Disproof of the Riemann Hypothesis

Let’s deﬁne S ( x ) = ϑ ( x ) − x , where ϑ ( x ) is the Chebyshev function. We prove that the Riemann Hypothesis is false when (cid:82) ∞ x S ( y ) × (1 + log y ) y 2 × log 2 y dy ≥ S ( x ) 2 x 2 × log x is satisﬁed for some number x ≥ 121. In addition, we demonstrate that the previous inequality is satisﬁed when S ( x ) ≥ 0 for some number x ≥ 121. It is known that S ( x ) changes sign inﬁnitely often. In this way, we show that the Riemann Hypothesis is indeed false.


Introduction
In mathematics, the Riemann Hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1  2 [1].Let N n = 2 × 3 × 5 × 7 × 11 × • • • × p n denotes a primorial number of order n such that p n is the n th prime number.Say Nicolas(p n ) holds provided q|N n q q − 1 > e γ × log log N n .
The constant γ ≈ 0.57721 is the Euler-Mascheroni constant, log is the natural logarithm, and q | N n means the prime number q divides to N n .The importance of this property is: Theorem 1.1.[2], [3].Nicolas(p n ) holds for all prime numbers p n > 2 if and only if the Riemann Hypothesis is true.
In mathematics, the Chebyshev function ϑ(x) is given by where p ≤ x means all the prime numbers p that are less than or equal to x.We know this: Theorem 1.2.[4].
Besides, Rosser and Schoenfeld derived a remarkable identity: We define another function: Putting all together yields the proof that the inequality (x) > u(x) is satisfied for a number x ≥ 3 if and only if Nicolas(p) holds, where p is the greatest prime number such that p ≤ x.
In this way, we introduce another criterion for the Riemann Hypothesis based on the Nicolas criterion and deduce some of its consequences.One of these consequences is that the Riemann Hypothesis is indeed false.

Results
Theorem 2.1.The inequality (x) > u(x) is satisfied for a number x ≥ 3 if and only if Nicolas(p) holds, where p is the greatest prime number such that p ≤ x.
Proof.We start from the inequality: which is equivalent to Let's add the following formula to the both sides of the inequality, q≤x log( q q − 1 ) − 1 q and due to the theorem 1.6, we obtain that q≤x log( q q − 1 ) Let's distribute it and remove B from the both sides: If we apply the exponentiation to the both sides of the inequality, then we have that which means that Nicolas(p) holds, where p is the greatest prime number such that p ≤ x.The same happens in the reverse implication.
Theorem 2.2.The Riemann Hypothesis is true if and only if the inequality (x) > u(x) is satisfied for all numbers x ≥ 3.
Proof.This is a direct consequence of theorems 1.1 and 2.1.
Theorem 2.3.If the inequality (x) ≤ 0 is satisfied for some number x ≥ 3, then the Riemann Hypothesis should be false.
Proof.This is an implication of theorems 1.7, 2.1 and 2.2.
Theorem 2.4.If the inequalities δ(x) ≤ 0 and S (x) ≥ 0 are satisfied for some number x ≥ 3, then the Riemann Hypothesis should be false.
Proof.If the inequalities δ(x) ≤ 0 and S (x) ≥ 0 are satisfied for some number x ≥ 3, then we obtain that (x) ≤ 0 is also satisfied, which means that the Riemann Hypothesis should be false according to the theorem 2.3.
Proof.We know that lim x→∞ (x) = 0 for the limits lim x→∞ δ(x) = 0 and lim x→∞ ϑ(x) x = 1.In this way, this is a consequence from the theorems 1.9 and 1.2.
Theorem 2.6.Under the assumption that is satisfied for some number x ≥ 121, then the Riemann Hypothesis should be false.
Proof.Under the assumption that for some number x ≥ 121, then we can deduce that according to the theorem 1.10.Using the theorem 1.5, then we obtain that q≤x 1 q ≤ log log ϑ(x) + B due to x ≥ 121.However, that would mean and therefore, the Riemann Hypothesis should be false because of the theorem 2.3.
Proof.By the theorem 1.3, we have that: for a number x ≥ 121.In this way, we have that 0 ≤ S (x) x < 1 when S (x) ≥ 0. Consequently, we obtain that S (x)  x ≥ S (x) 2 x 2 under the assumption that S (x) ≥ 0, since for every real number 0 ≤ z < 1, the inequality z ≥ z 2 is always satisfied.For that reason, we will have that S (x)  x×log x ≥ S (x) 2 x 2 ×log x .However, we know that for a constant C. In conclusion, the theorem is true for some number x ≥ 121 when S (x) ≥ 0.
Proof.This is a direct consequence of theorems 1.4, 2.6 and 2.7.

S 1 +
(y) × (1 + log y) y 2 × log 2 y dy = S (x) x × log x + such that could be a positive value which depends mostly on the number x ≥ 121 and S (x) ≥ 0. Certainly, we know that (1 + log y)y 2 × log 2 y dy = − 1 y × log y + Cwhere C is a constant.In addition, we have that∞ x S(y) × (1 + log y) y 2 × log 2 y dy = S (y) × ( (1 + log y) y 2 × log 2 y dy) log y) y 2 × log 2 y dy) dy after of applying the rule of Integration by Parts [10].In this way, we will have that ∞ x S (y) × (1 + log y) y 2 × log 2 y dy = S (x) x × log x + y × log y dy could be a positive value for some number x ≥ 121 when S (x) ≥ 0 and 1 x × log x dx = log log x + C