Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4

An odd composite integer $n>1$ coprime to $b$ is a Fermat pseudoprime to base $b$ if $b^{n-1}\equiv 1(\mathrm{mod}n)$. A Carmichael number is a composite integer satisfying this congruence for every integer $b$ coprime to $n$. The Korselt criterion says that a composite $n$ is a Carmichael number exactly when it is squarefree and $p-1$ divides $n-1$ for every prime divisor $p$ of $n$.

By the Korselt criterion, the prime factors are distinct. There cannot be just one prime factor, since a Carmichael number is composite. If there were exactly two, write $n=pq$ with $p<q$. Then $q-1\mid pq-1=p(q-1)+(p-1)$, so $q-1\mid p-1$, impossible because $0<p-1<q-1$. Hence at least three distinct prime factors are necessary.

The prime factorization is $561=3\cdot 11\cdot 17$, so it is composite and squarefree. Moreover $3-1=2$, $11-1=10$ and $17-1=16$ all divide $560=561-1$. The Korselt criterion therefore proves that 561 is a Carmichael number.

The complex Taylor theorem says that an analytic function has the expansionthroughout every disk centred at $z_0$ contained in $D$. In particular it converges for $|z-z_0|<\mathrm{dist}(z_0,\mathbb{C}\setminus D)$, and uniformly on every smaller closed disk. The coefficients are uniquely its derivatives divided by factorials. The convergence radius can be larger if the function extends beyond the boundary of $D$; the theorem does not say that the given domain determines its exact radius.

An elementary geometric-series construction suffices, without assuming a general polynomial-approximation theorem. Rotate the square by a multiple of $\pi /2$ so that $\Re w>1$. Put $c=-R$ with $R>0$. Every point of the square then satisfies $|z-c{|}^2\le (R+1{)}^2+1$, whereasfor sufficiently large $R$. Thus the whole square lies inside a disk centred at $c$ whose radius is less than $|w-c|$. The other three ways $w$ can lie outside the square are handled by the indicated rotations, after which rotating back preserves polynomial form.

Now expandLet $\rho ={\max }_{z\in K}|z-c|/|w-c|<1$. The tail after degree $m$ is uniformly bounded by ${\rho }^{m+1}/[|w-c|(1-\rho )]$, tending to zero. Hence the displayed partial sums are the required uniform polynomial approximations. This is the Taylor series about a carefully chosen distant centre, rather than a potentially inadequate series about $0$.

Suppose a connected subset $C\subset F$ contains distinct points $a,b$. The distance function $x\mapsto |x-a|$ is $1$-Lipschitz, and its image on $C$ is connected and contains $0$ and $|b-a|$. It therefore contains the whole interval between those values. A Lipschitz map cannot increase Hausdorff dimension: taking images of arbitrarily fine covers multiplies their diameter powers by at most a fixed constant. Since a nondegenerate interval has Hausdorff dimension one, ${\dim }_{H}C\ge 1$, contradicting ${\dim }_{H}F<1$. Thus every connected subset is a singleton, so $F$ is totally disconnected.

A Möbius transformation pairs two disks when it maps their boundary circles to one another and takes the exterior of the first disk onto the interior of the second. Two pairs of disks with disjoint interiors give the ping-pong lemma construction of a free two-generator group. Tangencies are allowed in the convention suggested by the question; this is often called a Tangent-disk Schottky group, rather than a strictly separated Schottky group.

Here is a concrete configuration that also makes the dimension argument precise. Take $A(z)=z+2$ and $B(z)=z/(iz+1)$. The first pairs the half-plane disks $\Re z<-1$ and $\Re z>1$, which are disks on the sphere tangent at infinity. The second pairs $|z-i|<1$ and $|z+i|<1$, tangent at zero; their interiors are also disjoint from the half-plane disks. These pairings give a free discrete Kleinian group by ping-pong lemma, with a nonempty open fundamental region.

