If $a=0$, the splitting field is $K$ and its Galois group is trivial. Otherwise choose a root $\alpha$ and a primitive $n$th root of unity $\zeta \in K$. Every root is ${\zeta }^j\alpha$, so $L=K(\alpha )$. Each automorphism is determined by $\sigma (\alpha )/\alpha \in {\mu }_n$. This is an injective group homomorphism, since the roots of unity are fixed by every automorphism. A subgroup of the cyclic group ${\mu }_n$ is cyclic. Thus $\mathrm{Gal}(L/K)$ is cyclic, with order dividing $n$.

This assertion is always true. Ordinal addition obeys $\alpha +\beta \ge \alpha$ and $\alpha +\beta \ge \beta$. Thus both summands are at most ${\omega }_1$. If $0<\beta <{\omega }_1$ and $\alpha <{\omega }_1$, both ordinals are countable, and their disjoint ordered union is countable, so their sum cannot be ${\omega }_1$. If instead $\alpha ={\omega }_1$ and $\beta >0$, appending a nonempty second block gives a sum strictly larger than ${\omega }_1$. These exhaust the possibilities with $0<\beta <{\omega }_1$; hence $\beta =0$ or $\beta ={\omega }_1$.

This assertion is false. Set $\alpha =1$ and $\beta ={\omega }_1$. The limit recursion givesEvery term is countable, and they are cofinal in the first uncountable ordinal. Yet $\alpha =1$ is neither zero nor ${\omega }_1$.

This assertion is false, even for nonzero limit ordinals. Take $\alpha =\omega$ and $\beta ={\omega }^2$. Then $\omega +{\omega }^2={\omega }^2$, while ${\omega }^2+\omega >{\omega }^2$. The first identity follows from the limit recursion, since ${\sup }_{n<\omega }(\omega +\omega n)={\omega }^2$; the second sum strictly extends the initial block of type ${\omega }^2$.

This assertion is always true. Synthetically, either bracketing places three disjoint blocks of order types $\alpha ,\beta ,\gamma$ in that order. The identity on these blocks is an order isomorphism, so both sums have the same order type. This proves associativity of ordinal addition for successor and limit ordinals alike.

For ${\Omega }_0>1$, introduce conformal time by $dt/d\tau =a$. Since $a^{\prime }=a\dot{a}$, the constraint becomesOn the initial expanding branch, separate variables and choose $a=t=\tau =0$ at the Big Bang. With $\alpha =H_0\sqrt{{\Omega }_0-1}$,The second formula follows by integrating $dt/d\tau =a$. The model reaches maximum size at $\alpha \tau =\pi /2$ and recollapses at $\alpha \tau =\pi$.

Today $a=1$ and $H_0>0$ selects the expanding branch, soSubstitution yields the requested ageIn particular the numerator is $\sqrt{{\Omega }_0}-1$, not $\sqrt{{\Omega }_0-1}$; the PDF displays the former. The flat-model limit is $1/(2H_0)$.

The fluid conservation equation is $\dot{\rho }+3H(\rho +P/c^2)=0$. With the radiation equation of state, it becomes $\dot{\rho }+4H\rho =0$, giving $\rho a^4={\rho }_0$. Define the critical density ${\rho }_{\mathrm{crit}}=3H^2/(8\pi G)$ and cosmological density parameter $\Omega =\rho /{\rho }_{\mathrm{crit}}$. Since $a_0=1$,Evaluating the Friedmann equation today gives $H_0^2+k{c}^2=H_0^2{\Omega }_0$, hence

In the Newtonian dust calculation, use a comoving spherical shell of physical radius $R(t)=a(t)\chi$. Homogeneity gives its enclosed mass $M=4\pi \rho R^3/3$, and the shell theorem removes the gravitational force of exterior shells. Assume pressureless matter, conserved enclosed mass, no cosmological constant, and no shell crossing. Its specific conserved mechanical energy isHomogeneity permits the energy per ${\chi }^2$ to be a constant, written $-k{c}^2/2$. Dividing by $a^2$ givesThis is a Newtonian derivation for dust. For radiation the enclosed gravitating mass is not conserved, so this same mechanical-energy argument cannot simply be reused; general relativity supplies the same Friedmann constraint with the total energy density divided by $c^2$, together with pressure-dependent fluid conservation.

