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Suppose continuity at $y$ failed. The negation of the relative-domain epsilon-delta definition gives an ${\epsilon }_0>0$ such that for every $\delta >0$ there is an $x\in [a,b]$ with $|x-y|<\delta$ but $|f(x)-f(y)|\ge {\epsilon }_0$. Choosing $\delta =1/n$ produces a sequence $x_n\to y$ whose image does not converge to $f(y)$, contradicting the hypothesis. Thus $f$ is continuous at $y$.

This proves the sequential characterization of continuity in metric spaces. At $a$ or $b$, the same argument uses points only in $[a,b]$, giving the appropriate one-sided continuity automatically.

Set $\theta =0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot{r}(0)=-4\cos (\pi /3)=-2$, while the tangential speed is $4\sin (\pi /3)=2\sqrt{3}$. HenceSubstitution in the Binet equation gives $12(u^{\prime \prime }+u)=3+9u$, or $u^{\prime \prime }+u/4=1/4$. This is an orbit equation for combined inverse-square and inverse-cube attraction. Solving the constant-coefficient equation and applying both initial conditions givesOn $0\le \theta \le 2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta =2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u^{\prime }(2\pi )=-1/\sqrt{3}$, so its radial velocity is now $+2$, rather than the original $-2$. The return is not a periodic return in position and velocity.

After that revolution, the first zero of $u$ is at $\theta =8\pi /3$, where $\sin (\theta /2)=-\sqrt{3}/2$. The physical branch therefore has $r\to \infty$ as $\theta \uparrow 8\pi /3$; it must not be continued into negative $u$. Since $dt/d\theta =1/(l{u}^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific potential energy is $\Phi (r)=-3/r-9/(2r^2)$ and the conserved energy per unit mass is $4^2/2+\Phi (1)=1/2$, giving escape speed $1$ at infinity, also obtained from $-l{u}^{\prime }(8\pi /3)=1$.

A central force has no angular component, so the supplied polar acceleration formula and Newton's second law give $d(r^2\dot{\theta })/dt=0$. Thus $l=r^2\dot{\theta }$ is the conserved specific angular momentum. For a nonradial orbit, $l\ne 0$ allows $\theta$ to be used as a parameter. With $u=1/r$ and a prime denoting $d/d\theta$,Also $r{\dot{\theta }}^2=l^2{u}^3$. The radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ therefore becomes the Binet equationFor a purely radial trajectory $l=0$, the angular parameter is unavailable, and one instead uses the direct radial equation. The orbit in part (b) is nonradial.

If ${\log }_{2}3=a/b$ were a rational number, choose positive integers $a,b$ because ${\log }_{2}3>0$. Exponentiation gives $2^a=3^b$, but the left side is even and the right side is odd. ThusThis is the basic irrational logarithm criterion: a rational logarithm would force positive integer powers of its base and argument to coincide.

For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

The exponential function series gives Euler's number $e={\sum }_{k=0}^{\infty }1/k!$. Suppose $e=a/b$ with integers $a,b$ and $b>0$. Choose $n\ge \max (b,2)$. Since $b$ divides $n!$, the numberis an integer. But its positive tail satisfiesThe inequality is strict because the factors after the first exceed $n+1$ when $j\ge 2$. No integer lies strictly between zero and one. This proves the irrationality of e:

If $q=\sqrt{2}+\sqrt{3}$ were a rational number, then $q^2=5+2\sqrt{6}$ would make $\sqrt{6}$ rational. Write $\sqrt{6}=a/b$ in lowest terms. Then $a^2=6b^2$, which is impossible by unique factorization: the exponent of the prime $2$ is even on the left and odd on the right. Therefore

It can be false even on a two-element finite set. The same constant maps $f=g=0$ on $\{0,1\}$ satisfy $fg=f$, while $g\ne {\mathrm{id}}_X$. Indeed, a constant $f$ makes $fg=f$ for every $g$. If $f$ were an injection, one could cancel it pointwise to conclude $g(x)=x$, but finiteness alone supplies no such hypothesis. As before, the assertion is trivial on the empty set or a singleton.