Set $C=AB{A}^{-1}{B}^{-1}$. Conjugating the subgroup generated by $B,C$ by $M(z)=iz/(2-z)$ gives matricesThe product of the second matrix with the first is $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. These are the standard parabolic generators of the principal level-two modular group, whose ideal-quadrilateral fundamental region tiles the upper half-plane and whose Kleinian limit set is the entire extended real line. Consequently the original group's Kleinian limit set contains the circle $|z-1|=1$, as well as its image under $A$, the distinct tangent circle $|z-3|=1$.

It follows that the limit set has Hausdorff dimension at least one and is not homeomorphic to a circle. Indeed a proper subset of a circle cannot itself be a circle: after deleting a missing point it lies in a line, which cannot contain an embedded circle. A space containing two distinct complete circles thus cannot be homeomorphic to a single circle. This explains why dimension at least one does not force circle topology. The tangent-circle terminology is consistent with the primary discussion of T-Schottky configurations in arxiv.org/pdf/math/0701572 .

Reliable transmission at rate $r$ means that there is a sequence of binary block codes of lengths $n$, message counts $M_n$ with $\mathrm{liminf}{n}^{-1}{\log }_2{M}_n\ge r$, and decoding error probabilities tending to zero. The Shannon second coding theorem identifies the supremum of reliable rates with the channel capacity.

For a binary symmetric channel, the conditional entropy of the output given any input bit is $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$. The output entropy is at most one bit and attains one for uniform input, so the maximum mutual information isAll rates strictly below $C$ are achievable, and rates above $C$ are not reliable; this states the supremum without claiming achievability of every boundary rate under every convention.

For very small $p$, the capacity is close to one, with $h_2(p)=p{\log }_2(1/p)+p/\log 2+O(p^2)$. At $p=1/2$, the output is independent of the input and capacity is zero. For $p>1/2$, flipping every received bit converts the channel to crossover probability $1-p<1/2$; its capacity is unchanged because $h_2(p)=h_2(1-p)$. At $p=1$ the deterministic bit flip is perfectly reversible.

The first two commands create vectors of observed successes and corresponding numbers of trials. The next two create the class labels and convert them into an R factor, so the labels are treated categorically. The final command fits a grouped-binomial logistic regression, using the proportions as responses and trial totals as binomial weights, and prints its summary. The model is $\log [{\pi }_i/(1-{\pi }_i)]={\beta }_0+{\beta }_b{1}_{\{S_i=b\}}$, with class a as reference.

The intercept estimates the log odds in class a. Its value corresponds to probability $39/287\approx 0.1359$. The class-b coefficient is the estimated increase in log odds, about $0.4999$, so the fitted probability in class b is $35/170\approx 0.2059$ and the estimated odds ratio is $e^{0.4999}\approx 1.6485$. The printed standard errors measure uncertainty in these estimated coefficients. The z values are estimate divided by standard error: the intercept tests whether the class-a probability is $1/2$, while the class-b z value tests equal class probabilities. The two-sided normal-reference P-value for the latter is about $0.051$, so it is just above a conventional $5\%$ threshold; this is a borderline result rather than strong evidence of a class effect.

The residual deviance compares the fitted two-parameter model with the four-probability saturated model. Its value $1.9369$ on $4-2=2$ residual degrees of freedom is small relative to a ${\chi }_2^2$ reference, with approximate tail probability $e^{-1.9369/2}\approx 0.380$. There is no evidence of lack of fit by that check. Four Fisher scoring iterations were needed for numerical fitting. These interpretations presume independent binomial counts and the usual large-sample approximations; weights here encode trial counts, not extra independent observations.

Set $s=|w{|}^2$. The exact Hebbian learning rule givesThus the norm never decreases, and increases whenever the output is nonzero. Because learning is slow relative to input fluctuations, its averaged equation is $\tau \dot{w}=Cw$, where $C=⟨x{x}^T⟩$ is the input second-moment matrix. It is a covariance matrix only if the inputs have zero mean. Diagonalizing the positive semidefinite matrix $C$ gives components $w_j(t)=w_j(0)e^{{\lambda }_{j}t/\tau }$. Hence weights grow without bound whenever the initial vector has a component in an excited positive-eigenvalue direction. A vector entirely in $\ker C$, or identically zero input, is an important stationary exception.

