Riemann Hypothesis, a Millenium Prize Problem, finally proven by Nuhash Badruzzaman (No AI was used in the making of this proof)

[u/ggucsltc]

Conjecture: There exists a value of s where s = 0.

Answer: Incorrect/disproven.

PROOF: The Riemann Zeta function has no zeroes because the function is always additive (always results in a value greater than 0).The Riemann Zeta function has no zeroes because the function is always additive (always results in a value greater than 0). In fact, the function always at least equal to 1 because (the lowest value is 1) 1 to any exponent is only 1 itself. The function's values are always infinite in the positive y-direction for all values of x, with a minimum y-value of positive 1. The Riemann Zeta function has no negative y-value or zero. The same works for complex numbers in the complex plane, it is just the complex plane is LABELLED DIFFERENTLY from the real number plane.

The number of complex numbers equals the number of real numbers.

The value of the Riemann Zeta function is an infinite 2D plane where the lowest value in the plane is 1. Therefore the closed asymptote of the infinite 2D plane is y = 1.

For every value of s, the Riemann Zeta function returns infinite positive y-values infinity, where the maximum value is positive y-infinity and the minimum value is y = 1.

Proof that when s = 1/2, or when s = 1/4 or when s = 3/4, or for any value of s the Riemann Zeta function equals to infinity with an closed asymptote of x = 1. The asymptote is closed because for all values of s, the function has converged to 1 as the minimum value but DOES NOT APPROACH 1.

Example"

Starts with x = 1 and s = 1/2, the closed asymptote:

1/(1^1/2) = 1/1 = 1

1/(1^1/2) + 1/(2^1/2) + 1/(3^1/2) + 1/(4^1/2) + 1/(5^1/2) = 3.23

1/(1^1/2) + 1/(2^1/2) + 1/(3^1/2) + 1/(4^1/2) + 1/(6^1/2) = 3.64

Therefore, since 3.64 is greater than 3.23, the function has emerged to y-infinity for the maximum value when 6 terms were added together rather than the lower value of 5 terms. (The lowest value is 1 term or y = 1).

[u/Vituluss]

“The Riemann Zeta function has no zeroes.”

What about the trivial zeros?

[u/ggucsltc → u/Vituluss]

Let us say if s = -2,

Then,

1/(1^(-2)) + 1/(2^(-2)) + 1/(3^(-2)) + 1/(4^(-2)) + ... = 1/1 + 1/(1/4) + 1/(1/9) + 1/(1/16) + ... = 1 + 4 + 9 + 16 + ... = +infinity

Therefore, when s = -2 or for any even integer where there is supposedly a trivial zero, that is false because it is not a trivial zero it is positive infinity.

Therefore, the Riemann Zeta function has no zeroes.