It can be false even on a two-element finite set. On $X=\{0,1\}$, let $f$ and $g$ both be the constant map with value $0$. Then $fg=g$, but $f(1)=0\ne 1$. In general the equation says exactly that $f$ fixes the image $g(X)$; it says nothing about other points. If $g$ happens to be a surjection, it does force $f$ to be the identity map, but finiteness alone does not ensure this. The empty set and a singleton have only one self-map, so the assertion is trivially true on those sets.

Take $fg=f\circ g$. The assertion can fail on an infinite set. On $X={\mathbb{N}}_0$, set $g(n)=n+1$, $f(0)=0$, and $f(n)=n-1$ for $n\ge 1$. Then $f(g(n))=n$ for every $n$, but $g(f(0))=1\ne 0$. Thus a right inverse need not be a left inverse.

On a finite set it is always true. From $fg={\mathrm{id}}_X$, $g$ is an injection: equality of two $g$-values gives equality after applying $f$. An injective self-map of a finite set is a bijection. Composing $fg={\mathrm{id}}_X$ with $g^{-1}$ shows $f=g^{-1}$, and hence $gf={\mathrm{id}}_X$. This includes the empty set.

The union need not be an equivalence relation, because transitivity can mix the two relations. Take $X=\{1,2,3\}$. Let $R$ have equivalence classes $\{1,2\}$ and $\{3\}$, and $S$ have classes $\{1\}$ and $\{2,3\}$. Then $1W2$ and $2W3$, but neither component relates $1$ to $3$, so $1W3$ fails. The union of equivalence relations is reflexive and symmetric; it is precisely transitivity that may fail.

The intersection is always an equivalence relation. Both component relations contain $(x,x)$, so $V$ is a reflexive relation. If $xVy$, both relations relate $x$ to $y$, and symmetry in each gives $yVx$. If $xVy$ and $yVz$, transitivity in each component gives both $xRz$ and $xSz$, hence $xVz$. This is the intersection of equivalence relations; the same argument works for any family of such relations.

Write $x=3+18k$. The second modular congruence becomes $18k\equiv -1(\mathrm{mod}23)$. The Bezout identity above says that $9$ is the inverse of $18$ modulo $23$, so $k\equiv -9\equiv 14(\mathrm{mod}23)$. ThereforeThe checks are $255=18\cdot 14+3=23\cdot 11+2$. The Chinese remainder theorem explains uniqueness modulo $18\cdot 23=414$, since the moduli are coprime integers.

Use the Euclidean algorithm: $23=18+5$, $18=3\cdot 5+3$, $5=3+2$, and $3=2+1$. Back-substitution gives the Bezout identity $1=9\cdot 18-7\cdot 23$. Multiplying by $101$ gives one solution of the linear Diophantine equation, and shifting the coefficients by multiples of $(23,-18)$ reduces it toIndeed, $18\cdot 12+23(-5)=216-115=101$. More generally all solutions are $x=12+23k$, $y=-5-18k$, $k\in \mathbb{Z}$: subtracting this particular solution gives $18(x-12)=-23(y+5)$, and the coprime integers $18,23$ force $23\mid (x-12)$.

For $u>0$, take the initial electron direction as the longitudinal axis and put $\beta =u/c$. Incoming four-momentum conservation givesThe transverse components must cancel. In the noncollinear case the two transverse directions are opposite, and ${\nu }_1\sin {\theta }_1={\nu }_2\sin {\theta }_2$. Hence the ratio of longitudinal momentum times $c$ to energy isWriting $s=({\theta }_1+{\theta }_2)/2$ and $d=({\theta }_1-{\theta }_2)/2$, the last ratio is $\cos s/\cos d$. It also equals $(1+\cos ({\theta }_1+{\theta }_2))/(\cos {\theta }_1+\cos {\theta }_2)$ whenever that denominator is nonzero. Since ${\gamma }^2{\beta }^2={\gamma }^2-1$, the angular relation for two-photon electron-positron annihilation isThe positive root follows from $u>0$. For noncollinear emission the positive momentum-to-energy ratio ensures $\cos s>0$, and $|d|<\pi /2$ ensures the stated denominator is positive.

