Arithmetic

Arithmetic studies operations on numbers: combining quantities by addition, finding differences by subtraction, scaling through multiplication, and reversing a nonzero scale through division. Its usefulness goes beyond producing an answer. A calculation can express how resources are shared, how a measurement changes with scale, or why two different descriptions of a quantity agree.

The allowed numbers matter. Start with the natural numbers $0,1,2,\dots$. Counting collections gives $2+3=5$, but removing three from two calls for the integer $-1$. Sharing three units equally between two recipients gives the rational number $3/2$. These operations motivate larger number systems: integers supply additive inverses, and rational numbers supply division by nonzero integers. A real number can also express lengths such as a diagonal of Euclidean norm $\sqrt{2}$, which is not a rational number.

The beauty is that bookkeeping, geometry, and algebra can check one another. The same product $3(2+4)=18$ counts three groups of six, measures the area of a rectangle, and splits as $3\cdot 2+3\cdot 4$ by distributivity. A number line makes negative integers visible as positions; a fraction records an exact share even when its decimal representation does not terminate. An arithmetic rule becomes easier to remember when these interpretations fit together.

A numeral is a written expression representing a number; different numerals can represent the same integer.

For distinct positive real numbers $a,b$, their logarithmic mean is $L(a,b)=(b-a)/(\log b-\log a)$, with continuous extension $L(a,a)=a$.

Numbers are mathematical objects used to express quantities and carry out operations. Familiar families include the natural numbers, integers, rational numbers, real numbers, and complex numbers; specifying the family determines which operations and solutions are available.

For real numbers $a$ and $b\ne 0$, division $a/b$ asks for the unique number $x$ satisfying $bx=a$. Equivalently, $a/b=a{b}^{-1}$ uses the multiplicative inverse of $b$. This description separates the role of the denominator from the result: the quotient undoes a specified multiplication.

For example, $3/2$ is the number whose double is three, so it is exactly $1.5$. A second interpretation asks how many portions of size $2$ fit into an amount $3$; the answer is one and a half portions. Equal sharing and measuring with a specified portion size lead to the same quotient, although the units assigned to the result can differ.

In the natural numbers, exact division need not be possible. Euclidean division instead keeps a remainder: $17=5\cdot 3+2$, with $0\le 2<5$. Thus seventeen objects form three complete groups of five with two left over. As a rational number, the same quotient is $17/5=3+2/5=3.4$. The remainder and the fractional part describe the same unfinished group in different ways.

Division by zero cannot retain the unique-solution rule. For $a\ne 0$, the equation $0x=a$ has no solution; for $a=0$, every real number $x$ solves $0x=0$, so there is no unique quotient. This explains both failures through multiplication, rather than assigning $a/0$ a value that would contradict ordinary arithmetic. ^[ref]

Division of integers that returns an integer quotient rather than an exact fraction. A rounding convention is required for negative operands. For $n\ge 0$ and $d>0$, the quotient is $q=\lfloor n/d\rfloor$ and the remainder is $r=n-qd$ with $0\le r<d$.

For an $n$-bit unsigned dividend $a$ and positive divisor $b$, read the dividend bits from most to least significant. If the processed prefix is $p$, maintain $p=bQ+R$ with $0\le R<b$. Appending bit $d$ gives trial remainder $T=2R+d$; append quotient bit $[T\ge b]$ and replace the remainder by $T-b[T\ge b]$. This preserves the quotient–remainder identity.

The old dividend and new quotient can share one $n$-bit processor register: after $k$ iterations its value is $(a{2}^k\mathrm{mod}{2}^n)+Q_k$, where $Q_k<2^k$. The shifted dividend occupies the upper positions and the accumulated quotient occupies the lower positions, so no information is overwritten. The high processor register holds $R_k$. For an $n$-bit dividend the next trial remainder is no larger than the next prefix, hence fits in $n$ bits through the final iteration. This last bound is specific to the dividend width; it must not be assumed for unrestricted multiword dividends.

For signed two's complement integers, a conditional positive bias converts the rounding behaviour of an arithmetic right shift into C integer division. It can be formed without a conditional branch by turning the sign bit into either zero or a low-$k$-bit bit mask.

Proof. Nonnegative $a$ needs no adjustment. If $a<0$, write $-a=2^{k}q+r$, $0\le r<2^k$. Then $a+(2^k-1)=-2^{k}q+(2^k-1-r)$, whose quotient rounded down is $-q$, exactly the quotient truncated towards zero. The bias fits safely for $1\le k<n$ in an $n$-bit signed word: it is added only to negative inputs, and the largest possible adjusted input is $2^k-2$. The corresponding remainder is $((a+b)\mathrm{mod}{2}^k)-b$, where $b=(2^k-1)1_{a<0}$.

