbach conjecture) immediately follows from the stronger version (binary Goldbach conjecture) proved above.

Twin prime theorem.

Theorem 3. There are infinitely many primes p such that p + 2 is also prime.

Proof 2. As the above result log N ln(N )

N (1 + N ) = log N ln(N ) N + log N ln(N )
(1 + N ) shall be always greater than 2 there shall be infinitely many twin primes with a prime gap of 2.Hence Twin prime conjecture stands proved and it can be called as Twin prime theorem.

Sophie Germain's prime theorem.

Theorem 4. There are infinitely many prime numbers of the form 2P + 1 Proof 3. As there shall be infinitely many twin primes with a prime gap of 2, there shall also be infinitely many prime numbers of the form 2P + 1 with prime gap of P + 1 . Hence Sophie Germain conjecture stands proved and it can be called as Sophie Germain's prime theorem.

6.21. Polignac's prime theorem.

Theorem 5. For every natural number k, there are infinitely many primes p such that p + 2k is also prime.

Proof 4. As there shall be infinitely many twin primes with a prime gap of 2, there shall also be infinitely many prime numbers of the form P + 2k with prime gap of 2k . Hence Polignac's conjecture stands proved and it can be called as Polignac's prime theorem.

Bertrands prime theorem.

Theorem 6. There is at least one prime between N and 2N.

Proof 5. According to PNT N ln(N ) shall be number of primes up to N. As the result log N ln(N )

2N = log N ln(N ) 2 + log N ln(N )
N shall be always greater than 1, there shall be at least one prime between N and 2N. Hence Bertrands postulates stands proved and it can be called as Bertrands theorem. This proof is most elementary compared to Chebyshev, Paul Erdos's proof.

Legendre's prime theorem.

Theorem 7. There is always a prime number between n 2 and (n + 1) 2 provided that n ̸ = -1, 0. Proof 6. Let N be a arbitrarily large number. Sum of squares of all the natural numbers up to N shall be N (N +1)(2N +1)

6

. Double of the sum shall be N (N +1)(2N +1)

3

. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). We can visualise N ln(N ) as a prime number itself and we can allow the prime gaps to change equivalently and complete the number sequence. Now if we take logarithm of N (N +1)(2N +1) 3 with respect to the base of N ln(N ) the result shall give us the lower bound of powers that can comfortably be applied on that prime to reach double of the sum of squares of all the natural numbers up to N i.e. N (N +1)(2N +1)

3

. In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of Legendre's primes (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set of equations

P g ≤ log N ln(N ) N (N +1)(2N +1) 3 , P 1 + P g = P 2 , P log N ln(N ) N (N +1)(2N +1) 3 1 ≤ N (N +1)(2N +1) 3
. Similarly replacing sum of N 2 by sum of (N + 1) 2 we get

P g ≤ log N ln(N ) (N +1)(N +2)(2N +3) 3 , P 1 + P g = P 3 , P log N ln(N ) (N +1)(N +2)(2N +3) 3 1 ≤ (N +1)(N +2)(2N +3)

3

. If our resultant exponent is greater than 2 both the cases (ideally it should be greater than or equal to 2 as we have ensured all primes are summed up 2 times) then that will be greater than the lower bound of prime gaps 2 which eventually ensures continuity of the pattern up to infinity. Clearly due to infinitude of primes, both the results

log N ln(N ) N (N + 1)(2N + 1) 3 = log N ln(N ) N + log N ln(N ) (N + 1) + log N ln(N ) (2N + 1) -log N ln(N ) 3 and log N ln(N ) (N +1)(N +2)(2N +3) 3 = log N ln(N ) (N + 1)((N + 1) + 1)((2N + 1) + 2) = log N ln(N ) (N + 1)+log N ln(N ) ((N + 1 log N ln(N ) ((2N + 2) + 1))-log N ln(N )
3 are greater than 2.And due to complete pattern of extra little quantity of +1 another prime can occur in the interval meaning that the lower limit of number of primes in the interval between n 2 and (n + 1) 2 . So the limit of Legendre's primes would be greater than 1. Thus there shall be at least one prime between n 2 and (n + 1) 2 as Legendre conjectured. Hence Legendre's prime conjecture stands proved and it can be called as Legendre's theorem.

Landau's prime theorem.

Theorem 8. There are infinitely many prime numbers of the form N 2 + 1.

Proof 7. We know P 1 + P 2 = Even number followed from Goldbach theorem 2N + 1 = Odd Prime if N= Odd prime followed from Sophie Germain theorem Now lets consider this following statement:

(N + 1) 2 + 1 = N 2 + 2N + 1 + 1 = N 2 + 1 + 2N + 1 = Even number obviously if N= Even
For infinite number of cases 2N + 1 = N=Odd Sophie Prime in the above equation , N 2 + 1 must have infinite number of prime solution applying Goldbach theorem.

Proof 8. Let N be a arbitrarily large number. Sum of squares of all the natural numbers up to N shall be N (N +1)(2N +1)

6

. Double of the sum shall be N (N +1)(2N +1)

3

. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). We can visualise N ln(N ) as a prime with respect to the base of N ln(N ) the result shall give us the lower bound of prime powers that can comfortably be applied on that prime to reach double of the sum of squares of all the natural numbers up to N i.e. N (N +1)(2N +1)

3

. In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of Landau's primes (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set of equations

P g ≤ log N ln(N ) N (N +1)(2N +1) 3 , P 1 + P g = P 2 , P log N ln(N ) N (N +1)(2N +1) 3 1 ≤ N (N +1)(2N +1) 3
. If our resultant exponent is greater than 2 (ideally it should be greater than or equal to 2 as we have ensured all primes are summed up 2 times) then that will be greater than the lower bound of prime gaps 2 which eventually ensures continuity of the pattern up to infinity. Clearly due to infinitude of primes, the result log N ln(N )

N (N +1)(2N +1) 3 = log N ln(N ) N + log N ln(N ) (N + 1) + log N ln(N ) (2N + 1
) is greater than 2. As exponent 2 is always lesser than the exponent as calculated above, there shall be infinitely many prime numbers of the form N 2 + 1. Hence Landau's prime conjecture stands proved and it can be called as Landau's prime theorem.

6.25. Brocard's prime theorem. Theorem 9. With the exception of 4, there are always at least four primes between the square of a prime and the square of the next prime.

Proof 9. Let N be a arbitrarily large number. Sum of squares of all the natural numbers up to N shall be N (N +1)(2N +1)

6

. Double of the sum shall be N (N +1)(2N +1)

3

. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). We can visualise N ln(N ) as a prime number itself and we can allow the prime gaps to change equivalently and complete the number sequence. Now if we take logarithm of N (N +1)(2N +1) 3 with respect to the base of N ln(N ) the result shall give us the lower bound of prime powers that can comfortably be applied on that prime to reach double of the sum of squares of all the natural numbers up to N i.e. N (N +1)(2N +1)

3

. In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of primes between two successive squares (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set

of equations P g ≤ log N ln(N ) N (N +1)(2N +1) 3 , P 1 + P g = P 2 , P log N ln(N ) N (N +1)(2N +1) 3 1 ≤ N (N +1)(2N +1)

3

. If our resultant exponent is greater than 2 (ideally it should be greater than or equal to 2 as we have ensured all primes are summed up 2 times) then that will be greater than the lower bound of prime gaps 2 which eventually ensures continuity of the pattern up to infinity. Clearly due to infinitude of primes, the result log N ln(N )

N (N +1)(2N +1) 3 = log N ln(N ) N + log N ln(N ) (N + 1) + log N ln(N ) (2N + 1
) is greater than 2. In case of interval between two consecutive primes the above limit get raised to the power of its own value meaning that there shall be at least 4 primes between the square of a prime and the square of the next prime. Hence Brocard's prime conjecture stands proved and it can be called as Brocard's prime theorem.

6.26. Opperman's prime theorem.

Theorem 10. For every integer x > 1, there is at least one prime number between x(x -1) and x 2 ,and at least another prime between x 2 and x(x + 1) Proof 10. Let N be a arbitrarily large number. Sum of square of all the natural numbers up to N shall be N (N +1)(2N +1)

6

. Double of the sum shall be N (N +1)(2N +1)

3

. Sum of all the natural numbers up to N shall be N (1+N ) 2 . Double of the sum shall be N (1 + N ) which shall include double of sum of all the primes up to N too. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). We can visualise N ln(N ) as a prime number itself and we can allow the prime gaps to change equivalently and complete the number sequence. Subtracting N (1 + N ) from N (N +1)(2N +1) 3 we get N (N + 1). 2N -2 3 . Now if we take logarithm of N (N + 1). 2N -2 3 with respect to the base of N ln(N ) the result shall give us the lower bound of prime powers that can comfortably be applied on that prime to reach N (N + 1). 2N -2 3 . In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of primes between x 2 and x 2 -x (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set of equations

P g ≤ log N ln(N ) N (N +1)(2N -2) 3 , P 1 + P g = P 2 , P log N ln(N ) N (N +1)(2N -2) 3 1 ≤ N (N +1)(2N -2)

3

.

Clearly due to infinitude of primes, the result log N ln(N )

N (N + 1). 2N -2 3 = log N ln(N ) N + log N ln(N ) (1 + N ) + log N ln(N )
2N -2 3 shall be greater than 2 meaning that there shall be at least one prime between x(x -1) and

x 2 . Again adding N (1 + N ) with N (N +1)(2N +1) 3 we get N (N + 1). (2N +4)

3

. Now if we take logarithm of N (N + 1). (2N +4) 3 with respect to the base of N ln(N ) the result shall give us the lower bound of prime powers that can comfortably be applied on that prime to reach N (N +1). (2N +4)

3

. In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of primes between x 2 +x and x 2 (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set of equations P 1 ≤ N ln(N ) to reach N (N + 1). (2N +4) 3 satisfying the following set of equations P g ≤ log N ln(N )

N (N +1)(2N +4) 3

, P 1 + P g = P 3 , P

log N ln(N ) N (N +1)(2N +4) 3 1 ≤ N (N +1)(2N +4) 3
. Clearly due to infinitude of primes, the result log N ln(N )

N (N + 1). (2N +4) 3 = log N ln(N ) N + log N ln(N ) (1 + N ) + log N ln(N ) (2N +4) 3
shall be greater than 2 meaning that there shall be at least one prime between x 2 and x(x + 1). Altogether Opperman's conjecture stands proved and it can be called as Opperman's theorem. Theorem 12. The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially. That smallest i such that ai = 1 is called the total stopping time of n. The conjecture asserts that every n has a well-defined total stopping time. If, for some n, such an i doesn't exist, we say that n has infinite total stopping time and the conjecture is false. If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound.

Proof 12. Collatz conjectured operations on any number (i.e. halving the even numbers or simultaneously tripling and adding 1 to odd numbers) may blow up to infinity or come down to singularity or may get stuck in a loop in between. Tripling and adding 1 to odd numbers will always land on an even number. Now to end the game we just need to step upon an even number which is of the form 2 n . Will that happen always up to infinity when odd primes are tripled and added to 1? Lets call n as Collatz exponent.

We have seen that odd primes are kind of descendants of sole even prime 2 via zeta zeros which again appear infinitely in cycles of 3 divisions. Multiplying by 3 simply connects it back to the sequence up to infinity. This small bias turns the game of equal probability into one sided game. Let N be a arbitrarily large number. Sum of squares of all the natural numbers up to N shall be N (N +1)(2N +1)

6

. Double of the sum shall be N (N +1)(2N +1)

3

. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). We can visualise N ln(N ) as a prime number itself and we can allow the prime gaps to change equivalently and complete the number sequence. Now if we take logarithm of

N (N +1)(2N +1) 3
with respect to the base of N ln(N ) the result shall give us the lower bound of prime powers that can comfortably be applied on that prime to reach double of the sum of squares of all the natural numbers up to N i.e. N (N +1)(2N +1)

3

. In other words if we consolidate the average prime gaps into a relatively large prime having approximate value of P 1 ≤ N ln(N ) then that will lead us also to lower limit of the collatz exponent (as our push action shall have squeeze reaction on the prime gaps) which will satisfy the following set of equations P g ≤ log N ln(N ) N (N +1)(2N +1) 3

, P 1 + P g = P 2 , P

log N ln(N ) N (N +1)(2N +1) 3 1 ≤ N (N +1)(2N +1)
3 . If our resultant exponent is greater than 2 (ideally it should be greater than or equal to 2 as we have ensured all primes are summed up 2 times) then that will be greater than the lower bound of prime gaps 2 which eventually ensures continuity of the pattern up to infinity. Clearly due to infinitude of primes, the result log N ln(N )

N (N +1)(2N +1) 3 = log N ln(N ) N + log N ln(N ) (N + 1) + log N ln(N ) (2N + 1
) is greater than 2 . As resultant exponent is always greater than 2 Collatz conjecture neither blows up to infinity nor it get stuck in a loop, it always lands on an even number of the form 2 n and one last step before the final whistle bring it down to singularity 1 as Collatz conjectured. Hence Collatz conjecture stands proved and it can be called as Collatz prime theorem.

Fermat's last theorem.

