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Exact arithmetic on rational numbers represented by integer numerators and nonzero integer denominators. Common factors can be removed with the greatest common divisor. Exact rational arithmetic avoids rounding error when deciding equality or integrality.

The additive group ${\mathbb{Q}}_{\mathrm{sq}}$ consists of the rational numbers whose reduced denominators are square-free integers. If $p_0,p_1,\dots$ lists the primes without repetition, then

For every additive function $f:\mathbb{R}\to \mathbb{R}$, every real number $x$, and every rational number $q$, one has $f(qx)=qf(x)$. This is the scalar multiplication part of being a linear map over the rational numbers.

First $f(0)=f(0)+f(0)$ gives $f(0)=0$, and $f(x)+f(-x)=f(0)$ gives $f(-x)=-f(x)$. Repeated addition gives $f(nx)=nf(x)$ for each positive integer $n$, and the negative case follows from the sign identity. If $q=m/n$ with $m\in \mathbb{Z}$ and $n>0$, thenNo continuity was used. In particular $f(1)=3$ fixes $f(-2/3)=-2$, $f(5/6)=5/2$, and every other rational value, even for a discontinuous additive function.

The rational homotopy group is obtained by tensoring a homotopy group with the rational numbers. It retains the free part and discards torsion.

The indicator of the rational numbers in the real numbers is discontinuous everywhere because both the rationals and irrationals are dense. The discontinuity theorem for Baire class one functions therefore shows that it is not Baire class one.

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 row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.

On $[0,1]$, use reduced representations of rational numbers with positive denominator. This reverses the nonzero values of the Thomae function, while leaving its values at irrational numbers zero. Every interval of positive length contains rational numbers of arbitrarily large reduced denominator: there are only finitely many fractions in $[0,1]$ with denominator at most a fixed $M$, whereas any open interval contains infinitely many rational numbers. Thus this function taking finite real number values is unbounded on every nondegenerate subinterval.

A row echelon form is in reduced row echelon form when every pivot element is one and is the only nonzero entry in its column.^[ref] This extra reduction displays each pivot unknown directly in terms of the free variables in a linear system. The zero matrix also satisfies the definition: it has no pivot elements.

Every matrix has exactly one reduced row echelon form under elementary row operations, with its size and column order held fixed.^[ref] Ordinary row echelon form is not unique: multiplying a nonzero row by two over the rational numbers keeps the staircase but changes the matrix. The reduced conditions remove that freedom.

For a fixed number of columns, each row space has a unique reduced matrix after zero rows are removed; at a fixed full size, the number of appended zero rows is also determined. Thus reduced row echelon form represents a vector subspace independently of the starting basis. This is the uniqueness used by Gaussian binomial coefficients to count vector subspaces over a finite field.

The fixed column order matters. Elementary column operations can change the row space, the kernel and the resulting reduced row echelon form. For instance, swapping the columns of $(12)$ over the rational numbers gives $(21)$, whose reduced row echelon form is $(11/2)$ instead of $(12)$. When solving a linear system, a column interchange also requires tracking the corresponding interchange of unknowns.

A planar orbit whose radius oscillates as its polar angle increases, producing successive lobes around the force center. For nonconstant $r$ with fundamental angular radial period $2\pi /\nu$ and nonzero specific angular momentum, the full position and velocity repeat precisely when $\nu$ is a rational number, provided the radius is bounded above and away from zero. If $\nu =p/q$ in lowest terms, $p$ radial periods advance the angle by $2\pi q$; $dt=r^{2}d\theta /h$ then gives a finite time period. This proves closure rather than merely a spatial self-intersection.

Work over the rational numbers, allowing unknowns in the real numbers when interpreting the geometry. Consider four equations in three unknowns:The third row of $A$ is the sum of its first two rows, and the fourth is their difference, second minus first. The same relations must hold on the right: consistency requires $\tau =1+4=5$ and $3=4-1$. The latter condition already holds. Gaussian elimination makes this compatibility test explicit while exposing the remaining freedom.

Apply the elementary row operations in this order, using each row's current value:then $R_3\leftarrow R_3-R_2$ and $R_4\leftarrow R_4-R_2$. The resulting augmented matrix is in row echelon form for every $\tau$:When $\tau \ne 5$, the third row has its pivot element in the augmented column. It says $0=\tau -5$, so the linear system has no solution. When $\tau =5$, that entire row vanishes and only two pivot elements remain. More equations than unknowns therefore do not force a unique solution: some equations may repeat information.

The final operation $R_1\leftarrow R_1-2R_2$ reduces the coefficient block toThe product of all six elementary matrices isDirect matrix multiplication verifies both displayed transformations. Its determinant is one, so this transformation is reversible. In particular, the third row of $E$ extracts the discrepancy between the third equation and the sum of the first two.

