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In a homogeneous isotropic elastic half-space $Y>0$, a Rayleigh wave with factor $e^{ik(X-ct)}$ has two depth factors $e^{-b_{L}Y}$ and $e^{-b_{T}Y}$. Substitution into the Navier-Cauchy equation factors the amplitude determinant into longitudinal and transverse factors, giving the displayed exponents. Strict exponential decay requires $0<c<\min (c_L,c_T)$. The longitudinal constituent has $(A,B)\propto (ik,-b_L)$; the transverse constituent has $ikA-b_{T}B=0$. The free-surface traction conditions select a combination of both constituents.

For a Rayleigh wave at a free surface of a homogeneous isotropic elastic half-space, define $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling the two surface tractions between the constituents with Rayleigh wave decay exponents gives the displayed equation with $0<s<1$. Equivalently $(b_T^2+k^2{)}^2=4k^2{b}_L{b}_T$. Squaring gives a factor $s$ times $P(s)=s^3-8s^2+(24-16\eta )s-16(1-\eta )$. For stable Lamé parameters, $0<\eta <3/4$, so $P(0)<0$, $P(1)=1$, and $P^{\prime \prime }(s)=6s-16<0$ on $(0,1)$. Its concavity ensures exactly one crossing in this interval. Both sides of the unsquared equation are positive, so this root gives the physical speed. The zero root is a degenerate static cancellation, not a propagating Rayleigh wave.

Let $D$ be a rational vector subspace of the real numbers and let $t\notin D$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$. For any prescribed real number $a$, the rule $h(d+qt)=f(d)+qa$ extends a rational-linear map $f$ on $D$ and remains rational-linear. The uniqueness follows because a nonzero rational multiple of $t$ cannot lie in $D$.

Let $t$ be transcendental over a field $k$ and let $\tau$ be the field automorphism sending $t$ to $1/t$. The rational function $u=t+1/t$ is fixed, and $t$ satisfies $X^2-uX+1=0$. Thus the extension over $k(u)$ has degree at most two. It is not one because the nontrivial automorphism $\tau$ fixes $k(u)$. The fixed field of $\tau$ also has index two by the fixed-field theorem, so it equals $k(u)$.

No nonconstant polynomial in $k[t]$ is fixed by inversion. For a polynomial of positive degree, substitution of $1/t$ creates negative powers, whereas the original has positive powers. Uniqueness of the coefficients of a Laurent polynomial rules out equality. Consequently this fixed field cannot have a generator that is a polynomial in the original variable, even though it has the simple rational-function generator above. The statement is valid in characteristic two as well: $t$ and $1/t$ remain distinct transcendental rational functions.

A resting shallow water layer has depth $h_0$ and wave speed $c_0=\sqrt{g{h}_0}$. A vertical wall receding into negative $x$ at speed $0<V<2c_0$ creates a simple wave with invariant $u-2c=-2c_0$. Inside the expansion fan $x/t=u+c$, proving the displayed profiles. The wall-adjacent state is $u=-V$, $c=c_0-V/2$; the fan occupies $c_0-3V/2<x/t<c_0$, with the resting state ahead. The wall lies at $x/t=-V$, strictly to the left of the fan tail because $c_0-V/2>0$. Thus the assumption on wall speed is exactly the condition for positive depth and continued contact between fluid and wall. The depth follows from $h=c^2/g$.

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.

The unrotated variants differ by $2\delta e_3\otimes e_1$, giving normal $e_1$. The rotation about $e_2$ with $\cos \theta =(1-{\delta }^2)/(1+{\delta }^2)$ and $\sin \theta =2\delta /(1+{\delta }^2)$ gives $U_{+}-\overline{R}{U}_{-}=[2\delta /(1+{\delta }^2)](e_1+\delta e_3)\otimes e_3$, giving the second normal $e_3$. If these are the only two compatible rotations, the only gradients in $SO(3)U_{+}$ compatible with $\overline{R}{U}_{-}$ are $U_{+}$ and $\overline{R}{U}_{+}$. Left multiplication by a rotation preserves rank, so the second classification follows directly from the first.

Let $A=R(k-1,\ell )$ and $B=R(k,\ell -1)$. From one vertex of any red-blue colored $K_{A+B}$, either at least $A$ incident edges are red or at least $B$ are blue. In the first case, apply the definition of the Ramsey number to its red neighbors: a red $K_{k-1}$ extends with the chosen vertex, and a blue $K_{\ell }$ is already enough. The blue case is analogous. This proves the recursion, with boundary values $R(k,2)=k$ and $R(2,\ell )=\ell$.

