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For integers $0\le r\le n$, the binomial coefficient isthe number of $r$-element subsets of an $n$-element set. It is zero outside that range. To obtain Pascal's identity, distinguish one element of an $(n+1)$-element set. An $r$-element subset either omits it, giving $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ choices, or contains it, leaving $\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$ choices. These cases are disjoint and exhaustive, so

If $p$ is a prime number and $0<r<p$, thenThe right side is divisible by $p$, while $r!$ is coprime to $p$. Cancellation modulo $p$ proves the prime-row binomial coefficient divisibilityApply the binomial theorem in the polynomial ring modulo $p$. The intermediate coefficients just proved to vanish giveMultiplying by $(1+X{)}^k$ yieldsFor $0\le r\le k<p$, the $X^p$ factor cannot contribute to the coefficient of $X^r$. Comparing coefficients proves the binomial coefficient congruence across a prime rowThis also covers $r=0$, when both sides are one.

A prime number is an integer $p>1$ whose only positive divisors are $1$ and $p$. Every integer greater than one has a prime factorization; in particular it has a prime divisor.

To prove infinitely many primes congruent to minus one modulo six, suppose a finite list $p_1,\dots ,p_n$ contains all such primes, and put $P=p_1\cdots p_n$; the empty product is one. The integer $N=6P-1>1$ is coprime to $6$, so each prime factor is congruent to $1$ or $-1$ modulo $6$. If all its prime factors, counted with multiplicity, were congruent to $1$, their product would also be congruent to $1$. But $N\equiv -1(\mathrm{mod}6)$. Hence some prime divisor $q$ satisfies $q\equiv -1(\mathrm{mod}6)$. None of the listed primes divides $N$, since $N\equiv -1(\mathrm{mod}{p}_i)$. This contradicts completeness of the list.

For infinitely many primes congruent to one modulo six, suppose instead that $p_1,\dots ,p_n$ lists every prime congruent to $1$ modulo $6$, and useIt is odd. Since each listed prime is coprime to $3$, $P^2\equiv 1(\mathrm{mod}3)$, so $N\equiv 1(\mathrm{mod}3)$ and $3$ does not divide it. Any prime divisor $q$ is therefore greater than $3$ and satisfies the modular congruenceThe allowed quadratic-congruence result gives $q\equiv 1(\mathrm{mod}6)$. Yet $N\equiv 3(\mathrm{mod}{p}_i)$, and $p_i>3$, so no listed prime divides $N$. Again there is a new prime in the required residue class. Thus both residue classes $6k-1$ and $6k+1$ contain infinitely many primes.

With the prescribed magnetic-only Lorentz force $m\ddot{\mathbf{r}}=q\dot{\mathbf{r}}\times \mathbf{B}$, the component equations areDifferentiate the squared planar speed:Thus $V=\sqrt{{\dot{x}}^2+{\dot{y}}^2}$ is constant, since the Lorentz force is perpendicular to the velocity.

For $V>0$, introduce the velocity angle $\theta$ by $\dot{x}=V\cos \theta$, $\dot{y}=V\sin \theta$. The two planar equations imply $\dot{\theta }=-q{B}_z(t)/m$. Therefore the velocity rotation in a time-dependent axial magnetic field has phaseEquivalently, the complex velocity satisfies $\frac{d}{dt}(\dot{x}+i\dot{y})=-iq{B}_z(t)(\dot{x}+i\dot{y})/m$, giving the same integrated phase. Integrating the velocities and $\ddot{z}=0$ provesThe constants $x_0,y_0,z_0$ are the initial position coordinates, $v_z$ is the constant velocity parallel to the field, $V\ge 0$ is the initial planar speed, and $\varphi$ is the initial planar-velocity direction modulo $2\pi$. Together they specify the initial position and velocity. If $V=0$, the planar position stays fixed and the angle $\varphi$ is immaterial.

