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The multiplicative group of integers modulo $n$ consists of the residue classes coprime to $n$. It has order $\phi (n)$.

If ${\mathcal{O}}_K=\mathbb{Z}[\theta ]$ and the minimal polynomial of $\theta$ factors modulo a rational prime $p$ asthenand the prime ideal $(p,g_i(\theta ))$ has norm $p^{\mathrm{deg}{g}_i}$.

This gives an explicit instance of unique factorization of ideals in a number field.

For three independent points drawn from rotationally symmetric distributions in the plane, with nonzero radii almost surely, the probability that their triangle contains the origin is $1/4$. The radial distributions can differ; independence and uniform independent angles are the essential ingredients.

Except for angle coincidences and antipodal pairs, which have probability zero, the origin is outside the triangle exactly when some line through the origin places all three points on the same side. Equivalently, all three angles lie in a semicircle. If this occurs, one of the three sample angles is the unique first endpoint of an oriented semicircle containing the other two. For any specified endpoint, the other independent uniform angles both fall in its next half-circle with probability $(1/2{)}^2=1/4$. The three endpoint events are disjoint almost surely, so the probability of containment in a semicircle is $3/4$.

The complementary probability is therefore $1/4$. This applies to uniform area samples in a disk and to independent uniform points on its boundary. The event depends on directions, so replacing one nonzero radius by another does not change it.

If distinct keys transform every ciphertext into equally probable plaintexts through a probability-preserving source symmetry, key equivocation never vanishes and unicity distance is infinite.

For a plaintext alphabet of size $m$ and entropy rate $h$ in bits per symbol, put $d={\log }_{2}m-h$. In the random-cipher approximation, the unicity distance scale iswhere $H(K)$ is cryptographic key Shannon entropy. Uniform cryptographic keys on a finite set $\mathcal{K}$ give $H(K)={\log }_2|\mathcal{K}|$. For correlated plaintext, its one-symbol Shannon entropy cannot replace the entropy rate without justification.^[ref]

For example, a 40-bit uniform cryptographic key and redundancy two bits per symbol suggest about twenty symbols under this model. This estimate does not bound computation time or guarantee unique recovery for each ciphertext. At zero redundancy it supplies no finite threshold; a probability-preserving source symmetry can also invalidate the random-cipher uniqueness intuition.

Key equivocation is the conditional entropy $H(K\mid C^n)$ remaining about a key after observing $n$ ciphertext symbols.

This theorem characterizes truth in an ultraproduct. Łoś's theorem says that a first-order formula holds in an ultraproduct exactly when it holds in a set of factors belonging to the ultrafilter.

The method of types combines two facts: there are few possible empirical distributions, and strings within each type class are equiprobable under an independent and identically distributed source. Information entropy determines how many strings share a frequency profile; relative entropy determines how unlikely that profile is under a specified source law. The connection is useful both for lossless source coding and for understanding rare observations.

The following formulas use natural logarithms, so information entropy and relative entropy are measured in nats. DefineHere $p$ is the source probability mass function and $q$ is an attainable empirical distribution. Use $0\ln 0=0$; first assume every $p(a)>0$. Base-two logarithms give the equivalent formulas with $2^{n{H}_2(q)}$ and $2^{-n{D}_2(q\Vert p)}$.

For a rational homology three-sphere $Y$, the linking form is the nonsingular pairingReversing the orientation of $Y$ negates this form.

Here the paired elements lie in the torsion subgroup of first homology.

For a sheet moving with $u(x,0)=\alpha x$ through otherwise stationary fluid, the layer thickness is $\delta =\sqrt{\nu /\alpha }$. Withthe boundary-layer equations reduce toThe exact solution is $f=1-e^{-\eta }$.

The twisted cubic is a nondegenerate projective variety, so every hypersurface containing it has degree at least two. At least two hypersurfaces are needed to cut out a curve in ${\mathbb{P}}^3$, making their degree product at least four rather than the curve's degree three.

A projective variety is nondegenerate when it is contained in no hyperplane of its ambient projective space.