For representable integer quotients with nonzero divisor, C integer division discards the fractional part towards zero and defines the remainder by $a=bq+r$. Thus a nonzero remainder has the dividend's sign. For example, $-13=5(-2)+(-3)$, so division by five returns quotient $-2$ and remainder $-3$. This convention differs from division rounded down, which would return quotient $-3$ and nonnegative remainder $2$. Division by zero and an unrepresentable quotient are excluded.^[ref]

A ratio compares two quantities $a,b$, written $a:b$ and corresponding to division $a/b$ when $b\ne 0$. Equal ratios of corresponding lengths characterize the scaling of similar triangles.

For example, $6:4$ and $3:2$ specify the same ratio because $6/4=3/2$. Multiplying both entries by the same nonzero scale preserves the quotient: $(ca)/(cb)=a/b$. This is useful when comparing mixtures or scaling a recipe; doubling both ingredient amounts preserves their ratio, while adding the same amount to both generally does not. For quantities with the same physical units, their ratio is dimensionless, connecting this elementary calculation with dimensional analysis.

A percentage expresses a ratio in hundredths: $p$ as a fraction is $100p\%$. For example, $18/60=0.3=30\%$. Comparing two percentages requires specifying whether the change is an absolute difference in percentage points or a relative change divided by the initial value.

A percentage point is one hundredth on a probability or proportion scale. A change from $20\%$ to $25\%$ is $5$ percentage points, while the relative increase is $(25-20)/20=25\%$. These answer different questions and have different denominators.

An inequality compares mathematical quantities using an ordering, such as $a<b$ or $a\le b$ for real numbers. A strict inequality excludes equality; a non-strict inequality allows it. For example, $4<5$ is strict, while $4\le 4$ is true because equality is allowed. Adding the same real number preserves either comparison; multiplying by a negative real number reverses its direction.

An ordered comparison such as $a<b$, excluding equality. In a sorted search, strict comparisons distinguish regions that can be discarded from boundary positions that may contain an equal target.

A linear inequality compares a linear functional of a vector of unknowns with a constant, for example $a^{T}x\le b$ or $a^{T}x<b$ over the real numbers. For $a\ne 0$, its solution set is a closed or open half-space, respectively. The inequality $x+y\le 1$ allows all points on one side of the line $x+y=1$, including the line. Multiplication by a negative real number reverses the inequality, so $x\ge 1$ is equivalently $-x\le -1$.

A system of linear inequalities requires several linear inequalities to hold simultaneously. The notation $Ax\le b$ compares coordinates componentwise, giving one inequality per row of the matrix $A$. Its solution set is the intersection of the corresponding half-spaces. For non-strict inequalities it is a linear polyhedron, which may be empty. For example, $x\ge 1$, $y\ge 1$, $x+y\le 1$ are inconsistent: adding their equivalent forms $-x\le -1$, $-y\le -1$, $x+y\le 1$ gives $0\le -1$. A Farkas certificate for linear inequalities generalizes this cancellation.

A sum of squares adds the squares of numbers $a_i^2$ for specified real numbers $a_i$. It is nonnegative and vanishes exactly when every term is zero. For the coordinates of a vector, it is the squared Euclidean norm; for regression residuals, it gives the residual sum of squares minimized by ordinary least squares.

A figurate number is an integer illustrated by a regular geometric arrangement of points. Triangular numbers and square numbers are basic examples: growing a prescribed shape turns a counting problem into a sequence of integers.

The sum of squares of the first $n$ positive integers: $S_n={\sum }_{k=1}^n{k}^2=n(n+1)(2n+1)/6$. It counts a square-based stack with layers containing $1^2,2^2,\dots ,n^2$ objects. Differences $S_b-S_{a-1}$ give sums over consecutive squares.

The number of objects in a square-based pyramid with successive square layers of sizes $1^2,2^2,\dots ,n^2$. If $F(n)=n(n+1)(2n+1)/6$, direct algebra gives $F(n)-F(n-1)=n^2$ and $F(0)=0$. Mathematical induction therefore proves the displayed summation formula. Its leading term is $n^3/3$, so this sum grows cubically, even though its largest individual summand grows quadratically.

For integers $s\ge 3$ and $n\ge 1$, the $n$th $s$-gonal number is $P_{s,n}=((s-2)n^2-(s-4)n)/2$. Its successive differences form an arithmetic progression. The cases $s=3,4,5,6$ give triangular numbers, square numbers, pentagonal numbers, and hexagonal numbers.