Theorem 13. There cannot be any integer solution for n > 2 which satisfies x n + y n = z n Proof 13. Let there be some value n > 2 which satisfies x n + y n = z n . Case n = 2 is known since Greeks (Pythagoras) if not earlier since Indians (Boudhayana). Now to search for greater values where from the search should begin. Obviously it should be 3 because if it discontinues at 3 then we should stop the search. At zero x 3 + y 3 = z 3 should satisfy the special case x 3 + y 3 = (x + y) 3 which can also be written as x 3 +y 3 = (x+y)(x 2 +xy+y 2 ). Replacing (x 2 +y 2 ) = 1 will bring zeta function into the picture analogous to Euler's sum to product form involving primes in the form of x 3 + y 3 = (x + y)(1 + xy).The idea is factoring out this (1+ something) terms one side and take their products to relate it universal zeta zeros related to primes. Similarly every higher order Diophantine equations of this form can be equated to zeros of higher order polynomials. Expanding those polynomials by repeated Newton's binomial expansion we shall get terms like (x 2 + y 2 ) = 1. For example x 4 + y 4 = (x 2 + 2xy + y 2 ).(x 2 + 2xy + y 2 ), replacing (x 2 + y 2 ) = 1 which can be written x 4 + y 4 = (1 + 2xy).(1 + 2xy). Even higher order cases will not be an exception. Now this (1+ something) form is reserved for primes in Euler's sum to product form of zeta function. If we try to unify these terms arising from zeros of all higher order polynomials we fall into a serious conflict as the first term (1 + xy) itself suggesting xy cannot be prime. Therefore there cannot be any integer solution for any value of n > 2 which satisfies x n + y n = z n as Fermat claimed. Fermat's last theorem holds good. He was not bluffing when he said he had the proof after all his claims was found to be true. Andrew Wiles has given 500+ pages proof. Proving Fermat's last theorem in more elementary way is a subject of academic interest now. I am presenting my signature proof which not to be construed as an attempt to undermine Andrew Wiles's work. His work has given the world of mathematics something new called Taniyama-Shimura theorem ( I don't know what it is all about). It's always good to keep on adding new theorems as they always proves to be handy in attacking novel problems. Let z n be a arbitrarily large number. Let two numbers x > ln(z n ) and y > ln(z n ) also satisfies x n + y n = z n . Now to find the limit n = e ln (z n ) e ln (z n ) we can look into its zeros in the d-unit circle which satisfies 1 = 1 n .e ln (n) . The only value n = 2 satisfies such conditions, therefore there cannot be any solution for any value of n > 2 which satisfies x n + y n = z n as Fermat claimed. We can work out the same proof using square roots also. I guess Fermat's original motivation was using square roots. Proof 14. In the proof of Fermat's last theorem we already got hint that x, y, z > 2 cases solution to above Diophantine equation will involve some composite numbers. Now if Beal theorem is true it will obey the limit z xy = 1 2 , 2 or xy z = 2, 1 2 which can only be true when AB C = 2, 1 2 or C AB = 1 2 , 2 respectively inside the d-unit circle. In all the cases A, B, and C have a common prime factor which is 2. So along the real numberline A x + B y = C z will be true for x, y, z > 2 if A, B, and C have a common prime factor.

abc theorem.

Theorem 15. For every positive real number ε , there exist only finitely many triples (a, b, c) of coprime positive integers, with a + b = c, such that c > rad(abc) 1+ε .

Proof 15. If we observe mimutely we will see infinity of abc conjecture triples (a, b, c) would have been true if and only if a + b -c = 0 = abc. But that is impossible because abc=0 has not completed the cycle although a+b-c=0 has completed a semicircle. The proven zeros of d-unit circle make infinitude of triples (a, b, c) impossible because 1 + 1 = 2 ̸ = rad(1.1.2) 1+ε , hence abc conjecture stands proved. We can conclude that for every positive real number ε , there exist finitely many triples (a, b, c) of coprime positive integers, with a + b = c, such that c > rad(abc) 1+ε .

Part 5. On the Unification of Complex Numbers

Let us begin with a short Introduction of complex numbers

The history of complex numbers dates back to ancient times, with the Greeks and the Chinese investigating equations with negative numbers. However, the concept of a complex number as we know it today was not fully developed until the 16th century, when Italian mathematicians like Niccolo Fontana Tartaglia and Gerolamo Cardano used complex numbers to solve cubic equations. The term "imaginary" for these quantities was coined by René Descartes in 1637. Later, in the 18th century, Swiss mathematician Leonhard Euler introduced the symbol "i" to represent the imaginary unit, and contributed significantly to the development of complex analysis. A more abstract formalism for the complex numbers was further developed by the Irish mathematician William Rowan Hamilton, who extended this abstraction to the theory of quaternions a hypercomplex number that involve more than one imaginary unit. Hypercomplex Numbers introduce additional dimensions beyond the real and imaginary axes found in the complex plane. There are several types of hypercomplex numbers, including quaternions, octonions, sedenions and trigintaduonions. Today, complex numbers are used in a wide range of fields, including physics, engineering, and computer science.However, complex numbers come with certain challenges and problems.While complex numbers can be geometrically interpreted as points on a plane and quaternions can be related to rotations in three-dimensional space, octonions lack a straightforward geometric interpretation. The reduced ability to visualize and geometrically interpret Hyper Complex numbers can make them more abstract and difficult to work with. Unlike real numbers, Quaternions do not commute under multiplication. This means that the order in which Quaternions are multiplied can affect the result.Sedenions do not possess the property of associativity under multiplication. Sedenions contain zero divisors. Zero divisors are elements that multiply to zero but are themselves non-zero. This means that not all sedenions have a multiplicative inverse. This lack of commutativity, associativity can make computations and algebraic manipulations more challenging and less intuitive.

If all complex numbers could be turned into real numbers

If all complex numbers could be turned into real numbers, it would imply a significant reduction in the mathematical structure and properties of complex numbers.The imaginary unit "i" is an essential component of complex numbers. It is defined as the square root of -1 and plays a fundamental role in various mathematical and scientific applications. If complex numbers were reducible to reals, the concept of the imaginary unit would cease to exist. The complex plane, where complex numbers are represented as points with real and imaginary coordinates, would no longer have meaning. The complex plane provides a geometric interpretation of complex numbers and facilitates visualizations of their properties, such as modulus, argument, and transformations. Removing complex numbers would eliminate the need for a two-dimensional plane to represent numbers.Complex analysis is a branch of mathematics that deals with functions defined on complex numbers. It has numerous applications in physics, engineering, and other fields. If complex numbers were reduced to reals, complex analysis would not exist in its current form, and the insights and techniques developed in complex analysis would be lost.Complex numbers are instrumental in solving polynomial equations. The Fundamental Theorem of Algebra states that every non-constant polynomial equation with complex coefficients has at least one complex root. If complex numbers were reduced to reals, the ability to find solutions to certain equations would be significantly hindered.Complex numbers have deep connections to various branches of mathematics and physics. They are used extensively in fields like quantum mechanics, signal processing, electrical engineering, and control systems. If complex numbers were reduced to reals, these areas would experience a loss of the mathematical tools and frameworks that complex numbers provide. Although we cannot get rid of Complex Numbers completely, In the present article I will try or atleast lay down the foundation stone towards the same direction. I could do this with the help of the insights I got while attempting to prove Riemann Hypothesis. Although it is a monumental accomplishment in the world of mathematics, I know it will take a lot of time before it get accepted all across the globe. After extensive research and countless hours dedicated to this endeavor, I believe that I have uncovered a compelling proof that demonstrates the validity of the Riemann Hypothesis. This proof could potentially have significant implications for our understanding of prime numbers, the distribution of their zeros, and the broader implications for various branches of mathematics.Given the immense impact that the Riemann Hypothesis holds for number theory and its connections to other mathematical fields, I am keen to share my findings with the broader mathematical community. If any of my readers find merit in my proof and believe that it holds great importance for mathematics, I kindly seek their assistance in advancing its recognition within the mathematical community. Their endorsement in whatsoever manner you may deem fit would provide a significant additional layer of credibility to the proof and its implications.

Let The definition of imaginary number iota be natural logarithm of 2

Before we dive into the definition of imaginary number i let us recall Complex Numbers are composed of a real part and an imaginary part, they are written in the form x + iy. x denotes the real part and iy denotes the imaginary part. Complex numbers can be represented on an Argand Diagram. An Argand diagram is similar to the Cartesian coordinate system except that the Real axis and Imaginary axis replace the X and Y axis respectively which you would usually expect see on the Cartesian system. The

ℜ{z} ℑ{z} -3 -3i -2 -2i -1 -1i 1 1i 2 2i 3 3i 2 + 3i
hypercomplex numbers are a generalization of the complex numbers. They were created in an attempt to describe certain geometric operations in spaces with a dimension higher than the 2-dimensional plane. Operations on the complex numbers can be used to describe many of the geometric operations on the plane. For instance multiplication by a real number corresponds to a scaling of the plane.

Multiplication by complex numbers with a modulus of 1 corresponds to a rotation of the plane. Adding complex numbers corresponds to translation of the plane. To turn complex numbers into reals we need to define the imaginary number in such a uniqe way that the definition set brimgs back the missing order amongst the complex numbers. The unit circle will be the best place to start with.Of course there were circles of radius 1 since the ancient Greek geometers, but the term "unit circle" could not have been described until Fermat and Descartes invented coordinates and developed analytic geometry in the 1630s.Later, Euler parametrized it and also studied it using complex numbers.The unit circle provides an easy way to define the sine and cosine functions, since for an arbitrary angle in standard position (i.e. vertex at origin, initial side the positive x-axis), the terminal side cuts the unit circle at the point whose x-coordinate is cosine of the angle and whose y-coordinate is sine of the angle.In mathematics, the concept of ordering is closely related to the idea of a number line. Real numbers can be represented on a number line, where each point corresponds to a specific real number. On the number line, we can clearly see the order of real numbers: we can say that one real number is greater than another by its position on the number line.However, when we introduce complex numbers, things become more intricate. Complex numbers have both a real part and an imaginary part, and they cannot be represented on a simple number line. In the complex plane, the horizontal axis represents the real part of the complex number, and the vertical axis represents the imaginary part. Each complex number corresponds to a unique point in the complex plane.Unlike the real numbers on the number line, complex numbers do not have a natural ordering based on their positions in the complex plane.Since the real and imaginary parts of complex numbers can vary independently, it is not meaningful to say that one complex number is greater or less than another until we evaluate the complex number into reals.A direct one-to-one mapping of complex numbers onto the real number line can be possible if somehow we can define some real value for imaginary number i. In our proof of Riemann Hypothesis we saw that inside d-unit circle four dimensional mathematics turns the imaginary number into a completely real one. The very first real solution to imaginary number i is natural logarithm of 2. Therefore let The definition of imaginary number iota be natural logarithm of 2.

Conceiving The Simplex Numbers

Having set the definition of imaginary number iota equal to be natural logarithm of 2 we are set to embark for a new world of new number system. Although complex numbers are named so to actually mean a complex (something like housing complex) of real and imaginary numbers, frequently it is misunderstood to mean a number system with complexity. Indeed complex operations and complex analysis can involve increasing level of complexity as we introduce higher number of dimensions.Now in our newly discovered number system which will turn every complex number into a real one based on the definition of imaginary number iota equal to be natural logarithm of 2 is going to simplify the whole process of complex operations. It will be justified to call this new namber system to be Simplex Numbers. Similar to complex number a typical Simplex number can also be expressed as (a + bi), where "a" represents the decimal part and "bi" represents the complex part. The decimal part can take any real value, while the complex part combines the imaginary unit "i" to be finally evaluated as equal to the natural logarithm of 2 with a real coefficient.The Simplex number system is a novel mathematical construct that combines the properties of both decimal numbers and complex numbers. It offers a unique framework for representing and manipulating numerical quantities, blending the familiar decimal system with the rich algebraic structure of complex numbers. Real part can also have multiple parts but they need to be simplified according to BODMAS / PEMDAS rule before evaluating any Simplex number. In the Simplex number system, the place value are similar to the rules followed in the decimal system we commonly use with an exception that it starts with base 10 and the power increases in opposite direction i.e. left to right. The complex part has got different place value than the real part. Similar to Decimal numbers the real part represents the normal decimal value representing decimal numbers of base 10. However the complex part has got hundreds as place value multiplier i.e. second power of 10, and the next has got thousands i.e. third power of 10 as place value multiplier, and so on. The decimal part functions similarly to the decimal system we are accustomed to, where digits from 0 to 9 represent different place values. The complex parts, on the other hand, follows the principles of complex numbers, where the each imaginary unit is introduced to represent natural logarithm of 2 at its successive natural logarithms.just like complex numbers can be represented geometrically on the complex plane, Simplex numbers can be visualized on a modified complex plane. The real axis represents the decimal part, while the imaginary axis represents the complex part.The Simplex number system allows for decimal precision, similar to the decimal system. This means that decimal parts can be represented with as many decimal places as needed, providing flexibility for precise calculations and measurements.

The Simplex Operations

The Simplex number system inherits the algebraic structure of complex numbers. This means that Simplex numbers can be added, subtracted, multiplied, and divided using rules similar to those used for complex and decimal numbers combined.Now we need to set the rules for Simplex Operation which will turn complex numbers into real one.just like in the decimal system, arithmetic operations like addition, subtraction, multiplication, and division can be performed on Simplex numbers. For instance, addition and subtraction are carried out by separately operating on the decimal and complex parts. Addition of Simplex numbers is self explanatory and is done in the same way as we would add other algebraic terms. The same applies for subtraction.These operations incorporate both the decimal and complex aspects of Simplex numbers, providing a comprehensive toolset for mathematical analysis and problem-solving.The introduction of the Simplex number system opens up new avenues for exploring mathematical concepts and solving real-world problems. Its hybrid nature allows for the representation of quantities that possess both decimal and complex characteristics, finding potential applications in fields such as engineering, physics, and computer science.While the Simplex number system is a theoretical construct at this stage, its development showcases the ongoing quest to expand and enrich mathematical frameworks. As mathematics continues to evolve, the integration of diverse number systems may lead to exciting discoveries and novel insights into the nature of numbers and their applications.Multiplication and division in the Simplex number system involve utilizing the non-Commutative, non-Associative, non-Distributive properties of Simplex numbers. The product of two Simplex numbers (a + bi) and (c + di) is calculated by combining like terms, resulting in a new Simplex number. Division follows a similar procedure, where the division result is calculated by dividing like terms.

