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A rational-linear map is a linear map when its domain and codomain are regarded as vector spaces over the rational numbers. An additive map on a rational vector subspace of the real numbers is automatically rational-linear: integer homogeneity follows by repeated addition, and division by a positive integer gives rational homogeneity of an additive function.

A rational matrix is a matrix whose entries are rational numbers. Multiplying by a common denominator produces an integer matrix and preserves its homogeneous solution set.

For a nonzero polynomial $g\in \mathbb{Q}[t]$, choose a positive integer $n$ clearing its denominators and set $c(g)=c(ng)/n$, where the polynomial content of an integer-coefficient polynomial is the positive greatest common divisor of its coefficients. This is independent of $n$: if $f/n=h/m$, then $mf=nh$, hence $mc(f)=nc(h)$.

The integer-coefficient product formula from Gauss lemma for polynomials givesMoreover $g/c(g)$ is an integer-coefficient primitive polynomial: if $g=f/n$, then $g/c(g)=f/c(f)$. For example,This normalization proves a useful divisibility descent. If $f\in \mathbb{Z}[t]$ is primitive, $g\in \mathbb{Z}[t]$ and $g=fq$ for $q\in \mathbb{Q}[t]$, then $c(q)=c(g)$ is an integer. Since $q/c(q)\in \mathbb{Z}[t]$, it follows that $q\in \mathbb{Z}[t]$.

A positive-integer labeling with $a⇝b\Rightarrow \ell (a)\le \ell (b)$ and $a/⇝b\Rightarrow \ell (a)>\ell (b)$ exists exactly when the condensation graph is totally ordered by reachability. Mutual paths force equal labels, and equal labels force mutual paths. Two incomparable components would require strict inequalities in both directions. Conversely number the components in topological order; total comparability forces every earlier component to reach every later one, proving both conditions. Compute the labeling in linear time. Reachability here includes length-zero paths.

A computation that retains pending operations while further calls run, then completes those operations as results return. A recursive procedure can instead describe an iterative process: passing an updated accumulator onward may leave no arithmetic pending. Avoiding growing call stack storage in that case depends on the implementation. A fixed number of variables also does not imply fixed storage in bits when their integer values grow.^[ref]

A reduced fraction $a/b$ has coprime integers $a,b$ and positive denominator $b$. Distinct reduced fractions with denominators at most $Q$ differ by at least $Q^{-2}$, since their difference has a nonzero integer numerator over a denominator at most $Q^2$.

For an integer $a$ and positive integer $b$, the remainder in Euclidean division is the unique integer $r$ in $0\le r<b$ for which $a=bq+r$ with an integer $q$. It is not the full quotient $a/b$.

For $|r|<1$, let $S_N$ contain the first $N$ terms, with $S_0=0$. The remainder after those terms isThis is the exact absolute error. A convenient bound replaces the denominator by $1-|r|$, giving $|R_N|\le |a||r{|}^N/(1-|r|)$ through the triangle inequality. The exact denominator is preferable when the sign or complex phase of $r$ is known.

For $a>0$ and $0<r<1$, the omitted fraction of the entire sum is $R_N/(a/(1-r))=r^N$. Capturing at least $99$ percent therefore means $r^N\le 0.01$. Taking logarithms and applying the ceiling function gives the smallest integer $N$:This gives $7,21,90,$ and $459$ terms for $r=0.5,0.8,0.95,$ and $0.99$, respectively. The difference is a concrete reason to care about the common ratio, even when all four series converge.

An absolute target has a different scale. For $a=1$, $r=0.8$, and required absolute error at most $0.01$, the total sum is $5$ andThe first acceptable term count is $N=28$: the remainder is approximately $0.0096714$, while at $N=27$ it is approximately $0.0120893$. Counting terms and checking the tail avoids an ambiguous instruction to stop when a term merely looks small.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

A repdigit whose digits are all 1 in a specified positional notation of base $b$. Its value is the geometric series $R_b(k)=1+b+\cdots +b^{k-1}=(b^k-1)/(b-1)$. In base ten, appending one digit gives $R_{10}(k+1)=10R_{10}(k)+1$, so its remainder modulo $n$ can be computed by a recurrence relation without constructing the full integer.

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.

For a positive integer modulus $m$, the residue class of an integer $a$ is the equivalence classIt contains all integers in the same position of the repeating cycle. For example,The residue class is this whole infinite collection; its canonical representative is the least nonnegative residue. Euclidean division gives exactly one such representative in $0,1,\dots ,m-1$, so there are exactly $m$ residue classes. When $m=1$, there is just one, $[0{]}_1=\mathbb{Z}$.

Define addition and multiplication of residue classes byThese definitions must give the same answer whichever integers represent the classes. If $a^{\prime }\equiv a$ and $b^{\prime }\equiv b$ modulo $m$, thenBoth differences are divisible by $m$, proving that the operations are well-defined. For example, $[9{]}_{12}+[5{]}_{12}=[2{]}_{12}$ and $[9{]}_{12}[5{]}_{12}=[9{]}_{12}$, since $45=3\cdot 12+9$. Replacing nine by $-3$ gives the same answers with smaller intermediate integers.

With these operations the residue classes form the finite commutative ring $\mathbb{Z}/m\mathbb{Z}$, a quotient ring. It need not be a field: $[6{]}_{12}[2{]}_{12}=[0{]}_{12}$ although neither factor is the zero class. These zero divisors explain why division requires a unit modulo n. Reducing after each multiplication is especially useful in modular exponentiation, where the unreduced powers can become enormous.