The polygonal number $P_{8,n}=n(3n-2)$ for a positive integer $n$.

The polygonal number $P_{7,n}=n(5n-3)/2$ for a positive integer $n$.

The $n$th hexagonal number is $H_n=n(2n-1)$ for a positive integer $n$. Since $H_n=T_{2n-1}$, every hexagonal number is also a triangular number.

The $n$th pentagonal number is $P_n=n(3n-1)/2$ for a positive integer $n$, giving $1,5,12,22,\dots$. A positive integer $x$ is pentagonal exactly when $24x+1$ is a perfect square and $(1+\sqrt{24x+1})/6$ is an integer, by the defining quadratic equation.

A value $k(3k-1)/2$ with $k$ any integer, including negative values and zero. For $k=\pm 1,\pm 2,\dots$, positive values occur as pairs $j(3j-1)/2$ and $j(3j+1)/2$. These are the indices in the Euler pentagonal identity.

A triangular number counts rows of $1,2,\dots ,n$ objects, with $n$ a nonnegative integer:The first values are $0,1,3,6,10,15,21,28,36,45,\dots$. The empty arrangement contributes zero, by the empty sum convention. Adding one more row gives the triangular number recurrence $T_n-T_{n-1}=n$ for $n\ge 1$. The row lengths form an arithmetic progression, while their running totals are triangular numbers.

Pair the rows with a reversed copy: $1+n=2+(n-1)=\cdots =n+1$. There are $n$ such pairs, so two copies contain $n(n+1)$ dots. Dividing by two proves the formula. This rectangle picture explains the factor one-half without requiring a formula for a long summation. For example, one hundred rows contain $T_{100}=5050$ dots.

The same count appears whenever every pair of different objects must be considered once. Five participants make ten possible handshakes; five sensors have ten different pairwise comparisons; a complete graph on five vertices has ten edges. These are the triangular number pair-counting identity, with one fewer row than objects. The distinction prevents a common error: $T_5=15$ counts all pairs from six objects, whereas five objects give $T_4=10$.

The triangular number square-complement identity says that two neighboring triangular numbers fill a square. Combining that identity with the one-row difference makes the difference of squares $T_n^2-T_{n-1}^2$ equal to $n^3$. It leads to Nicomachus's theorem: the sum of the first $n$ cubes of numbers is the square of a number $T_n^2$. A change in the way one groups objects exposes a relation between summation, multiplication and geometry.

This geometric view is already visible in Nicomachus's Introduction to Arithmetic by Nicomachus (Hahn teacher edition): Book II, Chapter 8 describes growing triangular rows, Chapter 12 combines successive triangular counts into squares, and Chapter 20 groups successive odd integers into cubes of numbers.^[ref] The linked edition uses Martin Luther D'Ooge's translation, revised with modern teaching material by Nathanael Hahn. Our empty-row index $T_0=0$ is a modern convention; the historical passage starts from unity and calls three the first actual triangle.

For a positive integer $n$, adding the row of length $n$ to the previous arrangement gives $T_n=T_{n-1}+n$. Thus the first differences of the triangular numbers are $1,2,3,\dots$. The triangular number recurrence provides incremental computation: after $T_6=21$, adding seven gives $T_7=28$.

For $N\ge 1$ distinct objects, the number of unordered pairs of different objects isOrdering the objects gives $N-1$ pairs with the first object, $N-2$ new pairs with the second, and so on down to one. Alternatively, $N(N-1)$ ordered pairs count every unordered pair twice. The binomial coefficient therefore counts the edges of a complete graph on $N$ vertices. This is also consistent with the handshaking lemma: each vertex has degree of a vertex $N-1$, and the degree sum counts every edge twice.

For $n\ge 1$,An $n$-by-$n$ array of cells has $n^2$ cells. One side of the diagonal, including the diagonal itself, contains $1+2+\cdots +n=T_n$ cells; the other contains $1+2+\cdots +(n-1)=T_{n-1}$. For $n=5$, the two counts are fifteen and ten, giving twenty-five. The identity is stated in Introduction to Arithmetic by Nicomachus (Hahn teacher edition) (Book II, Chapter 12, §§1–2, printed p. 70).

A nonnegative integer $M$ is a triangular number exactly when $8M+1$ is a perfect square. Its square root is then an odd integer, and the row index isCompleting the square in $M=n(n+1)/2$ proves necessity; reversing the calculation proves sufficiency. For example, $8\cdot 21+1=169=13^2$, giving $n=6$. In contrast, $8\cdot 20+1=161$ lies strictly between $12^2$ and $13^2$, so twenty is not a triangular number. The criterion turns an apparently open-ended search through a sequence into one exact perfect square check.