Example of Multiplication:

(3 + 4i)(-4 + 7i) = -12 + 28i (3 + 0i)(-4 + 7i) = -12 + 0i = -12
Example of Division:

4 + 7i 3 -2i = 4 3 + 7i -2 = 4 3 - 7i 2
The Simplex number system extends beyond basic arithmetic operations. Exponentiation, logarithms, and trigonometric functions can also be defined and applied to Simplex numbers. These operations consider both the decimal and complex aspects of the numbers, providing a comprehensive mathematical toolkit. Example of Exponentiation:

(3 + 4i) 2 (3 + 4i)(3 + 4i) = 9 + 16i
Example of Logarithm:

ln ((3 + 4i)(-4 + 7i))
= ln (-12) + ln 28.i = ln ( 1 12 ) + ln 28.i = -2.484 + 3.332i characteristics for example in electrical engineering, it could be used to model circuits that involve both decimal values (such as resistors) and complex values (such as capacitors or inductors). In finance, the Simplex number system might be utilized for analysing investments that have both real and complex components, such as options or derivatives. In physics, the Simplex number system might be utilized to study phenomena that involve both decimal measurements and complex quantities, such as oscillations or wave propagation.It's worth noting that the Simplex number system is a theoretical concept introduced here for illustrative purposes. Its practical implementation and development would require further consideration, refinement, and exploration by mathematicians, scientists, and researchers.The integration of the decimal and complex aspects in the Simplex number system represents a fascinating exploration of mathematical structures, showcasing the potential for new insights and applications in the realm of numbers and their properties.

12. Higher Dimensional Simplex Numbers Similar to how complex numbers extend beyond the realm of two-dimensional space, the Simplex number system can be further extended to higher dimensions. This opens up possibilities for applications in fields such as computer graphics, physics simulations, and signal processing.Quaternions are a fourdimensional extension of complex numbers and are represented as a + bi + cj + dk, where a, b, c, and d are real numbers, and i, j, and k are the quaternion units.To compute the product qp, we can use the quaternion multiplication rules:

i 2 = j 2 = k 2 = ijk = -1, ij = k, jk = i, ki = j, ji = -k, kj = -i, ik = -j.
We have just defined imaginary number i Now, let's multiply the quaternions under Simplex Operation:

(2

+ 3i + 4j + 5k)(-1 + 2i -3j + 6k) = -2 + 6i -12j + 30k
Similarly the Division result will be:

(2 + 3i + 4j + 5k) (-1 + 2i -3j + 6k) = -2 + 3 2 i - 4 3 j + 5 6 k
Hyperbolic quaternions are an extension of quaternions that incorporate hyperbolic numbers in each component.q = a + bi h + cj h + dk h is a hyperbolic quaternion when a, b, c, and d are real numbers, and i h , j h , and k h are the hyperbolic quaternion units has these products:

i h j h = k h = -j h i h , j h k h = i h = -k h j h , k h i h = j h = -i h k h and i 2 h = j 2 h = k 2 h = +1.
To turn the hyperbolic units into a real number we just need to to define i h = -i, j h = -j, k h = -k in Simplex Operation. Rest of the thing will remain the same. Now, let's multiply the hyperbolic quaternions under Simplex Operation:

(2 + 3i h + 4j h + 5k h )(-1 + 2i -3j + 6k) = -2 + 6i h -12j h + 30k h
Similarly the Division result will be:

(2 + 3i h + 4j h + 5k h ) (-1 + 2i -3j + 6k) = -2 + 3 2 i h - 4 3 j h + 5 6 k h
Now, let's multiply the 4 dimensional dual numbers under Simplex Operation:

(2 + 3ϵ 1 + 4ϵ 2 + 5ϵ 3 )(-1 + 2ϵ 1 -3ϵ 2 + 6ϵ 3 ) = -2 + 6ϵ 1 -12ϵ 2 + 30ϵ 3
Similarly the Division result will be:

(2 + 3ϵ 1 + 4ϵ 2 + 5ϵ 3 ) (-1 + 2ϵ 1 -3ϵ 2 + 6ϵ 3 ) = -2 + 3 2 ϵ 1 - 4 3 ϵ 2 + 5 6 ϵ 3

Evaluating The Simplex Numbers

Already things have started getting messy. There are more to go in the form of octonions(base-8), sedenions(base-16), trigintaduonions(base-32), babylonions(base-64)(I guess they are there and I am sure they do not violate any Laws of addition) , their Hyperbolic versions and their dual versions. Thankfully all these variants of hyper complex numbers will obey our definition for imaginary number i as we have chosen the unique one. We have seen some lack of inconsistencies above, to unify and simplify the notation ι c , ι h , ι d to denote complex number, perplex number, dual numbers respectively and use a subscript to denote the dimension. For example all the complex numbers up to 64 dimensions can be represented as

ι c1 , ι c2 , ι c3 , ι c4 , ι c5 , ι c6 , ι c7 , ι c8 , ι c9 , ι c10 , ι c11 , ι c12 , ι c13 , ι c14 , ι c15 , ι c16 , ι c17 , ι c18 , ι c19 , ι c20 , ι c21 , ι c22 ...
Similarly all the split complex/purplex/hyperbolic numbers upto 64 dimension can be represented as:

ι h1 , ι h2 , ι h3 , ι h4 , ι h5 , ι h6 , ι h7 , ι h8 , ι h9 , ι h10 , ι h11 , ι h12 , ι h13 , ι h14 , ι h15 , ι h16 , ι h17 , ι h18 , ι h19 , ι h20 , ι h21 , ι h22 ...
Similarly all the dual numbers numbers upto 64 dimension can be represented as:

ι d1 , ι d2 , ι d3 , ι d4 , ι d5 , ι d6 , ι d7 , ι d8 , ι d9 , ι d10 , ι d11 , ι d12 , ι d13 , ι d14 , ι d15 , ι d16 , ι d17 , ι d18 , ι d19 , ι d20 , ι d21 , ι d22 ...
However we should also allow/continue the practice if Greek letter iota is represented as English letter i. Let us now set the rules for evaluation of Simplex Numbers. To evaluate a complex number into a real number, we can multiply the imaginary part with the real part along with the pre-defined value of imaginary part.We have seen the imaginary number i can be defined to be equal to natural logarithm of 2, the imaginary hyperbolic numberi h can be defined to be equal to minus natural logarithm of 2 and finally the imaginary dual numberi d can be defined to be equal to minus natural logarithm of 2 squared. Now what shall be higher dimensional imaginary numbers for all these variants? Well to go for cycles in the context natural logarithm of 2 we can take repeated logarithm of 2, replacing minus 1 by 2 times of natural logarithm of 2 as and when they may appear in the iterative process. Now the question arise whether we shall consider the sign of the imaginary constants also? Answer shall be yes again as we are dealing with the power/indices plus sign means the Simplified imaginary constants (Simplex) part shall get multiplied with the real part and vice-versa.Did not got Why multiply? The default positive sign of complex numbers stands for positive power of exponents and which means multiplication and negative sign of complex numbers stands for negative power of exponents and which means division. But as we are flattening down the higher dimensions along the 1 dimensional number line we should also consider the sign it is bringing with it. Now lets discuss the scale factor for higher/lower dimensions to climb up and down. Why an extra scale factor? Because we use a number system based on powers of 10 for which there should be a default scale fator of 10 and some higher/lower multiples of 10 as we want to further ascend/descend to/from extra dimensions.What shall be the scale factor for imaginary number i . Starting with 10 for the real part, imaginary part i shall have a scale factor of 100. How come the decimal system be related to complex numbers,Wherefrom 100 is coming here? π 4 approximates to hundred. When higher dimensions are involved and pi is squared successively applying borrowing and carry over the decimal part, it can be rearranged to bring zeta function come into play and show that final sum of the decimal part equals to 1 as follows 1.10 n + 2.10 n-1 + 3.10 n-2 + 4.10 n-3 + 5.10 n-4 + ... 10 n when n tends to infinity out of infinite rotations among the unit circle, factoring out 10 n (1 + 1 2 + 1 3 + 1 4 + 1 5 + ...) 10 n finally giving us π 2 = 9 + ζ(1) = 9 + 1 = 10. Now Let us make a table of simplex constants of all the absolute real values for all the imaginary numbers up to 64 dimensions and its equivalent decimal scale factor. The scale factor chosen here is the natural one. However depending upon the application the scale factor may be adjusted to factors of 10 consistently.

SL. i c = ln(2), ln(ln(2))... i h = -ln(2), ln(-ln(2))... i d = -ln 2 (2),
i c ×(i h ) = -i c .i h similarly we can say i c ÷(i h ) = - i c i h
and for dual numbers no change in sign is required.

OR i h ×(i c ) = -i c .i h similarly we can say i h ÷(i c ) = - i h i c
and for dual numbers no change in sign is required.

Example of Cross Multiplication:

(

2 + 3i c1 + 4i c2 + 5i c3 )(2 + 3i h1 + 4i h2 + 5i h3 ) = 2.2 -3i c1 .3i h1 -4i c2 .4i h2 -5i c3 .5i h3 = 4 -9i c1 .i h1 -16i c2 .i h2 -25i c3 .i h3
Finally we are good to go for evaluation of Simplex numbers as follows:

= 4 -9i c1 -16i c2 -25i c3 = 4 × 16 × 0.3665129206 × 10 2 9 × 0.6931471806 × 25 × 0.3825728568 × 10 1 × 10 3 = 0.00393139907

Properties of Simplex logarithm

Thanks to Roger Cotes who first time used i in complex logarithm. Thanks to Euler who extended it to exponential function and tied i, pi and exponential function to unity in his famous formula. Now taking lead from both of their work and applying results of Zeta function and quaternion algebra we can define quaternion logarithm as follows. If q 1 = a 1 + ib 1 + jc 1 + kd 1 and q 2 = a 2 + ib 2 + jc 2 + kd 2 then simplified complex logarithm has the following property.

Theorem 16.

|ln (q 1 .q 2 )| = |ln (ℜ(q 1 )) + ln (ℜ(q 2 )) + i(ln (ℑ(q 1 )) + ln (ℑ(q 2 ))) ..| Proof 16.

|ln (q 1 .q 2 .q 3 .q 4 .q 5 .q 6 .q 7 .... = |ln (ℜ(q 1 )) + ln (ℜ(q 2 )) + +i(ln (ℑ(q 1 )) + ln (ℑ(q 2 ))+) + ...| Following Zeta functions analytic continuation, we can write :

|ln (q 1 .q 2 )| = |ln (ℜ(q 1 )) + ln (ℜ(q 2 )) + i(ln (ℑ(q 1 )) + ln (ℑ(q 2 ))) + ...| Corollary 1. ln ( q 1 q 2 ) = |ln (ℜ(q 1 )) -ln (ℜ(q 2 )) + i(ln (ℑ(q 1 )) -ln (ℑ(q 2 ))) + ...| Corollary 2.
e (q 1 +q 2 ) = e (ℜ(q 1 )) .e (ℜ(q 2 )) + i(e (ℑ(q 1 )) .e (ℑ(q 2 )) ) + j(e (ℑ(q 1 )) .e (ℑ(q 2 )) ) + ...

Corollary 3.

e (q 1 -q 2 ) = e (ℜ(q 1 )) e (ℜ(q 2 )) + i( e (ℑ(q 1 )) e (ℑ(q 2 )) ) + j( e (ℑ(q 1 )) e (ℑ(q 2 )) ) + ...

Corollary 4. |ln (q 1 + q 2 )| = |ln (ℜ(q 1 + q 2 )) + i(ln (ℑ(q 1 + q 2 ))) + ...| Corollary 5. |ln (q 1 -q 2 )| = |ln (ℜ(q 1 -q 2 )) + i(ln (ℑ(q 1 -q 2 ))) + ...| Theorem 17. If z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 then simplified Complex Logarithm has the following property. |ln (z 1 .z 2 )| = |ln (ℜ(z 1 )) + ln (ℜ(z 2 )) + i(ln (ℑ(z 1 ) + ln (ℑ(z 2 ))| Proof 17. |ln (z 1 .z 2 .z 3 .z 4 .z 5 .z 6 .z 7 ....)| = ln 1.2.3.4... + i ln 1.2.3.4... = ln (1) + ln (2) + ln (3) + ... + i ln ln (1) + ln (2) + ln (3) + ... = ln (ℜ(z 1 )) + ln (ℜ(z 2 )) + ln (ℜ(z 3 )) + ... + i ln (ℑ(z 1 )) + ln (ℑ(z 2 )) + ...
Following Zeta functions analytic continuation, we can write:

|ln (z 1 .z 2 )| = |ln (ℜ(z 1 )) + ln (ℜ(z 2 )) + i(ln (ℑ(z 1 ) + ln (ℑ(z 2 ))|
Example 3. Find the modulus of |ln ((5 + 13i).(12 + 17i))| using product to sum formula. And show that the result is same orders of magnitude that of actual product. In natural logarithmic scale both the values are of same orders of magnitude.

