Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4

The Wilson theorem states that for a prime $p$,Indeed, every nonzero residue modulo $p$ has a unique multiplicative inverse. The only residues equal to their own inverses solve $x^2\equiv 1(\mathrm{mod}p)$, hence are $1$ and $-1$. Pairing every other residue with its distinct inverse leaves

The Fermat little theorem states that for prime $p$,equivalently, if $p\nmid a$, then $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Wilson's theorem at $p=31$ givesso $28!\equiv 15(\mathrm{mod}31)$. Also $729\equiv 16(\mathrm{mod}31)$, and therefore

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A relation on $S$ is a subset of $S\times S$. An equivalence relation is reflexive, symmetric, and transitive. Its equivalence class at $x$ is $[x]=\{y\in S:x\sim y\}$.

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This is not an equivalence relation because it is not reflexive: $0\nsim 0$, since $0\cdot 0\ngtr 0$.

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The relation is reflexive and symmetric, but not transitive. For example,while $1\nsim -1$. Hence it is not an equivalence relation.

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This is an equivalence relation: $x-x=0\in \mathbb{Z}$; if $x-y\in \mathbb{Z}$, then $y-x\in \mathbb{Z}$; and integer differences add. Its classes arethe cosets of $\mathbb{Z}$ in $\mathbb{R}$.

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This relation is reflexive and transitive, but not symmetric: $1\sim 0$ whereas $0\nsim 1$. It is therefore not an equivalence relation.

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The parallel axis theorem says that for an axis through $x$, parallel to the centre-of-mass axis in direction $\hat{d}$,LetMoments about the axis through $x_C$ add, soUsing $M_A=M_C-M_B$ and solving giveswhen the denominator is nonzero. In the degenerate equal-distance case, this inertia equation alone does not determine $M_B$; the centre-of-mass equation $M_C{x}_C=M_A{x}_A+M_B{x}_B$ supplies the remaining information.

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For a lamina in the $xy$-plane,so the perpendicular axis theorem is immediately

For the uniform ellipse, put $x=au$, $y=bv$, where $u^2+v^2\le 1$. Symmetry of the unit disk givesand the same for $v^2$. Hence

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Set $c=1$. If $B$ is in the future lightcone of $A$ and $C$ in that of $B$, thenAdding and using the triangle inequality givesThus the statement is true.

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The statement is false. Takewith coordinates written as $(x,t)$. Both $B$ and $C$ lie on the future lightcone of $A$, but $B$ and $C$ are simultaneous and spatially separated, so neither is in the future lightcone of the other.

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The statement is true. The assumptions giveAdding and applying the triangle inequality provesso $B$ is in the future lightcone of $C$.

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The statement is false. With $(x,t)$ coordinates, takeBoth separations from $A$ are lightlike, but $B-C=(2,0)$ is spacelike.

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Again set $c=1$ and writeA boost of velocity $v$ givesDefineThe hypothesis says $D(0)<0$. Put $\xi =x_B-x_A$ and $\tau =t_B-t_A$. Causality gives $\tau \ge |\xi |$, while $T_B\ge |X_B|$, and $T_A=T_B+\tau$, $X_A=X_B+\xi$. ThereforeButHence at least one of $D(1),D(-1)$ is positive. By continuity, $D(v)>0$ for some physical velocity $|v|<1$ sufficiently close to that endpoint. In the corresponding inertial frame,as required.

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The Euler theorem states that if $\gcd (a,d)=1$, thenLet $r_1,\dots ,r_{\varphi (d)}$ be the reduced residue classes modulo $d$. Multiplication by $a$ permutes them, soTheir product is a unit and may be cancelled, proving the theorem.

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A decimal integer is congruent modulo $3$ to its digit sum. Cyclic permutation does not change that sum. Hence every rotation of a multiple of $3$ is again divisible by $3$.

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Divisibility by $2$ or $5$ depends only on the final digit. As the digits are cyclically rotated, every digit appears in the final position. Thus every rotation is divisible by $d\in \{2,5\}$ exactly when every digit is divisible by $d$.

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As printed, the claim is false: $70$ is 7-cyclic-divisible because its rotations are $70$ and $07=7$, but its digits are not all equal to $7$ and its length is not a multiple of $6$.

