Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3

Put $\alpha =\sqrt{N}$. Every positive solution satisfiesAfter reducing $x/y=p/q$, we still have $|\alpha -p/q|<1/(2q^2)$. Here is the continued fraction convergent criterion with its proof. If $p/q$ is reduced and obeys this inequality, it is a strict best approximation of the second kind: for any different $r/s$ with $0<s\le q$, the assumption $|s\alpha -r|\le |q\alpha -p|$ would givea contradiction. Choose consecutive continued fraction convergents $p_j/q_j,p_{j+1}/q_{j+1}$ with $q_j\le q<q_{j+1}$. Their determinant is $\pm 1$, their errors $e_j=q_j\alpha -p_j$ alternate in sign, and every integer pair $(r,s)$ is uniquely $a(p_j,q_j)+b(p_{j+1},q_{j+1})$. For $0<s<q_{j+1}$, either $b=0$, or $a,b$ have opposite signs: positive $a,b$ would make $s\ge q_{j+1}$, negative $a,b$ make $s<0$, and $a=0$ is impossible. Opposite signs make $a{e}_j$ and $b{e}_{j+1}$ have the same sign, so $|s\alpha -r|\ge |e_j|$. If $b=0$, reduction gives the $j$th convergent itself. Thus a strict best approximation must be a convergent. The determinant and alternating-error facts follow immediately from the recurrence $p_j=a_j{p}_{j-1}+p_{j-2}$, $q_j=a_j{q}_{j-1}+q_{j-2}$ and the positive continued-fraction tail formula. Therefore the reduced ratio $x/y$ is a convergent of $\sqrt{N}$. Without assuming $\gcd (x,y)=1$, the unreduced pair may be a common multiple of the convergent's numerator and denominator; for example $(10,2)$ for $N=24,M=4$ reduces to $5/1$.

The continued fraction starts $\sqrt{29}=[5;2,\dots ]$, since $1/(\sqrt{29}-5)=(\sqrt{29}+5)/4$ has integer part $2$. Its next continued fraction convergent is $5+1/2=11/2$. Direct substitution gives

Take the holomorphic function $f(z)=1/z$ on a neighborhood of the annulus. Every polynomial has integral zero around the unit circle by the Cauchy integral theorem, while ${\int }_{|z|=1}{z}^{-1}dz=2\pi i$. Uniform convergence on the annulus would imply convergence of these contour integrals, since their difference is bounded by $2\pi$ times the uniform error. Thus uniform polynomial approximation on the full annulus need not exist.

Choose the argument in $(0,2\pi )$ on the slit annulus and letThese compact sets increase to the slit annulus, and each has connected complement: the omitted angular corridor joins the inner disk to the outside of the outer disk. The polynomial Runge theorem says that a function holomorphic near a compact set with connected complement can be approximated uniformly there by polynomials. Choose $p_j$ with ${\sup }_{K_j}|p_j-f|<1/j$. Every compact subset $K$ of the slit annulus has positive distance from the removed positive ray, so $K\subseteq K_j$ for all sufficiently large $j$. Hence $p_j\to f$ uniformly on every such compact subset.

If the polynomials converge uniformly on the full annulus to $g$, their limit is continuous and equals $f$ on the slit annulus by part (ii). That subset is dense in the full annulus, so $g=f$ everywhere. The Cauchy integral theorem and uniform convergence on the unit circle then give

Let $H_k$ be the open hyperbolic half-plane beyond ${\gamma }_k$, opposite the central region $U$. These three half-planes have disjoint closures in the disk, and hyperbolic reflection $J_k$ exchanges $H_k$ and its complementary half-plane. In particular $J_k(U^{\circ })\subset H_k$ and $J_i(H_j)\subset H_i$ when $i\ne j$. A reduced nonempty reflection word therefore sends an interior point of $U$ into the half-plane corresponding to its first letter. It cannot be the identity. This ping-pong lemma argument proves that the reflection group isThe usual reflection tiling makes $U$ a fundamental domain for $W$: successive reflection across an offending side reduces a point to $U$, and the disjoint-side configuration yields locally finite tiles. Equivalently this is the disjoint-geodesic case of the reflection fundamental-domain theorem; there are no vertex relations because the sides never meet.