A material element is in local radiative equilibrium when its absorbed and emitted radiative powers balance, so radiation causes no net local heating. In a non-scattering grey atmosphere in local thermodynamic equilibrium, this reduces to equality of the angular mean specific intensity $J$ and the Planck function $B$. Integrating the radiative transfer equation over solid angle then shows that the radiative flux is constant through a plane-parallel atmosphere. The radiation can remain anisotropic; equilibrium constrains its angular mean, not every ray separately.

Let $T:X\to Y$ be a bounded linear operator with $X$ a Banach space. Suppose every $\Vert y\Vert <1$ has an approximation $Tx$ with $\Vert x\Vert <1$ and error less than $\delta \in (0,1)$. Repeatedly approximate the scaled residual to construct $x_n$ with $\Vert x_n\Vert <{\delta }^n$ and residual less than ${\delta }^{n+1}$. Completeness sums the series, and continuity gives $T\sum x_n=y$. Since $\Vert \sum x_n\Vert <(1-\delta {)}^{-1}$, scaling proves $B_Y(0,1-\delta )\subseteq T(B_X(0,1))$. Hence $T$ is a quotient operator, even without assuming completeness of $Y$.

One element that satisfies a relation simultaneously for every quantified input. A uniform witness also proves the corresponding pointwise statement, because it may be chosen for each particular input.

A witness whose existence is asserted separately for each input. The witness need not be the same element for different inputs, so this condition is weaker than existence of a uniform witness.

A projective triangle is an ordered triple of noncollinear projective points in a projective plane, together with their three joining projective lines. In the real projective plane this means that three chosen representative vectors are linearly independent. Pairing the vertices of two such triangles determines corresponding sides and the three corresponding-vertex joins.

For two projective triangles in a projective plane over a field, perspectivity from a point implies perspectivity from a line. The usual nondegenerate form assumes that the corresponding-side intersections are well defined and distinct. Its dual is the converse, by point-line duality. The result holds in the real projective plane; it is not an axiom valid in every abstract incidence plane.

One linear algebra proof explains the collinearity. Choose a representative $o$ of the center. Rescale the paired representatives to $p^{\prime }=p+\alpha o$, $q^{\prime }=q+\beta o$, and $r^{\prime }=r+\gamma o$, with nonzero coefficients in a nondegenerate configuration whose vertices differ from the center. The first corresponding-side intersection is represented by $x=\beta p-\alpha q=\beta p^{\prime }-\alpha q^{\prime }$. Similarly choose $y=\gamma q-\beta r$ and $z=\alpha r-\gamma p$. Then $\gamma x+\alpha y+\beta z=0$, so the three representatives are linearly dependent. Their projective points are consequently collinear.

In the real projective plane, represent a projective point by a nonzero vector and a projective line by its nonzero coefficient covector, identified with a vector in chosen coordinates. The join of distinct projective points represented by $p,q$ has coefficient vector $p\times q$, using the cross product. It is nonzero because the representatives are linearly independent, and its dot product with each representative vanishes. Dually, two distinct projective lines with coefficients $\ell ,m$ meet in the projective point represented by $\ell \times m$. Scaling an input scales the output and leaves the projective object unchanged. This makes joins, intersections, concurrency, and collinearity exact calculations with homogeneous coordinates.

For two triples of distinct points on two projective lines, the intersections $r_{ij}=p_i{q}_j\cap p_j{q}_i$ are collinear when the configuration is defined. Normalize a projective frame so $p_1=[0,1,0]$, $p_2=[1,1,0]$, $q_1=[0,0,1]$, $q_2=[1,0,1]$, and write $p_3=[1,\alpha ,0]$, $q_3=[1,0,\beta ]$. Then representatives of the three intersections are $(1,1,1)$, $(1,\alpha ,\beta )$, and $(\alpha \beta -1,\alpha (\beta -1),\beta (\alpha -1))$. The last is $\alpha \beta$ times the first minus the second, proving collinearity by linear dependence. This gives a coordinate proof over any field when the normalizations are available.

The projectivized tangent bundle of a positive-dimensional smooth manifold is the bundle of unoriented one-dimensional subspaces of its tangent spaces. A local tangent trivialization gives $P(TM){|}_U\cong U\times {\mathbb{RP}}^{n-1}$. If a base-chart transition is $F$, the line coordinates transform by $[v]\mapsto [D{F}_{x}v]$. In any projective chart, this is a ratio of smooth linear expressions with nonzero denominator. It therefore defines a smooth manifold of dimension $2n-1$.

For the two-sphere, a point consists of $x\in S^2$ and a line $L\subset x^{\perp }$. Choosing a unit vector $v\in L$ gives an oriented orthonormal basis $(x,v,x\times v)$. A unique rotation takes one chosen frame to another, so $S{O}_3$ acts transitively. Its stabilizer consists of the identity and the half-turn about $x$, which reverses $v$ while preserving its unoriented line. Thus $P(T{S}^2)\cong S{O}_3/C_2$.