If A and B are two non-empty sets, then the set of all ordered pairs (a, b) such that a ∈ A, b ∈ B is called the Cartesian Product of A and B, and is denoted by A×B . Thus, A×B = {(a,b) |a ∈ A,b ∈ B}.

**Cartesian Product**** **

Let us consider the
following two sets.

*A *is the set of 3
vegetables and *B *is the set of 4 fruits. That is,

*A *= {carrot, brinjal,
ladies finger} and *B *= {apple, orange, grapes, strawberry} What are the
possible ways of choosing a vegetable with a fruit? (Fig.1.2)

We can select them in
12 distinct pairs as given below.

(*c*, *a*),
(*c*, *o*), (*c*, *g*), (*c*, *s*), (*b*, *a*),
(*b*, *o*), (*b*, *g*), (*b*, *s*), (*l*, *a*),
(*l*, *o*), (*l*,*g*), (*l*, *s*)

This collection
represents the cartesian product of the set of vegetables and set of fruits.

If A and B are two
non-empty sets, then the set of all ordered pairs (*a*, *b*) such that *a *∈ *A*, *b *∈ *B *is
called the Cartesian
Product of A and B, and is denoted by *A*×*B *. Thus, *A*×*B *= {(*a*,*b*)
|*a *∈ *A*,*b *∈ *B*}.

·
*A *× *B *is the set of all possible ordered pairs between the
elements of *A *and *B *such that the first coordinate is an element
of *A *and the second coordinate is an element of *B*.

·
*B *× *A *is the set of all possible ordered pairs between the
elements of *A *and *B *such that the first coordinate is an element
of *B *and the second coordinate is an element of *A*.

·
If *a *= *b*, then (*a*, *b*) = (*b*, *a*).

·
The “cartesian product” is also referred as “cross product”.

Let A = {1, 2, 3} and B = {a, b}. Write *A* ×*B* and *B* ×*A* ?

*A *×*B *=
{1,2,3}×{*a*,*b*}* *=* *{(1,* a *),(1,* b *),(2,* a *),(2,* b *),(3,* a *),(3,* b *)}* *(as shown in Fig.1.3)

*B*×*A*
= {*a*,*b*}
× {1,2,3} = {(*a*,1), (*a*,2), (*a*,3),(*b*,1), (*b*,2), (*b*,3)} (as shown in Fig.1.3)

**Recall of standard infinite sets**

Natural Numbers **N** = {1, 2, 3, 4…} ;

Whole Numbers **W** = {0,1,2,3, ...};

Integers **Z** ={..., –2,–1,0,1,2,
...} ;

Rational Numbers

Real Numbers **R** = **Q** ∪ **Q****’** , where **Q’** is the set of all
irrational numbers.

For example, let
A be the set of numbers in the interval [3, 5] and B be the set of numbers in the
interval [2,3]. Then the
Cartesian product A×B corresponds to the
rectangular region shown in the Fig. 1.4. It consists of all points (*x, y*) within the region

**Progress check**

1. For any two
non-empty sets *A *and B, *A*×*B *is called as ______.

2. If *n*(*A*×
*B*) = 20 and *n*(*A*) = 5 then *n*(*B*) is ______.

3. If *A *= {-1,1} and *B *=
{−1,1} then geometrically describe the set of points of *A*×*B *.

4. If A, B are the
line segments given by the intervals (–4, 3) and (–2, 3) respectively,
represent the cartesian product of A and B.

**Note:**

The set of all points
in the cartesian plane can be viewed as the set of all ordered pairs (*x*,
*y*) where *x*, *y *are real numbers. In fact, **ℝ**×**ℝ**** **is the set of all
points which we call as the cartesian plane.

