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Associative property

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Property of a mathematical operation

"Associativity", "Associative", and "non-associative" redirect here. For other uses, seeAssociativity (disambiguation). For associative and non-associative learning, seeLearning § Types.

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Associative property
A visual graph representing associative operations; $(x\circ y)\circ z=x\circ (y\circ z)$
Type	Law,rule of replacement
Field	Elementary algebra Boolean algebra Set theory Linear algebra Propositional calculus
Symbolic statement	Elementary algebra $(x\,\,y)\,\,z=x\,\,(y\,\,z)\forall x,y,z\in S$ Propositional calculus $(P\lor (Q\lor R))\Leftrightarrow ((P\lor Q)\lor R)$ $(P\land (Q\land R))\Leftrightarrow ((P\land Q)\land R),$

Inmathematics, theassociative property^[1] is a property of somebinary operations that rearranging theparentheses in an expression will not change the result. Inpropositional logic,associativity is avalid rule of replacement forexpressions inlogical proofs.

Within an expression containing two or more occurrences in a row of the same associative operator, the order in which theoperations are performed does not matter as long as the sequence of theoperands is not changed. That is (after rewriting the expression with parentheses and in infix notation if necessary), rearranging the parentheses in such an expression will not change its value. Consider the following equations:

${\begin{aligned}(2+3)+4&=2+(3+4)=9\,\\2\times (3\times 4)&=(2\times 3)\times 4=24.\end{aligned}}$

Even though the parentheses were rearranged on each line, the values of the expressions were not altered. Since this holds true when performing addition and multiplication on anyreal numbers, it can be said that "addition and multiplication of real numbers are associative operations".

Associativity is not the same ascommutativity, which addresses whether the order of two operands affects the result. For example, the order does not matter in the multiplication of real numbers, that is,a ×b =b ×a, so we say that the multiplication of real numbers is a commutative operation. However, operations such asfunction composition andmatrix multiplication are associative, but not (generally) commutative.

Associative operations are abundant in mathematics; in fact, manyalgebraic structures (such assemigroups andcategories) explicitly require their binary operations to be associative. However, many important and interesting operations are non-associative; some examples includesubtraction,exponentiation, and thevector cross product. In contrast to the theoretical properties of real numbers, the addition offloating point numbers in computer science is not associative, and the choice of how to associate an expression can have a significant effect on rounding error.

Definition

[edit]

A binary operation ∗ on the setS is associative whenthis diagram commutes. That is, when the two paths from`S`×`S`×`S` toScompose to the same function from`S`×`S`×`S` toS.

Formally, abinary operation $\ast$ on asetS is calledassociative if it satisfies theassociative law:

(x\ast y)\ast z=x\ast (y\ast z)

, for all

x, y, z {\displaystyle x,y,z}

inS.

Here, ∗ is used to replace the symbol of the operation, which may be any symbol, and even the absence of symbol (juxtaposition) as formultiplication.

(xy)z=x(yz)

, for all

x, y, z {\displaystyle x,y,z}

inS.

The associative law can also be expressed in functional notation thus: $(f\circ (g\circ h))(x)=((f\circ g)\circ h)(x)$

Generalized associative law

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In the absence of the associative property, five factorsa,b,c,d,e result in aTamari lattice of order four, possibly different products.

If a binary operation is associative, repeated application of the operation produces the same result regardless of how valid pairs of parentheses are inserted in the expression.^[2] This is called thegeneralized associative law.

The number of possible bracketings is just theCatalan number, $C_{n}$ , forn operations onn+1 values. For instance, a product of 3 operations on 4 elements may be written (ignoring permutations of the arguments), in $C_{3}=5$ possible ways:

$((ab)c)d$
$(a(bc))d$
$a((bc)d)$
$a(b(cd))$
$(ab)(cd)$

If the product operation is associative, the generalized associative law says that all these expressions will yield the same result. So unless the expression with omitted parentheses already has a different meaning (see below), the parentheses can be considered unnecessary and "the" product can be written unambiguously as

a b c d {\displaystyle abcd}

As the number of elements increases, thenumber of possible ways to insert parentheses grows quickly, but they remain unnecessary for disambiguation.