Corollary 6.

ln (

z 1 z 2 ) = ln (ℜ(z 1 )) -ln (ℜ(z 2 )) + i ln (ℑ(z 1 )) -ln (ℑ(z 2 ))
Corollary 7.

e (z 1 +z 2 ) = e (ℜ(z 1 )) .e (ℜ(z 2 )) + i e (ℑ(z 1 )) .e (ℑ(z 2 ))

Corollary 8.

e (z 1 -z 2 ) = e (ℜ(z 1 )) e (ℜ(z 2 )) + i e (ℑ(z 1 )) e (ℑ(z 2 ))

Corollary 9.

|ln (z 1 + z 2 )| = ln (ℜ(z 1 + z 2 )) + i ln (ℑ(z 1 + z 2 ))
Corollary 10.

|ln (z 1 -z 2 )| = ln (ℜ(z 1 -z 2 )) + i ln (ℑ(z 1 -z 2 ))
Corollary 11.

|ln (z 1 + z 2 )| = ln (Re(z 1 )) + ln (Re(z 2 )) + i ln (Im(z 1 )) + ln (Im(z 2 ))

Corollary 12.

|ln (z 1 -z 2 )| = ln (Re(z 1 )) -ln (Re(z 2 )) + i ln (Im(z 1 )) -ln (Im(z 2 ))

Corollary 13.

|z 1 + z 2 | = e ln (z 1 +z 2 )
Corollary 14. One thing to notice is that pi is intricately associated with e. We view pi mostly associated to circles, what it has to do with logarithm? Can it also be a base to complex logarithm? Although pi based logarithm are not common, but they can be handy in complex logarithm. We know: Theorem 18. There cannot be any odd perfect numbers.

ln(2). π 4 = 1 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + • • • 1 - 1 3 + 1 5 - 1 7 + 1 11 - 1 13 + • • • = 1 + 1 3 - 1 5 + 1 7 -• • • + 1 + 1 2 + 1 4 + 1 6 + • • • -1 + 1 2 + 1 4 + • • • = 1 - i 3 3 + i 5 5 - i 7 7 -• • • + 1 - i 2 2 + i 4 4 - i 6 6 + • • • - 1 1 -1 2 = sin (i) + cos (i) -2
Minimal Proof 1. We have seen that the d-unit circle has even paity via 1 = 1 2 .e ln (2) , hence there cannot be any odd perfect numbers.

Are there infinitely many perfect numbers? Theorem 19. There are infinitely many even perfect numbers.

Minimal Proof 2. We have seen that the d-unit circle has harmonicity via 1 = 1 2 .e ln (2) , hence there shall be infinitely many even perfect numbers of the form 2 p (2 p -1) 2 where p is Mersenne prime.

Do any odd weird numbers exist?

Theorem 20. There cannot be any odd weird numbers.

Minimal Proof 3. We have seen that the d-unit circle has even paity via 1 = 1 2 .e ln (2) . Cyclically speaking weird numbers are quater perfect numbers, they exist by way of completing a cycle of three among the perfect numbers, semi perfect numbers, quater perfect or so called weird numbers. Hence weird numbers are not so weird and there cannot be any odd quater perfect numbers or odd weird numbers.

q = (2 n-m + 1) × 2n -1, r = (2 n-m + 1) 2 × 2 m+n -1,
where n > m > 0 are integers and p, q, and r are prime numbers, then 2 n × p) × q and 2 n ) × r are a pair of amicable numbers. Putting m=n-1 we get

p = 3 × 2 n-1 -1, q = 3 × 2 n -1, r = 9 × 2 2n-1 -1,
where n > 1 is an integer and p, q, and r are prime numbers, then 2 n × p) × q and 2 n ) × r are a pair of amicable numbers. As apparent from the completed cycle of three in the above equation Eulers form have got harmonicity form similar to the harmonicity 1 = 1 2 .e ln (2) in the d-unit circle. Hence there shall be infinitely many amicable numbers.

Are there any pairs of amicable numbers which have opposite parity? Theorem 25. There are no amicable numbers having opposite parity.

Minimal Proof 8. We have seen that the d-unit circle has even paity via 1 = 1 2 .e ln (2) . As we are talking about pairs, opposite parity will not sum up to even parity, hence there cannot be any opposite parity amicable numbers.

Are there any pairs of relatively prime amicable numbers? Theorem 26. There are no pairs of relatively prime amicable numbers.

Minimal Proof 9. Relatively prime pairs are not simply connected to each other so they can not cyclically descend to zero. Hence there can be no pairs of relatively prime amicable numbers.

Are there infinitely many betrothed / quasi amicable numbers? Theorem 27. There are infinitely many quasi amicable numbers.

Minimal Proof 10. We have seen that the d-unit circle has harmonicity via 1 = 1 2 .e ln (2) , hence there shall be infinitely many quasi amicable numbers the same way there are infinitely many amicable numbers.

Are there any pairs of quasi amicable numbers which have same parity? Theorem 28. There are no pairs of quasi amicable numbers which have same parity.

Minimal Proof 11. We have seen that the d-unit circle has even paity via 1 = 1 2 .e ln (2) . As we are talking about quasi amicable pairs same parity will not sum up to even parity, hence there cannot be any same parity quasi amicable numbers.

Minimal Proof 18. Fibonacci numbers that have a prime index p do not share any common divisors greater than 1 with the preceding Fibonacci numbers, due to the identity:gcd(F n , F m ) = F gcd(n,m) which implies the infinitude of primes. For n ≥ 3, F n divides F m if n divides m. If we suppose that m is a prime number p, and n is less than p, then it is clear that F p , cannot share any common divisors with the preceding Fibonacci numbers.

gcd(F p , F n ) = F gcd(p,n) = F 1
The above form have inbuilt cycle of three.Therefore there shall be infinitely many Fibonacci primes.

Are there infinitely many Lucas primes?

Theorem 36. There are infinitely many Lucas primes.

Minimal Proof 19. Lucas numbers are related to Fibonacci numbers following

L n = F n-1 + F n+1 = F n + 2F n-1 = F n+2 -F n-2 .
As Fibonacci primes have inbuilt cycle of three and the above relationship just amplifies the cycle of three into a double cycle of three there shall be infinitely many Lucas primes.

Are there infinitely many Pell primes?

Theorem 37. There are infinitely many Pell primes.

Minimal Proof 20. Pells equation x 2 -2y 2 = ±1 =⇒ 2y 2 ± 1 -x 2 = 0 have inbuilt cycle of three hence there shall be infinitely many Pell primes.

Are there infinite Newman-Shanks-Williams primes? Theorem 38. There are infinitely many Newman-Shanks-Williams primes.

Minimal Proof 21. P = (1+ √ 2) 2m+1 +(1- √ 2) 2m+1 2 =⇒ 1 + √ 2 2m+1 + 1 - √ 2 2m+1 -2P = 0 have
inbuilt cycle of three hence there shall be infinitely many Newman-Shanks-Williams primes.

Are there infinitely many Wieferich primes?

Theorem 39. There are infinitely many Wieferich primes.

Minimal Proof 22. p 2 |(2 p-1 -1) =⇒ 2 p-1 -1 p 2
= n have inbuilt cycle of three hence there shall be infinitely many Wieferich primes.

Are there infinitely many Wilson primes?

Theorem 40. There are infinitely many Wilson primes.

Minimal Proof 23. p 2 |((p -1)! + 1) =⇒ (p-1)!+1 p 2 = n have inbuilt cycle of three hence there shall be infinitely many Wilson primes. not get bogged down. Keep yourself tightly tied to zero, find the cycle of non-linear linear combination of three's, make it linear by finding its zeros you will catch them all. Once you complete the cycles you will feel to be on top of the world. You will not find anybody nearby to share your feeling that's sad part of the story. We can have infinite cycles like this, does that mean i shall have infinite number of real solutions: Yes, that's true but this infinites must be very smaller compared to the infinites of the number line itself. Why, why it will be so: Because this infinities are sitting in the exponent of e via Euler's formula e ix . Even smaller values of i will be growing exponentially fast. For example with i = ln 2 and a relatively smaller value for x = 271 will give us a number close to the number of particles in the observable universe. For higher values of i it will be growing even faster. Is there any general formula to calculate the cycles: Yes there can be as many formulas as many mathematicians on the earth can work out their own, all road goes to cycles. For n number of rotations, dimensions, etc. whatever name we call it the cycle will be completing near e i. ∞ n=1 (1+π n ) . I have used pi as the rotator because it's natural, it can be anything else as well. Also we can take partial sum or a truncated partial sum or a single term anywhere from the series to predict some particular cyclic behaviour. In case of a single iteration we can use e n 2 . For more accuracy we need to take smaller and smaller values of the rotator. For n=3 in the above series formula, it will be accurate enough to predict the cycle of big bounces happening periodically in cosmos. What will be the math: Refer 6.16 for the math behind the first cycle. What is so special about the number e , they are there in biology in the form of average number of heartbeats in human life time, they are there in cosmology in the form of Black hole alignment, they are everywhere. What is the next special cycle number, do that also appear in nature somewhere: Its e 4π-3 2π(π-3) = e 33.78 = 4.68 × 10 14 . Yes, they also appear. They are there in physics Speed of light × Planck Constant Graviational constant × Mass of electron ≈ e 2+ 4π-3 2π(π-3) ≈ e

(T) t R = l R c -2 = ℏ m R c 2 -2 = ℏG c 5 -2 2.906554422 × 10 89 s Planck charge Electric charge (Q) q R = 4πε 0 ℏc -2 = e √ α -2 3.517672633 × 10 36 C Planck temperature Temperature (Θ) T R = m R c 2 k B -2 = ℏc 5 Gk 2 B -2
2.007279736 × 10 -64 K

) t R = l R c -4 = ℏ m R c 2 -4 = ℏG c 5 -4 8.448058605 × 10 178 s Planck charge Electric charge (Q) q R = 4πε 0 ℏc -4 = e √ α -4 1.237402075 × 10 73 C Planck temperature Temperature (Θ) T R = m R c 2 k B -4 = ℏc 5 Gk 2 B -4
4.029171939 × 10 -128 K Here it goes( Refer Table 3).

The next one should be one fourth by guess:

No, its one third, remember zeta zeros lies on pi by 3 ( Refer Table 4). But the scale is not remaining quantum any more: Its the scale of collective quantum behaviour. algebra very neatly for example ζ(-1) + ζ(0) = 0 or 2ζ(-1) = 1 . To prove that every Hodge class on X is a linear combination with rational coefficients of the cohomology classes of complex subvarieties we just need compliance with addition laws of algebra and scalar multiplication which zeta function duly complies beyond any doubt. Therefore every Hodge class on X is algebraic. No need to mention that Mumford-Tate group is the full symplectic group.

Can we have some hints for BSD conjecture dealing elliptic curves. :

Here is the hint with minimal proof of numerical solution to BSD conjecture.

Z E,Q (s) = ζ(s)ζ(s -1) L(s, E)
ζ(s) is the usual Riemann zeta function and L(s, E) is called the L-function of E/Q. Kolyvagin showed that a modular elliptic curve E for which L(E, 1) is not zero has rank 0, and a modular elliptic curve E for which L(E, 1) has a first-order zero at s = 1 has rank 1. Hasse-Weil zeta function fails to throw some light on the rank of the abelian group E(K) of points of E at s = 1 as ζ(1) was known to be undefined . In the light of my proof of Riemann hypothesis and its generalisations we can now evaluate the rank easily. We set Hasse-Weil zeta function in left hand side to -1 and evaluate the right hand side putting ζ(1) = 1 which then give the average rank 1 2 including zero valued ranks . Similarly we can take harmonic conjugate of Hasse-Weil zeta function as follows:

Z * E,Q (s) = ζ(s).L(s, E) ζ(s -1)
Now setting it to -1 and at s=0 putting ζ(-1) = 1 2 and ζ(-2) = -1 2 we get the analytic rank of elliptic curves E over Q with order s=1 L(E, s) > 1 which equals 1. Following Kolyvagin theorem the Birch and Swinnerton-Dyer conjecture holds for all elliptic curves E over Q with order s=1 L(E, s) > 1.Tate-Shafarevich group must be finite for all such elliptic curves. Can we have some hints for Navier-Stokes dealing turbulence in fluid mechanics. : Here is the hint with minimal proof of numerical solution to Navier-Stokes existence. The continuity equation reads:

∂ρ ∂t + ∂ (ρu x ) ∂x + ∂ (ρu y ) ∂y + ∂ (ρu z ) ∂z = 0.
When the flow is incompressible, ρ does not change for any fluid particle, and its material derivative vanishes and the continuity equation is reduced to:

∂u x ∂x + ∂u y ∂y + ∂u z ∂z = 0.
Now above equation will be true when we can show:

∇ = ∂ ∂x î + ∂ ∂y ĵ + ∂ ∂z k = 0
Can the cyclic numbers be projected further based on 2 values of i found so far, what shall be the rotators: Yes, I iterated the process taking different irrational number as rotators and the same trend line. I could project it up to e 33.94 . The table can be enlarged up to say 10 1000 . Who is going to take the pain of searching right rotators to build a decimal system based ladder kind of table? I leave it for computer programmers, software engineers. Refer Table 5. Continued up to infinity ... Table 6. Tabulated value of Grand unified scale in ascending order up to zero. However existence itself is the result of amplification of some implied quantities by way of squaring/ doubling it. The act of squaring gives the notion of one way time, however when we take logarithm, it becomes cyclic and the apparent straight line bends back to the origin. The universal entropy is always zero as Maxwell's Demon has automated the process of adjusting the entropy at universal level and it does not increase which is not a violation of second law of thermodynamics. CP violation occurs due to obey CPT symmetry which is 3 dimensional as nature. No there is no exception to the principal of causality. The only single possible past as well as the single possible future is zero or nothing. The present moment is physically distinct from the past and future as the notion of time is an emergent property of consciousness and a conscious mind neither can follow cycles nor it can comprehend the infinities in cycles. Time zero links the quantum arrow of time to the thermodynamic arrow.