The standard rotation argument proves the likely intended statement. If an $L$-digit integer and all its rotations are divisible by $7$, then for each leading digit $a$,is divisible by $7$. The multiplicative order of $10$ modulo $7$ is $6$. If $6\nmid L$, then $7\nmid 10^L-1$, forcing every digit $a$ to be divisible by $7$, hence to be either $0$ or $7$. Thus the valid conclusion is:The counterexample shows why “all its digits are equal to 7” cannot replace the first alternative.

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LetBy Fermat's little theorem, $l\mid p-1$. If the digit count $L$ is a multiple of $l$, then $10^L\equiv 1(\mathrm{mod}p)$. The cyclic decimal divisibility relation givesand hence also modulo $p$. Since $p>7$ is coprime to $10$, every $10^j$ is a unit modulo $p$. Thereforefor every $j$. Since $c_0(n)=n$, this proves the required equivalence.

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For $n=1$,If $C_n=F_{n+1}{F}_{n-1}-F_n^2$, then the recurrence givesInduction yields the Fibonacci determinant identity known as Cassini's identity:

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For fixed $m,l$, setAt $n=0$, $D_0=F_l{F}_m$. Applying the Fibonacci recurrence to every term and cancelling gives $D_{n+1}=-D_n$. Induction therefore proves

Applying this identity to the relevant pairs of indices and using $F_{r+1}=F_r+F_{r-1}$ givesandThus, for $j\ge k\ge 0$,and

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For a nonempty set $S\subset \mathbb{R}$ bounded above, $\sup S$ is its least upper bound: every $s\in S$ satisfies $s\le \sup S$, and every $u<\sup S$ fails to be an upper bound.

A sequence $x_n$ converges to $x$ if for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|x_n-x|<\epsilon$.

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Every convergent sequence is bounded. Conversely, let $(x_n)$ be increasing and bounded above, and set $L=\sup \{x_n:n\ge 1\}$. For any $\epsilon >0$, $L-\epsilon$ is not an upper bound, so some $x_N>L-\epsilon$. Monotonicity then givesfor all $n\ge N$. Hence $|x_n-L|<\epsilon$, proving the monotone bounded sequence criterion.

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LetFor $n\ge 3$, the binomial theorem givesConsequently $(1000n^2+4)/1.1^n\to 0$: given $\epsilon >0$, the displayed rational upper bound is below $\epsilon$ for all sufficiently large $n$. Thus, directly from the definition of convergence,

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Yes. Put $\alpha =\sup S$ and $\beta =\sup T$. Every $s+t\le \alpha +\beta$, so this is an upper bound. Given $\epsilon >0$, choose $s\in S$ with $s>\alpha -\epsilon /2$ and $t\in T$ with $t>\beta -\epsilon /2$. Then $s+t>\alpha +\beta -\epsilon$. Hence

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No. Take $S=T=\{-2,-1\}$. Thenbut $ST=\{1,2,4\}$ and therefore $\sup (ST)=4$.

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No, even when every element of $S$ is positive. TakeThen $S^T=\{1/2,1/4\}$, so

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A set is countable set if it is finite or admits an enumeration by natural numbers. If $A_n$ are countable, choose enumerations $a_{n,m}$. Since ${\mathbb{N}}^2$ is countable by diagonal enumeration, the array $(n,m)\mapsto a_{n,m}$ enumerates ${\bigcup }_n{A}_n$ after repetitions are removed. This proves that a countable union of countable sets is countable.

The integers are countable, so $\mathbb{Z}\times {\mathbb{N}}_{>0}$ is countable. Mapping $(p,q)$ to $p/q$ surjects onto $\mathbb{Q}$, hence $\mathbb{Q}$ is countable. The same diagonal argument shows that $A\times B$ is countable whenever $A,B$ are.

Finally, if the reals in $(0,1)$ had decimal expansions $x_1,x_2,\dots$, choose a decimal whose $n$th digit differs from the $n$th digit of $x_n$, avoiding digits $0$ and $9$ to remove expansion ambiguity. This number differs from every listed number. The Cantor diagonal argument proves that $\mathbb{R}$ is uncountable.

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No. For every $m\in \mathbb{R}$, the line $y=mx$ contains the origin. These lines are distinct, so the uncountable familyhas nonempty common intersection.