The printed quotient needs a qualification for exactly collinear photons. The allowed configuration ${\theta }_1=0$, ${\theta }_2=\pi$ has positive frequencieswhich obey both conservation equations, but the printed left side is $0/0$. Thus the quotient holds for noncollinear angles, or as its appropriate noncollinear limit. The undivided identityremains true in the collinear case, where both sides vanish. The limit $u=0$ similarly requires care because the initial direction is then undefined and the photons are antiparallel.

Use metric signature $(+---)$, as required by the sign of the scalar product in this question. For a massive particle, the four-momentum isFor a photon of positive frequency $\nu$ and direction $e$, the Planck constant gives $E=h\nu$ andA future-pointing timelike four-vector $P=(P^0,p)$ has $P^0>|p|$ and $P^0>0$ in this signature. The ordinary Euclidean Cauchy-Schwarz inequality yieldsThis is stronger than the requested nonnegative inequality. For an electron and a positron of rest mass $m>0$, their future-pointing timelike four-momenta have $P_1^2=P_2^2=m^2{c}^2$. Thuswhich cannot equal the null norm of a single photon. Four-momentum conservation forbids spontaneous photon decay into an electron-positron pair in vacuum. An external body or field able to exchange momentum would be a different system.

Measure a signed angle $\theta$ from the upward radius $A\to O$ in the direction of falling. With $I_A=3m{a}^2/2$, gravitational torque givesThere is an idealization in the release condition: exactly upright and exactly at rest is an unstable equilibrium, so the exact ideal solution stays there. Uniqueness of this smooth equation with $\theta (0)=\dot{\theta }(0)=0$ prevents spontaneous departure. The usual falling calculation means an infinitesimal disturbance, with its energy tending to that of the upright rest state. For that limiting separatrix motion, conservation of mechanical energy givesOn the branch with increasing angle, $\dot{\theta }=\sqrt{8g/(3a)}\sin (\theta /2)$ for $0<\theta <2\pi$; the opposite branch has the opposite angular velocity. The nontrivial limiting motion approaches the upright position only as $t\to -\infty$, because $dt/d\theta$ has a logarithmically divergent integral near zero.

Let $e_r$ point from $A$ toward $O$, and $e_{\theta }$ in the direction of increasing angle, so $e_r=\sin \theta e_x+\cos \theta e_z$ and $e_{\theta }=\cos \theta e_x-\sin \theta e_z$. The centre of mass acceleration on the limiting falling branch isGravity has components $-mg\cos \theta e_r+mg\sin \theta e_{\theta }$. Applying Newton's second law to the whole disc givesThe signed component parallel to the radius, positive from $A$ toward $O$, is the required $mg(7\cos \theta -4)/3$. This same radial formula is consistent at the exact stationary upright state, where $\theta =0$, $a_O=0$, and $R=mg{e}_z$; nonzero falling angles require the limiting-release interpretation just stated.

The moment of inertia of the uniform disc about its central perpendicular axis isThe parallel axis theorem gives $I_A=I_O+m{a}^2=3m{a}^2/2$. For angle $\beta$ from the downward resting position, the gravitational torque is $-mga\sin \beta$, so $I_A\ddot{\beta }=-mga\sin \beta$. Linearizing at $\beta =0$ gives $\ddot{\beta }+(2g/(3a))\beta =0$. Thus the uniform disc pivoted at its rim is a physical pendulum with period