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## Theorems of the month

### Prime number theorem

For $\displaystyle{ \pi(x) := |\{p \in {\mathbb{P}_{\le x}} : x \in {}^{\omega}{\mathbb{R}}\}| }$, it holds that $\displaystyle{ \pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_\epsilon}\omega\;{\omega}^{\tilde{2}}) }$.

#### Proof:

From intervals of fix length $\displaystyle{ y \in {}^{\omega}{\mathbb{R}_{\gt 0}}, \check{y} }$ 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. The Stirling formula suggests the prime gap $\displaystyle{ n = {\epsilon}^{\sigma} = \mathcal{O}({_\epsilon}(n!)) }$.

For induction basis $\displaystyle{ n = 2 }$ resp. 3, the hypothesis states the first interval to contain $\displaystyle{ x_n/{_\epsilon}x_n }$ primes for $\displaystyle{ n \in {}^{\omega}{\mathbb{N}_{\ge2}} }$ and $\displaystyle{ x_4 \in [2, 4[ }$. Stepping from $\displaystyle{ x_n }$ to $\displaystyle{ x_n^2 }$ finds $\displaystyle{ \pi(x_n^2) = \pi(x_n) \check{x}_n }$ primes only from $\displaystyle{ \pi(x_n) = x_n/{_\epsilon}x_n }$. The average prime gap is $\displaystyle{ {_\epsilon}x_n }$, the maximal one $\displaystyle{ {_\epsilon}x_n^2 }$ and the maximal $\displaystyle{ x_n^2 }$ to $\displaystyle{ x_n }$ behaves like $\displaystyle{ \omega }$ to $\displaystyle{ {\omega}^{\tilde{2}}.\square }$

### Gelfond-Schneider theorem

It holds $\displaystyle{ a^b \notin {}_{\omega}^{\omega} \mathbb{A}_\mathbb{C} }$ where $\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus \mathbb{B} }$ and infinitesimal $\displaystyle{ \varepsilon, b \in {}^{\omega}\mathbb{A}_\mathbb{C} \setminus {}_{\omega}^{\omega}\mathbb{R} }$.

#### Proof:

The minimal polynomials $\displaystyle{ p }$ (and $\displaystyle{ q }$) of $\displaystyle{ c^r }$ resp. $\displaystyle{ c^{r\pm\varepsilon} = a^b }$ for maximal $\displaystyle{ r \in {}_{\omega}^{\omega}\mathbb{R}_{\gt 0} }$ and $\displaystyle{ f = p\;(q) }$ lead to the contradiction $\displaystyle{ {}^1f(c^{r(\pm\varepsilon)}) \ne 0 = (f(c^r) - f(c^{r\pm\varepsilon})) / (c^r - c^{r\pm\varepsilon}) = {}^1f(c^{r(\pm\varepsilon)}).\square }$

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