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Theorem of the month

Universal multistep theorem

For [math]\displaystyle{ n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, {\downarrow}\overset{\rightharpoonup}{x} \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, }[/math] [math]\displaystyle{ f : [a, b]\times{}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\overset{\rightharpoonup}{x}) := g_{\acute{k}}(x) }[/math], and [math]\displaystyle{ g_0(a) = f(\overset{\leftharpoonup}{a}, y_0, ... , y_{\acute{n}}) }[/math], the TS of the initial value problem [math]\displaystyle{ y^\prime(x) = f(x, y((\rightharpoonup)^0 x), ... , y((\rightharpoonup)^{\acute{n}} x)) }[/math] of order [math]\displaystyle{ n }[/math] implies

[math]\displaystyle{ y(\overset{\rightharpoonup}{x}) = y(x) + {\downarrow}\overset{\rightharpoonup}{x}{\pm}_{k=1}^{p}{\left (g_{p-k}(\overset{\rightharpoonup}{x}){\LARGE{\textbf{+}}}_{m=k}^{p}{\widetilde{m!}\tbinom{\acute{m}}{\acute{k}}}\right )} + \mathcal{O}(({\downarrow}\overset{\rightharpoonup}{x})^{\grave{p}}).\square }[/math]

Goldbach’s theorem

Goldbach’s theorem: Every even whole number [math]\displaystyle{ > 2 }[/math] is the sum of two primes.

Proof:

For [math]\displaystyle{ \hat{m} + \hat{n} = p_{m+r,n-r} + q_{m+r,n-r} + r, r \in \{0, 2, … , \max(g(n))\} }[/math], it follows alike [math]\displaystyle{ \hat{m} + \hat{n} = p_{m+s,n-s} + q_{m+s,n-s} + s, }[/math] [math]\displaystyle{ s \in \{0, 2, … , \max(g(n)) + 2\} }[/math]. This implies [math]\displaystyle{ \hat{m} + \hat{n} + 2 = p_{\grave{m}+r,\grave{n}-r} + q_{\grave{m}+r,\grave{n}-r} + r, r \in \{0, 2, … , \max(g(\grave{n}))\} }[/math]. Induction yields then the claim by the previous theorem.[math]\displaystyle{ \square }[/math]

Foundation theorem

Only the postulation of the axiom of foundation that every nonempty subset [math]\displaystyle{ X \subseteq Y }[/math] contains an element [math]\displaystyle{ x_0 }[/math] such that [math]\displaystyle{ X }[/math] und [math]\displaystyle{ x_0 }[/math] are disjoint guarantees cycle freedom.

Proof:

Set [math]\displaystyle{ X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\} }[/math] and [math]\displaystyle{ x_{\acute{n}} := \{x_n\} }[/math] for [math]\displaystyle{ m \in {}^{\omega}\mathbb{N} }[/math] and [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 2}\} }[/math] .[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics