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Theorems of the month
Prime number theorem
For [math]\displaystyle{ \pi(x) := |\{p \in {}^{\omega}{\mathbb{P}} : p \le x \in {}^{\omega}{\mathbb{R}}\}| }[/math] holds [math]\displaystyle{ \pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_e}\omega{\omega}^{\tilde{2}}) }[/math].
Proof:
In the sieve of Eratosthenes, the number of prime numbers decreases almost regularly. From intervals of fix length [math]\displaystyle{ y \in {}^{\omega}{\mathbb{R}_{>0}}, \hat{y} }[/math] set-2-tuples of prime numbers are formed such that the first interval has the unchanged representative prime number density and the second interval is empty, then the interval with the second most prime number density is followed by the second least one etc.
For induction basis [math]\displaystyle{ n = 2 }[/math] resp. 3, the induction hypothesis is that the first interval contains [math]\displaystyle{ x_n/{_e}x_n }[/math] prime numbers for [math]\displaystyle{ n \in {}^{\omega}{\mathbb{N}_{\ge2}} }[/math] and arbitrary [math]\displaystyle{ x_4 \in [2, 4[ }[/math]. Then the induction step from [math]\displaystyle{ x_n }[/math] to [math]\displaystyle{ x_n^2 }[/math] by considering the prime gaps of prime [math]\displaystyle{ p\# /q + 1 }[/math] for [math]\displaystyle{ p, q \in {}^{\omega}\mathbb{P} }[/math] proves that there are [math]\displaystyle{ \pi(x_n^2) = \pi(x_n) \check{x}_n }[/math] prime numbers only from [math]\displaystyle{ \pi(x_n) = x_n/{_e}x_n }[/math]. The average distance between the prime numbers is [math]\displaystyle{ {_e}x_n }[/math] and the maximal [math]\displaystyle{ x_n^2 }[/math] to [math]\displaystyle{ x_n }[/math] behaves like [math]\displaystyle{ \omega }[/math] to [math]\displaystyle{ {\omega}^{\tilde{2}}.\square }[/math]
Gelfond-Schneider theorem
It holds [math]\displaystyle{ a^b \in {}^{\omega} \mathbb{T}_\mathbb{C} }[/math] where [math]\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C}^{*} \setminus \{1\}, Q := {}^{\omega} \mathbb{R} \setminus {}^{\omega} \mathbb{T}_\mathbb{R} }[/math] and [math]\displaystyle{ b, \varepsilon \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus Q }[/math].
Proof:
Where [math]\displaystyle{ b \in Q }[/math] puts the minimal polynomial [math]\displaystyle{ p(a^b) = p(c^q) = 0 }[/math], assuming [math]\displaystyle{ a^b = c^{q+\varepsilon} }[/math] for maximum [math]\displaystyle{ q \in Q_{\gt 0} }[/math] leads to the contradiction [math]\displaystyle{ 0 = (p(a^b) - p(c^q)) / (a^b - c^q) = p^\prime(a^b) = p^\prime(c^q) \ne 0.\square }[/math]