The orientation-preserving subgroup consists of even words. Every pair $J_i{J}_j$ can be expressed in $A=J_2{J}_1$ and $B=J_3{J}_2$: the other nontrivial pairs are their inverses and $J_3{J}_1=BA$, $J_1{J}_3=A^{-1}{B}^{-1}$. Thus it is exactly $G$. A fundamental set is $U\cup J_{2}U$, with one representative chosen on each paired side; its interior is a fundamental domain. To see there are no relations between $A,B$, the reflection-word tree has vertices the reduced words and edges joining $w$ to $w{J}_i$. The even subgroup acts freely on this tree; its quotient has two vertices joined by three edges. Choosing the $J_2$ edge as a spanning tree leaves two independent loops, represented by $A$ and $B$. Reduced loops in a graph form a free group, soThe quotient is the double of the central three-sided region along its geodesic sides: an orientable pair of pants, topologically a sphere with three points removed. Its three ends are hyperbolic funnels, not cusps, because the sides are ultraparallel and the peripheral transformations have positive translation lengths. The compact convex core has three geodesic boundary circles; the complete quotient itself has no boundary.

A linear code of length $n$ over ${\mathbb{F}}_q$ is a linear subspace $C\subseteq {\mathbb{F}}_q^n$. A parity-check matrix $P$ satisfies $C=\ker P$. Its minimum distance of a code is the least Hamming distance between distinct codewords, equivalently the least Hamming weight of a nonzero word, because differences remain in the code.

For the usual binary bar product of binary linear codes, take equal-length binary codes with $C_2\subseteq C_1$ and defineIt is linear, of length $2n$ and dimension $\dim C_1+\dim C_2$, since the parametrization is injective. If $v=0$ and the word is nonzero, its weight is $2\mathrm{wt}(u)\ge 2d(C_1)$. If $v\ne 0$, coordinatewise $\mathrm{wt}(u)+\mathrm{wt}(u+v)\ge \mathrm{wt}(v)\ge d(C_2)$. The words $(u,u)$ with $u$ of minimum nonzero weight and $(0,v)$ with $v$ of minimum weight achieve the two candidate bounds. ThusFor a word $(x,y)$, membership is equivalent to $P_{1}x=0$ and $P_2(x+y)=0$. Consequently a parity-check matrix of a bar product isThe same distance argument actually needs only equal block lengths and a common field, not nesting. Over a general field replace the second row block by $(-P_2,P_2)$, so it tests $y-x\in C_2$. The displayed binary construction uses the conventional nesting.

Assume $n>p\ge 1$, the ordinary nonsaturated Gaussian linear model. The log likelihood is, up to a constant,The normal equations and full column rank give the least-squares estimator $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Write $H=X(X^{T}X{)}^{-1}{X}^T$ and $Q=I-H$, an orthogonal projection of rank $n-p$. Maximizing in ${\sigma }^2$ yieldsSince $QX=0$, the residual sum of squares is ${\epsilon }^{T}Q\epsilon$. Expanding its expectation uses only $\mathbb{E}{\epsilon }_i{\epsilon }_j={\sigma }^2{\delta }_{ij}$:This does not use Cochran's theorem. It proves bias $-p{\sigma }^2/n$ and givesIn the saturated case $n=p$, residuals vanish identically, the likelihood is unbounded as the positive variance tends to zero, and the unbiased residual estimator is unavailable. Thus the usual variance-estimation request requires $n>p$.

With no chemotaxis, the stationary oxygen equation is $D_c{c}^{\prime \prime }=k\bar{n}$. Symmetry and the two prescribed surface values giveBy Fick's first law, the vertical oxygen flux is $j_c=-D_c{c}^{\prime }=-k\bar{n}z$. Thus each face supplies an inward flux $k\bar{n}d/2$ per unit area, and the total influx is $k\bar{n}d$, equal to the integrated consumption across the layer. The profile is physically nonnegative only if $c_0\ge k\bar{n}{d}^2/(8D_c)$; otherwise the constant-consumption approximation must be modified.

For bacteria retained in the fluid, impose zero bacterial flux at the faces. The stationary bacterial equation says that $j_n=-D_n{n}^{\prime }+\mu n{c}^{\prime }$ is constant, hence zero. Integrating $n^{\prime }/n=(\mu /D_n)c^{\prime }$ gives $n(z)=n_0\exp (\mu c(z)/D_n)$, with $n_0$ fixed by the total bacterial population. Substitution into the oxygen equation provesZero bacterial through-flux is the boundary assumption behind this equilibrium relation.

