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Suppose first that a simple continued fraction is eventually periodic. Its periodic tail $\xi$ is fixed by the Möbius transformation represented by the continued-fraction matrix of one full period:ThusThe infinite tail is not rational, since a rational simple continued fraction terminates. Hence $\xi$ is a quadratic irrational, and the finite initial block expresses the original number as another rational Möbius transform of $\xi$. Therefore every eventually periodic simple continued fraction represents a quadratic irrational.

For the purely periodic fraction $x=[7,7,7,\dots ]$,so $x^2-7x-1=0$. Positivity selects

Apply the continued-fraction algorithm to $\sqrt{23}$:The complete quotient then repeats, givingThe convergents through one term before the final $8$ areThe last of these satisfies the Pell equationso a positive solution is

Because $K$ has characteristic $p$, the prime field ${\mathbb{F}}_p$ lies in $K$, andfor every $c\in {\mathbb{F}}_p$. These are all $p$ roots. Moreover $f^{\prime }(t)=-1$, so $f$ is a separable polynomial.

Let $m(t)$ be the minimal polynomial of an algebraic element of $\alpha$ over $K$. Translation by $c\in {\mathbb{F}}_p$ sends an irreducible factor $g(t)$ of $f(t)$ to the irreducible factor $g(t-c)$. The additive group of ${\mathbb{F}}_p$ therefore acts on the irreducible factors, and every orbit has size $1$ or $p$. An orbit of size $p$ would already contribute $p\mathrm{deg}m$ to the degree-$p$ polynomial $f$, forcing $\mathrm{deg}m=1$. Then $\alpha \in K$, all roots $\alpha +c$ lie in $K$, and $L=K$, contrary to the hypothesis.

Thus $m(t-c)=m(t)$ for every $c\in {\mathbb{F}}_p$. If $1\le d=\mathrm{deg}m<p$, comparison of the coefficient of $t^{d-1}$ in $m(t-c)-m(t)$ for $c\ne 0$ gives $-dc$ times the nonzero leading coefficient, a contradiction. Hence $\mathrm{deg}m=p$, and since $m$ is a monic divisor of the monic degree-$p$ polynomial $f$, we have $m=f$. Thus $f$ is irreducible over $K$.

All roots $\alpha +c$ already belong to $K(\alpha )$, so the splitting field isAs the splitting field of a separable polynomial, $L/K$ is a Galois extension. For each $c\in {\mathbb{F}}_p$, irreducibility gives a distinct $K$-automorphismThese exhaust the $p=[L:K]$ automorphisms. This is an Artin–Schreier extension, and

The diagonal Ramsey number $R(s)$ is the least positive integer $N$ such that every red-blue edge colouring of the complete graph $K_N$ contains a monochromatic copy of $K_s$.

For the existence and bound, write $R(r,s)$ for the least $N$ forcing either a red $K_r$ or a blue $K_s$. Given a vertex $v$ of $K_{R(r-1,s)+R(r,s-1)}$, the pigeonhole principle says that at least $R(r-1,s)$ incident edges are red or at least $R(r,s-1)$ are blue. In the first case those red neighbours contain a red $K_{r-1}$, which joins to $v$ to make a red $K_r$, or a blue $K_s$; the second case is symmetric. HenceStarting from $R(1,s)=R(r,1)=1$ and using Pascal's identity gives the binomial upper bound for a Ramsey numberThereforewhich proves both existence and the required estimate.

To determine $R(3)$, color a cycle graph $C_5$ red and its complement graph, also a $C_5$, yellow. This colors $K_5$ without a monochromatic triangle, so $R(3)>5$. At any vertex of a red-yellow $K_6$, at least three of the five incident edges have one color. If an edge among their three other endpoints has that color, it completes a monochromatic triangle with the chosen vertex; if none does, those three endpoints themselves form a triangle of the other color. Thus $R(3)\le 6$, and

The same local argument proves uniqueness on $K_5$. No vertex can have three incident edges of one color: their other endpoints would either contain an edge of that color or form a triangle of the other color. Every vertex therefore has vertex degree at most two in each color, and because the two degrees sum to four, each is exactly two. Each color class is consequently a regular graph of degree two on five vertices, hence a single $C_5$. Up to relabelling, the coloring is exactly the cycle-and-complement coloring above.

Finally, the monochromatic-triangle counting formula gives, for a coloring of $K_6$ with red degrees $d_v$,The bound is attained by coloring all edges of $K_{3,3}$ red and the two complementary copies of $K_3$ yellow: there are no red triangles and exactly two yellow triangles. The least requested number is therefore

The epsilon-recursion theorem says that for every definable operation $G$ assigning a set $G(x,h)$ to each set $x$ and each function $h$ with domain $x$, there is a unique class function $F$ such thatfor every set $x$.