For $B_z(t)=\beta t$, the inner integral gives $\theta (t)=-a{t}^2+\varphi$, with $a=q\beta /(2m)>0$. Under the specified initial data,Rescale $u=\sqrt{a}s$. The given Fresnel integrals then giveHence the limiting position isThe negative $y$ coordinate follows from clockwise rotation of planar velocity for positive charge and positive $B_z$. The position converges through cancellation of increasingly rapid oscillations; its convergence does not mean the constant planar speed tends to zero.

Label the initially left-hand rod $A$ and the right-hand rod $B$. At impact their centres are $(0,\ell )$ and $(0,-\ell )$, and their contact points are both the origin. Each rod has total mass $2m$ and moment of inertiaLet the impulse on $B$ be $+J{\mathbf{e}}_x$, so the impulse on $A$ is $-J{\mathbf{e}}_x$. The impulsive form of Newton's second law givesThe contact lever arms from the centres are $(0,-\ell )$ for $A$ and $(0,+\ell )$ for $B$. Both angular impulses are $-\ell J{\mathbf{e}}_z$. Thus, taking counterclockwise rotation as positive,Equal and opposite linear impulses therefore produce equal clockwise rotations, not opposite rotations.

The restitution law uses the velocities of the contacting endpoints, including the rotational terms. For $A$ that velocity isand for $B$ it is $u_B=-V+J/m$. The approach speed before impact is $2V$. The coefficient of restitution $e$ therefore givesSubstituting into the centre and angular velocities yields the endpoint collision of two freely rotating equal rods:For the first, perfectly elastic case, $e=1$, soAll initial translational energy has become rotational energy. Indeed the total initial kinetic energy is $2m{V}^2$, while the final energy for general $e$ isThe loss is $m{V}^2(1-e^2)$, zero for an elastic collision. Conservation of total angular momentum about the origin provides another check: initially it is $-4m\ell V$; after collision the centre-motion contribution is $-2m\ell V(1-e)$ and the spins contribute $-2m\ell V(1+e)$, with the same total. The initial distance $d$ determines when the collision occurs but does not enter the impulsive velocities.

Take rightward velocities as positive just before the right-wall collision. The incoming particle velocity is $v$ and the wall velocity is $-V$. Their total momentum divided by total mass gives the centre of mass velocityIn an elastic collision, relative velocity reverses, so the particle's signed outgoing velocity is $v_{\mathrm{out}}=2v_{\mathrm{cm}}-v$. For $M>m$ it is negative, and its speed isConsequently the exact change in speed isFor $m/M\to 0$ with $v,V$ fixed, this proves the requested leading increment $v^{\prime }-v\simeq 2V$. Reflection gives the same increment at the left wall.

This approximation is not uniform when $v/V$ is allowed to become arbitrarily large. To neglect recoil relative to the increment $2V$ requiresThe inequalities $m\ll M$ and $v\gg V$ alone do not imply it: for example, $m/M=10^{-3}$ and $v/V=10^3$ give an exact zero speed increment. The subsequent slowly moving-wall approximation therefore uses the consistent joint range $1\ll v/V\ll M/m$, or treats the walls as effectively prescribed massive reflectors. The assumed slow change of wall speed over many impacts embodies the same negligible-recoil requirement.

Within this approximation, the separation derivative is the right-wall velocity minus the left-wall velocity:For a time window containing many crossings but little separation change, $\ell /v\ll \Delta t\ll \ell /V$, the flight time between successive wall collisions is approximately $\ell /v$. Thus their number and the cumulative speed change arePassing to the averaged differential equation givesHence the particle-speed invariant between slowly approaching walls isIt is an adiabatic invariant for the rapid elastic bouncing motion while the separation and wall velocities vary slowly on the flight-time scale.

Let $T(z)=(az+b)/(cz+d)$, with $ad-bc\ne 0$. If $c=0$, then $d\ne 0$ anda composition of a nonzero scaling and a translation. If $c\ne 0$, rearrange the fraction asThus every Möbius transformation is a composition, in order, of translation by $d/c$, reciprocal inversion, nonzero scaling by $(bc-ad)/c^2$, and translation by $a/c$. All formulas act on the Riemann sphere, including zero and infinity. The scaling parameter must be nonzero for that factor to be a Möbius transformation.

