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Database normalization organizes relational schemas using their dependencies to reduce redundant facts and update anomalies. If a table stores (student,course,instructor,office) and each instructor has one office, repeating the office in every enrolment creates multiple places to update. Splitting instructor-office facts into a separate relation can remove that redundancy. A decomposition must be checked for lossless join and, when needed, dependency preservation; merely creating more tables is not a correctness argument. Different normal forms impose different conditions on functional dependencies and other dependency types.

Retains the database tuples satisfying a condition.

Retains specified database attributes, eliminating repeated database tuples under set semantics.

Combines database tuples satisfying a matching condition. A natural join uses equality of all shared database attribute names.

For two database relations with compatible database attributes, retains the database tuples in the first and absent from the second.

Relational division expresses an all-required-matches query. Let $R\subseteq X\times Y$ be a database relation, $S\subseteq Y$ the required values and $C\subseteq X$ an explicit candidate database relation. The desired result isIt can be constructed using Cartesian product, relational projection and relational difference:The inner relational difference contains missing required pairs. Its relational projection identifies failing candidates; removing them from $C$ proves the formula.

For example, $R$ can record scientific instrument/test pairs and $S=\{\mathrm{heat},\mathrm{vibration}\}$ the required tests. An scientific instrument with both pairs passes; one with only the heat pair does not. This is a universal quantifier requirement, unlike an ordinary relational join asking for some match.

When $S$ is empty, vacuous truth gives $D=C$. Choosing $C={\pi }_X(R)$ is conventional but cannot recover candidates absent from $R$; use an independently stored scientific instrument database relation if those candidates must be included. Stating the candidate universe makes this edge case unambiguous.

Consider independent rejection sampling trials with proposal density $q$, accepted-value density $f$, acceptance probability $a\in (0,1)$ and rejection probability $r=1-a$. Splitting the proposal law by its accept/reject outcome gives $q=af+r{g}_R$, where $g_R$ is the rejected-proposal distribution.

If an algorithm permits $n$ trials, returns the first accepted value when one occurs, and otherwise returns the final rejected proposal, its output density is the displayed mixture. The event of $n$ failures has probability $r^n$ and gives the conditional law $g_R$; every successful stopping trial gives law $f$. Hence the forced return introduces bias unless $g_R=f$, and its contribution decays geometrically with the cap. Conditional on success before any specified trial, the output still has law $f$, since accepted value and successful-trial index are independent.

Stopping after a fixed number of rejection sampling proposals preserves the target law of accepted labels, conditional on the number accepted. Stopping instead after a fixed amount of elapsed computing time can change the finite-sample law if processing costs depend on the proposed state. The deadline may censor precisely the expensive states.

For an exact counterexample, let both target and proposal be uniform on two states, so every proposal is accepted. Let processing cost be one time unit for the first state and two for the second. Stop between one and two units and retain only finished proposals. Conditional on obtaining an output, the first state is the only possible value, despite its target probability being one half. This proves that a runtime cutoff need not be an unbiased stopping rule for the realized finite output. It does not imply asymptotic bias of the indefinitely growing completed sequence; it identifies a finite-horizon censoring effect.

For $n\ge 3$ cyclically ordered planar vertices, let $e_j=v_{j+1}-v_j$, with indices modulo $n$. They describe a convex regular polygon exactly when $e_0\ne 0$ and there is a single choice of sign for which every successive edge vector is obtained by a rotation through $2\pi /n$ in that orientation. Necessity follows from equal sides and equal exterior angles. Conversely, the edge vectors have the form $R^j{e}_0$. Their sum is zero because the corresponding geometric series runs through the $n$ distinct roots of unity. Summing its partial series shows that the vertices are a translated and scaled copy of those roots, so they form a convex regular polygon. A larger turn such as $4\pi /5$ can instead trace a star, which is why merely checking a constant turning angle is insufficient.

