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A local diffeomorphism is a differentiable map that restricts near every point to a diffeomorphism onto an open subset of its target. The local inverse functions need not combine into a single global inverse function: different parts of the domain of a function can map onto the same output region.

For continuously differentiable maps between open subsets of ${\mathbb{R}}^n$, being a local diffeomorphism is equivalent to having an invertible derivative at every point. The inverse function theorem proves one direction. For the other, the chain rule for each local inverse function gives two inverse linear maps. The periodic angle example under the inverse function theorem has this property everywhere, even though it has infinitely many inputs for every point of the punctured plane. The noninjective map with constant Jacobian determinant gives another useful distinction: even a constant nonzero Jacobian determinant need not give global injectivity.

Every local diffeomorphism is an open map. To see this, take an open subset of the source and intersect it with the neighbourhoods where the map is a diffeomorphism. Each local image is open, and their union is the full image. Thus nearby outputs remain available locally, despite the possible global ambiguity.

For a positive real base $b\ne 1$, the logarithm is the inverse function of the exponential function $t\mapsto b^t$:It asks which exponent produces a given positive number. Thus ${\log }_{2}8=3$ and ${\log }_2(1/8)=-3$. The base restrictions matter: $b=1$ would produce only one value, while a negative base does not define $b^t$ as a real-valued function for every real exponent.

The useful transformation isFor example, $8\cdot 16=128$ corresponds to $3+4=7$ in base two. Products become sums and ratios become differences. The logarithm does not generally convert a sum into a sum of logarithms. Base conversion uses the natural logarithm: ${\log }_{b}x=\ln x/\ln b$. It also gives $d({\log }_{b}x)/dx=1/(x\ln b)$, positive when $b>1$ and negative when $0<b<1$.

Historically, John Napier's 1614 Mirifici logarithmorum canonis descriptio presented logarithmic tables and their use, documented in the Smithsonian Libraries' digitized original. Tables made multiplication-heavy calculations more manageable by replacing products with addition and a lookup. The 1619 Constructio title page below belongs to a later publication from the same work on logarithms. These historical publications preceded the modern inverse function and integral presentation used here.

A real-valued function $f:D\to \mathbb{R}$, with domain of a function $D\subseteq \mathbb{R}$ in the usual order of the real numbers, is a monotonic function if it is either a nondecreasing function throughout $D$ or a nonincreasing function throughout $D$. The same direction must work for every pair $x<y$ in $D$. Being a strictly increasing function on one part and a strictly decreasing function on another does not make the whole function monotone: for example, $x^2$ on the real line changes direction at zero. A constant function is both a nondecreasing function and a nonincreasing function.

The word "increasing" is ambiguous across textbooks: some allow equal outputs and some require unequal outputs. Use nondecreasing function for the weak inequality and strictly increasing function for the strict inequality; use nonincreasing function and strictly decreasing function in the other direction. Neither continuity nor differentiability is part of the definition of a monotonic function.

Strict monotonicity gives injectivity, so an equation $f(x)=b$ has at most one solution and an inverse function exists on the image of a function. Weak monotonicity permits repeated outputs and jumps. In probability theory, a cumulative distribution function is a nondecreasing function; its quantile function uses a generalized inverse to handle those flat parts and jumps. Monotonicity alone does not guarantee continuity, even for a strictly increasing function.

The natural logarithm $\ln x$ is the logarithm with base Euler's number, and the inverse function of the real exponential function. A particularly geometric definition isFor $x>1$, it is the area under the hyperbola $y=1/t$ from $1$ to $x$. For $0<x<1$, the reversed integration limits give a negative value. The fundamental theorem of calculus gives $(\ln x{)}^{\prime }=1/x$, so the natural logarithm is increasing, while $(\ln x{)}^{\prime \prime }=-1/x^2$ gives concavity.

The change of variables $t=au$ shows ${\int }_a^{ab}dt/t={\int }_1^{b}du/u=\ln b$ for $a,b>0$. Thus the intervals from $1$ to $2$ and from $2$ to $4$ each contribute area $\ln 2$, explaining $\ln 4=2\ln 2$. Equal ratios give equal increments on a logarithmic scale, even though the ordinary differences grow. This makes the natural logarithm useful for comparing multiplicative changes and solving for time in exponential growth or exponential decay. The integral construction and inverse function properties are discussed in OpenStax, the natural logarithm as an integral.

