Counterexample of the Riemann Hypothesis

Under the assumption that the Riemann hypothesis is true, von Koch deduced the improved asymptotic formula θ ( x ) = x + O ( √ x × log 2 x ), where θ ( x ) is the Chebyshev function. We obtain a result which contradicts this asymptotic formula. By contraposition, we deduce that the Riemann hypothesis is false.


Introduction
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].On the other hand, h(x) + O( f (x)) denotes the collection of functions having the growth of h(x) plus a part whose growth is limited to that of f (x).Thus, expresses the same as g(x) − h(x) = O( f (x)).
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. Say Nicolas(p n ) holds provided The constant γ ≈ 0.57721 is the Euler-Mascheroni constant, log is the natural logarithm, and p n is the n th prime number.The importance of this property is: , [3].Nicolas(p n ) holds for all prime numbers p n > 2 if and only if the Riemann hypothesis is true.
We know the following properties for the Chebyshev function: Theorem 1.2.[4].If the Riemann hypothesis holds, then for all x ≥ 10 8 .
We also know that Theorem 1.4.[6].If the Riemann hypothesis holds, then for all numbers x ≥ 13.1.
Let's define H = γ − B such that B ≈ 0.2614972128 is the Meissel-Mertens constant [7].We know from the constant H, the following formula: For x ≥ 2, the function u(x) is defined as follows We use the following theorems: Theorem 1.6.[9].For x > −1: Theorem 1.7.[10].For x ≥ 1: Let's define: Definition 1.8.We define another function: Putting all together yields the proof that the inequality ϖ(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.2

Results
Theorem 2.1.The Riemann hypothesis is true if and only if the inequality ϖ(x) > u(x) is satisfied for all numbers x ≥ 3.
Proof.In the paper [3] is defined the function: We know that f (x) is lesser than 1 when Nicolas(p) holds, where p is the greatest prime number such that 2 < p ≤ x.In the same paper, we found that where U(x) = −ϖ(x) [3].When f (x) is lesser than 1, then log f (x) < 0. Consequently, we obtain that −ϖ(x) + u(x) < 0 which is the same as ϖ(x) > u(x).Therefore, this is a consequence of the theorem 1.1.
Theorem 2.2.If the Riemann hypothesis holds, then for all numbers x ≥ 13.1.
Proof.Under the assumption that the Riemann hypothesis is true, then we would have after of distributing the terms based on the theorem 1.4 for all numbers x ≥ 13.1.If we apply the logarithm to the both sides of the previous inequality, then we obtain that q≤x log( q q − 1 ) < γ + log log x + log 1 That would be equivalent to where we know that according to theorem 1.7 since 8×π× √ x 3×log x+5 ≥ 1 for all numbers x ≥ 13.1.We use the theorem 1.5 to show that q≤x log( q q − 1 ) − 1 q = H − u(x) which is the same as We eliminate the value of H and thus, Under the assumption that the Riemann hypothesis is true, we know from the theorem 2.1 that ϖ(x) > u(x) for all numbers x ≥ 13.1 and therefore, Hence, Suppose that θ(x) = ϵ × x for some constant ϵ > 1.Then, In addition, we know that using the theorem 1.6 since log ϵ log x > −1 when ϵ > 1. Certainly, we will have that Thus, .
If we add the following value of log x log θ(x) to the both sides of the inequality, then We know this inequality is satisfied when 0 < ϵ ≤ 1 since we would obtain that log x log θ(x) ≥ 1.Therefore, the proof is done.
Proof.If the Riemann hypothesis holds, then for all x ≥ 10 8 due to the theorem 1.2.Now, suppose there is a real number x ≥ 10 8 such that θ(x) > x + √ x × log 1.9 x.That would be equivalent to log θ(x) > log(x + √ x × log 1.9 x) and so, 1 log θ(x) < 1 log(x + √ x × log 1.9 x) for all numbers x ≥ 10 8 .Hence, log x log θ(x) < log x log(x + √ x × log 1.9 x) .
If the Riemann hypothesis holds, then is satisfied, then the Riemann hypothesis should be false.Let's define the function The Riemann hypothesis would be false when there exists some number y ≥ 10 8 such that for all x ≥ y the inequality υ(x) ≤ 0 is always satisfied.We ignore when 2 ≤ x ≤ 10 8 since θ(x) < x according to the theorem 1.3.We know that the function υ(x) is monotonically decreasing for every number x ≥ 10 8 .The derivative of υ(x) is negative for all x ≥ 10 8 .The derivative of υ(x) is x × log(x) + 1.9 × log 1.9 (x) + 0.5 × log 2.9 (x) x × ( √ x + log 1.9 (x)) × log 2 (x + √ x × log 1.9 (x)) + 1 x × log(x + √ x × log 1.9 (x)) .
Indeed, a function υ(x) of a real variable x is monotonically decreasing in some interval if the derivative of υ(x) is lesser than zero and the function υ(x) is continuous over that interval [11].
It is enough to find a value of y ≥ 10 8 such that υ(y) ≤ 0 since for all x ≥ y we would have that υ(x) ≤ υ(y) ≤ 0, because of υ(x) is monotonically decreasing.We found the value y = 10 8 complies with υ(y) ≤ 0. In this way, we obtain that υ(x) ≤ 0 for every number x ≥ 10 8 .Consequently, under the assumption that the Riemann hypothesis is true, then θ(x) < x + √ x × log 1.9 x for all x ≥ 10 8 .Hence, this implies that the Riemann hypothesis is false using the theorem 1.2.