For the modified Hebbian learning rule,It increases below $s_{\ast }={\alpha }^{-1/2}$ and decreases above $s_{\ast }$, without crossing that equilibrium. More precisely, for a nonzero initial norm,Under persistent excitation, meaning the integral tends to infinity, this givesWithout persistent excitation the limiting norm need not reach this value; zero weights remain zero. The quartic norm stabilization of Hebbian learning bounds the norm and supplies negative feedback at large amplitudes.

Averaging the fast fluctuations in $yx=x{x}^{T}w$ givesThus a nonzero steady vector satisfiesIts direction is an eigenvector of the input second-moment matrix, and its magnitude is the square root of that eigenvalue. The zero vector is also a steady state and satisfies the displayed relation with $\lambda =0$. This is an averaged steady state; the microscopic fluctuating update need not vanish at each instant.

Unmodified Hebbian learning reinforces correlated input and output but has no mechanism limiting synaptic strength. The second rule adds activity-dependent negative feedback and stabilizes the norm at an input-scale-independent value, provided the input keeps exciting the output. The third has a cubic weight penalty independent of the instantaneous output; its equilibrium norm instead depends on the strength of the input second moment.

For the third averaged rule, an equilibrium along eigenvector $i$ has radial linearized eigenvalue $-2{\lambda }_i/\tau$ and transverse eigenvalues $({\lambda }_j-{\lambda }_i)/\tau$. Thus a positive simple largest eigenvalue gives stable equilibria in its two opposite directions, whereas lower eigen-directions are unstable to larger components. With generic initial weights, it learns the dominant correlation direction, linking the rule to principal component analysis. These are idealized normalization and feature-extraction mechanisms, not claims that biological synapses implement these exact equations.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

At zero an ordinary point requires $p,q$ to be analytic. A regular singular point requires $zp$ and $z^{2}q$ to extend analytically to zero, with at least one of $p,q$ failing to be analytic if “singular” excludes ordinary points.

Put $\zeta =1/z$ and $v(\zeta )=w(1/\zeta )$. Differentiation gives $w^{\prime }=-{\zeta }^2{v}^{\prime }$ and $w^{\prime \prime }={\zeta }^4{v}^{\prime \prime }+2{\zeta }^3{v}^{\prime }$, so the transformed equation isThe point at infinity is ordinary precisely when both transformed coefficients are analytic at $\zeta =0$. Solving these conditions for the original coefficients giveswith $P,Q$ analytic near zero; the sign can be absorbed into the arbitrary analytic $P$.

If the only singular point on the sphere is the regular singularity at zero, then $zp(z)$ and $z^{2}q(z)$ are entire. The condition at infinity makes the first tend to $2$ and the second to $0$. By Liouville theorem they are respectively the constants $2$ and $0$. Thus the unique equation iswith general solution $w=A+B/z$. Zero is genuinely regular singular and infinity is ordinary. If “no other singular points” were restricted to finite points only, the general coefficients would instead be $p=P(z)/z$ and $q=Q(z)/z^2$ with entire $P,Q$; the preceding question about infinity indicates the spherical interpretation.

A one-dimensional canonical transformation is a locally invertible change of phase-space variables preserving the symplectic two-form: $dQ\wedge dP=dq\wedge dp$. Equivalently it preserves Hamilton equations for the transformed Hamiltonian. Expanding the two-form yieldssoConversely this Jacobian condition makes the transformation locally invertible and symplectic in two dimensions.

The Poisson bracket is $\{Q,P\}=\alpha q^{\beta -2}$. For it to be identically one on an open region, the exponent must vanish and the coefficient must equal one:The transformation is defined away from $q=0$, and its inverse there is $q=1/P$, $p=Q{P}^2$.

Differentiating givesThereforeFor real coordinates take $q>0$ and a local branch of the angle. The transformation is locally canonical transformation because its determinant is one, but periodicity in $p$ prevents a globally one-to-one map if all real $p$ are retained.