A pairing function is a computable function that gives a bijection from the Cartesian product ${\mathbb{N}}_0^2$ to ${\mathbb{N}}_0$, with two computable functions recovering its coordinates. Here ${\mathbb{N}}_0=\{0,1,2,\dots \}$ consists of the natural numbers, including zero. If $p$ is the pairing function and $L,R$ its decoding functions, the requirements areEvery ordered pair has exactly one code, every code represents an ordered pair, and decoding always terminates. For a bijection given by a total computable function, computability of the inverse function also follows by enumerating all ordered pairs until the one with code $z$ appears. A primitive recursive pairing function makes the stronger requirement that both encoding and coordinate decoding are primitive recursive.

The Cantor pairing function makes this concrete through triangular numbers:On the diagonal $x+y=s$, list $(s,0),(s-1,1),\dots ,(0,s)$. There are $s+1$ ordered pairs there, and $T(s)$ ordered pairs on earlier diagonals. Their codes therefore occupy precisely the consecutive natural numbers from $T(s)$ through $T(s+1)-1$. These intervals touch without overlapping and cover all natural numbers; within each interval the offset $y$ determines the ordered pair. This proves both injectivity and surjectivity, hence the bijection. It also explains why the infinite grid is a countably infinite set: every ordered pair is reached after finitely many steps. A row-by-row listing would never finish its first infinite row. Authored treatments give the same diagonal construction with the coordinates exchanged.^[ref]

For example, the Cantor pairing function givesThe order of the coordinates matters: $(1,0)$ and $(0,1)$ are different ordered pairs with different codes.

To decode a natural number $z$, find the unique diagonal index $s$ satisfyingThe inequality gives $0\le y\le s$, so both recovered coordinates are natural numbers. At $z=18$, $T(5)=15\le 18<T(6)=21$, hence $s=5$, $y=18-15=3$, and $x=5-3=2$. At $z=0$, $s=0$ and both coordinates are zero. Solving the quadratic inequality gives the equivalent floor function expressionThis is exact as a statement about the real numbers and their square roots. An implementation using floating-point arithmetic can round across a diagonal boundary, especially for large $z$. Use an exact integer square root instead: with $q=\lfloor \sqrt{8z+1}\rfloor$ computed using integer arithmetic, $s=\lfloor (q-1)/2\rfloor$. Alternatively, search for the diagonal using exact integer comparisons. The primitive recursive pairing function proof below uses bounded minimization and needs no approximate square root.

A pairing function is a reversible encoding: unlike a hash function mapping arbitrarily many inputs into a fixed finite range, it loses no coordinates. This works because the mathematical natural numbers have no upper bound. A machine word with $b$ binary digits has $2^b$ possible values, while two such words have $2^{2b}$ possible ordered pairs; a bijection into one word of the same width is impossible. Evaluating the Cantor pairing function in a fixed-width integer type can overflow, including in the intermediate product $(x+y)(x+y+1)$. Exact implementation requires sufficiently large integer storage or an explicit restricted input range. One mathematical code need not occupy fewer binary digits than its two inputs.

Iterating a pairing function gives a reversible encoding of tuples of any fixed positive length. With the left-nesting convention, the Cantor pairing function encodes $(2,3,4)$ asDecode the outer ordered pair first: $T(22)\le 257<T(23)=276$ gives $(18,4)$. Decode its first coordinate $18$ to recover $(2,3)$, giving the nested ordered pair $((2,3),4)$ and thus the original tuple $(2,3,4)$.

The tuple length and nesting convention must be known. The same code $257$ represents $(18,4)$ as a two-entry tuple and $(2,3,4)$ as a three-entry tuple; nesting alone does not make variable-length tuples self-describing. To encode variable-length tuples, supply their length or another explicit termination convention as part of the encoding. This simple change of representation is useful in computability theory and Gödel numbering: a function of several natural numbers can be studied through a single numerical input, while all coordinates remain recoverable. The striking fact is that a whole infinite grid can be serialized without omitting a point.