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Yes. The set ${\mathbb{Q}}^2$ is countable, hence the set of unordered pairs of distinct rational points is countable. Two distinct points determine at most one line. Mapping each such pair to its line therefore has countable image, and every line in the stated collection lies in that image. Thus the collection is countable.

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The particle moves on a horizontal circle of radiusIf $T$ is the tension in the light rod, vertical and radial balance giveDividing eliminates both $T$ and $m$, yielding the rotating conical pendulum with an offset pivot conditionThe ratio under the square root has units $T^{-2}$. The answer is independent of $m$; increasing $L$ increases the orbit radius and lowers the angular velocity required at fixed ${\varphi }_1$.

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The spring length is $a+x$, its tension is $kx$, and the orbit radius isVertical balance first givesRadial balance isUsing the vertical equation,Unlike the rigid-rod result, this depends on $m$: a larger mass stretches the spring farther and changes the radius of the orbit.

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A central force conserves angular momentumPut $u=1/r$. Since $\dot{\theta }=\ell u^2/m$,where primes now denote differentiation with respect to $\theta$. Substitution into the radial equation gives the Binet equation

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For $V=-km/r$, one has $dV/dr=km{u}^2$, soAfter choosing the angular origin, the solution isFor $0\le e<1$, using $r=\sqrt{x^2+y^2}$ and $r+ex=p$ givesThis is an ellipse, with the force centre at one focus, as described by a Kepler orbit.

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ForBinet's equation becomesA circular orbit has constant $u=u_c$ and therefore satisfiesThe right side has maximum $4/(27r_0)$ at $u_c=1/(3r_0)$. Thus there are two circular orbits when $0<K<4/(27r_0)$, one marginal orbit at equality, and none above it.

Writing $u=u_c+\eta$ and linearizing givesHence the outer orbit $u_c<1/(3r_0)$, equivalently $r_c>3r_0$, is stable; the inner orbit is unstable. These are the circular-orbit branches of the shifted inverse-radius potential.

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To first order in $r_0$,The expanded orbit equation is thereforeIts required family of solutions iswith any $0<A<1$. Successive periapses occur when $\omega \theta$ changes by $2\pi$, so their angular separation isThe periapsis advance per revolution is consequently $2\pi K{r}_0+O(r_0^2)$.

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With total mass $M={\sum }_i{m}_i$, defineThe angular momentum about the centre of mass isExpanding, using ${\sum }_i{m}_i(x_i-R)=0$ and $P=M\dot{R}$, gives

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Every particle is initially at distance $R$ from the axis, so

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Each freely moving particle has position, relative to its own initial radial and tangential unit vectors,Thus all particles lie on a circle of radiusTheir angular momenta are constant and sum toSince $I(t)=mR(t{)}^2$, the angular velocity isWith $\theta (0)=0$,This is the geometry of a freely expanding rotating particle ring.

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For uniform surface density $m_0/(\pi R^2)$,

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At time $t$, the remaining water has angular momentumDuring $dt$, the ejected mass is $-dm$, and its signed tangential speed in the inertial frame is $R\omega +u$. Conservation of angular momentum gives, to first order,After simplification, the angular rocket equation isHere $u$ is signed relative to the direction of rotation; reversing the spray changes its sign.

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For $u=0$, the equation becomesso $\omega /m$ is constant. With $m(t)=m_0-\mu t$,It reaches zero when the water is exhausted, at

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The four-momentum of a massive particle iswhile for a massless particleFor $P_1=P_2+P_3$, invariance givesIn the rest frame of particle 1, $P_1=(m_{1}c,0)$, sowith $E_1=m_1{c}^2$. Equal and opposite final momenta are real exactly whenthe threshold condition for a relativistic two-body decay.

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At step $n$, the parent and daughter rest masses areThe decay is allowed when $M\ge m$, which holds because $M=\lambda m$ with $\lambda >1$. Hence all $N$ stages occur:The initial rest mass is $NM+m_0$ and the final rest mass is $Nm+m_0$. Thusof rest mass is converted into kinetic energy, corresponding to energy $Nm(\lambda -1)c^2$.

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Momentum conservation is $P_A+P_B=P_C+P_D$. The diagonal bilinears are $P_a^2=m_a^2{c}^2$. The six off-diagonal ones areandThese express every bilinear invariant through the Mandelstam variables and masses. Expanding $s+t+u$ and using momentum conservation cancels all mixed products, leaving