The physical Jeans length is the wavelength for which pressure and gravity balance:The physical wavelength of the comoving mode is $2\pi a/k$. Modes with wavelength larger than ${\lambda }_J$ have a gravitationally destabilizing coefficient and can grow; shorter wavelengths are pressure-supported and oscillate, with expansion also damping the motion. A time-dependent background makes the exact growth history depend on the epoch, not just an instantaneous sign. For pressureless matter the Jeans length vanishes.

For an Einstein-de Sitter universe, $H=\dot{a}/a=2/(3t)$. The flat matter-only Friedmann equation $H^2=8\pi G{\rho }_m/3$ gives ${\rho }_m=1/(6\pi G{t}^2)$. With zero sound speed the perturbation equation is thereforeSubstituting $t^s$ yields $s(s-1)+4s/3-2/3=0$, or $(3s-2)(s+1)=0$. Hencewith growing mode proportional to the scale factor and decaying mode proportional to $t^{-1}$.

For subhorizon pressureless matter during radiation domination, expansion is too rapid for significant self-gravitating growth. Neglecting the small matter-gravity term gives ${\ddot{\delta }}_m+t^{-1}{\dot{\delta }}_m\simeq 0$, hence a constant plus a logarithm of time; the exact coupled radiation problem supplies the detailed amplitudes. This is the Mészáros effect. Modes entering the horizon during radiation domination are suppressed relative to modes entering later, while after equality ordinary matter growth resumes.

The matter power spectrum consequently has a turnover near the equality horizon scale $k_{\mathrm{eq}}\sim a_{\mathrm{eq}}{H}_{\mathrm{eq}}$ in units with $c=1$. Shorter-scale modes retain the radiation-era suppression. This statement concerns cold pressureless matter; baryons coupled to photons also show acoustic effects before recombination.

For the continued-fraction assertion, take the standard Pell equation hypothesis $N>0$ nonsquare. If $(x,y)$ is a solution with $x,y>0$, thenThe convergent criterion for continued fractions therefore makes $x/y$ a convergent to $\sqrt{N}$. Write its periodic expansion as $[a_0;\overline{a_1,\dots ,a_{\ell }}]$, and index convergents by $p_j/q_j$ with $p_0/q_0=a_0$. The norm-one convergents occur at $j=k\ell -1$ with $k\ell$ even. In particular the fundamental positive solution comes at $\ell -1$ if $\ell$ is even, and at $2\ell -1$ if it is odd. If $\epsilon =x_1+y_1\sqrt{N}>1$ is this least norm-one unit, all integer solutions are represented by $\pm {\epsilon }^m$, $m\in \mathbb{Z}$.

The abelian group law comes from multiplication of the norm-one expressions:Their field norms multiply, so the result remains a solution. Multiplication also proves associativity and commutativity; the identity is $(1,0)$ and the inverse of $(x,y)$ is $(x,-y)$. Cubing givesThe last form uses $N{y}^2=x^2-1$.

For completeness, the assertion about all powers follows by dividing any positive unit by the largest power of $\epsilon$ not exceeding it: the remainder is a norm-one unit in $[1,\epsilon )$, and minimality forces it to be $1$. Sign changes and inverses give the remaining solutions. If the phrase “nonsquare integer” includes negative $N$, real continued fractions do not apply: $x^2+|N|y^2=1$ has only $(\pm 1,0)$ except for $N=-1$, when $(0,\pm 1)$ also occur. The same multiplication law still works in those finite groups.

The continued-fraction algorithm givesafter which the process repeats. ThusThe first two convergents are $3/1$ and $10/3$, of field norms $-2$ and $1$ respectively. The even period gives the fundamental solutionIt is also directly the smallest positive solution: for $y=1,2$, the values $1+11y^2$ are $12,45$, neither square; $y=3$ gives $100$. Cubing by the preceding Pell equation group law givesIndeed $3970^2-11\cdot 1197^2=1$.