Use binary expansions ending in zeros at dyadic rationals. The doubling map shifts the expansion left:In ordinary coordinates it is $2x$ on $[0,1/2)$ and $2x-1$ on $[1/2,1)$. The branches are continuous; the left limit at $1/2$ is $1$ while its value and right limit are $0$. This is its only discontinuity in the half-open interval.

For every $N\ge 2$, the repeating binary block $0^{N-1}1$ givesIts least period is $N$: a shorter repeating block would make more than one digit equal to one in a length-$N$ period. The orbit avoids the discontinuity, and its multiplier is $(F^N{)}^{\prime }=2^N>1$, so these periodic orbits are repelling and unstable. The same multiplier argument applies to every periodic orbit, with the endpoint fixed point treated one-sidedly.

In the two-branch horseshoe formulation of Glendinning chaos, disjoint subintervals are each mapped onto the same interval by an iterate. Here $J=(0,1)$, $K_0=(0,1/2)$ and $K_1=(1/2,1)$ already satisfy $F(K_0)=F(K_1)=J$. Their inverse branches $x\mapsto x/2$ and $x\mapsto (x+1)/2$ realize every binary itinerary. This is a horseshoe for a piecewise continuous map; on the circle the endpoint jump disappears. The symbolic explanation also shows dense periodic points, by repeating any prescribed finite prefix, and topological transitivity, because an iterate maps any sufficiently small dyadic cylinder onto $(0,1)$. Finally every neighborhood contains points sharing an arbitrarily long prefix but with different subsequent digits; an appropriate iterate separates their images by a fixed amount, for instance more than $1/4$. This proves sensitive dependence on initial conditions as well. The binary shift has full two-symbol chaotic dynamics.

For $x>0$, differentiate the absolutely convergent geometric series to obtainIf $\sigma =\Re z>1$, the sum of integrals of absolute values isThus Fubini theorem permits termwise integration, even for complex $z$. The substitution $u=mx$ and the definition of the Gamma function giveThe missing $m=1$ terms cancel in the difference of the two Riemann zeta functions.

The quadratic first integrals of the torque-free Euler equations for a torque-free rigid body areDifferentiation and substitution of all three Euler equations make each derivative vanish: the coefficients respectively sum to $(I_2-I_3)+(I_3-I_1)+(I_1-I_2)=0$ and $I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)=0$. At either endpoint these constants are $2E=I_2{\Omega }^2$, $L^2=I_2^2{\Omega }^2$, proving $L^2=2E{I}_2$.

Eliminating the other squared components gives, with $s={\omega }_2/\Omega$,Hence $|s|\le 1$ and the second Euler equation gives $ds/d\tau =\pm (1-s^2)$, where $\tau =\Omega t\sqrt{(I_2-I_1)(I_3-I_2)/(I_1{I}_3)}$ increases with $t$. The specified orbit goes from $s=1$ to $s=-1$, so its sign must be negative. The sign in the original PDF's reduced equation is therefore inconsistent with its endpoint conditions. The corrected heteroclinic satisfiesIntegrating by separation gives this family for $|s|<1$; the integration constant is a time translation. The printed positive-sign equation instead has the increasing solutions $s=\tanh (\tau -{\tau }_0)$ and cannot meet the stated endpoints. For completeness its other real solutions are $s=\pm 1$ and the branches $s=\coth (\tau -{\tau }_0)$ on intervals avoiding their pole. These latter branches are excluded by the conserved-energy constraint $|s|\le 1$. The scalar heteroclinic is unique up to translation. For the full angular-velocity vector there are two sign-related branches: simultaneously reversing ${\omega }_1$ and ${\omega }_3$ preserves all three Euler equations and both endpoints. Full-vector uniqueness therefore also requires choosing one of those branches.

