Every nontrivial extensional property of partial computable functions or computably enumerable languages is undecidable.

Order vertices by nonincreasing positive degrees $d_i$, and let $b_i$ count preceding neighbors. The Madiman-Tetali inequality applied to preceding-neighbor sets and $b_i$ copies of singleton $i$ bounds the entropy of a uniformly chosen independent set. In each local term, all-zero neighbor assignment permits two choices at $i$, while every other assignment permits at most one. The maximum-entropy bound gives ${\log }_2(2^{b_i+1}-1)$ for that term. Complete bipartite graphs $K_{d,d}$ attain equality when one part is ordered before the other.

For a graph whose vertices are partitioned into parts, an independent transversal is an independent set containing exactly one vertex from each part. If a cycle's parts each have twelve vertices, choose one vertex uniformly and independently per part. A cycle edge joining distinct parts is selected at both ends with probability $1/144$. Each event depends on at most $47$ others, since each part meets at most $24$ cycle edges. The Lovász local lemma applies because $48e/144<1$, proving existence for an arbitrary such partition.

For every infinite set $X$ there is an independent family of sets consisting of $2^{|X|}$ subsets of $X$. One construction uses a set $D$ of pairs $(s,H)$, where $s\subseteq X$ is finite and $H\subseteq \mathcal{P}(s)$. For each $A\subseteq X$, let $I_A=\{(s,H):A\cap s\in H\}$. Given distinct $A_1,\dots ,A_m$, choose a finite $s$ distinguishing all their traces and choose $H$ to contain precisely the traces prescribed to have value one. This realizes every finite Boolean choice. With the axiom of choice, $|D|=|X|$, so a bijection transports this family to $X$. Its independent assignments give $2^{2^{|X|}}$ distinct ultrafilters.

Expand every term in finitely many generators as a finite nonempty sum of monomials, and associate its exponent support with the upward-closed subset of ${\mathbb{N}}^r$ that it generates. The reverse-inclusion well-quasi-ordering of monomial ideals supplies an earlier support upset containing a later one. Every later monomial is then a multiple of an earlier monomial, hence is absorbed by it under $a+ab=a$. The earlier sum absorbs the later sum. Thus every infinite term sequence has an earlier element preceding a later one in the reverse natural order. Reflexivity, antisymmetry and transitivity follow directly from commutative idempotent addition.

For an otherwise outbred pedigree, each of the two shared grandparents contributes $(1/2{)}^{2+2+1}=1/32$ to the kinship coefficient of first cousins. Thus their child has inbreeding coefficient $1/16$. Equivalently, cousin kinship is one quarter of the $1/4$ kinship of their sibling parents.

// Target: biology.bigb

For a biallelic locus with frequencies $p=1-q$ and $q$, and inbreeding coefficient $F$, ancestral-copy identity with probability $F$ gives genotype probabilities $P(AA)=p^2+Fpq$, $P(Aa)=2pq(1-F)$ and $P(aa)=q^2+Fpq$. In the IBD component both copies equal a single ancestral draw; in the remaining component the two ancestral alleles are independent biological population draws. A fully penetrant recessive disease therefore has risk $q^2+Fq(1-q)$.

// Target: biology.bigb

For $U$ uniform on $(0,1)$, $Y$ has density $q(y)=2y^{-2}$ on $y\ge 2$. The importance sampling weight for the standard Cauchy distribution tail is $W=f(Y)/q(Y)$, and its expectation is the target tail probability. Writing $A=\arctan (1/2)$ gives $\mathbb{E}W=A/\pi$ and $\mathbb{E}{W}^2=(A+2/5)/(4{\pi }^2)$ by integrating $y^2/(1+y^2{)}^2$. The weight varies only between $2/(5\pi )$ and $1/(2\pi )$, making its variance much smaller than that of a rare-event indicator. This is a comparison of sampling variances, not a deterministic ordering of every pair of empirical variance estimates.

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt is a subset of the original domain of a function. The notation does not require an inverse function: the preimage of a singleton set can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The preimage of a singleton set is also its fiber of a function.