A squared triangular number is the square of a number $T_n$, where $T_n$ is a triangular number. Beginning at $n=0$, the values are $0,1,9,36,100,225,441,\dots$. Nicomachus's theorem gives them a second description as cumulative sums of cubes of numbers.

For every nonnegative integer $n$,For example, $1^3+2^3+3^3+4^3=1+8+27+64=100=T_4^2$. A short proof uses the triangular number recurrence and the triangular number square-complement identity:Summing this difference of squares identity gives a telescoping sum; all intermediate squared triangular numbers cancel. Nicomachus describes a related grouping of the odd integers: take one, then the next two, then the next three, producing $1,3+5,7+9+11,\dots$, which are $1^3,2^3,3^3,\dots$. This historical grouping is recorded in Introduction to Arithmetic by Nicomachus (Hahn teacher edition) (Book II, Chapter 20, §5, printed p. 75).

A square triangular number is simultaneously a triangular number and a perfect square. For example, $36=T_8=6^2$. This differs from a squared triangular number: $T_4^2=100$ is not itself triangular, by the recognition criterion for triangular numbers. Writing $T_n=m^2$ gives $(2n+1{)}^2-8m^2=1$, a Pell equation that explains why this intersection of two counting patterns has additional arithmetic structure.

The geometric mean replaces a list of nonnegative real numbers by one repeated value that preserves their product of numbers. For $n\ge 1$ inputs,For two positive inputs, $G=\sqrt{x_1{x}_2}$ is their positive square root. The geometric mean of $1$ and $9$ is three: replacing $1\cdot 9$ by $3\cdot 3$ retains the same product. In contrast, their arithmetic mean five retains the sum $1+9=5+5$. The geometric mean is zero if any input is zero; the logarithm formula below requires every input to be positive. ^[ref].

Its usefulness becomes immediate for successive growth factors. Suppose a hypothetical positive quantity starts at $Q_0=1000$, is multiplied by $3/2$, then by $2/3$. It goes $1000\to 1500\to 1000$, so the geometric mean factor is $\sqrt{(3/2)(2/3)}=1$: a constant factor of one for both steps gives the same endpoint. The arithmetic mean factor is $13/12$; repeating it would give $1000(13/12{)}^2\approx 1173.61$, a different endpoint. A mean of factors summarizes multiplicative change; subtracting one afterwards gives the equivalent change per step. Equal-duration steps are necessary to interpret this as change per common time interval.

For positive inputs, logarithms turn multiplication into addition, soThus the geometric mean is an arithmetic mean in logarithmic coordinates, followed by the exponential function. For positive measurements with a common unit, one can use dimensionless ratios to a fixed reference $x_0$: $G=x_0\exp (n^{-1}{\sum }_i\log (x_i/x_0))$. Changing the common unit rescales $G$ by the same factor as the inputs. As a check, doubling one input and halving another retains $G$ while usually changing the arithmetic mean.

The arithmetic-geometric mean inequality gives $G\le A$. Apply that same inequality to the multiplicative inverses $1/x_i$ to obtain $H\le G$, where $H$ is the harmonic mean. Consequentlyfor positive inputs, with equality throughout exactly when all inputs agree. For $30$ and $60$, the three values are $H=40$, $G=30\sqrt{2}\approx 42.4264$, and $A=45$. They preserve, respectively, the sum of inverses, the product, and the sum. No one of these summaries preserves all three for unequal inputs.

For positive quantities measured at successive steps, the growth factor $g_i=Q_i/Q_{i-1}$ is a dimensionless ratio. It can be below one when the quantity shrinks. Repeated multiplication gives $Q_n=Q_0{\prod }_{i=1}^n{g}_i$, so the common factor that reproduces the endpoint after $n$ identical steps is their geometric mean. A factor $g$ corresponds to a relative change $g-1$; factors must be averaged before subtracting one when summarizing compounded change.

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For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted geometric mean isIt takes a weighted mean of logarithms. A zero weight excludes that input; equal weights give the ordinary geometric mean. If $x_i$ occurs $m_i$ times in a list of total length $M$, its correct weight is $w_i=m_i/M$. For the list $(12,3,3,3)$,and replacing all four entries by $G_w$ preserves the product $324$.

There is also an optimization interpretation. A logarithmic transformation turns multiplicative discrepancies into additive ones. For a positive representative $c$, the weighted mean minimizes squared deviations identity givesThus $c=G_w$ uniquely minimizes weighted quadratic loss in logarithmic coordinates. The weighted arithmetic mean minimizes squared discrepancies in the original coordinates. These two objectives explain why different choices of what constitutes an error select different summaries.