At the threshold $P_{\ast }=3+2\sqrt{2}$, the determinant quadratic has a double zero atConsequently the scaled disturbance length in $\cos (s/L)$ isThe physical length is ${\ell }_u{L}_{\mathrm{crit}}=\sqrt{{\ell }_u{\ell }_v}$ at threshold; its full spatial wavelength is $2\pi$ times this length, not this length itself. For much shorter lengths, the large negative diffusion terms smooth both components. For much longer lengths, diffusion contributes little and the stable homogeneous reaction feedback, especially the inhibitor response, suppresses disturbances. The selected length lies between those two stabilization mechanisms.

Assume $Q>1$, so $T(k)<0$ for every wavenumber. A Turing instability therefore occurs exactly when $D(k)<0$ for some $k$. Put $z=k^2\ge 0$. The upward quadratic $1+(1-P)z+P{z}^2$ has a negative minimum on this half-line precisely whenCombining these givesThe unstable band is $z_{-}<k^2<z_{+}$, whereThe degradation-limited diffusion distances are proportional to ${\ell }_u=\sqrt{D_u/{\beta }_u}$ and ${\ell }_v=\sqrt{D_v/{\beta }_v}$. Thus $P=({\ell }_v/{\ell }_u{)}^2$ and the criterion says the inhibitor must travel more than $(1+\sqrt{2})$ times as far as the activator before degradation. Short-range activation and longer-range inhibition permit a local peak to reinforce itself. On a finite domain the allowed wavenumbers must also intersect the unstable band; the displayed condition assumes continuously available modes.

Set $u=1+\hat{u}(t)\cos (s/L)$ and $v=1+\hat{v}(t)\cos (s/L)$ and retain linear terms. With $k=1/L$, the Fourier mode has second derivative $-k^2\cos (ks)$. HenceThe trace and determinant of this linearization are

A nonzero spatially constant steady state satisfies $v=u$ and $v=u^2$, hence $(u,v)=(1,1)$. The first derivatives of $u^2/v-u$ there are $1,-1$, while those of $Q(u^2-v)$ are $2Q,-Q$. ThusSince $Q>0$, both eigenvalues have strictly negative real part exactly when $Q>1$. For $0<Q<1$ the equilibrium is unstable. At $Q=1$ its linear eigenvalues are $\pm i$; this is the marginal boundary and does not give asymptotic stability from linearization. The positive-trace criterion in the hint tests positive real parts; stability requires applying it to $-J$.

Choose the nondimensionalizationThe chain rule gives ${\partial }_{\tau }={\beta }_u{\partial }_t$ and ${\partial }_x^2=({\beta }_u/D_u){\partial }_s^2$. Dividing the first equation by ${\beta }_u$ and the second by ${\beta }_v$ gives the required scaled reaction–diffusion system. Positive degradation and diffusion constants are assumed.

First interpret the decimal expansions as fractional expansions $0.d_1{d}_2\cdots$, with every $d_j\in \{4,5\}$. The resulting set $X_0$ is compact: a diagonal subsequence argument selects successively stabilized digits from any sequence in $X_0$, and the tail bound ${\sum }_{j>m}9\cdot 10^{-j}=10^{-m}$ proves convergence to the corresponding expansion.

It has empty interior. At level $m$ its points lie in $2^m$ decimal intervals, each of length $10^{-m}$, whose total length is $(1/5{)}^m$. An interval of positive length could not lie in all these covers. Thus $X_0$ is closed and nowhere dense. Each translate $X_0+q$ is also nowhere dense, and the rational numbers are countable. The Baire category theorem applied to $\mathbb{R}$ yieldsAny point in this complement has the required property. If the printed description also permits integer digits before the decimal point, there are only countably many such integer parts, and negative expansions merely add reflected translates. This makes $X$ a countable union of translates or reflections of $X_0$. The same countable-union argument therefore still proves that some real $t$ cannot be written as $x+q$.