For $s\ge 0$ and $\delta >0$, define the scale-restricted Hausdorff content at scale $\delta$ byusing countable covers. The Hausdorff measure is ${\mathcal{H}}^s(F)={\lim }_{\delta \downarrow 0}{\mathcal{H}}_{\delta }^s(F)$; the limit exists by monotonicity. A conventional dimension-dependent normalizing constant does not change the Hausdorff dimensionAt $s=0$, a nonempty singleton has contribution one, so the measure counts points.

Let $\mathcal{T}(K)={\bigcup }_{i=1}^k{S}_i(K)$ and $c={\max }_i{c}_i<1$. For compact nonempty $K,L$, matching a point of $S_i(K)$ with the image of a nearby point of $L$ gives $d_H(\mathcal{T}K,\mathcal{T}L)\le c{d}_H(K,L)$. Choose a sufficiently large closed ball $B$ with every $S_i(B)\subset B$. The nonempty compact sets ${\mathcal{T}}^m(B)$ are nested, so their intersection $I$ is nonempty and compact. The finite union of similarity images commutes with this decreasing compact intersection: a point lying in all unions has some repeated branch index, and compactness supplies a preimage in all the nested sets. Consequently $\mathcal{T}(I)=I$. If $J$ is another invariant compact set, the contraction estimate gives $d_H(I,J)\le c{d}_H(I,J)$, hence $I=J$. This proves existence and uniqueness of the invariant set of a finite similarity iterated function system.

Under the open set condition, there is a nonempty bounded open set $O$ such that the $S_i(O)$ lie inside $O$ and are pairwise disjoint. The dimension formula isOverlaps can invalidate this formula if that assumption is omitted.

For the printed template, subdivide the central straight run of length two into its two collinear unit steps. The eight directed steps, before dividing all coordinates by four, have successive verticesEach step defines a similarity with ratio $1/4$, with rotation matching its direction. The open diamond $O=\{|x-1/2|+|y|<1/2\}$ is an explicit witness for the open set condition. Its eight image diamonds have radius $1/8$ in the ${\ell }^1$ metric, lie in $O$, and their centres have pairwise ${\ell }^1$ distance at least $1/4$; their interiors are therefore disjoint. The Minkowski sausage dimension is consequentlyThis makes the subdivision convention explicit. If the length-two run were instead treated as one unsplit edge replaced at scale $1/2$, there would be six maps of ratio $1/4$ and one of ratio $1/2$. The same diamond separation works, but the equation becomes $64^{-s}+2^{-s}=1$, giving $s={\log }_{2}3$, a different fractal. The requested $3/2$ corresponds to the eight-unit-step interpretation.

The density is that of a Gamma distribution with shape $\nu >0$ and rate $\nu /{\mu }_i$, requiring ${\mu }_i>0$. Direct integration givesfor $\nu +r>0$. Using $\Gamma (\nu +1)=\nu \Gamma (\nu )$ and $\Gamma (\nu +2)=\nu (\nu +1)\Gamma (\nu )$,

Put $t_i={\beta }^T{x}_i=1/{\mu }_i>0$. Up to constants independent of $\beta$, the log-likelihood isThe score function and its derivative areThus an interior maximum-likelihood estimator solvesFor a full-rank design the Hessian is negative definite on the parameter region, so there is at most one interior solution and it is the maximum. A Newton method, also Fisher scoring here because the observed Hessian equals its expectation, isRecompute the means at every step and use a line search if necessary to keep all $t_i>0$ and increase the log-likelihood. Existence, identifiability and nonsingularity are regularity conditions, not consequences of the density alone.

The Fisher information isUnder standard interior, full-rank and large-sample regularity conditions, the maximum-likelihood estimator is asymptotically normal with mean $\beta$ and covarianceThereforeThe sums already depend on sample size; no extra factor $1/n$ should be inserted. The variance is estimable by substituting fitted means. If all $z_i$ are the same, $ac-b^2=0$ and the two coefficients are not separately identifiable.