The strictly concave function $\log x$ and the reversed Jensen's inequality give $G_w\le A_w$, where $A_w={\sum }_i{w}_i{x}_i$ is the weighted arithmetic mean. Apply the same reasoning to the multiplicative inverses to get $H_w\le G_w$ for the weighted harmonic mean. Thus $H_w\le G_w\le A_w$; equality holds exactly when all positive-weight inputs agree. For this list, $H_w=48/13$, $G_w=3\sqrt{2}$, and $A_w=21/4$. The weights specify which inputs are represented and must stay the same across the comparison.

Starting from $0<a_1\le b_1$, replace the lower endpoint by the geometric mean and the upper endpoint by the arithmetic mean. The arithmetic-geometric mean inequality gives $a_n\le a_{n+1}\le b_{n+1}\le b_n$. The bounded monotone sequence theorem gives limits, and the arithmetic recurrence forces them to coincide. The limiting value is the arithmetic-geometric mean of the initial pair.

The natural numbers are the nonnegative integers $0,1,2,\dots$.

The minimum excludant of a finite set of nonnegative integers is the smallest nonnegative integer absent from the set. For example, $\mathrm{mex}\{0,1,3\}=2$.

For a set of positive integers, its minimum excluded positive integer is the least positive integer absent from the set. This is the positive-indexed version of the minimum excludant, which normally allows zero.

Every nonempty subset of the natural numbers has a least element in their usual order. Applying this principle to positive integer linear combinations proves Bézout's identity. This principle for the usual natural-number order is distinct from the general well-ordering theorem asserting that every set can be equipped with some well-order.

Assume that a proposed statement fails for some natural number, and choose the least failing index using the well-ordering principle. Prove the necessary base cases and reduce the least failure to a smaller index that must satisfy the statement. If that smaller case implies the alleged failure also satisfies it, there is a contradiction. Reductions by a step larger than one require enough consecutive base cases to cover every possible residue class.

A real number is an element of the complete ordered field $\mathbb{R}$.

The condition that a real number is at least zero. A nonnegative flow in a directed flow network follows the specified orientation or has zero value; a negative algebraic solution would describe the opposite orientation instead.

For a real number $x$, its cube root is the unique real number $y$ satisfying $y^3=x$. The function $y\mapsto y^3$ is strictly increasing, so negative inputs also have a unique real cube root.

The integer cube root of a nonnegative integer $n$ is $\lfloor \sqrt[3]{n}\rfloor$, the unique integer $r$ satisfying $r^3\le n<(r+1{)}^3$. It can be computed by binary search using integer cube comparisons.

A number line represents real numbers as points on a line, with zero as origin and a chosen positive direction and unit scale. Addition becomes a shift and multiplication becomes a rescaling about zero.

A linear scale places a value $x$ at a position $u=ax+c$, where $a>0$ and $c$ are constants. Equal differences in $x$ occupy equal distances because $u_2-u_1=a(x_2-x_1)$. This contrasts with a logarithmic scale, which assigns equal distances to equal ratios.

The Archimedean property says that the natural numbers are unbounded in the real numbers. In particular, for every positive real number $a$ there is a positive integer $n$ with $1/n<a$. Otherwise a purported upper bound for all natural numbers contradicts the least-upper-bound property by translating that bound by one.

A binary number is a number represented in base two.

After adding $k$ least-significant binary digits, the numerical sum of the two processed input suffixes equals the written $k$ result digits plus $carry{2}^k$, with $carry\in \{0,1\}$. Initially both suffixes are empty. For the next column, write $(x_k+y_k+carry)\mathrm{mod}2$ and pass $\lfloor (x_k+y_k+carry)/2\rfloor$ to the next column. Multiplying these column quantities by $2^k$ preserves the equation. After all columns, the remaining carry supplies the extra leading digit, proving correctness without converting either full input to a machine integer.

For an $n$-bit binary string $A=a_{n-1}\cdots a_0$, the unsigned interpretation assigns positive positional weights $2^i$ to its bits. Its values range from zero to $2^n-1$. The same stored bits have two's complement value $\mathrm{bin}(A)-a_{n-1}{2}^n$, so interpreting a machine word is a separate choice from storing its bits.

The number of binary digits in the usual representation of a positive integer $n$: $\lfloor {\log }_{2}n\rfloor +1$. The bit length of zero is conventionally zero in integer programming interfaces.

Every real number in $[0,1]$ is a sum ${\sum }_{j\ge 1}{\epsilon }_j{2}^{-j}$ with binary digits ${\epsilon }_j\in \{0,1\}$.