The Baire category theorem says that in a nonempty complete metric space, a countable intersection of open dense subsets is dense. Equivalently, such a space cannot be a countable union of nowhere dense sets: a set is nowhere dense when the interior of its closure is empty. Indeed the complements of those closures are open dense; their intersection is dense and in particular nonempty, so a countable union of the original sets cannot exhaust the space. The same argument shows that the complement of such a union is dense.

Within the prescribed uniform-density approximation, $n=3M/(4\pi m_p{R}^3)$. Treating the given approximate pressure coefficient as one, the degeneracy pressure and specified kinetic-energy relation giveFor a uniform sphere, $m(r)=M(r/R{)}^3$, so shell assembly givesThusAt fixed mass, this energy has its minimum at $R_{\ast }=2\alpha /\beta$, with $E^{\prime \prime }(R_{\ast })=2\alpha /R_{\ast }^4>0$. Expansion beyond this radius raises the energy, while strong compression costs positive energy growing as $R^{-2}$, faster than the attractive $R^{-1}$ term. This explains radial stability of the model white dwarf. The numerical coefficient in $\alpha$ inherits the approximate pressure law; a relativistic degeneracy law would change this stability argument.

Let the surface be at $R$ with vanishing external pressure. Multiply hydrostatic equilibrium by $4\pi r^3$ and integrate:The other side is $-{\int }_0^{R}4\pi Gm(r)\rho (r)rdr=-{\int }_0^{M}Gm/r(m)dm$. Building the star one shell at a time identifies this as its negative gravitational potential energy $E_{\mathrm{grav}}$. ThereforeIf the star has nonzero surface pressure $P_s$, the integration instead gives $3⟨P⟩V-3P_{s}V=-E_{\mathrm{grav}}$; the requested form assumes $P_s=0$.

At a positive equilibrium of the bounded Hill model, the mutual repression stability criterion becomesThus $mn>1$ is necessary. Because the same $\lambda$ is imposed in both synthesis functions, it is not sufficient for arbitrary unequal positive exponents. The additional equilibrium constraint isHere is an exact criterion for all the printed exponent choices, including nonintegers. If $m>1$ and $n>1$, multistability is possible: for large $\lambda$ the intermediate equilibrium satisfies $x\sim {\lambda }^{(m-1)/(mn-1)}$, $y\sim {\lambda }^{(n-1)/(mn-1)}$, so both increase without bound and $K\to mn>1$. These asymptotics follow by balancing $x{y}^m\sim \lambda$ and $y{x}^n\sim \lambda$; the limiting two algebraic equations have a nonsingular log-Jacobian when $mn\ne 1$, so a nearby exact equilibrium exists for large $\lambda$. If both exponents are at most one, $K<1$ everywhere and multistability is impossible.

For the remaining case, interchange $x,y$ if necessary so $0<m\le 1<n$. For each $x>0$, let $Y(x)>0$ be the unique solution ofwhich exists because the left side is strictly decreasing; for $m=1$ the right side always exceeds one. DefineThen the exact answer for this remaining range isThis is a one-variable test depending only on $m,n$, with no unknown $\lambda$. To prove it, parameterize all equilibria by $\lambda (x)=x(1+Y(x{)}^m)$. Implicit differentiation givesThe denominator is positive, and $\lambda (x)$ goes from zero to infinity as $x$ goes from zero to infinity. A negative-slope portion therefore forces two folds and an interval of three equilibria, with two attracting branches. If $K\le 1$ everywhere, $\lambda$ is strictly increasing apart from possible isolated stationary points, so there is only one equilibrium for each $\lambda$. This is the common-strength Hill repression criterion.