Interpretation of quantum mechanics: How does the quantum description of reality, which includes elements such as the superposition of states and wave function collapse or quantum decoherence, give rise to the reality we perceive? Another way of stating this question regards the measurement problem: What constitutes a "measurement" which apparently causes the wave function to collapse into a definite state? Unlike classical physical processes, some quantum mechanical processes (such as quantum teleportation arising from quantum entanglement) cannot be simultaneously "local", "causal", and "real", but it is not obvious which of these properties must be sacrificed, or if an attempt to describe quantum mechanical processes in these senses is a category error such that a proper understanding of quantum mechanics would render the question meaningless. Can a multiverse resolve it?:

Wave function never collapses, we can restore wave pattern just by detecting the spin simultaneously in both of the detector in the famous double slit experiment. We will see the wave pattern comes back. The act of measurement do not play any role in a scalable experiment, still we need little bit approximations corresponding to desired level of accuracy due to uncertainties involved. Quantum entanglement is local and not subject to causality as the event itself is its cause. We can do an experiment to prove its locality. 3 lab situated in 3 cities anywhere in the globe will entangle pair of photons among themselves and make a triangle as such. Now a fourth lab will not be able to entangle any further photons with any of the three above. Which will prove entanglement is local also do not violate causality, but its real. Multiverse will not be required to explain entanglement.

Yang-Mills theory: Given an arbitrary compact gauge group, does a non-trivial quantum Yang-Mills theory with a finite mass gap exist? : Yes it exists. See the Math chapter.

Color confinement: Quantum chromodynamics (QCD) color confinement conjecture is that color charged particles (such as quarks and gluons) cannot be separated from their parent hadron without producing new hadrons. There is not yet an analytic proof of color confinement in any non-abelian gauge theory.: Yes, its true if they are combination of 3 only. Mesons although hadrons are exceptions to this rule. Physical information: Are there physical phenomena, such as wave function collapse or black holes that irrevocably destroy information about their prior states? How is quantum information stored as a state of a quantum system?:

Information cannot be destroyed, they may undergo physical changes cyclically but they reappear elsewhere coming in contact with its conjugal pair. Dimensionless physical constant: At the present time, the values of the dimensionless physical constants cannot be calculated; they are determined only by physical measurement. What is the minimum number of dimensionless physical constants from which all other dimensionless physical constants can be derived? Are dimensional physical constants necessary at all?:

For each cycle there can be different dimensionless physical constants. Even in the planck cycle alpha is not enough, we need to half it to describe the cyclicity among physical events. Dimensional physical constants are not necessary at all to describe reality in simple terms as nature can play with its infinity card, to honour the mathematics it cycles through one single constant which is natural logarithm of 2.Yet, when it comes to detailing, physical constants are required to depict the classical universe.

Fine-tuned universe: The values of the fundamental physical constants are in a narrow range necessary to support carbon-based life. Is this because there exist other universes with different constants, or are our universe's constants the result of chance, or some other factor or process? In particular, Tegmark's mathematical multiverse hypothesis of abstract mathematical parallel universe formalized models, and the landscape multiverse hypothesis of spacetime regions having different formalized sets of laws and physical constants from that of the surrounding space -require formalization.:

The apparent fine tuning is due to renormalising infinities arising out of math to some finite result to match our experiment or observation. Doing so we miss the cycles of physical events and we settle with a narrow bound which is not the reality. The reality is everywhere its infinite and within that infinity rising of carbon based life forms is not at all a miracle. Zeta results and numerical relativity coming out of zeta results directly proves all the four levels of Tegmark's mathematical multiverse hypothesis. We need to formally accept it. Quantum field theory: Is it possible to construct, in the mathematically rigorous framework of algebraic QFT, a theory in 4-dimensional spacetime that includes interactions and does not resort to perturbative methods?: Yes, once we bring cycles in our math and theories our math for algebraic QFT will not fail.

Cosmology and general relativity.

Short answer to unsolved problems in physics 2. Problem of time: In quantum mechanics time is a classical background parameter and the flow of time is universal and absolute.

In general relativity time is one component of four-dimensional spacetime, and the flow of time changes depending on the curvature of spacetime and the spacetime trajectory of the observer. How can these two concepts of time be reconciled?:

They can be reconciled at T=0. There from local quantum time line can begin giving the impression that local time is flowing. But after completion of a half time cycle the local time will start flow backward through cyclic changes and that will cause Big Bounce which will again go back to T=0 meaning "Time does not exist".

Cosmic inflation: Is the theory of cosmic inflation in the very early universe correct, and, if so, what are the details of this epoch? What is the hypothetical inflaton scalar field that gave rise to this cosmic inflation? If inflation happened at one point, is it self-sustaining through inflation of quantum-mechanical fluctuations, and thus ongoing in some extremely distant place?: Yes, its mathematically plausible but cyclic maths do not allow it so fast as Alan guth brought it to fix the Big Bang model. Once it happens, always it happens, therefore eternal inflation which is relatively slow across the various parts of the universe is more correct picture.

Horizon problem: Why is the distant universe so homogeneous when the Big Bang theory seems to predict larger measurable anisotropies of the night sky than those observed? Cosmological inflation is generally accepted as the solution, but are other possible explanations such as a variable speed of light more appropriate?:

The local anisotropies gets exponentially smoothened out by eternal inflation so the homogeneousness. Variable speed of light has greater domain to work. Einstein realised that in a rotating frame of reference speed of light may not remain constant following Mach principal, he took the relativity route instead to keep the speed of light constant. His choice was wise. Now we just need a scale factor to scale up or scale down while moving between two rotating frame of reference. Overall speed of light remain constant.

Origin and future of the universe: How did the conditions for anything to exist arise? Is the universe heading towards a Big Freeze, a Big Rip, a Big Crunch, or a Big Bounce? Or is it part of an infinitely recurring cyclic model?:

Universe existed and it will exist forever, Big bounce keep on happening cyclically across the universe meaning every structures that forms in an universe has got an age but the universe itself is timeless.

Size of universe: The diameter of the observable universe is about 93 billion light-years, but what is the size of the whole universe?: Its truly infinite even bigger than 10 500 multiverses as string theories suggest. Yet It can be unified into an eternal omniverse having zero entropy in total thermal equilibrium which can be seen as one boltzman brain where every event whether possible or impossible can happen over infinite time cycles. Baryon asymmetry: Why is there far more matter than antimatter in the observable universe?:

Because antimatters are moving backward in time through cycle of changes, Its like one particle universe.

Cosmological constant problem: Why does the zero-point energy of the vacuum not cause a large cosmological constant? What cancels it out?: Because that energy runs infinite time cycles, infinite multiverses. Ln(2) is the cosmological constant and its huge enough to bind infinite multiverses and run the show for infinite time cycles. Dark matter: What is the identity of dark matter? Is it a particle? Is it the lightest superpartner (LSP)? Or, do the phenomena attributed to dark matter point not to some form of matter but actually to an extension of gravity?: Failed combinations like Quaterquarks, Pentaquarks, or moderately successful Hexaquarks, Dodecaquarks and so on all acting like BEC, not interacting with enough strength to get amplified enough and get detected by our telescopes, sounds fit as dark matter candidates.

Dark energy: What is the cause of the observed accelerated expansion (de Sitter phase) of the universe? Why is the energy density of the dark energy component of the same magnitude as the density of matter at present when the two evolve quite differently over time; could it be simply that we are observing at exactly the right time? Is dark energy a pure cosmological constant or are models of quintessence such as phantom energy applicable?: Observed accelerated expansion (de Sitter phase) of the universe is due to non linear stages or phases or epochs in cycle of changes. The energy density of the dark energy component remain always of the same magnitude as of the density of matter through cycle of changes. There is nothing called right time as time itself does not exist. Dark energy apart from being a cosmological constant is also a mathematical constant having value of natural logarithm of 2. There can be infinite cycles, so the universal cosmological constant may also be infinitely small as we keep on dividing spaces into infinite partitions.

Dark flow: Is a non-spherically symmetric gravitational pull from outside the observable universe responsible for some of the observed motion of large objects such as galactic clusters in the universe?:

No, The universe has got zero gaussian curvature meaning at large scale space-time is flat but at smaller scales it can have both positive or negative curvature. Due to some local non zero curvature observable universe may feel an outside pull. But if we are able to sum up the curvatures meticulously then we will see change of cycles are finally making everything flat. Axis of evil: Some large features of the microwave sky at distances of over 13 billion light years appear to be aligned with both the motion and orientation of the solar system. Is this due to systematic errors in processing, contamination of results by local effects, or an unexplained violation of the Copernican principle?: No, this is not an error, actually this is so. The first cyclic number is 4.33 billion. Three cycles of that comes around 13 billion. So beyond 13 billion light years its more or less same cycle so the same synchronised motion, same orientation.

Shape of the universe: What is the 3-manifold of comoving space, i.e. of a comoving spatial section of the universe, informally called the "shape" of the universe? Neither the curvature nor the topology is presently known, though the curvature is known to be "close" to zero on observable scales. The cosmic inflation hypothesis suggests that the shape of the universe may be unmeasurable, but, since 2003, Jean-Pierre Luminet, et al., and other groups have suggested that the shape of the universe may be the Poincaré dodecahedral space. Is the shape unmeasurable; the Poincaré space; or another 3-manifold?: 3-manifold of comoving space is spherical. As we take snapshots over different time we cannot follow the time cycles so it appears as a flat like a straight line. If we project it further to infinities it will give us zero curvature meaning rays of straight lines are coming out from the origin in all directions and it can have all possible shapes over infinite time.

The largest structures in the universe are larger than expected. Current cosmological models say there should be very little structure on scales larger than a few hundred million light years across, due to the expansion of the universe trumping the effect of gravity. But the Sloan Great Wall is 1.38 billion light-years in length. And the largest structure currently known, the Hercules-Corona Borealis Great Wall, is up to 10 billion light-years in length. Are these actual structures or random density fluctuations? If they are real structures, they contradict the 'End of Greatness' hypothesis which asserts that at a scale of 300 million light-years structures seen in smaller surveys are randomized to the extent that the smooth distribution of the universe is visually apparent.: Again this is due to scale gap. Why we are settling with one single bang. Series of bangs across the universe have bounded the smaller structures into a greater structure. Random density fluctuations although not necessary but allowed once to explain the first bang, thereafter it does not remain random, there is a frequent pattern even among the random numbers such as primes. Due to that pattern small series of bangs keep on happening every now and then. There is no end of greatness yet the hypothesis is true as beyond every 300 million light years the the randomness just differs in plus/minus sign which finally smoothens out to zero in visuals. From an infinitely small patch of sky to the whole omniverse everything is connected to every other parts through cycles and zeros.

Quantum gravity.

Short answer to unsolved problems in physics 3. Vacuum catastrophe: Why does the predicted mass of the quantum vacuum have little effect on the expansion of the universe?:

Because the predicted mass is expanding another unit of space outside right now, still it won't tear apart the universe as universe has no boundary. Quantum gravity: Can quantum mechanics and general relativity be realized as a fully consistent theory (perhaps as a quantum field theory)? Is spacetime fundamentally continuous or discrete? Would a consistent theory involve a force mediated by a hypothetical graviton, or be a product of a discrete structure of spacetime itself (as in loop quantum gravity)? Are there deviations from the predictions of general relativity at very small or very large scales or in other extreme circumstances that flow from a quantum gravity theory?: No , We should not mix Gravity with QM, we can unify quantum fields to make a QFT but it will not mix with GR except at singularities. Even if we try it will give us a zero everywhere, what we shall do with zeros. I fear we need to carry both, to make sense of everything. Space time is continuous by means of cycles of discrete quanta. We do not need Graviton as Gravity is not real force, its curvature of space time. GR is a generalised concept, if there is a requirement for scale adjustment we are allowed to do so in GR and the suggested scale factors in this paper will bridge the gap between GR and QM.

Black holes, black hole information paradox, and black hole radiation: Do black holes produce thermal radiation, as expected on theoretical grounds? Does this radiation contain information about their inner structure, as suggested by gauge-gravity duality, or not, as implied by Hawking's original calculation? If not, and black holes can evaporate away, what happens to the information stored in them (since quantum mechanics does not provide for the destruction of information)? Or does the radiation stop at some point leaving black hole remnants? Is there another way to probe their internal structure somehow, if such a structure even exists?: Yes, Black holes exists but the singularity arises not only due to infinite gravity but also may be due to zero gravity at least at the centre of the black hole it must be zero. Black holes do not consume information. Whatever information falls inside the event horizon are radiated back in scrambled form over a huge scale of time by Hawking Radiation. Those scrambled information from all the black holes can get assembled into some useful information again and again. A size zero black hole remnants can always exist and there will be no physics inside unless we donate some physics to it. We wont spent any of our scarce physics time (worlds best minds already spent three decades) as we know there are no more hairs inside, why should we make it hairy again? Extra dimensions: Does nature have more than four spacetime dimensions? If so, what is their size? Are dimensions a fundamental property of the universe or an emergent result of other physical laws? Can we experimentally observe evidence of higher spatial dimensions?: Ultimate reality is nature have zero dimensions. Nature does not care whether somebody somewhere is drawing a line through it and then doing whatever other geometries through it, finally all gets erased by somebody else and the cycle continues. Another reality is that nature is 3 dimensional through cycles of threes. Four space-time dimensions represents cycles of changes which connects both of this reality. Now there can be infinite number of cycles, then why it should be 10+1 dimensional alone as string theories require. Those extra dimensions if seen closely are nothing but extra degrees of freedom that we can borrow from zero to return it back elsewhere. So far our observation suggest space-time is flat and the curvature is experimentally observed at shorter distances does not suggest presence of any extra dimensions. If that was the case we could have lost all the starlight in those extra dimensions coming towards us from a distant star from edge of the universe in contrary to the results found in famous Michelson-Morley experiment. If still we insist that we will be calling those extra degrees of freedom as extra dimensions only because we are accustomed to do so then I must say that there are infinite number of extra dimensions none of which can be detected because of their cyclic orientation.