A historical reference for comparing sets through bijections is Georg Cantor's paper Ein Beitrag zur Mannigfaltigkeitslehre, published in 1878 in volume 84 of Journal für die reine und angewandte Mathematik, pages 242–258.^[ref]

Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility. The derivative $U^{\prime }$ is continuous and strictly decreasing: concavity makes it nonincreasing, and equality at two distinct points would make $U$ affine between them, contradicting strict concavity. The Inada conditions therefore make $U^{\prime }$ a bijection from $(0,\infty )$ onto $(0,\infty )$. For each $y>0$, the derivative of $U(x)-xy$ is positive below $I(y)$ and negative above it. Consequently its unique global maximum is finite andThere is no need to differentiate $I$ to find the first derivative of this utility conjugate. Comparing the maxima at $y$ and $y+h$ gives, for $h>0$,Continuity of $I$, together with the corresponding inequalities for $h<0$, givesThus $V$ is strictly decreasing and strictly convex, since its derivative is strictly increasing. This proves all the asserted properties except the second differentiability, which does not follow from the printed hypotheses.

Here is an explicit Inada utility with vanishing curvature. PutThe positive integral defining $p$ makes $U$ strictly increasing. Also $p^{\prime }=-w$ is negative except at one point, so $p$ is strictly decreasing and $U$ is strictly concave. Near zero $w(s)$ is asymptotic to $s^{-2}$, and at infinity it is asymptotic to $s^{-2}$; thus $p(0+)=\infty$ and $p(\infty )=0$. This smooth utility function satisfies every printed positive-domain assumption, but $U^{\prime \prime }(1)=0$. At $y_0=p(1)=3-4\log 2$, differentiability of $I$ would imply, by differentiating $p(I(y))=y$,which is impossible. Hence $V^{\prime }=-I$ has no finite derivative there. More explicitly, $p(x)-p(1)=-(x-1{)}^3/12+o((x-1{)}^3)$, so its inverse function has a cube-root singularity. The twice-differentiable conclusion is false as printed. With the additional assumption $U^{\prime \prime }(x)<0$ everywhere, the inverse function theorem proves the intended dual differentiability with nonvanishing utility curvature:

For the financial assertions, use the positive-wealth admissible trading strategy class appropriate to the positive-domain utility function, and expectations for which the expressions are defined. A normalized terminal state-price density satisfies $Z_N>0$, $\mathbb{E}{Z}_N=1$, and defines an equivalent martingale measure $Q$ by $dQ=Z_{N}dP$. A self-financing portfolio obeyswhere the holdings are measurable before the next price increment. Under $Q$ the gains are a martingale when integrable; for the usual lower-bounded admissible class, localization gives a supermartingale. In particular the state-price budget constraint is $\mathbb{E}[Z_N{X}_N]\le x$. The positive-domain qualification matters: assumptions only on $U$ at positive arguments do not control its values at nonpositive wealth.

The definition of the utility conjugate gives the pointwise inequalityTaking expectations, using the state-price budget constraint, and then taking the supremum over strategies gives $u(x)\le \mathbb{E}V(y{Z}_N)+yx$ for every $y>0$ and every state-price density. Taking the two infima proves the utility duality with martingale deflators upper bound:If an expectation is infinite, use the same inequality in the extended-real sense whenever the negative parts make it well defined; no subtraction of two infinities is needed.

A complete market is one in which every terminal contingent claim in the specified claim class is attainable by a self-financing portfolio. Bounded terminal claims suffice for the uniqueness argument. In the standard finite-horizon discrete-time market, absence of arbitrage gives existence of an equivalent martingale measure by the fundamental theorem of asset pricing. To prove uniqueness, let $Z_N$ and ${\tilde{Z}}_N$ be two normalized state-price densities. Every bounded claim $H$ has a replicating price $p(H)$, and its replication has the same price under either equivalent martingale measure:Take $H=1_A$ for every terminal event $A$. The integrable difference $Z_N-{\tilde{Z}}_N$ has zero integral over every event, hence vanishes almost surely: its positive and negative sets would otherwise give a nonzero integral. The normalized terminal state-price density is unique up to null sets. Its density process is then uniquely $Z_n=\mathbb{E}[Z_N\mid {\mathcal{F}}_n]$. This uses the usual pricing replication class, where bounded claims have bounded replicating value processes; arbitrary doubling strategies are not an admissible definition of completeness.

