# Associative property

(Redirected from Associative operation)

In mathematics, the associative property[1] is a property of some binary operations. In propositional logic, associativity is a valid rule of replacement for expressions in logical proofs.

Within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. That is, rearranging the parentheses in such an expression will not change its value. Consider the following equations:

$(2+3)+4=2+(3+4)=9 \,$
$2\times(3\times4)=(2\times3)\times4=24 \,$

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

Associativity is not to be confused with commutativity, which addresses whether a × b = b × a.

Associative operations are abundant in mathematics; in fact, many algebraic structures (such as semigroups and categories) explicitly require their binary operations to be associative.

However, many important and interesting operations are non-associative; some examples include subtraction, exponentiation and the vector cross product. In contrast to the theoretical counterpart, the addition of floating point numbers in computer science is not associative, and is an important source of rounding error.

## Definition

Formally, a binary operation $*$ on a set S is called associative if it satisfies the associative law:

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

Here, $*$ is used to replace the symbol of the operation, which may be any symbol, and even the absence of symbol like for the multiplication.

$(xy)z=x(yz) = xyz \qquad\mbox{for all }x,y,z\in S.$

The associative law can also be expressed in functional notation thus: $f(f(x,y),z) = f(x,f(y,z))$.

## Generalized associative law

If a binary operation is associative, repeated application of the operation produces the same result regardless how valid pairs of parenthesis are inserted in the expression.[2] This is called the generalized associative law. For instance, a product of four elements may be written in five possible ways:

1. ((ab)c)d
2. (ab)(cd)
3. (a(bc))d
4. a((bc)d)
5. a(b(cd))

If the product operation is associative, the generalized associative law says that all these formulas will yield the same result, making the parenthesis unnecessary. Thus "the" product can be written unambiguously as

abcd.

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

## Examples

The addition of real numbers is associative.

Some examples of associative operations include the following.

• The concatenation 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).
$\left. \begin{matrix} (x+y)+z=x+(y+z)=x+y+z\quad \\ (x\,y)z=x(y\,z)=x\,y\,z\qquad\qquad\qquad\quad\ \ \, \end{matrix} \right\} \mbox{for all }x,y,z\in\mathbb{R}.$
Because of associativity, the grouping parentheses can be omitted without ambiguity.
• Addition and multiplication of complex numbers and quaternions is associative. Addition of octonions is also associative, but multiplication of octonions is non-associative.
$\left. \begin{matrix} \operatorname{gcd}(\operatorname{gcd}(x,y),z)= \operatorname{gcd}(x,\operatorname{gcd}(y,z))= \operatorname{gcd}(x,y,z)\ \quad \\ \operatorname{lcm}(\operatorname{lcm}(x,y),z)= \operatorname{lcm}(x,\operatorname{lcm}(y,z))= \operatorname{lcm}(x,y,z)\quad \end{matrix} \right\}\mbox{ for all }x,y,z\in\mathbb{Z}.$
$\left. \begin{matrix} (A\cap B)\cap C=A\cap(B\cap C)=A\cap B\cap C\quad \\ (A\cup B)\cup C=A\cup(B\cup C)=A\cup B\cup C\quad \end{matrix} \right\}\mbox{for all sets }A,B,C.$
• If M is some set and S denotes the set of all functions from M to M, then the operation of functional composition on S 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 sets M, N, P and Q, with h: M to N, g: N to P, and f: P to 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.
• Consider a set with three elements, A, B, and C. 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 mapping is not commutative.

## Propositional logic

### Rule of replacement

In standard truth-functional propositional logic, association,[3][4] or associativity[5] are two valid rules of replacement. The rules allow one to move parentheses in logical expressions in logical proofs. The rules are:

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

and

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

where "$\Leftrightarrow$" is a metalogical symbol representing "can be replaced in a proof with."

### Truth functional connectives

Associativity is a property of some logical connectives of truth-functional propositional logic. The following logical equivalences demonstrate that associativity is a property of particular connectives. The following are truth-functional tautologies.