For illustration and a closed test on the boundary $m=1$, $Y=x/(1+x^n-x)$ andobtained at $x^n=(n+1)/(n-1)$. In particular $(m,n)=(1,2)$ has $mn>1$ but maximum $3\sqrt{3}/8<1$, so it cannot be multistable. For positive integer exponents the allowed choices simplify to both exponents at least two, or one equal to one and the other at least four. For arbitrary real exponents the preceding exact criterion is needed; replacing it by $mn>1$ would discard the common-strength constraint.

The unbounded power laws behave differently. Combining their equilibrium equations gives $x^{1-mn}={\lambda }^{1-m}$. If $mn\ne 1$, there is exactly one positive equilibrium,Its slope product is exactly $mn$, so it is attracting if $mn<1$ and a saddle if $mn>1$, but there are never two isolated attracting equilibria. If $mn=1$, equilibria exist only when ${\lambda }^{1-m}=1$; when this holds there is a whole curve of equilibria, with a zero tangent eigenvalue and transverse eigenvalue $-2$. This includes $m=n=1$, $xy=\lambda$, for every $\lambda$. A neutrally degenerate continuum is not bistability between isolated attractors. Thus these power laws do not generate ordinary multistability; boundedness was essential to the outer-root argument in part (a).

The nullclines are $x=f(y)$ and $y=g(x)$, both decreasing curves in the positive quadrant. To their left/right the horizontal velocity is respectively positive/negative; below/above the second nullcline the vertical velocity is positive/negative. Bounded positive synthesis functions give an inward-pointing sufficiently large rectangle, while the axes point into the quadrant. More directly, put $H(x)=f(g(x))-x$: $H(0)>0$, and $H(x)<0$ for large enough $x$. Continuity therefore gives a positive equilibrium.

At an equilibrium, the Jacobian matrix is $J=\left(\begin{array}{cc}-1& f^{\prime }(y_{\ast })\\ g^{\prime }(x_{\ast })& -1\end{array}\right)$, with eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its logarithmic-slope product isbecause $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$. Thus a product exceeding one makes this equilibrium a saddle and gives $H^{\prime }(x_{\ast })>0$. Immediately on its left $H<0$, and immediately on its right $H>0$. Since $H(0)>0$ and $H$ is eventually negative, there must be other roots on both sides, crossing in the attracting direction. For simple roots those have $H^{\prime }<0$, hence two negative Jacobian eigenvalues: the saddle separates at least two stable steady-state alternatives. The divergence is $-2$, so periodic attractors cannot replace these equilibria in a bounded planar region. Nongeneric flat crossings or intervals of equilibria require a nonlinear stability interpretation; the strict saddle inequality alone does not guarantee that every other root is hyperbolic. This is the usual mutual repression stability criterion behind bistability.

The conclusion also has a nonlinear trapping justification. Choose $l<u$ on the left of the saddle with $H(l)>0>H(u)$; such endpoints exist by the signs just proved. The rectangle $[l,u]\times [g(u),g(l)]$ is forward invariant: on the vertical edges, $f(y)-l\ge f(g(l))-l>0$ and $f(y)-u\le f(g(u))-u<0$, while the horizontal-edge velocities point inward or are tangent. The same construction on the right of the saddle gives a second disjoint trapping rectangle. Each contains a nonempty compact attracting invariant set, obtained by intersecting its nested forward images; strict vertical inwardness also rules out invariant boundary trajectories. If a downward crossing is isolated, endpoints can be taken arbitrarily close to it, proving nonlinear stability, and the absence of periodic orbits makes nearby trajectories converge to that equilibrium. Intervals of equilibria instead give attracting alternatives consisting of families, so bounded monotone functions without nondegeneracy do not necessarily produce two isolated hyperbolic attractors.

For the Fourier mode $u_m^n=g^n{e}^{im\xi }$, substitution givesThe denominator is never zero for real $\mu ,\xi$, and numerator and denominator are complex conjugates, so $|g|=1$. Parseval's identity therefore makes each step norm-preserving on square-summable grid data. The method is von Neumann stable for every $\mu >0$, with no upper Courant restriction. It is an implicit nondissipative scheme; unconditional stability does not remove phase error or guarantee accuracy at a large time step.