For a comoving fluid volume, $V\propto a^3$ and $E=\rho c^{2}V$. The first-law relation $dE=-PdV$ gives $c^2(Vd\rho +\rho dV)=-PdV$. Since $\dot{V}/V=3\dot{a}/a$, the cosmological perfect-fluid continuity equation isFor the Newtonian energy argument, assume negligible pressure, spherical homogeneity, Newtonian gravity and constant enclosed mass. A test particle at physical radius $r=a\chi$ has constant specific energyDivide by $a^2{\chi }^2/2$ and set $k=-2ϵ/(c^2{\chi }^2)$ to obtain the Friedmann equationHomogeneity makes $k$ independent of the chosen comoving shell. In the relativistic interpretation it specifies the spatial-curvature sign: positive, zero or negative $k$ gives closed spherical, flat or open hyperbolic spatial geometry, with a conventional rescaling allowing $k=1,0,-1$. Newtonian mechanics itself derives the energy constant, not the spatial metric. When pressure is appreciable, the general relativistic equations still have the displayed Friedmann form, but the constant-mass Newtonian test-particle derivation cannot be applied without that qualification.

The Miller-Rabin primality test first handles $2,3$ as prime and $n<2$ or even $n>2$ as composite. For odd $n>3$, write $n-1=2^{s}d$ with $d$ odd and choose a base $2\le a\le n-2$. A nontrivial $\gcd (a,n)$ proves compositeness. Otherwise compute $x=a^d\mathrm{mod}n$: the round passes if $x=1$ or $x=-1$. If neither holds, square successively, at most $s-1$ times, and pass if a value $-1$ occurs; otherwise declare composite. An occurrence of $1$ before $-1$ exhibits a nontrivial square root of one and is also a failure. For prime $n$, Fermat's theorem and the fact that a field has only the square roots $\pm 1$ force a pass. The standard strong-liar bound says that an odd composite passes for at most one quarter of the bases; independent random rounds therefore have error probability at most $4^{-r}$ after $r$ rounds. Passing is a probable-prime verdict, not a proof of primality. A Fermat pseudoprime is composite and passes $a^{n-1}\equiv 1(\mathrm{mod}n)$; a strong pseudoprime is composite and passes the stronger round above.

Since $p\ge 5$ is odd, $2^p\equiv 2(\mathrm{mod}3)$ andis a product of two odd integers greater than one. Thus $N$ is odd and composite. Also $2^{2p}=4^p\equiv 1(\mathrm{mod}N)$. Fermat's theorem modulo $p$ gives $p\mid 4^{p-1}-1$, and $p\ne 3$, so $p\mid N-1=4(4^{p-1}-1)/3$. This number is even; hence $2p\mid N-1$. Thereforeproving that $N$ is a base-two Fermat pseudoprime.

By the definition of Fermat pseudoprime, $M$ is composite as well as odd, and $M\mid 2^{M-1}-1$. Put $L=2^M-1$. A proper factorization $M=ab$ makes $2^a-1$ a proper divisor of $L$, so $L$ is composite. NowSince $2^M\equiv 1(\mathrm{mod}L)$, $2^d\equiv 1(\mathrm{mod}L)$. This is the first passing condition of the Miller-Rabin primality test, with $s=1$. Thus $2^M-1$ is a base-two strong pseudoprime. If “pseudoprime” were used to include primes, the compositeness conclusion would fail, for instance $M=3$ gives $L=7$; the standard composite definition matters.

For a continuous loop in $\mathbb{C}\setminus \{0\}$, choose a continuous argument lift $\theta :[0,1]\to \mathbb{R}$ with $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift is constructed by subdividing the interval into pieces whose images lie in an argument chart and matching the arguments at their endpoints. The winding number of a continuous closed path isChanging the initial lift adds one constant multiple of $2\pi$ and does not change this integer.

A loop homotopy is a continuous $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ with $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$ and $H(s,0)=H(s,1)$ for every $s$. Compactness gives a positive lower bound $r$ for $|H|$ and uniform continuity. Partition the $s$ interval finely enough that neighboring loops $\eta ,\delta$ satisfy $|\eta (t)-\delta (t)|<r$. Their quotient lies in $|z-1|<1$, and has a single-valued continuous argument with equal endpoint values. Its winding number is zero. Arguments of a product add, so $w(\eta )=w(\delta )+w(\eta /\delta )=w(\delta )$. Applying this at every step provesThis proves homotopy invariance of winding number directly, including continuous curves without differentiability.

The principal argument is continuous and single-valued on $\mathbb{C}\setminus (-\infty ,0]$. The given loop stays in this slit plane, and its initial and final values have the same principal argument. The definition therefore gives $\boxed{w(\gamma ;0)=0}$.