For subsets $U,V\subseteq Y$, preimages preserve both unions and intersections:For the intersection, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output subsets.

In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See Lebl, Proposition 0.3.15: preimages of unions, intersections and complements.

This makes preimages useful for transporting conditions. The solution set of $g(x)=0$ is $g^{-1}(\{0\})$, and a zero set is a preimage of zero. Requiring two output conditions takes the intersection of their preimages. In topology, a continuous function pulls every open set back to an open set; a measurable function pulls the specified measurable sets in its codomain back to measurable sets in its domain of a function. These definitions use preimages because their set operations remain consistent even when the function is noninjective.

The lux is the unit of illuminance: one lumen of luminous flux uniformly incident on one square metre gives one lux. It measures illumination arriving at a surface, rather than the surface's luminance or the source's luminous intensity.

For a small source of luminous intensity $I_v$, normal incidence at distance $r$ gives $E_v=I_v/r^2$ when the source can be treated as a point and intervening losses are negligible. Thus a $1\mathrm{cd}$ source gives $1\mathrm{lx}$ at $1\mathrm{m}$ and $0.25\mathrm{lx}$ at $2\mathrm{m}$. A meter's sensor must be oriented consistently because tilting changes the projected collecting area.

Mayer's relation states that the difference between heat capacity at constant pressure and heat capacity at constant volume for a fixed-composition ideal gas isHere both capacities are total heat capacities in $\mathrm{J}{\mathrm{K}}^{-1}$, $n$ is amount of substance, $N$ is the number of particles, $R$ is the molar gas constant and $k_B$ is the Boltzmann constant. In other normalizations the same statement isThe first is a difference of molar heat capacities, in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the second uses specific heat capacities, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Writing $C_p-C_V=R$ with total capacities would omit the sample's amount of substance. The relation holds pointwise even when the capacities depend on temperature. ^[ref]

The proof outline is short: enthalpy is $H=U+pV=U+nRT$, and internal energy of an ideal gas is independent of volume at fixed temperature. Differentiating at fixed composition and amount of substance therefore gives $C_p=dH/dT=dU/dT+nR=C_V+nR$. It uses the ideal gas law, rather than assuming any particular number of molecular degrees of freedom. At fixed pressure, the additional heat supplies expansion work, $pdV=nRdT$. The two processes can have the same temperature endpoints and equal $\Delta U$ while requiring different heat inputs.

For one mole of the rigid-rotation diatomic ideal gas model, $C_V=5R/2$ and $C_p=7R/2$. Heating by $10\mathrm{K}$ with $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $Q_V\simeq 207.86\mathrm{J}$ in a rigid sealed vessel and $Q_p\simeq 291.01\mathrm{J}$ under a frictionless piston at fixed pressure. Both gain $207.86\mathrm{J}$ of internal energy, while the piston delivers approximately $83.14\mathrm{J}$ of work. This is a calculated model budget. If the capacities vary with temperature, the corresponding finite statement is stillunder the fixed-pressure and fixed-volume heating assumptions in heat capacity. Their final volume values generally differ. Combining Mayer's relation with heat capacity ratio also gives $C_V=nR/(\gamma -1)$ and $C_p=\gamma nR/(\gamma -1)$ locally; constant values cannot be extrapolated across intervals in which $\gamma (T)$ changes.

For a smooth stable single thermodynamic phase of a general simple compressible system, the broader Maxwell relation identity iswhere $\alpha$ is the coefficient of thermal expansion and ${\kappa }_T$ the isothermal compressibility, with composition fixed throughout. Substituting the ideal gas law gives $nR$; a real gas need not do so. This distinction matters near a phase transition or when interactions appreciably change the equation of state. It explains why Mayer's relation is useful: thermal measurements under different constraints become linked to a mechanical response, while the difference retains a particularly simple value for an ideal gas.^[ref]

