Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.

Count the pairs $(g,\omega )$ for which $g$ fixes $\omega$. Counting first by group elements gives ${\sum }_g\mathrm{Fix}(g)$, while counting first by points gives ${\sum }_{\omega }|\mathrm{Stab}(\omega )|$. On a group orbit of size $r$, orbit-stabilizer theorem makes each stabilizer have size $|G|/r$, so that group orbit contributes $|G|$ to the second sum. Dividing by $|G|$ proves Burnside's lemma:A tetrahedron has four vertices. A rotational symmetry is determined by the image of one vertex, with four possibilities, and the cyclic orientation of the three vertices around it, with three possibilities. All these choices are realizable, so its rotational symmetry group of a tetrahedron has order $12$. More explicitly, its elements are the identity, the two nontrivial rotational symmetries around each of four vertex-to-opposite-face axes, and the half-turn about each of three axes through opposite edge midpoints. Thus there is one element of order 1, eight of order 3, and three of order 2. The induced permutations of vertices are the even permutations, identifying the group with $A_4$.

Apply the group action to the set of edge colourings. A colouring is fixed by a rotational symmetry precisely when its colour is constant on each cycle of the induced edge permutation. The identity has six edge cycles and fixes $4^6$ colourings. A vertex-axis rotational symmetry has two cycles of length three, fixing $4^2$. An opposite-edge half-turn fixes the two edges on its axis as sets and exchanges the other four in two pairs, giving four cycles and $4^4$ fixed colourings. ThereforeReflections are excluded because the equivalence in the source uses rotational symmetries.

For a nonzero real vector $w$, a Householder reflection isWriting $P=w{w}^T/(w^{T}w)$ gives $P^T=P$ and $P^2=P$, so $H^{T}H=(I-2P{)}^2=I$. Thus a Householder matrix is an orthogonal matrix, and it reflects across the hyperplane perpendicular to $w$.

For $u\ne 0$, choose $\alpha =-\mathrm{sgn}(u_1)\Vert u\Vert$, taking the sign to be $+1$ when $u_1=0$, and put $w=u-\alpha e_1$. This $w$ is nonzero. Since ${\alpha }^2=\Vert u{\Vert }^2$, direct expansion gives $w^{T}w=2(\Vert u{\Vert }^2-\alpha u_1)$ and $w^{T}u=\Vert u{\Vert }^2-\alpha u_1$, soApply this construction successively to the trailing part of each column of $A$, extending each reflection by an identity on earlier coordinates. Previously created zeros are preserved. If a trailing column vector is zero, use the identity there. The resulting product $Q^T$ is orthogonal and $Q^{T}A=R$ is upper triangular, yielding the QR factorization $A=QR$ even when $A$ is singular.

To solve $Ax=b$, multiply by $Q^T$ and use back substitution on $Rx=Q^{T}b$ when $R$ is nonsingular. If $A$ is singular, $\det R=\det (Q^T)\det A=0$; an upper triangular determinant is its diagonal product, so at least one diagonal entry is zero. In that case consistency and free variables require separate treatment.

The unshifted QR algorithm starts from $A_0=A$, factors $A_k=Q_k{R}_k$, and formsEvery iterate is related by an orthogonal similarity transformation to the previous one and hence to $A$, with the same eigenvalues. Symmetry is preserved. For symmetric matrices with appropriate spectral separation, the iterates approach diagonal form: the off-diagonal entries decrease and the diagonal exposes the eigenvalues. Practical shifted QR accelerates convergence and uses deflation when off-diagonal couplings become negligible. With a shift ${\mu }_k$, factor $A_k-{\mu }_{k}I=Q_k{R}_k$ and set $A_{k+1}=R_k{Q}_k+{\mu }_{k}I$; the same similarity identity still holds.

Universal diagonal convergence of unshifted QR would be too strong. For example $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ admits the standard factorization $Q=A$, $R=I$, so the iteration remains at $A$ rather than becoming diagonal, although its eigenvalues are $\pm 1$. Such equal-modulus cases require a convergence qualification or treatment of a retained block. Similarity is unconditional; convergence to a useful limiting form is an additional algorithmic property.

The Lagrange interpolation polynomialhas degree at most $n$ and takes the required value at each node, proving existence. If two such polynomials existed, their difference would have $n+1$ distinct roots despite degree at most $n$, so it would vanish identically. This proves uniqueness.

Define $P(x)=[(x-x_0)r(x)-(x-x_n)q(x)]/(x_n-x_0)$. It has degree at most $n$. At $x_0$ its value is $q(x_0)=f_0$, and at $x_n$ it is $r(x_n)=f_n$. At every interior node both $q$ and $r$ equal $f_i$, giving $P(x_i)=f_i$. Uniqueness therefore givesIf the data come from $x^m$ with $m\le n$, uniqueness gives $p=x^m$. If $m<n$, that same polynomial has degree at most $n-1$ and satisfies each smaller interpolation problem, so $q=r=x^m$.