For an integer $d>1$, the proportion of proper fractions $1/d,\dots ,(d-1)/d$ already in lowest terms. There are $\phi (d)$ such fractions, so $R(d)=\phi (d)/(d-1)$. The prime factorization of $d$ gives $R(d)=\frac{d}{d-1}{\prod }_{p\mid d}(1-1/p)$.

A positive integer whose sum with its digit reversal has only odd numerical digits, with leading zeros forbidden in both the original and reversed representations. For example, $36+63=99$. Studying the arithmetic carries of mirrored digit pairs allows whole lengths to be counted without enumerating all integers of that length.

On the lateral surface of an infinite circular cylinder of radius $R$, the induced metric is $R^{2}d{\theta }^2+d{z}^2$. Lifting a curve to the unrolled plane $(x,z)=(R\theta ,z)$ preserves its arc length. If the endpoints differ by angular coordinate $\Delta \theta$ and height $\Delta z$, the possible lifted endpoint displacements are $(R(\Delta \theta +2\pi k),\Delta z)$ for integers $k$. Each straight segment gives a geodesic, and the shortest of their lengths isChoose the representative of $\Delta \theta$ with smallest absolute value modulo $2\pi$; if its absolute value is $\pi$, the two angular directions tie. This minimum is attained because the candidate lengths tend to infinity as $|k|\to \infty$. The formula concerns travel on the lateral surface; it does not allow a shortcut through the cylinder's interior or across end caps.

For $R=1$, take $p=(1,0,0)$ and $q=(0,1,3)$ in surrounding Euclidean space. The angular difference is $\pi /2$ and height difference is $3$. The $k=0$ route has length $\sqrt{9+{\pi }^2/4}\approx 3.386$. The $k=-1$ route goes the other way around and has length $\sqrt{9+9{\pi }^2/4}\approx 5.586$. Both are among the geodesics on a circular cylinder, but only the first is a minimizing geodesic between these endpoints.

This is a global consequence of intrinsic flatness of a circular cylinder: locally everything is Euclidean, yet the periodic angular coordinate leaves several competing routes. A shortest-route problem on the curved surface becomes a family of ordinary planar distance calculations.

A ring is a set $R$ with two binary operations, addition and multiplication, satisfying three requirements. Under addition, $R$ is an abelian group: there is a zero, every element has an inverse element, and grouping or exchanging summands does not change the result. Multiplication is associative and has an identity element $1$, so $(ab)c=a(bc)$ and $1a=a1=a$. Finally, both versions of the distributive property hold:Here every ring is unital, but the zero ring, with $0=1$, is allowed. Commutativity of multiplication is an additional condition defining a commutative ring. These conventions agree with ^[ref].

The integers form a ring with their usual operations. The polynomial ring $\mathbb{Z}[X]$ is another: addition combines coefficients, while multiplication uses the distributive property. For example,The same rules justify symbolic calculations in both rings; this is why one definition can organize apparently different kinds of arithmetic. In every ring, $a0=0a=0$, and $a(-b)=-(ab)$. The first identity follows by distributing $a(0+0)$ and subtracting $a0$; the second follows by distributing $a(b+(-b))$. Thus familiar sign rules are consequences of the structure.

Order still matters in some rings. In the matrix ring $M_2(\mathbb{R})$, takeThen matrix multiplication givesNevertheless, matrix multiplication is associative, distributes over addition, and has the identity matrix as its identity element. This is a valid ring. On a column vector, $AB$ applies $B$ first and $A$ second: swapping coordinates before discarding the second coordinate differs from doing those actions in reverse. The failure of commutativity records a meaningful difference in function composition.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

For every positive integer $n$, Herbert Robbins's bounds giveEquivalently, the remainder in $\log (n!)=\log S_n+r_n$ satisfies $1/(12n+1)<r_n<1/(12n)$. This both proves that $S_n$ underestimates $n!$ and squeezes $n!/S_n$ to one, establishing Stirling's formula. Robbins published the bounds in “A Remark on Stirling's Formula” in 1955. ^[ref]

For $n=10$, the relative error of $S_{10}$ lies strictly between $1-e^{-1/121}$ and $1-e^{-1/120}$, approximately $0.823\%$ and $0.830\%$. A bound specifies the accuracy of an approximation before using it in another calculation.

This instance of Rodrigues' formula constructs the Legendre polynomial for every nonnegative integer $l$. Put $v=(x^2-1{)}^l$; then $(x^2-1)v^{\prime }=2lxv$. Differentiating $l+1$ times by the Leibniz rule gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$ for $y=v^{(l)}$, which is the Legendre differential equation after reversing signs. The leading term of the normalized derivative is $(2l)!x^l/[2^l(l!{)}^2]$, proving its polynomial degree is exactly $l$. At $x=1$, the product $(x+1{)}^l(x-1{)}^l$ contributes only the term with all $l$ derivatives on the second factor, giving $v^{(l)}(1)=2^{l}l!$. Thus the normalization is $P_l(1)=1$.

The rounded Bernstein polynomial replaces each coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)$ in the unnormalized basis $x^k(1-x{)}^{n-k}$ by its floor function. It therefore has integer coefficients in the ordinary monomial basis. If $f(0),f(1)\in \mathbb{Z}$, only interior rounding errors remain andTo prove this, bound each interior rounding error by one. On $[0,1/4]$, sum the resulting powers as a geometric series of ratio at most $1/3$ and maximize $x(1-x{)}^{n-1}$; reflect for $[3/4,1]$. On the middle interval each degree-$n$ product is at most $(3/4{)}^n$. integer endpoints are necessary for this convergence: a noninteger endpoint produces a fixed nonzero rounding error there.