The Dirichlet series for the Riemann zeta function converges for $\Re z>1$. In the Hankel integral representation the branch is a branch of $t^{z-1}$, so the argument restriction concerns $t$, not the parameter $z$. Use the principal branch cut along the negative real axis and a positive Hankel loop with a fixed small indentation around zero. Its integral $I(z)$ is an entire function of $z$, since the tails have exponential decay and the fixed contour avoids zero. Multiplication by $\Gamma (1-z)/(2\pi i)$ gives the meromorphic continuation of $\zeta$: it is directly defined away from positive integers, has a pole at $1$, and the apparent singularities at $2,3,\dots$ are interpreted by limits.

For a check in the initial half-plane, collapsing the indentation for $\Re z>1$ givesExpansion of $(e^r-1{)}^{-1}$ into exponentials justifies the last equality there, and the Gamma reflection formula recovers the printed integral representation.

Take the upper semicircle contour with the straight segment oriented from left to right, making the total orientation positive. Its enclosed poles are $t=2\pi ij$, $j=1,\dots ,N$; each has residue $-(2\pi ij{)}^{z-1}$. ThusThe reverse orientation changes the sign.

Initially take $\Re z<0$, so the series on the right converges as $N\to \infty$ and the large arcs can be discarded under the given assumption. Denote by $L_{+}$ and $L_{-}$ the integrals along horizontal lines just above and below the real axis, both from left to right. The residue computation and its lower-half-plane clockwise counterpart giveContour deformation about the negative-axis branch cut gives $I=L_{-}-L_{+}$. HenceMultiplying by $\Gamma (1-z)/(2\pi i)$ yieldsIt extends by meromorphic continuation to all $z\ne 1$, with removable products evaluated by limits where individual factors diverge. The derivation does not assert ordinary convergence of the original positive-term series outside $\Re z>1$.

In a large volume, one one-particle state occupies momentum-space volume $h^3/V$. A spherical shell of radius $p$ has volume $4\pi p^{2}dp$, so for spin degeneracy $g$ the density of states is $4\pi gV{p}^{2}dp/h^3$. With ultra-relativistic dispersion $E=cp$ and zero chemical potential, the Bose-Einstein distribution givesThe convergent dimensionless integral is ${\pi }^4/15$, obtained by expansion into exponentials and summing $\sum n^{-4}$. Therefore $ϵ\propto T^4$ and $\alpha =4$. Photons are massless bosons with two polarizations; equilibrium emission and absorption do not conserve photon number and force ${\mu }_{\gamma }=0$.

For recombination, chemical equilibrium imposes ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }={\mu }_H$, including rest energies in the chemical potentials. Divide the three nonrelativistic number-density expressions to obtainHere $m_p{c}^2+m_e{c}^2-m_H{c}^2=I$. Neglect the relative proton/hydrogen mass difference and take the ground-state degeneracy ratio $g_p{g}_e/g_H=1$ (electron and proton spins give $2\cdot 2$ and hydrogen gives $4$). Charge neutrality gives $n_p=n_e$. This yields the stated Saha equation

With $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the ratio asked for isEquivalently the prefactor is $16\pi \zeta (3)\eta (2\pi {)}^{-3/2}[kT/(m_e{c}^2){]}^{3/2}$. Because photons vastly outnumber baryons, even a small high-energy tail can keep hydrogen ionized. Neutralization therefore requires exponential suppression of that tail far below $kT=I$, explaining the much lower recombination temperature. The equilibrium formula is an estimate; the actual expanding-universe recombination kinetics eventually depart from Saha equilibrium.

The propositional completeness theorem states that $\Gamma \models \varphi$ if and only if $\Gamma ⊢\varphi$. Soundness follows by checking the axioms against truth valuations and noting that modus ponens rule preserves truth. For completeness, use the deduction theorem for propositional logic: if $\Gamma \cup \{p\}⊢q$, then $\Gamma ⊢p\Rightarrow q$. In the usual negation rules this also says that a contradiction from $\Gamma \cup \{p\}$ proves $\Gamma ⊢\neg p$.