The cosmic censorship hypothesis and the chronology protection conjecture: Can singularities not hidden behind an event horizon, known as "naked singularities", arise from realistic initial conditions, or is it possible to prove some version of the "cosmic censorship hypothesis" of Roger Penrose which proposes that this is impossible? Similarly, will the closed timelike curves which arise in some solutions to the equations of general relativity (and which imply the possibility of backwards time travel) be ruled out by a theory of quantum gravity which unites general relativity with quantum mechanics, as suggested by the "chronology protection conjecture" of Stephen Hawking?: GR and QM cannot be unified except inside singularities. Singularities arising from experimental realistic initial conditions will not be scalable enough compared to the natural one therefore the hypothesis cannot be tested, yet it must be true as conceptually we can see naked/clothed singularities as an absolute zero. Possibility of backwards time travel can be ruled out. Chronology protection conjecture of Stephen Hawking will not be violated as within the cycles and across the cycles causality is not violated. Quantum entanglement is a local phenomena which will not pass the rectangle of entanglement test. A,B,C,D four labs anywhere in the globe cannot entangle a pair of particles among themselves and complete the rectangle or a square as a special case, although the same experiment will be successful in case of a triangle. Absence of Non-local phenomena implies nature do not violate causality.

High-energy physics/particle physics.

Short answer to unsolved problems in physics 4. Hierarchy problem: Why is gravity such a weak force? It becomes strong for particles only at the Planck scale, around 1019 GeV, much above the electroweak scale (100 GeV, the energy scale dominating physics at low energies). Why are these scales so different from each other? What prevents quantities at the electroweak scale, such as the Higgs boson mass, from getting quantum corrections on the order of the Planck scale? Is the solution supersymmetry, extra dimensions, or just anthropic fine-tuning?: Gravity is not a force, its curvature of space-time. Its weak because it is shared by all the infinite number

  So another equation of the points lying on unit circle |z| = 1 2 e ix = 1. Although both the equation are of unit circle, usefulness of |z| = 1 2 e ix = 1 is greater than |z| = |e ix | = 1 as |z| = 1 2 e ix = 1 bifurcates mathematical singularity 21 of 95 and introduces unavoidable mathematical duality particularly in studies of primes and zeta function. |z| = 1 2 e ix = 1 can be regarded as d-unit circle.

6. 30 .

 30 Beal theorem. Theorem 14. If A x + B y = C z where A, B, C, x, y, and z are positive integers with x, y, z > 2 then A, B, and C have a common prime factor.

  )| = |ln (ℜ(1.2.3.4.5.6.7...)) + i ln (ℑ(1.2.3.4.5.6.7...)) + ...| = |ln (1) + ln (2) + ln (3) + ... + i ln (ln (1) + ln (2) + ...) + ...|

  |ln ((5 + 13i).(12 + 17i))| = |ln (5) + ln (13)|+|ln (12) + ln (17)| = 3.02(approx+3.76(approx) = 6.79(approx) |ln ((5 + 13i).(12 + 17i))| = |ln (5) + ln (12) + i(ln (13 + ln (17)| = 6.77(approx)

|z 1 -

 1 z 2 | = e ln (z 1 -z 2 ) Corollary 15. |ln (z 1 .z 2 )| = |ln z 1 + ln z 2 | z 1 -ln z 2 | Corollary 17. |ln (z)| = ln (Re(z)) + i ln (Im(z)) 15. Pi based logarithm

=

  Lets set:π = sin (i) + cos (i) and replacing π -2 = ln (π) we can write ln e π π -ie we can write π -ie = -1 Part 6. The Minimal proofsLet us generalise the patternAre there any odd perfect numbers?

  = e22.19 = 4, 324, 402, 934 do that appear in nature somewhere: Yes, they are special. They are there in physics 2 × Mass of electron × Speed of light squared × Charles ideal gas constant Boltzmann constant ≈ e π (π-3)

  Locality: Are there non-local phenomena in quantum physics? If they exist, are non-local phenomena limited to the entanglement revealed in the violations of the Bell inequalities, or can information and conserved quantities also move in a non-local way? Under what circumstances are non-local phenomena observed? What does the existence or absence of non-local phenomena imply about the fundamental structure of spacetime? How does this elucidate the proper interpretation of the fundamental nature of quantum physics?:

  Proof 11. According to PNT we know that there shall be N ln(N ) number of primes up to N with an average prime gap of ln(N ). The same is also true in case of primes i.e. there shall be P ln(P ) number of primes up to P with an average prime gap of ln(P ). Now let us consider the following logarithmic expression.

	log P ln(P )	(P + 2) < log P ln(P )	P +	1 P	log P ln(P )	P
	Which we can show by the following steps
	log P ln(P )	(P + 2) < log P ln(P )	P +	1 P	log P ln(P )	P
	=⇒ log P ln(P )	(P + 2) < log P ln(P )	P + log P ln(P )	P	1 P
	=⇒ log P ln(P )	(P + 2) < log P ln(P )	P.P	1 P
	=⇒ log P ln(P )	(P + 2) < log P ln(P )	P 1+ 1 P
	=⇒ log Pn ln(Pn)	(P + 2) < log Pn ln(Pn)	P	1+ 1 Pn n
	P n → ∞ =⇒ P n+1 <P	1+ 1 n
	6.27. Firozbakht's prime theorem.
	1
	Theorem 11. p n+1 √ p n+1 < n √ p n Equivalently: p n+1 < p n n (where p n is the nth prime) is a strictly decreasing function of n, i.e., 1+ 1 n n for all n ≥ 1

n Hence Firozbakht's prime conjecture stands proved and it can be called as Firozbakht's prime theorem. 6.28. Let us prove Collatz conjecture.

Table 1 :

 1 ln(-ln 2 (2))... SL. i c = ln(2), ln(ln(2))... i h = -ln(2), ln(-ln(2))... i d = -ln 2 (2), ln(-ln 2 (2))... SL. i c = ln(2), ln(ln(2))... i h = -ln(2), ln(-ln(2))... i d = -ln 2 (2), ln(-ln 2 (2))... Real value for imaginary part & Scale Factors So far we have not done anything towards unification. Lets start the unification process. Complex numberi c , Dual numberi d and hyperbolic numberi h can be unified with the following relationship in simplex operation:

	SF

  35.72 ≈ 3.27 × 10 15 , they are there in biology in the form of average number of cell divisions in human life time, they are there in cosmology in the form of Large number hypothesis, Multiverse they are everywhere. But Large number hypothesis if of 10 43 orders of magnitude: Its 3 cycles again e 100 ≈ 10 43 .

	Name	Dimension	Expression		Value (SI unit)
	Planck length	Length (L)	l R =	ℏG c 3	-2	2.612280807 × 10 70 m
	Planck mass	Mass (M)	m R =	ℏc G	-2	4.736869309 × 10 16 kg
	Planck time	Time

Table 2 .

 2 Tabulated value of squared Planck UnitsIs there a third: Here goes the third quantisation from the Schrödinger equation ( Refer Table2).

		Ĥ |Ψ⟩ -E|Ψ⟩ = 0 =⇒	Ĥ |Ψ⟩ E|Ψ⟩	= 1
	Name	Dimension	Expression			Value (SI unit)
	Planck length	Length (L)	l R =	ℏG c 3	-4	6.824007974 × 10 140 m
	Planck mass	Mass (M)	m R =	ℏc G		-4	2.243793085 × 10 33 kg
	Planck time	Time (T

Table 3 .

 3 Tabulated value of double squared Planck Units

	Is this scale cosmic:
	Yes, that's true.
	But how the physics equations will balance, it cannot be one sided change, what should
	the quantum relativistic version of it:

  11.78 e 5.89 e 3.93 e 2.94 e 2.36 e 1.96 -1 e 11.81 e 5.90 e 3.94 e 2.95 e 2.36 e 1.96 22.19 e 11.09 e 7.4 e 5.55 e 4.44 e 3.7 33.78 e 16.89 e 11.26 e 8.45 e 6.76 e 5.63 + ϕ 2 e 33.94 e 16.97 e 11.31 e 8.48 e 6.78 e 5.65

	SL	Formula	i/j	g 1	g 2	g 3	g 4	g 5	g 6
	1	e	iπ gp	ln (2)	e 2.18 e 1.09 e 0.73 e 0.54 e 0.44 e 0.36
	2	π	je gp	1 ln (2)	e 4.49 e 2.24	e 1.5 e 1.12 e 0.9 e 0.75
	3	π	je gp iπ	e e-2
	4 π + ϕ 5 e gp iπ gp 1 e π-3 6 π je gp e 2	e 22.99 e 11.50 e 7.66 e 5.75 e 4.60 e 3.83
	7	e	iπ gp	4π-3 2π(π-3)
	8	π	je gp	e 2

e e e

Example of Trignometric functions: sin((3 + 4i)(-4 + 7i)) = sin (-12 + 28i) = sin(-12) cosh(28) + cos(28) sinh (-12)i = 388011186582.81 -78334.3584923298i Hyperbolic complex numbers are a two-dimensional extension of the real numbers that include the imaginary unit j, where j 2 = 1. The change of sign distinguishes the Hyperbolic-complex numbers from the ordinary complex ones.As we are in the process of unification, notation j may be confusing with higher dimensional quaternion number system. For better clarity, henceforth we shall use i h for hyperbolic complex numbers.To turn the hyperbolic unit i h into a real number we just need to define i h = -i = -ln(2) just before evaluation in Simplex Operation. Rest of the thing will remain the same. Example of Multiplication:

Dual numbers are a two-dimensional extension of the real numbers that include the imaginary unit ϵ ̸ = 0, where ϵ 2 = 0,. To turn the dual unit ϵ ̸ = 0 into a real number we just need to define ϵ ̸ = 0 = ln(2).ln(-2) equal to the product of imaginary number i and hyperbolic number j in Simplex Operation. Rest of the thing will remain the same. Example of Multiplication:

Example of Division:

The Simplex number system has the potential to find applications in various fields. For instance, in engineering, it could be useful in modelling and analysing systems that possess both real and complex

Do any Lychrel numbers exist?

Theorem 21. There are no Lychrel numbers.

Minimal Proof 4. A Lychrel number is a natural number that cannot form a palindrome through the iterative process of repeatedly reversing its digits and adding the resulting numbers. The process of repeatedly reversing digits of any number and adding the resulting digits to get another number is equivalent of cyclic harmonisation along the d-unit circle ultimatte which sends the number back to its origin.As all the numbers have their additive identity there cannot exist any Lychrel numbers. Candidate Lychrel numbers will form palendrom after a large number of iteration. For example the reverse-andadd process on196 will form palendrom somewhere in the range of e 196 = 10 85.12171845 following the harmonics of the identity 1 = 1

196 .e ln (196) .

Is 10 a solitary number?

Theorem 22. 10 has no friendly partner.

Minimal Proof 5. In number theory, friendly numbers are two or more natural numbers with a common abundancy index, the ratio between the sum of divisors of a number and the number itself (hereinafter called the friendly ratio). As 10 is not a prime ideally speaking it should have a friendly partner. But 10 is a harmonic number resulting its friendly ratio also to be harmonic. Any candidate friendly partner of 10 are always some harmonics away so they never come close enough to become real friends. Therefore 10 has no friendly partner.

Do any Taxicab(5, 2, n) exist for n > 1?

Theorem 23. There cannot be any Taxicab(5, 2, n) number for n > 1.

Minimal Proof 6. In mathematics, the generalized taxicab number Taxicab(k, j, n) is the smallest number which can be expressed as the sum of j kth positive powers in n different ways. For k = 3 and j = 2, they coincide with taxicab numbers.

Taxicab(3, 2, 2) = 1729 = 1 3 + 12 3 = 9 3 + 10 3 -famously stated by Ramanujan. Are there infinitely many amicable numbers?

Theorem 24. There are infinitely many amicable numbers.

Minimal Proof 7. Euler's rule for amicable numbers states that if

Are there infinitely many Wagstaff primes?

Theorem 29. There are infinitely many Wagstaff primes.

Minimal Proof 12. P = 2 q +1 3 =⇒ 3P = 2 q + 1 = N 2 + 1 have inbuilt cycle of three where q is an odd prime. Therefore there shall be infinitely many Wagstaff primes.

Are there infinitely many Pierpont primes? Theorem 30. There are infinitely many Pierpont primes.

Minimal Proof 13. For nonnegative integers u and v 2 u 3 v + 1 will generate infinitely many Pierpont primes as the form have inbuilt cycle of three in the form of 3 v .

Are there infinitely many Cullen primes?

Theorem 31. There are infinitely many Cullen primes.

For n = 1 3 the form have inbuilt cycle of three.Therefore there shall be infinitely many Cullen primes.

Are there infinitely many Woodall primes?

Theorem 32. There are infinitely many Woodall primes.

For n = 1 3 the form have inbuilt cycle of three.Therefore there shall be infinitely many Woodall primes.

Are there infinitely many Carol primes?

Theorem 33. There are infinitely many Carol primes.

three in the form of 2-1-1=0.Therefore there shall be infinitely many Carol primes.

Are there infinitely many Kynea primes?

Theorem 34. There are infinitely many Kynea primes.

three in the form of 2-1-1=0.Therefore there shall be infinitely many Carol primes.

Are there infinitely many Fibonacci primes?

Theorem 35. There are infinitely many Fibonacci primes?

Part 7. The Conclusion

Let us draw the conclusion

Riemann hypothesis stands proved in different ways primarily involving the concept of duality in dunit circle, the concept of harmonic conjugation in complex analysis, the concept quaternions in 4dimensional number system. It has given a lot of new mathematics to research further.