An example where this does not work is thelogical biconditional↔. It is associative; thus,A ↔ (B ↔C) is equivalent to(A ↔B) ↔C, butA ↔B ↔C most commonly means(A ↔B) and (B ↔C), which is not equivalent.

Examples

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The addition of real numbers is associative.

Some examples of associative operations include the following.

Theconcatenation of the three strings"hello"," ","world" can be computed by concatenating the first two strings (giving"hello ") and appending the third string ("world"), or by joining the second and third string (giving" world") and concatenating the first string ("hello") with the result. The two methods produce the same result; string concatenation is associative (but not commutative).
Inarithmetic,addition andmultiplication ofreal numbers are associative; i.e.,
$\left.{\begin{matrix}(x+y)+z=x+(y+z)=x+y+z\quad \\(x\,y)z=x(y\,z)=x\,y\,z\qquad \qquad \qquad \quad \ \ \,\end{matrix}}\right\}{\mbox{for all }}x,y,z\in \mathbb {R} .$
Because of associativity, the grouping parentheses can be omitted without ambiguity.
The trivial operationx ∗y =x (that is, the result is the first argument, no matter what the second argument is) is associative but not commutative. Likewise, the trivial operation $x\circ y=y$ (that is, the result is the second argument, no matter what the first argument is) is associative but not commutative.
Addition and multiplication ofcomplex numbers andquaternions are associative. Addition ofoctonions is also associative, but multiplication of octonions is non-associative.
Thegreatest common divisor andleast common multiple functions act associatively. $\left.{\begin{matrix}\operatorname {gcd} (\operatorname {gcd} (x,y),z)=\operatorname {gcd} (x,\operatorname {gcd} (y,z))=\operatorname {gcd} (x,y,z)\ \quad \\\operatorname {lcm} (\operatorname {lcm} (x,y),z)=\operatorname {lcm} (x,\operatorname {lcm} (y,z))=\operatorname {lcm} (x,y,z)\quad \end{matrix}}\right\}{\mbox{ for all }}x,y,z\in \mathbb {Z} .$
Taking theintersection or theunion ofsets: $\left.{\begin{matrix}(A\cap B)\cap C=A\cap (B\cap C)=A\cap B\cap C\quad \\(A\cup B)\cup C=A\cup (B\cup C)=A\cup B\cup C\quad \end{matrix}}\right\}{\mbox{for all sets }}A,B,C.$
IfM is some set andS denotes the set of all functions fromM toM, then the operation offunction composition onS is associative: $(f\circ g)\circ h=f\circ (g\circ h)=f\circ g\circ h\qquad {\mbox{for all }}f,g,h\in S.$
Slightly more generally, given four setsM,N,P andQ, withh :M →N,g :N →P, andf :P →Q, then $(f\circ g)\circ h=f\circ (g\circ h)=f\circ g\circ h$ as before. In short, composition of maps is always associative.
Incategory theory, composition of morphisms is associative by definition. Associativity of functors and natural transformations follows from associativity of morphisms.
Consider a set with three elements,A,B, andC. The following operation:
× A B C
A A A A
B A B C
C A A A
is associative. Thus, for example,A(BC) = (AB)C =A. This operation is not commutative.
Becausematrices representlinear functions, andmatrix multiplication represents function composition, one can immediately conclude that matrix multiplication is associative.^[3]
Forreal numbers (and for anytotally ordered set), the minimum and maximum operation is associative: $\max(a,\max(b,c))=\max(\max(a,b),c)\quad {\text{ and }}\quad \min(a,\min(b,c))=\min(\min(a,b),c).$

×	A	B	C
A	A	A	A
B	A	B	C
C	A	A	A

Propositional logic

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Transformation rules
Propositional calculus
Rules of inference (List)
Implication introduction / elimination (modus ponens) Biconditional introduction / elimination Conjunction introduction / elimination Disjunction introduction / elimination Disjunctive / hypothetical syllogism Constructive / destructive dilemma Absorption / modus tollens / modus ponendo tollens Modus non excipiens Negation introduction
Rules of replacement
Associativity Commutativity Distributivity Double negation De Morgan's laws Transposition Material implication Exportation Tautology
Predicate logic
Rules of inference
Universal generalization / instantiation Existential generalization / instantiation
v t e

Rule of replacement

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In standard truth-functional propositional logic,association,^[4]^[5] orassociativity^[6] are twovalid rules of replacement. The rules allow one to move parentheses inlogical expressions inlogical proofs. The rules (usinglogical connectives notation) are:

$(P\lor (Q\lor R))\Leftrightarrow ((P\lor Q)\lor R)$

and

$(P\land (Q\land R))\Leftrightarrow ((P\land Q)\land R),$

where " $\Leftrightarrow$ " is ametalogical symbol representing "can be replaced in aproof with".