For logarithmic utility, direct maximization gives $V(y)=-\log y-1$. In the complete market, write $Z=Z_N$. The unique-density utility conjugate bound is minimized at $y=1/x$, givingThe candidate terminal portfolio wealth is $H^{\ast }=x/Z$, whose cost is exactly $\mathbb{E}[Z{H}^{\ast }]=x$. If completeness includes nonnegative $Q$-integrable claims, it is replicable and attains the bound. There is also an approximation proof using only bounded-claim completeness. DefineThen $0<b_n\le 1$, $b_n\uparrow 1$, $H_n$ is positive and bounded, and $\mathbb{E}[Z{H}_n]=x$. Replicating $H_n$ therefore gives an admissible terminal claim of initial cost $x$. Since $\mathbb{E}(\log Z{)}^{+}\le \mathbb{E}Z=1$, the negative parts of $-\log (Z\vee n^{-1})$ have an integrable common bound. Monotone convergence after adding this lower bound givesConsequently the logarithmic terminal wealth in a complete market isThe value may be $+\infty$ if $\mathbb{E}(\log Z_N{)}^{-}=\infty$; the displayed formula remains meaningful, while attainment by $H^{\ast }$ depends on the precise admissible claim class.

Always true. If $(f\circ g)(x)=(f\circ g)(x^{\prime })$, injectivity of $f$ gives $g(x)=g(x^{\prime })$, and injectivity of $g$ gives $x=x^{\prime }$. Thus the function composition is injective. For any $y\in X$, since $f$ is surjective, choose $z$ with $f(z)=y$. Since $g$ is surjective, choose $x$ with $g(x)=z$. Then $(f\circ g)(x)=y$, proving that the composition is surjective. Consequently $f\circ g$ is a bijection; its inverse function is $g^{-1}\circ f^{-1}$.

An inverse function of $f:X\to Y$ is a function $h:Y\to X$ satisfying bothThese are two-sided inverse identities. If $f(x)=f(x^{\prime })$, apply $h$ to get $x=x^{\prime }$, so $f$ is injective. For every $y\in Y$, the identity $f(h(y))=y$ supplies a preimage, so $f$ is surjective. Thus $f$ is a bijection.

Conversely, if $f$ is a bijection, each $y\in Y$ has exactly one preimage. Define $h(y)$ to be that preimage. Then $f(h(y))=y$ and $h(f(x))=x$, giving both identities. Hence a function has a two-sided inverse exactly when it is a bijection. This also makes its inverse function unique.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

LetThe clock $A$ is an absolutely continuous function and is strictly increasing: its derivative is positive away from the Brownian zero set, which has zero Lebesgue measure. It also tends to infinity. This is immediate for $\delta =1$; for $\delta >1$, recurrence and the Strong Markov property imply that the Brownian occupation time of, for example, $\{x:1<|x|<2\}$ is unbounded, while the integrand is bounded below there by ${\delta }^2$.

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.

A primitive recursive pairing function is a pairing function whose encoding function and both coordinate functions of its inverse function are primitive recursive functions. Thus the bijection can be used in either direction inside a construction by primitive recursion and function composition in recursion theory. A decoder that is merely a computable function would not establish this stronger closure property. Sutner's definition of a primitive recursive pairing system explicitly includes the decoding functions; his more general systems need only be injective, whereas a pairing function here is a bijection.^[ref]

The Cantor pairing function is a primitive recursive pairing function. For its encoding, construct the triangular numbers directly by primitive recursion:Since addition is primitive recursive, so are $T$ and $\pi$ by primitive recursion and function composition in recursion theory. This establishes the claim without treating division by two as an unexplained primitive operation.