For uniqueness, suppose $F$ and $F^{\prime }$ both satisfy the recursion. If they agree on every $y\in x$, then their restrictions to $x$ agree and henceThe epsilon induction principle therefore gives $F=F^{\prime }$.

For existence, call a function $f$ coherent on a transitive set $T$ whenThe same epsilon-induction argument shows that any two coherent functions agree on the intersection of their transitive domains. Assuming coherent functions have been constructed below every member of $x$, their compatible union defines all values below $x$; adjoining the value prescribed by $G(x,\cdot )$ extends the construction to $x$. Epsilon induction establishes this construction for every $x$. The compatible local functions therefore unite to the required class function $F$.

A relation $r$ on $x$ is well-founded when every nonempty subset of $x$ has an element with no $r$-predecessor in that subset. It is extensional when

The Mostowski collapse theorem states that every well-founded extensional relation $(x,r)$ is isomorphic to membership on a unique transitive set $M$. Define recursivelyWell-founded recursion makes $\pi$ well defined. Epsilon-style induction along $r$, together with extensionality, shows that $\pi$ is injective. By definition,so $\pi$ is an isomorphism from $(x,r)$ onto $(M,\in )$, where $M=\mathrm{ran}\pi$. If $z\in \pi (a)\in M$, then $z=\pi (b)$ for some $bra$, so $z\in M$; hence $M$ is transitive. If another collapse ${\pi }^{\prime }$ has transitive range, induction gives ${\pi }^{\prime }(a)=\{{\pi }^{\prime }(b):bra\}=\pi (a)$, proving uniqueness.

Finally, suppose $x$ is finite. Its collapse $M$ is a finite transitive set, so every member of $M$ is hereditarily finite. Therefore every well-founded extensional relation on $x$ collapses to a transitive subset of $V_{\omega }$.

Conversely, let $x$ be infinite. Choose the initial infinite ordinal $\kappa$ equinumerous with $x$. The successor ordinal $\kappa +1$ has the same cardinality as $\kappa$. Pulling its membership relation back along a bijection $x\to \kappa +1$ gives a well-founded extensional relation on $x$ whose collapse is $\kappa +1$. But $\kappa \in \kappa +1$ is not hereditarily finite, so $\kappa +1⊈V_{\omega }$. Consequently,

Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for ${\dot{q}}_i(q,p,t)$, and the Hamiltonian is the Legendre transformIts differential isbecause the $p_{i}d{\dot{q}}_i$ terms cancel. The Euler-Lagrange equation identifies $\partial \mathcal{L}/\partial q_i={\dot{p}}_i$, so coefficient comparison gives Hamilton's equations

For the symmetric top, the momenta areThusThe Legendre transform gives

Both $\varphi$ and $\psi$ are cyclic coordinates, so $p_{\varphi }$ and $p_{\psi }$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\varphi ,\psi$,under the Poisson bracket. Hence three independent first integrals in involution are

The $u^2/v$ term makes $u$ locally autocatalytic, while increasing $v$ suppresses this production. Thus $u$ is the activator and $v$ is the inhibitor.

At a positive spatially homogeneous equilibrium,Hence $u=2bv$ and $v=u^2$, givingThe reaction Jacobian matrix there isBecause $b>0$, the trace-determinant stability criterion shows that the spatially uniform reaction kinetics are stable exactly when

For a perturbation proportional to $e^{\lambda t+ikx}$, put $q=k^2$. Its growth rates are the eigenvalues ofThe trace is $2b-1-(d+1)q<0$ in the reaction-stable regime, while the determinant isA Turing instability occurs exactly when this upward-opening quadratic is negative for some $q>0$. By the two-species diffusion-driven instability criterion, this requiresTogether these reduce toThus the unstable region in the $(b,d)$-plane lies below the straight line $d=(6-4\sqrt{2})b$ and to the left of $b=1/2$.

Inside this region, the unstable squared-wavenumber band isAt the bifurcation boundary the two roots coincide, soSubstituting $d=(6-4\sqrt{2})b$ yields

The set $K$ is compact, and its complement is connected: two disjoint closed disks do not enclose any bounded complementary component. Since $f$ is holomorphic on the open neighbourhood $\Omega$ of $K$, the polynomial Runge theorem gives, for each $n$, a polynomial $p_n$ such thatHence

Define a function on the disconnected open set $\Omega$ byIt is holomorphic on $\Omega$. Applying the first part produces polynomials $P_n$ with

For an obstruction, takeThese are disjoint nonempty bounded closed sets. If polynomials $Q_n$ converged uniformly to zero on $K_1$, the maximum modulus principle on a bounded domain would giveThey therefore cannot also satisfy $Q_n(0)\to 1$. Thus this pair $K_1,K_2$ has the requested non-approximation property.