Truth functional connectives

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Associativity is a property of somelogical connectives of truth-functionalpropositional logic. The followinglogical equivalences demonstrate that associativity is a property of particular connectives. The following (and their converses, since↔ is commutative) are truth-functionaltautologies.^{[citation needed]}

Associativity of disjunction: $((P\lor Q)\lor R)\leftrightarrow (P\lor (Q\lor R))$
Associativity of conjunction: $((P\land Q)\land R)\leftrightarrow (P\land (Q\land R))$
Associativity of equivalence: $((P\leftrightarrow Q)\leftrightarrow R)\leftrightarrow (P\leftrightarrow (Q\leftrightarrow R))$

Joint denial is an example of a truth functional connective that isnot associative.

Non-associative operation

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A binary operation $*$ on a setS that does not satisfy the associative law is callednon-associative. Symbolically,

$(x*y)*z\neq x*(y*z)\qquad {\mbox{for some }}x,y,z\in S.$

For such an operation the order of evaluationdoes matter. For example:

Subtraction: $(5-3)-2\,\neq \,5-(3-2)$
Division: $(4/2)/2\,\neq \,4/(2/2)$
Exponentiation: $2^{(1^{2})}\,\neq \,(2^{1})^{2}$
Vector cross product: ${\begin{aligned}\mathbf {i} \times (\mathbf {i} \times \mathbf {j} )&=\mathbf {i} \times \mathbf {k} =-\mathbf {j} \\(\mathbf {i} \times \mathbf {i} )\times \mathbf {j} &=\mathbf {0} \times \mathbf {j} =\mathbf {0} \end{aligned}}$

Also although addition is associative for finite sums, it is not associative inside infinite sums (series). For example, $(1+-1)+(1+-1)+(1+-1)+(1+-1)+(1+-1)+(1+-1)+\dots =0$ whereas $1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+\dots =1.$

Some non-associative operations are fundamental in mathematics. They appear often as the multiplication in structures callednon-associative algebras, which have also an addition and ascalar multiplication. Examples are theoctonions andLie algebras. In Lie algebras, the multiplication satisfiesJacobi identity instead of the associative law; this allows abstracting the algebraic nature ofinfinitesimal transformations.

Other examples arequasigroup,quasifield,non-associative ring, andcommutative non-associative magmas.

Nonassociativity of floating-point calculation

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In mathematics, addition and multiplication of real numbers are associative. By contrast, in computer science, addition and multiplication offloating point numbers arenot associative, as different rounding errors may be introduced when dissimilar-sized values are joined in a different order.^[7]

To illustrate this, consider a floating-point representation with a 4-bitsignificand:

(1.000₂×2⁰ + 1.000₂×2⁰) +1.000₂×2⁴ = 1.000₂×2¹ + 1.000₂×2⁴ = 1.001₂×2⁴

1.000₂×2⁰ + (1.000₂×2⁰ +1.000₂×2⁴) = 1.000₂×2⁰ + 1.000₂×2⁴ = 1.000₂×2⁴

Even though most computers compute with 24 or 53 bits of significand,^[8] this is still an important source of rounding error, and approaches such as theKahan summation algorithm are ways to minimize the errors. It can be especially problematic in parallel computing.^[9]^[10]

Notation for non-associative operations

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Main article:Operator associativity

In general, parentheses must be used to indicate theorder of evaluation if a non-associative operation appears more than once in an expression (unless the notation specifies the order in another way, like ${\dfrac {2}{3/4}}$ ). However,mathematicians agree on a particular order of evaluation for several common non-associative operations. This is simply a notational convention to avoid parentheses.