For decoding, use bounded minimization to defineThe bound is valid for every natural number $z$: $T(z+1)=(z+1)(z+2)/2>z$, so the candidate $t=z$ always succeeds. The search includes zero; when $z=0$, it immediately returns $s(0)=0$. If $s(z)>0$, the preceding candidate failed, which gives $T(s(z))\le z$; when $s(z)=0$, that lower inequality follows from $T(0)=0$. Thus this search finds exactly the diagonal in the Cantor pairing function inverse, withThe comparison is primitive recursive, and bounded minimization of a primitive recursive predicate is primitive recursive. Therefore $s$ is a primitive recursive function. Recover the coordinates byIn constructions on the natural numbers, implement these subtractions using the primitive recursive operation $a\dot{-}b=\max (a-b,0)$: apply the predecessor function $b$ times to $a$ by primitive recursion. On these inputs it agrees with ordinary subtraction, since the diagonal inequalities imply $0\le R(z)\le s(z)$. Consequently $L$ and $R$ are primitive recursive functions, and $\pi (L(z),R(z))=z$. Zimmermann proves the relevant closure under bounded minimization and the primitive recursive character of the Cantor encoding and coordinate decoding; his convention adds the first coordinate, so its coordinates are exchanged relative to $\pi$ here.^[ref]

For a worked bounded minimization, take $z=18$. Candidates $t=0,1,2,3,4$ give $T(t+1)=1,3,6,10,15$, none exceeding $18$. Candidate $t=5$ gives $T(6)=21>18$, so $s(18)=5$, $R(18)=3$, and $L(18)=2$. The search bound was $18$; it terminates much earlier in this example. The bound proves membership in the primitive recursive functions; it is not a claim that scanning all candidates is the fastest implementation.

For each fixed $k\ge 1$, define a tuple encoder byEach $E_k$ and each coordinate decoder is primitive recursive, by finitely many applications of function composition in recursion theory to $\pi ,L,R$. For the three-entry tuple encoded as $257$, applying $R$ returns $4$, applying $L$ returns $18$, and then $L(18)=2$, $R(18)=3$. The tuple length remains part of the agreed convention, as explained for a general pairing function.

This makes primitive recursive pairing functions useful for encoding several evolving values as one state in primitive recursion. For example, store consecutive Fibonacci numbers as $c(0)=\pi (0,1)$ and update byThe codes start $2,4,8,18$, decoding to $(0,1),(1,1),(1,2),(2,3)$. Every step uses primitive recursive encoding, decoding and addition; a single numerical state therefore carries the two values needed for the next step. This is also the reason a suitable Gödel numbering should permit sufficiently simple decoding: representation should preserve the class of functions being studied.

The quantile function converts a cumulative probability level into a value. For a cumulative distribution function $F$ and $0<p<1$, use the generalized inverseIt selects the first threshold reaching the requested level. The infimum handles jumps and flat portions, so this construction does not require an ordinary inverse function. Other endpoint conventions are possible at $p=0$ or $p=1$; the open probability interval avoids those issues. NIST introduces the inverse-direction question as the percent point function. ^[ref].

For the finite example under probability distribution,Thus $Q_X(0.1)=0$, $Q_X(0.5)=1$, $Q_X(0.9)=3$. The value at $p=0.2$ is still zero, because $F_X(0)=0.2$ already reaches that level. For $F_Y(y)=y^2$ on $[0,1]$, the quantile function is $Q_Y(p)=\sqrt{p}$, giving $Q_Y(0.5)=1/\sqrt{2}\approx 0.70710678$.

A median is a central threshold; $Q(1/2)$ supplies one, though a median need not be unique. The same quantile function becomes a sampling algorithm through inverse transform sampling. Jumps in $F$ become flat bands in $Q$, precisely the mechanism needed to assign positive probability to one outcome.

For a function $f:X\to Y$ and a subset $A\subseteq X$, its restriction $f{|}_A$ keeps the same values but allows only inputs in $A$. Restricting the domain of a function can remove collisions and permit an inverse function, as the nonnegative quadratic function example shows. This changes the allowed inputs; it is different from keeping the inputs and changing the codomain.