Part 8. The question answer sessions

The Doubt Clearence sessions Frequently appeared Questions and Answers 1. Here goes list of some frequently appeared questions related to Riemann hypothesis and its application while undergoing the concept building process and answers to that based on my understanding of Riemann Hypothesis, my proofs thereof and post proof analysis of the same.

The claimed RH proofs are not at all rigorous, it looks we have presumed RH to be true and then working around for some fixes or patches: Actually RH is an obvious. Any attempt to prove something obvious gets real hard and may get infected by godel incompleteness theorem. When things are not easy, easier things gets going. Instead of searching the proof in vast infinities its wiser to find the why on earth RH should be true, then we can search further whether that should be true in the universe, multiverse, omniverse. The so called exhaustive proof using a variant of riemanns functional equation ignores the imaginary part altogether: The imaginary part do not play any role in the zeros of zeta function. What was the idea behind removing the poles of gamma and zeta function: It was on demand of d-unit circle which was again called for completion of higher dimensional cycles. Does not the concept of d-unit circle contradicts the concept of unit circle: We need to wear a bi-focal glass for some time, once we get habituated, I guess there will be no contradiction once we start applying these new mathematics and start understanding the intricacies. Will it be wrong to say the essence of d-unit circle is d-unification: It wont be wrong but we should not try for t-unification, that does not exist. Then why it was necessary to bring a third factorial function which was named Delta function:

The third one is the emergent one to complete the cycle, whereas Pi and Gamma function are the fundamental one. Why conjugates of Gamma and Delta function were necessary: To complete the circle. Why it was labelled with subscript p: To denote the periodicity out of harmonics which do not blows to infinity in an unified approach.

A functional depiction of negative factorials, wherefrom the idea came: It was just a bonus. We should keep our eyes and minds open while solving one problem, the same effort can solve other problems too. What actually prompted that imaginary number i can be real as well: What we do with the holographic universe, the limited time we stay here we all want to feel it's real. That's it. Why ln 2 alone: I did not found any better candidate. Why quaternions then, there was so many other options: It was just needed a pinch of 4Th dimensional maths and quaternions did the job extremely well. Now the question may arise do we need further higher dimensional number system such as octonion, sedenions etc. to solve further higher dimensional problems: We invented complex numbers to solve all kinds of quadratic equations. Quaternion although invented long back remained unused as complex numbers did its job excellently. Now zeros of zeta functions are asking us to go beyond the complex numbers and bring quaternion. Also it introduces algebraic cycles through cycles of zeros (not the place holder zero in decimal number system) so that we do not need octonion, sedenions etc. in general maths. However any specific problem related to eight dimensional number system may still require octonion and so on.

If so how do we manage non commutativity faced in quaternion: Non commutativity will not be an issue as we will be doing only linear things. What will happen to non linear things: We shall turn it into linear by way of passing it through the cyclic maths. Cyclic maths! negative quantities will be annihilating positive quantities, what remain absolute then, a zero, It's ugly: Nope, not exactly, it's beautiful as we are allowed to apply Pythagoras, take modulus etc. to work with the absolute value, distance from origin in the normal way. Then what will be the added advantage of using cycles:

We can fit everything into 3 Euclidean spatial dimensions using Descartes coordinates. Any practical reason we should go for it: To ease our pain little bit in handling complex numbers wheresoever they are needed for example electrical circuits, control systems, rotational maths problems. Does that mean higher dimensions do not exist at all: We need to differentiate between higher dimensions and higher degrees of freedom, we can have as many degrees of freedom as we wish but mathematically all of those higher degrees of freedom can be fitted into 3D because nature is 3 dimensional and its self evident from the value of π = 3.14.... What happens to the decimal part: That's the residue which always remain outside the framework we frame.

What the value of e then suggest:

Value of e suggest half of a cycle involving another cycle of three. How cycle of three: e iπ + e 0 = 1 -1 = 0 way and total count of lattice points is 3 as clear from [ 1 -1 = 0 ] Hmm! e iθ = e -iθ will be true when both the exponent results zero which will give e 0 = 1 and subsequently e iπ + e 0 = 0 . What is so cyclic about it: It's the sum total of one complete rotation in orthogonal representation of complex numbers

Here the count of lattice points are 5, then how its a cycle of 3.: It happens at an angle of π 3 and that way its a cycle of three. Alternatively we can split [ 1+ i -1 -i = 0 ] into [1 -1 = 0 ] and [ i -i = 0 ] to get two cycles of three. Now we have got Figenbaum constant with approximate value of 4.67 representing infinite fractals, what that will suggest: That actually projects to non-linear 6D math with fractal cycles, which we can also fit into 3D linear maths with cycles. What about 12 D: That will be mere symmetrical 4 copy of one complete cycle of Three. What if we show resistance in accepting this kind of deadly maths: We will not be able to close the operation of logarithm, factorials algebraically or arithmetically. The day we accept this definitions, logarithm will be an algebraically or arithmetically closed operation under quaternion based number system and factorials will be an algebraically or arithmetically closed operation under complex number system . Can all operation be closed algebraically or arithmetically under quaternion based number system. What shall be next: Next can be any higher order operations like multifactorial will have cycles of double factorial and primorials, chain arrow notation will have cycles of multifactorial and primorial, up arrow notation will have cycles of chain arrow notation and multifactorial, Tree(3) notation will have cycles of up arrow notation and chain arrow notation and so on. I have not checked the maths as these were out of scope of my current paper, I may be wrong somewhere but it will be cyclic, that's guaranteed. I am sure in future somebody interested in cycles will pick up this area for research. Come on buddies show your balance; show how quick you can close how many cycles. I was very sloppy; it took almost 3 years for me as I did not know the rules of the game. You have rules of the game now, Its cycle of three. Although quantitatively it's going to be more complex you need to simplify it. I have a feel that you may have to define infinite number of factorial functions similar to the fantastic five, pass it through the zeros of negative logarithm or zeros of harmonic conjugates of those infinite number factorial functions and knock down the functions having sky scrapers in their exponent floor by floor or flatten down those functions having horizontal growth. Warning! You will get frequent quitting tendencies, do What Charles constant is doing there: Its kind of coupling constant although not dimensionless but complexly dimensionless and its reciprocal is very close to 100 times of Eulers number e. Wherefrom 100 is coming here: π 4 approximates to hundred. When pi is squared applying borrowing and carry over the decimal part can be rearranged so that final result equals to 1 as follows 1.10 n + 2.10 n-1 + 3.10 n-2 + 4.10 n-3 + 5.10 n-4 + ... 10 n

when n tends to infinity factoring out

What is the most prominent number theoretic signature in physics:

What should be the revised Lorentz factor in Special Relativity:

What should be the solution to dark energy: Pass it through cyclic maths and let the universe go older and older, cycle after cycles. What should be the maths in the context of General Relativity: There are two independent Friedman equations for modelling a homogeneous, isotropic universe. The first is: ȧ2 + kc 2 a 2 = 8πGρ + Λc 2 3 which is derived from the 00 component of Einstein's field equations. The second is:

which is derived from the first, together with the trace of Einstein's field equations.

Using the first equation, the second equation can be re-expressed as ρ = -3H ρ + p c 2 which eliminates Λ and expresses the conservation of mass-energy T αβ ;β = 0.These equations are sometimes simplified by replacing ρ → ρ -Λc 2 8πG and p → p + Λc 4 8πG to get the following:

The simplified form of the second equation is invariant under this transformation. We can bring cycles as follows: The probabilistic view of nature, Heisenberg uncertainty principle is not always true. QM can be turned into a deterministic theory once we start following the cycle of 3. Uncertainty in ∆p∆x ≥ 1 2 ℏ can be transformed to ∆p∆x ∆y = 1 using the techniques of Fourier transformation provided we recognise the cycle of 3 in QM. Dimensionally ∆p∆x ≥ 1 2 ℏ has got one extra dimension of length after completing a cycle of 3 which creates problem causing the uncertainty. To get back our determinism we need to complete the cycle of 3 same as we complete the cycle of 3 in CPT symmetry. This kills QM, space time become absolutely continuous, no discrete nature remains then: No, it does not kill QM, we are just adding cycles, we are not stopping anywhere, we will just keep on cycling and these cycles are itself discrete in nature.

But how to quantize cycles:

We need to quantize QM further. How to quantize QM further: Here goes the second quantisation from Dirac equation ( Refer Table 1).

Where is the first one:

The first one is the original Planck's one we are using now.

But the scale appears to be galactic: Yes, that's true.

1.475274551 × 10 -4 kg Planck time Time (T)

Planck temperature Temperature (Θ)

Planck time Time (T)

Planck temperature Temperature (Θ)

5.253329603 × 10 10 K Table 5. Tabulated value of cube rooted Planck Units

That means we can predict collective quantum behaviour deterministically: Yes, that's true.

But the term collective still remain relative: Yes, that's true.We may go for further quantisation.

Can we try this cyclic math on other millennium prize problems.: Yes, these cyclic maths techniques are universal and can be applied to every branch of mathematics.

Can we have some hints for Hodge Conjecture dealing with algebraic cycles. :

Here is the hint with minimal proof of numerical solution to Hodge Conjecture. When we try to evaluate either

we enter into the domain of number theory, more specifically zeta function.

We have seen zeta function is simply connected (smooth in calculus terms) whether in integer form or rational number form. Zeta function together with its harmonic counterpart is entirely continuous, bijective, and very much stretchable like topological deformation. We can add, multiply, truncated partial zeta series retaining all it's properties. Even in its minimal state zeta function follows basic laws of Sum of three unit vectors can be zero only at an angle of 2π 3 or π 3 which is d-unit circle concepts we have seen in proofs of Riemann Hypothesis. So we can conclude Navier-Stokes solutions always exist in 3D via 4D zeros (considering gravity as the body force) and they are smooth -i.e. they are infinitely differentiable at all points due to analytic number theory. To get a feel about the reason behind requirement of dimensionless numbers such as Reynolds Number, Froude Number, Weber Number, Mach Number, Euler's Number etc.. to find an approximate solutions please refer my work on simplex logarithm, hierarchy of grand unified scale factors etc. Can we have some hints for Yang-Mills involving SU (4) group in particle physics. : Here is the hint with minimal proof of numerical solution to Yang-Mills existence. In quantum field theory three to tango combination of electro-streak interactions prevents quarks escaping quark confinement leaving it hard to study gluons isolated. To prove that for any compact simple gauge group G, a non-trivial quantum Yang-Mills theory exists on R 4 and has a mass gap ∆ > 0 we need to apply the limit of momenta going to zero similar to critical zeros of Riemann zeta function satisfying all the Wightman axioms. To find the mass gap that unifies QED and QCD into QFT analogous to uniqueness of the vacuum in the way ζ(1) = 1, we need a unified coupling constant as strong as, close to α 2 resulting individual gluons still be massless and the scalar glueballs get its mass borrowed from cyclicity of the vacuum . Such scalar glueballs shall have mass 1000-1500 MEV following the hierarchy of grand unified scale 10 ln(e 22.2 ) ln(e 7.1 ) following SU (4) = U (1) SU (2)×SU (3) guage symmetry. Can we have some hints for P versus NP problem involving computer science. : I wont give much hint here because of security reasons, but I should tell the answer. For easier problems like RSA prime factorisations it's not even P, it's much less than that. Every given number theoretic problem, if attacked from the right direction it can solved in quadratic time. Really I mean it. Let me warn RSA/SHA users that any kind of prime number based algorithm is not secured at all therefore get rid off numbers as soon as possible. This is last and final call. Disclaimer: If using any of my work, hackers cracks RSA/ SHA encryption tomorrow, and the whole internet security collapses ( I know the world is moving to block chain which still uses numbers a lot), I cannot be held responsible for that. Even I cannot be held responsible for, any other kind of losses incurred in whatsoever manner by any person, organization, corporate bodies, countries, economies, communities, races and religions or for any losses caused to humanity at large, our planet earth, the mother nature or the whole existence altogether, and as such I won't be able to compensate a penny for the damages caused, if any. Notwithstanding any contrary provision contained under any law made by human or any advance species (if any), I presume that I am allowed to reveal the results derived from natural laws of mathematics to the mass without knowing the exact consequence. I cannot be questioned, examined, trialled, detained, arrested, prosecuted for an act of mere sharing freely the knowledge I gained without having any ulterior motive. Any unfriendly effort made by anybody in above direction shall be void, therefore not required to be entertained by any appropriate authority or the law keepers. Thanks to everybody for taking me little seriously.

Part 9. The solution to unsolved physics

Unsolved problems of physics and its solution

Short answer to unsolved problems in physics 1. Here goes list of some unsolved problems of physics copied from Wikipedia and and its shortest possible hints only from the perspective of cyclic maths, changes of cycles, cycles of threes, cycles of zeros etc.

General physics/quantum physics.