Aleft-associative operation is a non-associative operation that is conventionally evaluated from left to right, i.e.,

$\left.{\begin{array}{l}a*b*c=(a*b)*c\\a*b*c*d=((a*b)*c)*d\\a*b*c*d*e=(((a*b)*c)*d)*e\quad \\{\mbox{etc.}}\end{array}}\right\}{\mbox{for all }}a,b,c,d,e\in S$

while aright-associative operation is conventionally evaluated from right to left:

$\left.{\begin{array}{l}x*y*z=x*(y*z)\\w*x*y*z=w*(x*(y*z))\quad \\v*w*x*y*z=v*(w*(x*(y*z)))\quad \\{\mbox{etc.}}\end{array}}\right\}{\mbox{for all }}z,y,x,w,v\in S$

Both left-associative and right-associative operations occur. Left-associative operations include the following:

Subtraction and division of real numbers^[11]^[12]^[13]^[14]^[15]: $x-y-z=(x-y)-z$; $x/y/z=(x/y)/z$
Function application: $(f\,x\,y)=((f\,x)\,y)$

This notation can be motivated by thecurrying isomorphism, which enables partial application.

Right-associative operations include the following:

Exponentiation of real numbers in superscript notation: $x^{y^{z}}=x^{(y^{z})}$
Exponentiation is commonly used with brackets or right-associatively because a repeated left-associative exponentiation operation is of little use. Repeated powers would mostly be rewritten with multiplication:; $(x^{y})^{z}=x^{(yz)}$
Formatted correctly, the superscript inherently behaves as a set of parentheses; e.g. in the expression $2^{x+3}$ the addition is performedbefore the exponentiation despite there being no explicit parentheses $2^{(x+3)}$ wrapped around it. Thus given an expression such as $x^{y^{z}}$ , the full exponent $y^{z}$ of the base $x {\displaystyle x}$ is evaluated first. However, in some contexts, especially in handwriting, the difference between ${x^{y}}^{z}=(x^{y})^{z}$ , $x^{yz}=x^{(yz)}$ and $x^{y^{z}}=x^{(y^{z})}$ can be hard to see. In such a case, right-associativity is usually implied.
Function definition: $\mathbb {Z} \rightarrow \mathbb {Z} \rightarrow \mathbb {Z} =\mathbb {Z} \rightarrow (\mathbb {Z} \rightarrow \mathbb {Z} )$; $x\mapsto y\mapsto x-y=x\mapsto (y\mapsto x-y)$
Using right-associative notation for these operations can be motivated by theCurry–Howard correspondence and by thecurrying isomorphism.

Non-associative operations for which no conventional evaluation order is defined include the following.

Exponentiation of real numbers in infix notation^[16]: $(x^{\wedge }y)^{\wedge }z\neq x^{\wedge }(y^{\wedge }z)$
Knuth's up-arrow operators: $a\uparrow \uparrow (b\uparrow \uparrow c)\neq (a\uparrow \uparrow b)\uparrow \uparrow c$; $a\uparrow \uparrow \uparrow (b\uparrow \uparrow \uparrow c)\neq (a\uparrow \uparrow \uparrow b)\uparrow \uparrow \uparrow c$
Taking thecross product of three vectors: ${\vec {a}}\times ({\vec {b}}\times {\vec {c}})\neq ({\vec {a}}\times {\vec {b}})\times {\vec {c}}\qquad {\mbox{ for some }}{\vec {a}},{\vec {b}},{\vec {c}}\in \mathbb {R} ^{3}$
Taking the pairwiseaverage of real numbers: ${(x+y)/2+z \over 2}\neq {x+(y+z)/2 \over 2}\qquad {\mbox{for all }}x,y,z\in \mathbb {R} {\mbox{ with }}x\neq z.$
Taking therelative complement of sets: $(A\backslash B)\backslash C\neq A\backslash (B\backslash C)$ .
(Comparematerial nonimplication in logic.)

History

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William Rowan Hamilton seems to have coined the term "associative property"^[17] around 1844, a time when he was contemplating the non-associative algebra of theoctonions he had learned about fromJohn T. Graves.^[18]

Relationship with commutativity in certain special cases

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In general, associative operations are not commutative. However, under certain special conditions, it may be the case that associativity implies commutativity. Associative operators defined on an interval of the real number line are commutative if they are continuous and injective in both arguments.^[19] A consequence is that every continuous, associative operator on two real inputs that is strictly increasing in each of its inputs is commutative.^[20]