For $\tau =5$, $[R\mid Eb]$ is the reduced row echelon form of the full augmented matrix. For $\tau \ne 5$, only its coefficient block is reduced: scale the third row by $1/(\tau -5)$ and clear the entries above its augmented pivot element. The full reduced row echelon form is thenThe last column is part of the matrix when testing the echelon conditions. Calling $[R\mid Eb]$ fully reduced before clearing that column would miss this distinction.

A real number is simply normal in base $K$ if each individual digit in its positional notation has limiting digit block frequency $1/K$. This is the length-one part of the definition of a normal number. It places no equal-frequency requirement on longer words.

For example, $x=0.010101{\dots }_2=1/3$ is a simply normal number in base two: each binary digit has frequency $1/2$. But $00$ and $11$ never occur, and $01$ and $10$ each have frequency $1/2$. Thus $x$ is not a normal number in base two. Even a sequence that looks more complicated, such as the deterministic Thue–Morse sequence, must be tested against the actual digit block frequency definition; visual complexity is no substitute for it.

The decimal representation $y=0.01234567890123456789\dots$ makes the same point in base ten. Every digit has frequency $1/10$, yet the only adjacent pairs areEach has limiting digit block frequency $1/10$; the other ninety pairs have frequency zero. A normal number in base ten requires $1/100$ for every one of the hundred pairs. This rational number is a simply normal number in base ten, but it is not a normal number in that base.

The solution set of a mathematical condition $P$ in a specified domain $D$ collects every allowed value that satisfies it:It describes the answer without committing to the steps used to find it. For the equation $x^2=1$ over the real numbers, the solution set is $\{-1,1\}$; restricting to positive real numbers leaves $\{1\}$. For $2x=1$, the solution set is $\{1/2\}$ over the rational numbers and the empty set over the integers. The domain is part of the question.

An equation can have one solution, several solutions, no solutions, or an entire domain of solutions. Over the real numbers, $2x+3=9$ gives the singleton set $\{3\}$, $x^2+1=0$ gives the empty set, and the mathematical identity $(x+1{)}^2=x^2+2x+1$ gives the whole real line. Changing that second equation's domain to the complex numbers gives $\{-i,i\}$. The elementary definition by values that satisfy an equation and the classification by solution sets appear in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

Requiring several conditions simultaneously takes the intersection of their solution sets. The squared equation $x=(x-2{)}^2$ has solution set $\{1,4\}$. Adding the restriction $x\ge 2$ giveswhich is also the solution set of $\sqrt{x}=x-2$. The extraneous solution example explains why the sign restriction is needed. Requiring at least one of several conditions instead takes their union. This turns logical combinations of constraints into reusable operations on sets.

For several variables, a solution is an ordered assignment, usually represented by a tuple. The real equation $x+y=1$ has solution set $\{(t,1-t):t\in \mathbb{R}\}$, a straight line in the Euclidean plane with coordinates $(x,y)$. Adding $x-y=0$ intersects that straight line with another, leaving the single tuple $(1/2,1/2)$. A system of linear inequalities has no solution exactly when the intersection of its individual solution sets is the empty set. For optimization constraints, this common solution set is the feasible region.

Because a solution set is a set, the order in which its members are listed is irrelevant. The multiple root of $(x-1{)}^2=0$ is still the one solution $1$; its multiplicity of a root is additional polynomial information, rather than a second occurrence inside the solution set. Equivalent equations preserve the solution set on the stated domain, even if their algebraic forms differ.

A spreadsheet tabular model assigns a consistent meaning to each row and column. For an original example, let rows 2–4 describe three items, with quantity in column A, unit price in column B, subtotal in column C, and total including tax in column D. Put the single shared tax rate in F1. The chosen rate $1/5$ is an illustrative assumption. Quantities, unit prices and F1 are inputs; columns C and D hold spreadsheet formulas and display their computed values as outputs.

Enter decimal inputs 2.5, 1.75, 4.5 for the unit prices and 0.2 for the tax rate. The following table expresses those values as exact fractions so that the arithmetic is easy to check. It does not require typing a fraction into an input spreadsheet cell or assume that all software calculates with exact rational numbers.

Start with C2 containing =A2*B2 and D2 containing =C2*(1+$F$1). Here * requests multiplication, and 1+$F$1 adds the tax fraction to one. Copy these two spreadsheet formulas down into rows 3 and 4, a fill operation. The relative spreadsheet references change with the destination row: C3 becomes =A3*B3 and D3 becomes =C3*(1+$F$1); C4 becomes =A4*B4 and D4 becomes =C4*(1+$F$1). The absolute spreadsheet reference $F$1 continues to read the shared rate. Copying spreadsheet formulas adjusts relative coordinates while retaining absolute coordinates. ^[ref]

The same mathematical expression applies to every item. Writing $q_i$ for quantity, $p_i$ for unit price and $t$ for the shared tax rate givesFor row 2, the exact arithmetic is $(3\cdot 5/2)(1+1/5)=(15/2)(6/5)=9$. The subtotals add to $47/2$, so the totals add to $(47/2)(6/5)=141/5$, agreeing with $9+42/5+54/5$. The spreadsheet formulas encode the relationship; the displayed numbers are its results for these particular inputs.