Let $K\subset S^3$ be a smooth knot and let $m\ne 0$ be an integer. The exterior of its $m$-twist-spun knot in $S^4$ is a fiber bundle over a circle, with fiber the $|m|$-fold cyclic branched covering of $S^3$ along $K$ with an open three-ball removed. For the open complement, use the corresponding branched cover with a point removed. Its monodromy comes from the cyclic deck action.^[ref]

For $m=\pm 1$, the branched cover is $S^3$, the compact fiber is a three-ball, and the resulting two-dimensional surface knot is unknotted. Zero-twist spinning is ordinary spinning and is excluded from this nonzero-twist fibration theorem; a zero-fold branched cover is not its fiber.^[ref]

Let ${\Delta }^n$ be an $n$-dimensional simplex, $n\ge 1$, with barycentric coordinates $x_0,\dots ,x_n$. Suppose closed sets $F_0,\dots ,F_n$ cover it and each $F_i$ contains the opposite face of a simplex $\{x_i=0\}$. Then their common intersection is nonempty. This is the higher-dimensional opposite-face version of the triangle KKM lemma.

If the common intersection were empty, normalized distance functions to the $F_i$ would give a continuous map $g:{\Delta }^n\to \partial {\Delta }^n$ carrying every face of a simplex into itself. On the boundary, the straight interpolation $(1-t)x+tg(x)$ stays in the same face of a simplex, so $g{|}_{\partial {\Delta }^n}$ is homotopic to the identity map. It therefore induces the identity on reduced top homology. Yet it extends across the contractible simplex, whose reduced homology is zero, a contradiction.

Conversely, any such face-preserving boundary-valued continuous map would produce a counterexample by $F_i=\{x:g_i(x)=0\}$. These are closed, cover the simplex, contain their opposite faces of a simplex and have empty common intersection, because the $g_i$ sum to one. Thus the covering assertion and the absence of that map are equivalent.

Let closed subsets $F_0,F_1,F_2$ of a triangle have empty common intersection. Their distance functions $d_i(x)=\mathrm{dist}(x,F_i)$ have positive sum at every point, soare continuous barycentric coordinates. If one $F_i$ is empty it may be assigned the constant positive function one instead. The denominator condition is essential; the formula is undefined on a common intersection.

If the $F_i$ cover the triangle, at least one $d_i(x)$ vanishes everywhere, placing $g(x)$ on its boundary. If $F_i$ also contains the side opposite the $i$th vertex, then $g_i=0$ on that side: the map carries each side into itself. The no-retraction covering principle rules out this boundary-valued continuous map, proving the triangle KKM lemma.

An equilateral triangle has three equal side lengths.

An acute triangle has all three interior angles strictly smaller than a right angle.

An anti-trapped surface is the time reverse of a trapped surface: both future-directed orthogonal null expansions are positive.

A one-dimensional scalar conservation law describes the balance of a scalar density $u$ and its conservation law flux $f(u)$:The second identity follows from the first for smooth $u$: the quantity inside a fixed interval changes only through its boundary fluxes. For a classical solution, the chain rule also gives the quasilinear partial differential equation $u_t+f^{\prime }(u)u_x=0$, with characteristic speed $f^{\prime }(u)$.

The conservative form still defines a discontinuous weak solution. With initial condition $u_0$, it requiresfor every smooth compactly supported test function on $[0,\infty )\times \mathbb{R}$. For $f(u)=u^2/2$, a jump from $1$ to $0$ travels at speed $1/2$ by the Rankine-Hugoniot condition, between incoming characteristic speeds $1$ and $0$. Reversing those states still gives a conservative jump of speed $1/2$, but entropy solution admissibility selects a rarefaction wave. Thus conservation determines jump balance while an entropy condition selects the continuation after characteristic crossing.^[ref]

A linear transport equation has prescribed coefficients and is linear in the unknown:Assume the coefficients and initial condition $u_0$ are smooth enough for the following characteristic curves and classical solution. With $\dot{X}(t,\alpha )=b(t,X(t,\alpha ))$ and $X(0,\alpha )=\alpha$, set $C(t)=c(t,X(t,\alpha ))$ and $F(t)=f(t,X(t,\alpha ))$. The chain rule reduces the equation to $\dot{U}+CU=F$. The integrating factor givesDifferentiating the formula verifies the ordinary differential equation. For constant $c=\gamma >0$, zero $f$ and constant $b$, it becomes $u(t,x)=e^{-\gamma t}{u}_0(x-bt)$: translation with exponential decay. With $c=\mathrm{div}b$, it instead describes the conservative continuity equation. The difference of two solutions with the same $f$ satisfies the homogeneous equation, the form of superposition useful for uniqueness.^[ref]