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Theorems of the month
Greatest-prime Criterion
If a real number may be represented as an irreducible fraction [math]\displaystyle{ \widehat{ap}b \pm \hat{s}t }[/math], where [math]\displaystyle{ a, b, s }[/math], and [math]\displaystyle{ t }[/math] are natural numbers, [math]\displaystyle{ abst \ne 0 }[/math], [math]\displaystyle{ a + s > 2 }[/math], and the (second-)greatest prime number [math]\displaystyle{ p \in {}^{\omega }\mathbb{P}, p \nmid b }[/math] and [math]\displaystyle{ p \nmid s }[/math], then [math]\displaystyle{ r }[/math] is [math]\displaystyle{ \omega }[/math]-transcendental.
Proof:
The denominator [math]\displaystyle{ \widehat{ap s} (bs \pm apt) }[/math] is [math]\displaystyle{ \ge 2p \ge 2\omega - \mathcal{O}({_e}\omega\sqrt{\omega}) > \omega }[/math] by the prime number theorem.[math]\displaystyle{ \square }[/math]
Transcendence of Euler's Constant
For [math]\displaystyle{ x \in {}^{\omega }{\mathbb{R}} }[/math], let be [math]\displaystyle{ s(x) := \sum\limits_{n=1}^{\omega}{\hat{n}{{x}^{n}}} }[/math] and [math]\displaystyle{ \gamma := s(1) - {_e}\omega = \int\limits_{1}^{\omega}{\left( \widehat{\left\lfloor x \right\rfloor} - \hat{x} \right)dx} }[/math] Euler's constant, where rearranging shows [math]\displaystyle{ \gamma \in \; ]0, 1[ }[/math].
If [math]\displaystyle{ {_e}\omega = s(\hat{2})\;{_2}\omega }[/math] is accepted, [math]\displaystyle{ \gamma \in {}^{\omega }\mathbb{T}_{\mathbb{R}} }[/math] is true with a precision of [math]\displaystyle{ \mathcal{O}({2}^{-\omega}\hat{\omega}\;{_e}\omega) }[/math].
Proof:
The (exact) integration of the geometric series yields [math]\displaystyle{ -{_e}(-\acute{x}) = s(x) + \mathcal{O}(\hat{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx }[/math] for [math]\displaystyle{ x \in [-1, 1 - \hat{\nu}] }[/math] and [math]\displaystyle{ t(x) \in {}^{\omega }{\mathbb{R}} }[/math] such that [math]\displaystyle{ |t(x)| < {\omega} }[/math].
After applying Fermat's little theorem to the numerator of [math]\displaystyle{ \hat{p}(1 - 2^{-p}\,{_2}\omega) }[/math] for [math]\displaystyle{ p = \max\, {}^{\omega}\mathbb{P} }[/math], the greatest-prime criterion yields the claim.[math]\displaystyle{ \square }[/math]