This separation is useful for forecasting and sensitivity analysis. Keeping quantities and unit prices fixed while changing only F1 from $1/5$ to $1/4$ gives exact totals $75/8$, $35/4$, and $45/4$. More generally, a rate change $\Delta t$ changes each total byThis follows by expanding the two mathematical expressions and cancelling $s_i(1+t)$. Here $\Delta t=1/20$, so the increases are $3/8$, $7/20$, and $9/20$, adding to $47/40$. One shared input makes the assumption visible and consistent across all rows. Overwriting a computed total with a number would instead remove that spreadsheet formula; its value would cease to follow the model when the rate changes.

// Target: computer-science.bigb

The square class of $a\in K^{\ast }$ is its image in the square-class group of a field $K^{\ast }/K^{\ast 2}$. Two nonzero elements have the same square class exactly when their quotient is a square. Every element of this quotient has order dividing $2$. Over the rational numbers, a square class has a unique signed squarefree integer representative.

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

An ultrametric space is a metric space whose metric satisfies the ultrametric inequality. The rational numbers with $d_p(x,y)=|x-y{|}_p$, and their P-adic number completion, are examples. Their geometry organizes points into nested clusters rather than measuring separation along a number line.

For integers $a,b$ and $k\ge 0$, the definition of the P-adic valuation givesThus a closed ball of radius $p^{-k}$ in the integers is a congruence class modulo $p^k$. In the p-adic integers the same closed ball is $a+p^k{\mathbb{Z}}_p$. More shared low-order digits mean a smaller p-adic absolute value of the difference.

The binary tree first splits the sixteen integers into even and odd congruence classes, then refines modulo $4$, $8$, and $16$. Each node fixes one more low-order digit. The leaves $5$ and $13$ share the class $5$ modulo $8$, then separate modulo $16$, so their distance is exactly $1/8$. The leaves are deliberately ordered by their low-order digits; horizontal spacing and edge lengths are layout choices, not the metric. The singleton leaves describe this finite sample; a congruence class modulo $16$ in all p-adic integers remains infinite.

The ultrametric inequality also explains the unusual behavior of balls. Every point of an open ball can serve as its center: if $d(a,b)<r$, then $d(a,x)<r$ if and only if $d(b,x)<r$. Two open balls that intersect are nested, since a point of their intersection serves as a common center and their radii can be compared. The same nesting rule holds for closed balls and for a pair consisting of one open ball and one closed ball.

Every open ball is a clopen set. For a point $y$ outside $B(a,r)$, the open ball $B(y,d(a,y))$ stays outside $B(a,r)$: if $d(y,z)<d(a,y)$, the ultrametric inequality forces $d(a,z)=d(a,y)\ge r$. Hence the complement is an open set. Also every closed ball of positive radius is an open set, because $B(b,r)$ lies in the closed ball of radius $r$ around $a$ whenever $d(a,b)\le r$. A radius-zero closed ball is a singleton and need not be an open set. Distinct points can therefore be separated by a clopen set, proving that an ultrametric space is totally disconnected. These ball properties turn distance thresholds into a hierarchy of compatible clusters. ^[ref]

Consider an original abstract data type whose values are rational numbers. Store an ordered pair $(n,d)$ of integers, interpreted as a numerator and a denominator. Use exact, unbounded integer arithmetic in this model. The storage space, representation invariant and abstraction function areThe representation invariant does not require a positive denominator or a reduced fraction. Consequently $(1,2)$, $(2,4)$ and $(-1,-2)$ are three different concrete states with the same abstract rational number $1/2$. Similarly, $(0,1)$ and $(0,-3)$ both represent zero. The abstraction function is a surjection: every rational number has an integer numerator and a nonzero integer denominator, so every abstract value has a representation.

Both $(1,0)$ and $(0,0)$ violate the representation invariant; neither has a value under this abstraction function. By contrast, $(2,4)$ is completely valid. Simplifying it to the reduced fraction $1/2$ may be a useful implementation choice, but is not a requirement of this representation. Strengthening the representation invariant to require a positive denominator and a greatest common divisor of one would choose a unique stored representative for each rational number; every constructor and operation would then need to establish that stronger condition.

Equality of abstract values is therefore different from equality of stored ordered pairs. For valid $(a,b)$ and $(c,d)$,This defines an equivalence relation on $R_{\mathrm{valid}}$: each equivalence class contains all representations of one rational number. An API equality operation can compare these cross-products using exact integer arithmetic. Comparing the two fields individually would incorrectly declare $(1,2)$ and $(2,4)$ unequal as rational numbers. The general reason is that an abstraction function may map several concrete values to one abstract value.^[ref]