Here the target $C$ of $g$ is the unit circle. The boundary loop $g(e^{2\pi it})$ extends through the disk, so the radial homotopy $g(s{e}^{2\pi it})$ contracts it to a constant loop. Thus its winding number is zero. On the boundary put $h(z)=g(z)/z^n$, also circle-valued. Adding argument lifts gives $w(h)=0-n=-n$. If $z^n+g(z)$ never vanished, $h$ would avoid $-1$, which is the only negative real point of the unit circle. Part (iii) would give $w(h)=0$, contradicting $n\ne 0$. Therefore there is a boundary point with $z^n+g(z)=0$. The circle-valued hypothesis is essential; it is not a claim about arbitrary complex-valued $g$.

Introduce dimensionless quantities $\tilde{S}=S/S_0$, $\tilde{I}=I/S_0$, $\tilde{x}=x\sqrt{r{S}_0/D}$ and $\tilde{t}=r{S}_{0}t$. Substitution and cancellation of $r{S}_0^2$ give the stated spatial SIR model withDrop tildes. A travelling wave with $z=x-ct$ obeysAt the leading edge $S=1,I=0$, the infective equation linearizes to $I^{\prime \prime }+c{I}^{\prime }+(1-\lambda )I=0$. For an invading epidemic the susceptible level must exceed the removal threshold, $\lambda <1$. A nonoscillatory decaying tail has exponentsIf the discriminant is negative, a nonzero real tail oscillates in sign, violating nonnegative infective density. Thus for a right-moving waveThis is a necessary admissibility condition from the leading-edge calculation, not by itself a global existence proof.

To derive the final size relation for an epidemic, use $I=c{S}^{\prime }/S$ and integrate the infective ODE across the wave. The tails have $I,I^{\prime }\to 0$, givingThereforeFor a nontrivial epidemic $0<\sigma <1$. The function $s-\lambda \log s$ is strictly convex, minimized at $s=\lambda$. For $0<\lambda <1$ it has the trivial root $1$ and one further root $0<\sigma <\lambda$. For $\lambda \ge 1$, it exceeds one for every $s<1$, so only the no-epidemic root remains; this also proves the necessity of $\lambda <1$. On the nontrivial branch,It rises from $\sigma \sim e^{-1/\lambda }$ as $\lambda \downarrow 0$ to $\sigma =1-2(1-\lambda )+O((1-\lambda {)}^2)$ as $\lambda \uparrow 1$. The sketch distinguishes this branch from the always-present trivial root.

For parameter-dependent centre manifold theorem, append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the eigenvalues are $0,-1,0$: the $y$ direction is strongly stable and $(x,\mu )$ span the extended centre directions. An extended centre manifold for a parameter is an invariant graph of a function $y=h(x,\mu )$ tangent to $y=0$; nearby trajectories approach its reduced dynamics along stable fibers. The stable eigenvalue remains separated from zero, so local bifurcation and stability are governed by this reduction. This is the parameter-extension content of the requested stable/centre-manifold discussion.

The centre-manifold invariance equation isReflection symmetry permits an invariant graph of a function with $h(-x,\mu )=h(x,\mu )$ and $h(0,\mu )=0$. At quadratic order the left side vanishes and the right side is $-h-x^2$, giving $h=-x^2+O(|\mu |x^2+x^4)$. More explicitly its first mixed coefficient is $2\mu x^2$, found by equating the $\mu x^2$ terms. The reduced equation isThus the origin is locally asymptotically stable for $\mu <0$, becomes a saddle for $\mu >0$, and creates two stable equilibria $x=\pm \sqrt{\mu /2}+O({\mu }^{3/2})$, $y=-\mu /2+O({\mu }^2)$. This is a supercritical pitchfork bifurcation. The exact nearby nonzero branches provide a check: the equilibrium equations givewhich are real and nonzero for small positive $\mu$. The omitted branch $y=1+\sqrt{1+\mu }$ is not near the origin. At $\mu =0$ the leading $-2x^3$ also attracts, although at an algebraic rather than exponential rate.