A regular heptagon is a regular polygon with seven sides. Its unit-circle vertices have complex coordinates $e^{2\pi ik/7}$ for $k=0,\dots ,6$. It cannot be constructed from a unit segment by a straightedge and compass construction: the first coordinate $\cos (2\pi /7)$ has degree three over $\mathbb{Q}$, since the seventh cyclotomic field has degree six and degree two over its real subfield. Every finite constructible coordinate extension has power-of-two degree, which cannot contain a degree-three subextension by the tower law.

For $(L_0+ϵL_1)u=f(ϵ)$ and $Bu=g(ϵ)$, expand all three quantities in powers of $ϵ$. The order-$n$ problem has $L_0{u}_n=f_n-L_1{u}_{n-1}$ and $B{u}_n=g_n$, with $u_{-1}=0$. In particular, Taylor series boundary coefficients are generally nonzero even when the differential operator depends linearly on the parameter. If the homogeneous realization of $L_0$ is invertible and boundary data admit smooth lifts, the same inverse solves each successive inhomogeneous boundary value problem. Induction establishes unique coefficients at every order. This formal conclusion does not alone assert convergence of the full power series; convergence needs additional bounds on the inverse and the perturbation.

For a fixed symbol $a$ and a regular language $L$, inserting exactly one $a$ at an arbitrary position preserves regularity. Make two copies of a recognizing finite-state automaton, one before and one after the insertion. Preserve its ordinary automaton transitions in each copy, and add a transition consuming $a$ from each before-state to its matching after-state. Accept only in the after-copy. Every accepting run makes exactly one of the additional moves.

Regular languages are closed under finite intersection, but not countably infinite intersection. Indeed, for any formal language $K\subseteq {\Sigma }^{\ast }$ over a finite alphabet, let $L_i$ contain every word of length at least $i$, together with the members of $K$ of length less than $i$. Each $L_i$ is a regular language, since its exceptional short part is a finite language, and ${\bigcap }_{i\ge 1}{L}_i=K$. Thus every formal language, including one that is not a regular language, can arise in this way.

For a nonzero real normal vector $n$, orthogonal reflection in $n^{T}x=d$ has the displayed affine map. It is $x\mapsto Hx+t$ with $H=I-2n{n}^T/(n^{T}n)$ and $t=2dn/(n^{T}n)$. In homogeneous coordinates its matrix isEvery point on the hyperplane is fixed, and the signed normal displacement changes sign. Choosing any point $p$ on the hyperplane gives $t=p-Hp$, so translating by $-p$, applying the reflection matrix $H$ and translating back gives the same transformation.

An alkaline copper(II)–tartrate reagent used in a Fehling's test.

A test named after Hermann von Fehling. Heating with Fehling's reagent can produce copper(I) oxide from reducing sugars; several different substances can give a positive result.

For a node $v$ of a red-black tree, count black nodes strictly below $v$ on any path to an absent-child sentinel, including that black sentinel. The red-black invariants make this count independent of the path. Other conventions omit the sentinel or count $v$; specify the convention when using numerical values. Equality of the counts, rather than their common additive offset, is what the balancing condition requires.

A red-black tree with no red stored nodes is a perfect binary tree. Indeed, every root-to-sentinel path has equal length because every real node is black; a missing branch at an earlier level would make its path shorter. Thus every real level is full. Its size is $2^h-1$ for some number $h$ of real levels. Conversely, colouring every stored node of a perfect binary tree black satisfies every red-black invariant. This proves both directions of the size criterion.

In the usual bottom-up algorithm, insert a red node by the binary search tree rule. If its parent is black, no repair is needed. If its parent and uncle are red, recolour both black and recolour the grandparent red, then continue upward. If the uncle is black, an inner configuration first rotates the parent to become an outer configuration; rotating the grandparent and exchanging its colour with the parent's resolves the red-red conflict. Apply the left/right mirror cases as needed and finish with a black root. The tree rotations preserve key order, and the colour changes preserve equal black height. Different valid insertion conventions can produce different valid red-black trees, so a worked trace must identify its convention.

A formal language whose complement of a set, relative to the set of all input words, is a recursively enumerable language. A recognizer for the complement detects every negative instance eventually and need not halt on positive instances. A formal language is a decidable language exactly when it and its complement are both recursively enumerable languages.