For three distinct finite points, define the cross-ratio normalizationIt sends $z_1,z_2,z_3$ to $0,1,\infty$ respectively. If $z_3=\infty$, use $(\zeta -z_1)/(z_2-z_1)$; if $z_1=\infty$, use $(z_2-z_3)/(\zeta -z_3)$; if $z_2=\infty$, use $(\zeta -z_1)/(\zeta -z_3)$. These are the corresponding limits of the same formula and cover all possible triples. Form the analogous normalization $F_w$ of the target triple. Thenhas all three specified images, proving existence. For uniqueness, the quotient of any two such maps fixes the three source points. Conjugating by $F_z$ gives a Möbius transformation fixing $0,1,\infty$. Fixing zero and infinity forces it to be $\lambda \zeta$, and fixing one forces $\lambda =1$. The quotient is identity, so the two maps are equal.

The Möbius transformations permuting three points are therefore in one-to-one correspondence with permutations of the triple. For the normalized set the six maps and their ordered images of $(0,1,\infty )$ areThey realize all six permutations. The action map to the symmetric group is a homomorphism, is injective by the three-point uniqueness proof, and is onto by this list. Hence

Since $\sigma$ is a group homomorphism, $\sigma (0)=I$, $\sigma (x+y)=\sigma (x)\sigma (y)$, and $\sigma (-x)=\sigma (x{)}^{-1}$. The product is closed on $\mathbb{R}\times {\mathbb{R}}^2$. Its two associative bracketings giveand coincide by the homomorphism property. The identity element is $(0,0)$, and the two-sided inverse isFor example, multiplying on the right gives $v-\sigma (x)\sigma (-x)v=0$, and multiplying on the left gives $-\sigma (-x)v+\sigma (-x)v=0$. Thus this law defines a group, specifically a semidirect product.

If it is an abelian group, comparing $(x,0)\ast (0,w)$ and $(0,w)\ast (x,0)$ gives $\sigma (x)w=w$ for every $w$. Hence $\sigma (x)=I$ for every $x$. Conversely this trivial action makes the product ordinary vector addition. Therefore the unique action giving a commutative group isFor the final condition, compare $(0,a{e}_1)\ast (x,w)$ and $(x,w)\ast (0,a{e}_1)$. They are $(x,a{e}_1+w)$ and $(x,w+\sigma (x)a{e}_1)$ respectively. They coincide for every $a,x,w$ exactly when $\sigma (x)e_1=e_1$ for every $x$. The first column of $\sigma (x)$ is then $(1,0{)}^{\mathsf{T}}$; since it lies in the special linear group, its determinant is one and its other diagonal entry is also one. ThusMatrix multiplication gives $r(x+y)=r(x)+r(y)$. Conversely any such additive function gives a homomorphism into $SL(2,\mathbb{R})$ fixing $e_1$ and therefore makes these translations central. This proves the full characterization of the central translations in a linear semidirect product:No continuity or measurability is assumed, so it would be unjustified to restrict the answer to $r(x)=cx$.

A group homomorphism sends identity to identity. For $\theta (g),\theta (h)$ in its image,which is still in the image; hence the image of a group homomorphism is a subgroup of $H$. Let $N=\ker \theta$. It is a subgroup by the same identity, and for $k\in N$ and $g\in G$,Thus $gN{g}^{-1}\subseteq N$, and replacing $g$ by $g^{-1}$ gives equality, proving that $N$ is a normal subgroup.