For a nonrelativistic particle, $E(p)=m{c}^2+p^2/(2m)$. The Gaussian integral yieldsIndeed ${\int }_0^{\infty }{p}^2{e}^{-p^2/(2m{k}_{B}T)}dp=\sqrt{\pi }(2m{k}_{B}T{)}^{3/2}/4$, which combines with $4\pi /h^3$ to give the prefactor. For the two species this givesChemical equilibrium imposes ${\mu }_a+{\mu }_{\alpha }={\mu }_b+{\mu }_{\beta }$. If the two massless bath species have zero chemical potentials, then ${\mu }_a={\mu }_b$; their masses alone do not imply this chemical-potential condition.

Weak reactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$, with leptons treated as relativistic, interconvert neutrons and protons. Negligible lepton chemical potentials give approximately $n_n/n_p=\exp [-(m_n-m_p)c^2/(k_{B}T)]$, since their degeneracies and masses are nearly equal. Equilibrium favors protons increasingly as temperature falls. But the weak reaction rate decreases faster than the expansion rate and eventually falls below it: cosmological weak freeze-out prevents the ratio from following its equilibrium exponential to zero. Free neutrons subsequently decay, reducing the frozen ratio further.

In Big Bang nucleosynthesis, cooling eventually makes deuterium survive photodissociation. Nuclear reactions then build helium through deuterium and the light intermediate nuclei, locking almost all surviving neutrons into stable helium-four. Neglecting binding-energy corrections and taking $r=n_n/n_p\le 1$, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$ andFor example $r=1/7$ gives a helium mass fraction $1/4$, with the remaining baryons predominantly hydrogen.

The cumulative hierarchy is defined by transfinite recursion:The rank of a set is $\mathrm{rk}(x)={\sup }_{y\in x}(\mathrm{rk}(y)+1)$, equivalently the least $\alpha$ such that $x\subseteq V_{\alpha }$. The standard hierarchy properties give $\mathrm{rk}(V_{\alpha })=\alpha$, so every ordinal occurs as a set rank; the ordinal $\alpha$ itself is another example of rank $\alpha$.

Let $x$ be transitive of rank $\alpha$, and fix $\beta <\alpha$. The supremum formula gives an element of $x$ whose rank is at least $\beta$. Choose such an element $y$ of least possible rank. If $\mathrm{rk}(y)>\beta$, the same supremum formula gives $z\in y$ with rank at least $\beta$. Transitivity gives $z\in x$, and membership strictly decreases rank, contradicting the choice of $y$. Thus $x$ contains an element of every rank below $\alpha$.

A nonempty finite set has rank the maximum of finitely many successor ordinals, hence a successor ordinal; the empty set has rank zero. Conversely, for any $\beta$, the singleton $\{V_{\beta }\}$ has rank $\beta +1$. ThusA finite transitive set of rank $\alpha$ contains, by the proved property, distinct elements of every rank $\beta <\alpha$. Finiteness therefore forces $\alpha <\omega$. Every finite ordinal is itself a finite transitive set of that rank. Consequently

Take a uniformly random ordering of the vertices and select each vertex that precedes every neighbor. The selected set is independent: if two adjacent vertices were both selected, each would have to precede the other. Among the $d(v)+1$ vertices consisting of $v$ and its neighbors, each is equally likely to come first, so the selection probability is $1/(d(v)+1)$. Linearity of expectation gives expected size ${\sum }_{v}1/(d(v)+1)$. At least one ordering attains at least this average; its integer size is therefore at least the ceiling. This proves the Caro-Wei bound

Write $r=n/m$. The Cauchy-Schwarz inequality givesThe proved Caro-Wei bound therefore gives an independent set of at least $m$ vertices.

No. The disjoint union of $m$ copies of the complete graph $K_{n/m}$ has exactly the prescribed number of edges and has independence number $m$: an independent set can choose at most one vertex from each component, and choosing one from every component attains $m$.

Choose the radicals with positive imaginary part. The roots of unity areThe first comes from $z^2+z+1=0$. For the second, $t={\zeta }_5+{\zeta }_5^{-1}$ obeys $t^2+t-1=0$, so $t=(\sqrt{5}-1)/2$; solving $z^2-tz+1=0$ gives the displayed expression. The Kummer extension towers areEach step is quadratic and the base contains the second roots of unity. The second tower really has degrees $2,2$: its intermediate field is real, while the added radical is nonreal, and $\sqrt{5}=1+2({\zeta }_5+{\zeta }_5^{-1})$ belongs to the cyclotomic field.