The heat capacity ratio compares heat capacity at constant pressure $C_p$ with heat capacity at constant volume $C_V$ at the same equilibrium state and fixed composition:Here $C_p,C_V$ refer to the whole sample, in $\mathrm{J}{\mathrm{K}}^{-1}$; molar heat capacities $C_{p,\mathrm{m}}=C_p/n$ and $C_{V,\mathrm{m}}=C_V/n$ are per mole; specific heat capacities $c_p=C_p/m$ and $c_V=C_V/m$ are per unit mass. The amount of substance $n$ or mass $m$ cancels only when both capacities use the same normalization. Thus heat capacity ratio is dimensionless and independent of sample size. For a fixed-composition ideal gas with positive finite $C_V$, Mayer's relation giveswhere $R$ is the molar gas constant. More ways to store internal energy make a given temperature increase require more heat, bringing heat capacity ratio closer to unity. ^[ref]

The equipartition theorem predicts $C_{V,\mathrm{m}}=fR/2$ when $f$ independent quadratic terms in molecular energy are thermally active and their number stays fixed; consequently $\gamma =1+2/f$. A classical monatomic gas with no appreciable electronic excitation has $f=3$, giving $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. For an ideal gas of diatomic molecules with three translational and two rotational terms active, but its vibrational normal mode effectively frozen, $f=5$ gives $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. A fully classical harmonic vibration adds both kinetic energy and potential energy, giving $f=7$ and $\gamma =9/7$. These are model regimes, not universal constants of every gas: rotational freezing, electronic excitation, and changing composition invalidate the simple count.^[ref]

The mechanical significance of heat capacity ratio follows from the first law of thermodynamics. For a fixed amount of ideal gas in a reversible adiabatic process, $C_V(T)dT=-pdV$. Combining this with the ideal gas law gives the local relationThe partial derivative holds at fixed entropy and composition even when heat capacity ratio varies with temperature. The finite-path formulas $p{V}^{\gamma }=\mathrm{constant}$ and $T{V}^{\gamma -1}=\mathrm{constant}$ require the additional approximation of constant heat capacity ratio. A rapid compression also needs internal equilibration to approximate this path; rapid motion alone does not ensure reversibility. The same local response gives isentropic compressibility $1/(\gamma p)$ and, for small sound disturbances in this ideal gas regime, speed of sound $a=\sqrt{\gamma p/\rho }$, where $\rho$ is mass density. Thus thermal storage controls mechanical stiffness, sound propagation and the measurements in Rüchardt experiment and Clément-Desormes experiment.

A monatomic gas consists of individual atoms or ions rather than molecules. In the classical dilute ideal-gas regime, its three translational degrees of freedom give $U=3N{k}_{B}T/2$ by equipartition theorem. Thus $C_V=3N{k}_B/2$, while $C_P-C_V=N{k}_B$ follows by differentiating the ideal-gas enthalpy $H=U+PV=5N{k}_{B}T/2$. Consequently the adiabatic index is $\gamma =C_P/C_V=5/3$ and the specific internal energy is $e=3P/(2\rho )$. The same translational result applies to a fully ionized mixture when the particle numbers are fixed. This excludes additional excitation, ionization and radiation contributions to the heat capacity.

The ideal gas law is the equation of state relating the absolute pressure $P$, volume $V$, particle count $N$, and absolute thermodynamic temperature $T$ of a classical ideal gas:Here $k_B$ is the Boltzmann constant, $N_{\mathrm{A}}$ is the Avogadro constant, $R$ is the molar gas constant, and $n_{\mathrm{mol}}$ is the amount of substance in moles. The distinct symbol $n_{\mathrm{mol}}$ avoids confusing amount of substance with the number density $n=N/V$, measured in ${\mathrm{m}}^{-3}$: the same law can be written $P=n{k}_{B}T$. Use absolute pressure, relative to a vacuum, rather than an unconverted gauge pressure. ^[ref]

The exact SI definitions of the Avogadro constant and Boltzmann constant giveThis numerical value is exact; the applicability of the ideal gas model to a real gas is an approximation. ^{[ref] [ref]}

For $n_{\mathrm{mol}}=1\mathrm{mol}$, $T=300\mathrm{K}$, and $V=0.025{\mathrm{m}}^3$, direct substitution givesAt fixed temperature and particle count, doubling the volume to $0.050{\mathrm{m}}^3$ halves the pressure to approximately $49.887\mathrm{kPa}$: this is Boyle's law. At fixed volume and particle count, doubling the temperature from $300\mathrm{K}$ to $600\mathrm{K}$ doubles the pressure to approximately $199.547\mathrm{kPa}$. These ratios use kelvins, not readings in the degree Celsius scale: $300\mathrm{K}$ is $26.85{}^{\circ }\mathrm{C}$, and $600\mathrm{K}$ is $326.85{}^{\circ }\mathrm{C}$.