Define a map from the quotient group byIf $gN=hN$, then $h^{-1}g\in N$ and $\theta (g)=\theta (h)$, so the map is well defined. Conversely equality of the images implies $h^{-1}g\in N$, giving injectivity. It is onto by the definition of image, and $\Phi ((gN)(hN))=\theta (gh)=\theta (g)\theta (h)$. This proves the first isomorphism theorem for groups in this case:An orthogonal matrix $Q$ has $Q^{\mathsf{T}}Q=I$, so $(\det Q{)}^2=1$. The determinant is therefore a group homomorphism $O(3)\to \{1,-1\}$, with image both signs because $\mathrm{diag}(-1,1,1)$ has determinant $-1$. Its kernel is the special orthogonal group $SO(3)$. Hence $SO(3)$ is normal in $O(3)$, andFor the last request, the determinant-twist projection in odd dimension gives an explicit example:Indeed $\det \pi (Q)=(\det Q{)}^4=1$, and scalar factors commute, so $\pi (QR)=\pi (Q)\pi (R)$. If $\pi (Q)=I$, then $Q=(\det Q)I$, making $Q=I$ or $Q=-I$. Both lie in the kernel, which is therefore exactly $\{I,-I\}$, of order two. The map also fixes every member of $SO(3)$, so it is onto.

For the centralizer of a subset, the identity commutes with every $h\in A$. If $g,k$ commute with each member of $A$, then for every such $h$,Also $gh=hg$ implies $g^{-1}h=h{g}^{-1}$ after multiplication by $g^{-1}$ on both sides. Thus identity, products and inverses lie in $C(A)$, proving that it is a subgroup.

For the conjugation action, identity acts trivially, andThese are the defining laws of a left group action. Its orbits of a group action are the conjugacy classes. The stabilizer of $h$ is exactly its centralizer, since $gh{g}^{-1}=h$ if and only if $gh=hg$. The orbit-stabilizer theorem therefore givesA class has one element exactly when that element commutes with all of $G$, so the singleton classes are precisely the elements of $C(G)$, the centre of a group. Partitioning $G$ into all its conjugacy classes now proves the class equationFor $|G|=p^r$, Lagrange's theorem for finite groups makes every class size a power of $p$. A nonsingleton class has size at least $p$ and is therefore divisible by $p$. Reducing the class equation modulo $p$ yields $|C(G)|\equiv 0(\mathrm{mod}p)$, because $r>0$. The identity lies in $C(G)$, so its order is positive and divisible by $p$. ConsequentlyThis proves the nontrivial center of a finite p-group directly from the conjugation classes.

Use suffix notation, summing repeated indices, and write ${\partial }_i=\partial /\partial x_i$. For a twice continuously differentiable vector field, mixed derivatives commute. Thus the divergence of its curl satisfiesbecause the derivatives are symmetric in $i,j$, while the Levi-Civita symbol is antisymmetric in those indices. HenceFor the second identity, use the contraction of two Levi-Civita symbols:Rearranging each component proves the vorticity cross-product identityThe second identity needs first derivatives; the first additionally uses equality of the mixed second derivatives.

Take the unit normal to point outwards. The divergence of the field isThe divergence theorem turns the closed surface integral into a volume integral over $[0,1{]}^3$:For a direct check of the surface integral, on $x=1$ the outward component is $F_x=2y$, contributing ${\int }_0^1{\int }_0^{1}2ydydz=1$. On $x=0$ it vanishes. On both $y=0$ and $y=1$, $F_y=xy(y-1{)}^2=0$. On $z=1$, $F_z=0$, whereas on $z=0$ the field component is $F_z=-1$ and the outward normal is $-\hat{\mathbf{z}}$, giving a contribution $1$. The six face contributions are therefore $1,0,0,0,0,1$, again totaling 2. The positive bottom-face contribution is a consequence of its outward orientation.

Both subgroups contain the identity element, so $H\cap K$ is nonempty. If $x,y\in H\cap K$, then $x{y}^{-1}$ belongs to both $H$ and $K$, hence to their intersection. This gives closure under inverses by setting $x=e$, and closure under products by using the inverse of $y$. Therefore $H\cap K$ is a subgroup of $G$, and also of both $H$ and $K$.

Lagrange's theorem for finite groups states that the order of a subgroup divides the order of its finite ambient group; more precisely $|L|=[L:M]|M|$ for $M\le L$. Applying it twice givesSince $p$ and $q$ are coprime, their only positive common divisor is one. Thus