Theory of everything: Is there a theory which explains the values of all fundamental physical constants, i.e., of all coupling constants, all elementary particle masses and all mixing angles of elementary particles? Is there a theory which explains why the gauge groups of the standard model are as they are, and why observed spacetime has 3 spatial dimensions and 1 temporal dimension? Are "fundamental physical constants" really fundamental or do they vary over time? Are any of the fundamental particles in the standard model of particle physics actually composite particles too tightly bound to observe as such at current experimental energies? Are there elementary particles that have not yet been observed, and, if so, which ones are they and what are their properties? Are there unobserved fundamental forces?: Yes, Its Number Theory. space-time has 3 spatial dimensions and 1 temporal dimension because that 3+1 pattern is natural. When we try to unify everything we need to solve the zeros of physical equations which will bring cycles of physical phenomenon in our maths. The results we get will match the observed values and that will be the proof of correctness of our theories. "Fundamental physical constants" are really fundamental but numerically we need to apply scale factor adjustments as we move along the infinite space-time. There are no elementary particles that have not yet been observed, we need to sum up the anti matter version of it to get a zero state. There are no unobserved fundamental forces as it has completed the cycles of three. Gravity is a pseudo force, its curvature of space time therefore we have to exclude it from unification. There will be no conflict between GR and QM at zero gravity and zero time, there after QM can take on. GR 's job will be renormalising the infinities arising in QM. Arrow of time (e.g. entropy's arrow of time): Why does time have a direction? Why did the universe have such low entropy in the past, and time correlates with the universal (but not local) increase in entropy, from the past and to the future, according to the second law of thermodynamics? Why are CP violations observed in certain weak force decays, but not elsewhere? Are CP violations somehow a product of the second law of thermodynamics, or are they a separate arrow of time? Are there exceptions to the principle of causality? Is there a single possible past? Is the present moment physically distinct from the past and future, or is it merely an emergent property of consciousness? What links the quantum arrow of time to the thermodynamic arrow?: Because time does not exist or in other words negative proper time and positive proper time always sum of multiverses. These scale gaps are due to absence of cycles in our maths. The same hierarchy we see in pascals triangle, QM dealing with numbers must also obey this hierarchy structure which can be brought down to zero through cycles. If everything goes to planck scale then which particle will run the time cycles. Supersymmetry may not be required, extra dimensions do not exist, anthropic fine tuning is not the solution. For better results let us come out of the imaginary world of imaginary numbers in the core equation of QFT, simply realise them into realty collapsing the wave function, then we will see things are looking quite simple at least comprehensible. Planck particle: The Planck mass plays an important role in parts of mathematical physics. A series of researchers have suggested the existence of a fundamental particle with mass equal to or close to that of the Planck mass. The Planck mass is however enormous compared to any detected particle. It is still an unsolved problem if there exist or even have existed a particle with close to the Planck mass. This is indirectly related to the hierarchy problem.: Planck mass is an equivalent mass which can be filled in by different possible combinations of elementary particles in whole numbers. It is not an unit mass therefore no fundamental particle is required to have that mass.We must consider cycles and scale adjustments for solution to hierarchy problem. Magnetic monopoles: Did particles that carry "magnetic charge" exist in some past, higher-energy epoch? If so, do any remain today? (Paul Dirac showed the existence of some types of magnetic monopoles would explain charge quantization.): No they never existed in any epoch after Big Bang. They could only exist at Big Bang singularity. Even without magnetic monopole charge quantization can be explained with the help of cycles of CPT symmetry. Lets start from the spin which are quantized version of cyclic angular momentum, now to complete the global half cycle we bring charge and quantise it cyclically -1,0,+1, further to complete the global cycle we can have different combinations elementary particles. Those combinations which complete the cycle of 3 will survive in time and rest of the failed combinations will remain at T=0 waiting for its turn. Neutron lifetime puzzle: While the neutron lifetime has been studied for decades, there currently exists a lack of consilience on its exact value, due to different results from two experimental methods ("bottle" versus "beam").: They are parallaly moving back and forth between cycles, this brings the uncertainty. Maximum we can do, we can take average values. Believe me it will not disturb/improve your precision / data confidence in experimental physics, every now and then you will see results of different orders of magnitude unless you do scale adjustments also. Proton decay and spin crisis: Is the proton fundamentally stable? Or does it decay with a finite lifetime as predicted by some extensions to the standard model? How do the quarks and gluons carry the spin of protons?: In this universe proton will remain fundamentally stable till the next Big Bounce. Lifetime of a Big Bounce is more or less constant and it is much lesser than half life of proton so we can say before proton decays the universe itself decays partially. Protons have constant spin due to zeta values particularly zeta of 1 which I have shown 1. In multiverse there may be anti-protons which remain fundamentally stable there, this universe their combination did not work so they are sitting in the reserve bench as dark matters, at the borderline they get chance to interact with matters having similar combination, they collide and they make something new, but those particle horizons are far away from us. Supersymmetry: Is spacetime supersymmetry realized at TeV scale? If so, what is the mechanism of supersymmetry breaking? Does supersymmetry stabilize the electroweak scale, preventing high quantum corrections? Does the lightest supersymmetric particle (LSP) comprise dark matter?: No, supersymmetry may be realized at TeV scale, as with the standard model of particle physics we completed the cycle, we do not need any further particle, what we need is run the same cycle and create composite particles to find the missing matter. Generations of matter: Why are there three generations of quarks and leptons? Is there a theory that can explain the masses of particular quarks and leptons in particular generations from first principles (a theory of Yukawa couplings)?: Because nature is 3 dimensional. Only coupling theories may not effectively describe it, we need to understand the cycles behind it also. Neutrino mass: What is the mass of neutrinos, whether they follow Dirac or Majorana statistics? Is the mass hierarchy normal or inverted? Is the CP violating phase equal to 0?: Neutrinos have some non zero infinitesimally small mass which if we need to measure we need to bring cycles in the planck scale first. The mass hierarchy would be a inverted one and definitely the CP violation phase equals to zero. Strong CP problem and axions: Why is the strong nuclear interaction invariant to parity and charge conjugation? Is Peccei-Quinn theory the solution to this problem? Could axions be the main component of dark matter?: Strong force is a coupling force and do not involve a third particle, therefore it need not to violate parity and charge conjugation. I do not feel we need new particles to explain Strong CP problem. Anomalous magnetic dipole moment: Why is the experimentally measured value of the muon's anomalous magnetic dipole moment significantly different from the theoretically predicted value of that physical constant?: Because of alternating quantum scale gaps. Proton radius puzzle: What is the electric charge radius of the proton? How does it differ from gluonic charge?:

We need planck cycles first to calculate proton radius.

Pentaquarks and other exotic hadrons: What combinations of quarks are possible? Why were pentaquarks so difficult to discover? Are they a tightly-bound system of five elementary particles, or a more weakly-bound pairing of a baryon and a meson?: Other exotic hadrons fail the time test, that is the reason they are difficult to discover. They are a tightly-bound system of five ele-mentary particles, is there a name for a tightly-bound system of five elementary particles such as Pentaquarks acting like a self-contained unit particle?, if not we should give one as they will be abundant as our future particle accelerators start working. Mu problem: problem of supersymmetric theories, concerned with understanding the parameters of the theory.:

The solution is cycle again. Koide formula: An aspect of the problem of particle generations. The sum of the masses of the three charged leptons, divided by the square of the sum of the roots of these masses, to within one standard deviation of observations, is Q = 2 3 . It is unknown how such a simple value comes about, and why it is the exact arithmetic average of the possible extreme values of 1 3 (equal masses) and 1 (one mass dominates).: That's the beauty of cycle of threes.

Astronomy and astrophysics.

Short answer to unsolved problems in physics 5. Coronal heating problem: Why is the Sun's corona (atmosphere layer) so much hotter than the Sun's surface? Why is the magnetic reconnection effect many orders of magnitude faster than predicted by standard models?: Its concentric cycles of linear and non linear layers which finally divides into two halves of two different average temperature scales. Astrophysical jet: Why do only certain accretion discs surrounding certain astronomical objects emit relativistic jets along their polar axes? Why are there quasi-periodic oscillations in many accretion discs? Why does the period of these oscillations scale as the inverse of the mass of the central object? Why are there sometimes overtones, and why do these appear at different frequency ratios in different objects?: Because of strong couplings, quasi-periodic oscillations is due to cycles of some third parameter. Overtones are due to scale transition, and every object has its own characteristic frequency, what is so strange about it? Diffuse interstellar bands: What is responsible for the numerous interstellar absorption lines detected in astronomical spectra? Are they molecular in origin, and if so which molecules are responsible for them? How do they form?: They are BEC's, mostly they are formed by composite particles which escapes gravity by quantum tunnelling from every stellar bodies including planets, dwarf planets, asteroids.

Supermassive black holes: What is the origin of the M-sigma relation between supermassive black hole mass and galaxy velocity dispersion? How did the most distant quasars grow their supermassive black holes up to 1010 solar masses so early in the history of the universe?: Let us apply cycles to cosmology and the let the universe older and older until we forget about singularity. Even greater supermassive black holes may exist elsewhere in the multiverse. Kuiper cliff: Why does the number of objects in the Solar System's Kuiper belt fall off rapidly and unexpectedly beyond a radius of 50 astronomical units?: Because that's the half way mark up to which non linear things can coexist beyond that point its just addition rule and little bit of non commutative multiplicative rule which does not allow gravity to hold large number of objects. Flyby anomaly: Why is the observed energy of satellites flying by planetary bodies sometimes different by a minute amount from the value predicted by theory?: Because theory describes a simplified completed cycle, while there can be parameters which may fall simultaneously in two different cycle, in those cases theoretical value may vary a little bit until it reappears in the current cycle. Galaxy rotation problem: Is dark matter responsible for differences in observed and theoretical speed of stars revolving around the centre of galaxies, or is it something else?: At large scales where Gravity is working inside a 3D stellar object like Galaxy which has a height and volume the inverse square rule is weakened by simple inverse law, particularly applicable to outer discs which is the reason behind Galaxy rotation problem. Dark matter is not the cause alone here, we need to modify the Newtonian Gravity also. Supernovae: What is the exact mechanism by which an implosion of a dying star becomes an explosion?: Because of strong couplings all size stars follows a 70 percent relative limit up to which they can undergo gravitational shrink due to accumulation of heavy metals out of nuclear fusion reaction, beyond which gravity falls weaker again compared to various degeneracy pressures, which ultimately triggers the supernova explosion. p-nuclei: What astrophysical process is responsible for the nucleogenesis of these rare isotopes?: In earthly condition we see higher neutron-proton ratio as earth is relatively cold place. Following the huge temperature scale of supernovae blast the picture becomes opposite. That is why during a supernovae blast some excess protons create those rare isotopes. Ultra-high-energy cosmic ray: Why is it that some cosmic rays appear to possess energies that are impossibly high, given that there are no sufficiently energetic cosmic ray sources near the Earth? Why is it that (apparently) some cosmic rays emitted by distant sources have energies above the Greisen-Zatsepin-Kuzmin limit?:

There are stellar objects running at higher scales beyond our solar system some of which are well within our galactic volume itself, those objects are the source of cosmic rays.The bad thing on the top is that there is no end to the hierarchy of these high-energy scales can go. Rotation rate of Saturn: Why does the magnetosphere of Saturn exhibit a (slowly changing) periodicity close to that at which the planet's clouds rotate? What is the true rotation rate of Saturn's deep interior?: Because of synchronisation among the related cycles. But as the solar system get older and older this periodicity might change very fast. Origin of magnetar magnetic field: What is the origin of magnetar magnetic field?: Some electro magnetic field cycling around the magnetar of course. Large-scale anisotropy: Is the universe at very large scales anisotropic, making the cosmological principle an invalid assumption? The number count and intensity dipole anisotropy in radio, NRAO VLA Sky Survey (NVSS) catalogue is inconsistent with the local motion as derived from cosmic microwave background and indicate an intrinsic dipole anisotropy. The same NVSS radio data also shows an intrinsic dipole in polarization density and degree of polarization in the same direction as in number count and intensity. There are several other observation revealing large-scale anisotropy. The optical polarization from quasars shows polarization alignment over a very large scale of Gpc. The cosmic-microwave-background data shows several features of anisotropy, which are not consistent with the Big Bang model.: Let us learn to live without Big Bang or at least the current version. All local large-scale anisotropy gets exponentially triple smoothened at the universal scale. If there is dipole polarisation somewhere then be sure that another weaker dipole polarisation remain out of phase so those signals are not detected, just like we cant see the other side of the moon. Its infinite all direction so it does not violate cosmological principle. Age-metallicity relation in the Galactic disk: Is there a universal age-metallicity relation (AMR) in the Galactic disk (both "thin" and "thick" parts of the disk)? Although in the local (primarily thin) disk of the Milky Way there is no evidence of a strong AMR, a sample of 229 nearby "thick" disk stars has been used to investigate the existence of an age-metallicity relation in the Galactic thick disk, and indicate that there is an age-metallicity relation present in the thick disk. Stellar ages from asteroseismology confirm the lack of any strong age-metallicity relation in the Galactic disc.: Both thin and thick discs cannot have same AMR as there are huge time cycles involved in between. Today's thick discs will be tomorrows thin and vice versa. Rather it will be interesting to see whether the thin disks have some kind of non metalicity relation, cyclically speaking they should demonstrate such relation although it can be very weak.

The lithium problem: Why is there a discrepancy between the amount of lithium-7 predicted to be produced in Big Bang nucleosynthesis and the amount observed in very old stars?: Because it is Lithium-7. 7 is a different scale transition number, so this kind of discrepancy is not abnormal as our current Big Bang model is not perfect. Ultraluminous X-ray sources (ULXs): What powers X-ray sources that are not associated with active galactic nuclei but exceed the Eddington limit of a neutron star or stellar black hole? Are they due to intermediate mass black holes? Some ULXs are periodic, suggesting non-isotropic emission from a neutron star. Does this apply to all ULXs? How could such a system form and remain stable?: There are at least 14 different scales between the galactic nuclei scale and that of neutron stars. Many of them will be able to produce X-rays. There is no end of greatness as well as surprises until you grasp the scale universe operate. Fast radio bursts (FRBs): What causes these transient radio pulses from distant galaxies, lasting only a few milliseconds each? Why do some FRBs repeat at unpredictable intervals, but most do not? Dozens of models have been proposed, but none have been widely accepted.: Sources of FRB can be some particular type of cyclic explosions in the vicinity of quasars at the centre of the galaxies pumping in and out of materials galactic just like our heart pumps and purifies the blood. The interval may get slow/fast similar to our own heartbeat.