The curved slices show Boyle's law at three temperatures; the straight slice shows proportionality to absolute temperature at one fixed volume. They are relations among equilibrium states, rather than a prescribed process history. Compression could hold the temperature fixed through heat transfer, or raise it during adiabatic compression. To calculate a process's heat or work, supplement the ideal gas law with the process constraint, the first law of thermodynamics, and the relevant internal energy or heat capacity information. The graph's extrapolation toward $T=0$ is mathematical; it does not assert that a real gas remains classical or gaseous there.

For a stellar mixture with fixed mean molecular weight, the ideal gas law can be written $P=\mathcal{R}\rho T/\mu$, where $\rho$ is mass density and $\mathcal{R}$ is the stellar gas constant. It assumes negligible interaction corrections and a nondegenerate gas; excitation or ionization can change the particle count and thermodynamic response.

In his 1834 memoir on the motive power of heat, Émile Clapeyron combined gas laws into a relation between pressure, volume, and temperature while analysing heat engines (printed page 164). The equation above uses modern SI notation and absolute thermodynamic temperature; it is not a transcription of his historical temperature convention. ^[ref]

The molar gas constant is the product of the Avogadro constant and the Boltzmann constant. It converts the particle-number form of the ideal gas law, $pV=N{k}_{B}T$, into the amount-of-substance form $pV=nRT$, using $N=n{N}_A$ with $n$ measured in moles.

The partial pressure of a component of a gas mixture is the pressure it would exert alone at the same temperature and total volume. For an ideal mixture, $p_i=n_{i}RT/V$ and $p={\sum }_i{p}_i$. The chemical potential of an ideal gas uses $p_i$ for that component rather than the total pressure $p$.

If a number field of degree $n$ and signature $(r_1,r_2)$ has discriminant $d_K$, every ideal class contains an integral ideal $I$ with

For $s=\sqrt{-21}$, the integer ring is $\mathbb{Z}[s]$. The Minkowski bound for ideal classes is less than six. The ramified primes $(2,1+s)$ and $(3,s)$ give independent classes of order two; their product is the class of $(5,s-2)$ because $(3+s)$ factors as the product of these three prime ideals. Element norms $a^2+21b^2$ cannot equal 2, 3 or 6, so these classes are distinct. Ideals of norm at most five supply no further classes, proving the four-element group.

Put $\omega =(1+\sqrt{229})/2$, $P=(3,\omega )$ and $Q=(5,\omega -4)$. The Minkowski bound for ideal classes is $\sqrt{229}/2<8$, so the ideal class group is generated by primes above three and five; two and seven are inert. The prime ideal factorizations $(6+\omega )=PQ$ and $(15+2\omega )=P^3$ reduce it to a cyclic group of order dividing three. It is not trivial: the unit $ϵ=(15+\sqrt{229})/2$ has field norm minus one. If an algebraic integer had norm of absolute value three, multiplying by a power of this unit would make both real absolute values at most $\sqrt{3ϵ}$. For $a=x+y\omega$, this bounds $|y|\sqrt{229}=|a-a^{\prime }|\le 2\sqrt{3ϵ}<\sqrt{229}$, forcing $y=0$, which cannot have norm of absolute value three.

The Langevin function is the mean projection of a freely orientable three-dimensional unit vector weighted by $e^{\xi \cos \theta }$. Integrating over solid angle gives $Z_1=4\pi \sinh \xi /\xi$, and differentiation gives $⟨\cos \theta ⟩={\partial }_{\xi }\log Z_1=\coth \xi -1/\xi$. It behaves as $\xi /3$ at small argument and $1-1/\xi$ at large positive argument.