# What do these terms mean: commutative, associative, distributive

I am reading a book, and I am trying to understand what the writer really mean by the following terms. I would like to understand what these words mean in relation to the examples.

In regular algebra, addition and multiplication are commutative: $$A + B = B + A$$ $$A \times B = B \times A$$ they are also associative: $$A + (B + C) = (A + B) + C$$ $$A \times (B \times C) = (A\times B) \times C$$ And multiplication is said to be distributive: $$A \times (B + C) = (A \times B) + (A \times C)$$

In Boolean algebra, the $+$ operator is distributive over the $\times$ operator: $$W + (B \times F) = (W + B)\times (W + F)$$ $$W = \text{white}\qquad B = \text{black}\qquad F = \text{female}$$

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What part of this don't you understand? –  Thomas Andrews Dec 28 '11 at 20:46
There is a typo, when you say multiplication is said to be distributive, in symbols that should be $A\times (B+C)=A\times B+A\times C$. –  André Nicolas Dec 28 '11 at 20:48
The meaning of the words are the formulas that come right after the colon in each case -- no more, no less. –  Henning Makholm Dec 28 '11 at 21:28
@HananN. Are you asking how the mathematical concepts relate to the meaning of the words in the English language? –  Bill Dubuque Dec 28 '11 at 21:46
Commutative: the result is the same if operands are commutated. $x + y$ is the same as $y + x.$ Assosicative: $x + y + z$ is the same as $(x + y) + z$ as $x + (y + z)$. It doesn't matter if you assigned an order (also: associate) to $(x+y)$ before $(y + z)$ and vice versa. Distributive: results are the same when you distribute. –  user2468 Dec 29 '11 at 0:04

You've quoted the definitions right there; there's not much more to understanding beyond just getting a lot of experience with manipulating expressions using those rules to gain some intuition of their utility.

For instance, you may figure out that if you have an operation * which is commutative but not associative, then the following manipulations are valid:

$a * (b * c) = a*(c*b) = (b*c) * a = (c*b)*a$

but the following is not:

$a*(b*c) = b*(a*c)$

The effect of associativity is to allow you to drop the parentheses altogether without worrying about ambiguity of the expression. So there are 5 different interpretations of the expression $a*b*c*d$, namely, $(a*b)*(c*d), ((a*b)*c)*d, (a*(b*c))*d, a*((b*c)*d)$ and $a*(b*(c*d))$. Associativity guarantees that these are all the same.

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Let us examine the associative property first.

1. Associative property of addition is how numbers are associated or
grouped together. We need a convention on how to compute 1+(2+3). The
associative property shows that 1+(2+3) yields same value as (1+2)+3.

For instance,

      ( *   **)   ***     *   (**    ***)
(1 +  2 ) +  3   =  1 + (2   +  3 )
( A + B ) +  C   =  A + (B   +  C )

2. Consider the following diagram where we have three 4s or four 3s which depends on if we group by rows or by columns.

 * * * *
* * * *
* * * *


Thus commutative principle for multiplication shows that the order does not matter and 3 X 4 = 4 x 3 or the general form:

(A)(B)$\equiv$(B)(A)

3. For distributive principle of multiplication consider the following array of counters to represent (3)(12):

 * * * * * * * * * *  | * *
* * * * * * * * * *  | * *
* * * * * * * * * *  | * *


The rectangular array is divided into two rectangles such as the
first one is (3)(10) and the next one is (3)(2). The multiplication has been broken into two parts ie distributed. The product (3)(10+2) can be replaced as (3)(10) + (3)(2).

The full statement is:

Multiplication is distributive with respect to addition.

Symbolically, it can written as: (A) (B+C) $\equiv$ (A) (B) + (A) (C)

Multiplication is also distributive with respect to subtraction.

Thus, (A) (B-C) $\equiv$ (A) (B) - (A) (C)

Reference:

Lay, L.Clark. The Study of Arithmetic. The Macmillan Company. 1966.

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