For the final claim assume $n\ge 2$. Put $t=q-r$. Its node values are $t(x_0)=f_0-r(x_0)$, $t(x_i)=0$ for $1\le i\le n-1$, and $t(x_n)=q(x_n)-f_n$. The linear spline $s$ joining these values has a ramp on the first interval, is zero on all middle intervals, and has a ramp on the last. ThusOn each node interval $s^{\prime }$ is constant and $t-s$ vanishes at both ends. Hence $\int s^{\prime }(t^{\prime }-s^{\prime })dx=0$ on that interval. Integrating the supplied identity and summing giveswhich proves the claimed inequality and the minimum Dirichlet energy of a linear interpolant.

There is a source exception: with $n=1$, $q=f_0$ and $r=f_1$, so $t$ is constant and the left side of the asserted bound is zero. If $f_0\ne f_1$, its right side is $2(f_0-f_1{)}^2/(x_1-x_0)>0$. Thus the final inequality requires $n\ge 2$, although the preceding interpolation statements remain valid for $n=1$.

Let $f_0,f_1$ be densities or probability mass functions relative to a common measure, describing two simple hypotheses. A randomized hypothesis test is a measurable function $0\le \phi \le 1$, whose value is the conditional rejection probability. Fix $0<\alpha <1$. The Neyman-Pearson lemma states that a test ${\phi }_{\ast }$ with ${\mathbb{E}}_0{\phi }_{\ast }=\alpha$ is most powerful among tests with size at most $\alpha$ if it rejects where $f_1>c{f}_0$, accepts where $f_1<c{f}_0$, and randomizes as necessary on equality, for a threshold $c\ge 0$. Sets where $f_0=0<f_1$ are included in rejection. A likelihood-ratio quantile and boundary randomization supply such a test.

To prove optimality, let $\phi$ be any competitor with ${\mathbb{E}}_0\phi \le \alpha$. Pointwise, the threshold rule makesIntegrating and rearranging givesThus no competitor of the permitted size has larger statistical power, proving the lemma. Randomization permits exact size when discrete observations cannot supply it with a deterministic rejection region.

The velocity is $\mathbf{u}=\nabla \varphi$ because the flow is irrotational. Incompressible flow requires $\nabla \cdot \mathbf{u}=0$, hence the velocity potential satisfies Laplace's equation:The exact kinematic boundary condition at the free surface is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$. With constant atmospheric pressure, the dynamic condition from Bernoulli's principle is ${\varphi }_t+\frac{1}{2}|\nabla \varphi {|}^2+g\eta =0$, after choosing the time-dependent potential gauge. For small amplitudes, the products of perturbations are second order. Evaluating the remaining first-order terms at the undisturbed level $y=0$ givesSubstitution of $\varphi =f(y)\sin (kx-\omega t)$ into Laplace's equation gives $f^{\prime \prime }-k^{2}f=0$. For $k\ne 0$, decay as $y\to -\infty$ selects $f(y)=B{e}^{|k|y}$. The two boundary conditions for $\eta =A\cos (kx-\omega t)$ yieldFor a nonzero wave amplitude these are compatible precisely whenWith the conventional $k>0$ and positive frequency, the dispersion relation is $\omega (k)=\sqrt{gk}$. The exponential decay is the infinite-depth boundary condition; retaining a growing exponential would violate it.

Orient the body boundary $C$ counterclockwise. Its outward normal vector satisfies $\mathbf{n}ds=(dy,-dx)$, so the pressure force on the body has complex componentFor steady incompressible flow and irrotational flow, Bernoulli's principle gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant-pressure integral vanishes. Write $q=w^{\prime }=u-iv$. The impermeability condition makes velocity tangent to $C$, implying $qdz=\bar{q}d\bar{z}$. Thus $|q{|}^{2}d\bar{z}=q^{2}dz$ along the boundary, andThis proves the Blasius force theorem, with force measured per unit length perpendicular to the flow plane.

For a circle of radius $a$, Milne-Thomson's circle theorem replaces an incident analytic potential $f(z)$ by $f(z)+\overline{f(a^2/\bar{z})}$, making the stream function constant on the circle. Its reflected singularities must lie inside the circle; here the physical source lies outside. With $c=Q/(2\pi )$ and $d=a^2/b$, this yields, up to additive constants,The interior image source and sink have equal and opposite strengths, so there is no extra net source inside the body. No independent vortex term is added. The potential may have logarithmic branches, but the velocity and pressure are single-valued.

Only $d$ and $0$ lie inside $C$. The residues of $(w^{\prime }{)}^2$ areTheir sum is $-2c^2{a}^2/[b(b^2-a^2)]$. The residue theorem and the preceding force formula therefore giveThe force points towards the external source. Reflection symmetry across the horizontal axis independently checks the vanishing vertical component.