Associativity of disjunction:

$(P \or (Q \or R)) \leftrightarrow ((P \or Q) \or R)$
$((P \or Q) \or R) \leftrightarrow (P \or (Q \or R))$

Associativity of conjunction:

$((P \and Q) \and R) \leftrightarrow (P \and (Q \and R))$
$(P \and (Q \and R)) \leftrightarrow ((P \and Q) \and R)$

Associativity of equivalence:

$((P \leftrightarrow Q) \leftrightarrow R) \leftrightarrow (P \leftrightarrow (Q \leftrightarrow R))$
$(P \leftrightarrow (Q \leftrightarrow R)) \leftrightarrow ((P \leftrightarrow Q) \leftrightarrow R)$

## Non-associativity

A binary operation $*$ on a set S that does not satisfy the associative law is called non-associative. Symbolically,

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

For such an operation the order of evaluation does matter. For example:

$(5-3)-2 \, \ne \, 5-(3-2)$
$(4/2)/2 \, \ne \, 4/(2/2)$
$2^{(1^2)} \, \ne \, (2^1)^2$

Also note that infinite sums are not generally associative, for example:

$(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+\dots \, = \, 0$

whereas

$1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+\dots \, = \, 1$

The study of non-associative structures arises from reasons somewhat different from the mainstream of classical algebra. One area within non-associative algebra that has grown very large is that of Lie algebras. There the associative law is replaced by the Jacobi identity. Lie algebras abstract the essential nature of infinitesimal transformations, and have become ubiquitous in mathematics.

There are other specific types of non-associative structures that have been studied in depth. They tend to come from some specific applications. Some of these arise in combinatorial mathematics. Other examples: Quasigroup, Quasifield, Nonassociative ring.

### Nonassociativity of floating point calculation

In mathematics, addition and multiplication of real numbers is associative. By contrast, in computer science, the addition and multiplication of floating point numbers is not associative, as rounding errors are introduced when dissimilar-sized values are joined together.[6]

To illustrate this, consider a floating point representation with a 4-bit mantissa:
(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 a 24 or 53 bits of mantissa,[7] this is an important source of rounding error, and approaches such as the Kahan Summation Algorithm are ways to minimise the errors. It can be especially problematic in parallel computing.[8] [9]

### Notation for non-associative operations

In general, parentheses must be used to indicate the order of evaluation if a non-associative operation appears more than once in an expression. However, mathematicians agree on a particular order of evaluation for several common non-associative operations. This is simply a notational convention to avoid parentheses.

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

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

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

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

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

• Subtraction and division of real numbers:
$x-y-z=(x-y)-z\qquad\mbox{for all }x,y,z\in\mathbb{R};$
$x/y/z=(x/y)/z\qquad\qquad\quad\mbox{for all }x,y,z\in\mathbb{R}\mbox{ with }y\ne0,z\ne0.$
• Function application:
$(f \, x \, y) = ((f \, x) \, y)$
This notation can be motivated by the currying isomorphism.

Right-associative operations include the following:

$x^{y^z}=x^{(y^z)}.\,$
The reason exponentiation is right-associative is that a repeated left-associative exponentiation operation would be less useful. Multiple appearances could (and would) be rewritten with multiplication:
$(x^y)^z=x^{(yz)}.\,$
$\mathbb{Z} \rarr \mathbb{Z} \rarr \mathbb{Z} = \mathbb{Z} \rarr (\mathbb{Z} \rarr \mathbb{Z})$
$x \mapsto y \mapsto x - y = x \mapsto (y \mapsto x - y)$
Using right-associative notation for these operations can be motivated by the Curry-Howard correspondence and by the currying isomorphism.

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

$\vec a \times (\vec b \times \vec c) \neq (\vec a \times \vec b ) \times \vec c \qquad \mbox{ for some } \vec a,\vec b,\vec c \in \mathbb{R}^3$
• Taking the pairwise average of real numbers:
${(x+y)/2+z\over2}\ne{x+(y+z)/2\over2} \qquad \mbox{for all }x,y,z\in\mathbb{R} \mbox{ with }x\ne z.$
• Taking the relative complement of sets $(A\backslash B)\backslash C$ is not the same as $A\backslash (B\backslash C)$. (Compare material nonimplication in logic.)

2. ^ Durbin, John R. (1992). Modern Algebra: an Introduction (3rd ed.). New York: Wiley. p. 78. ISBN 0-471-51001-7. "If $a_1, a_2, \dots, a_n \,\, (n \ge 2)$ are elements of a set with an associative operation, then the product $a_1 a_2 \dots a_n$ is unambiguous; this is, the same element will be obtained regardless of how parentheses are inserted in the product"