**Example 1.1** If *A* =
{1,3,5} and *B* = {2,3} then (i) find *A* ×*B* and *B* ×*A*.

(ii) Is *A*
× *B* = *B* ×*A*? If not why? (iii) Show that *n*(*A*×*B*) = *n*(*B*×*A*)
= *n*(*A*)× *n*(*B*)

** Solution **Given that

(i)* A*×*B
*= {1,3,5} × {2,3} = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)} ...(1)

*B× A* = {2,3} × {1,3,5} = {(2,1),
(2,3), (2,5), (3,1), (3,3), (3,5)} ...(2)

(ii) From
(1) and (2) we conclude that *A* × *B* ≠ *B* ×*A* as (1, 2) ≠ (2, 1) and (1, 3) ≠ (3, 1), etc.

(iii)* n*(*A*)=3;* n *(*B*)
= 2.

From (1)
and (2) we observe that, *n (A×B) = n
(B×A)* = 6;

we see
that, *n* (*A*) ×*n* (*B*) = 3 × 2 = 6 and *n* (*B*) × *n* (*A*)
= 2×3 = 6

Hence, *n *(*A*×*B*) =*n*
(*B*×*A*) = *n*(*A*) × *n*
(*B*) = 6.

Thus, *n *(*A*×*B*) =*n*
(*B*×*A*) = *n*(*A*) × *n*
(*B*).

**Example 1.2 **If *A*×*B
=* {(3,2), (3,4), (5,2), (5,4)} then find *A* and *B*.

*Solution**A×B* ={(3,2), (3,4), (5,2), (5,4)}

We have *A* = {set of all first coordinates of
elements of *A × B* }. Therefore, A =
{3,5}

B = {set
of all second coordinates of elements of A *×
*B }. Therefore, B = {2,4}

Thus *A* = {3,5} and *B* = {2,4}.

**Example 1.3** Let *A* = {*x* ∈ **N
**| 1 < *x* <4} , *B* = {*x* ∈ **W**| 0 ≤ *x* < 2} and *C* = {*x* ∈ **N ***|
x*<3} .

Then verify that

(i) *A* *×* (*B* U *C*) = (*A* *×*
*B*) U (*A* *×* *C*)

(ii) *A* *×* (*B* Ո *C*) = (*A* × *B*) + (*A *×*
C*)

*Solution**A *=* *{*x *∈ **N**| 1 < *x* < 4} = {2, 3} , *B* = {*x* ∈**W** | 0 ≤ *x* < 2} = {0, 1},

*C*
= {*x*
∈ **N** | *x* < 3}= {1,2}

(i)* A *×* *(*B *∪*C*)* *=* *(*A
× B *)* *,* *(*A × C*)

*B *∪
*C
*=* *{0, 1}* *∪* *{1, 2}* *=* *{0, 1, 2}

*A *×* *(*B *∪*C*) = {2,
3} × {0, 1, 2} = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} ...(1)

*A *× *B *= {2, 3}
× {0, 1} = {(2,0),(2,1),(3,0),(3,1)}

*A *× *C *= {2, 3}
× {1, 2} = {(2, 1), (2, 2), (3, 1), (3, 2)}

(*A* × *B*) ∪ (*A* ×*C*) = {(2, 0), (2, 1), (3, 0), (3,
1)} , {(2, 1), (2, 2), (3, 1), (3, 2)}

= {(2, 0), (2, 1), (2, 2), (3,
0), (3, 1), (3, 2)} ...(2)

From (1) and (2), *A* × (*B* ∪*C* ) = (*A*×*B* ) ∪ (*A*×*C*) is verified.

(ii)* A *×* *(*B *∩*C*)* *=* *(*A *×*B*)* *∩* *(*A *×*C*)

*B *Ո
*C *= {0, 1} ∩ {1, 2} = {1}

*A *×* *(*B *∩*C*) = {2, 3} ×{1} = {(2,1),(3,1)} ... (3)

*A *× *B*= {2, 3} ×{0, 1} = {(2, 0),(2, 1),(3, 0),(3, 1)}

*A *× *C *= {2, 3} ×{1, 2} = {(2, 1),(2, 2),(3, 1),(3, 2)}

(*A* × *B*) ∩ (*A* ×*C*) = {(2, 0),(2, 1),(3, 0),(3, 1)}
∩ {(2, 1),(2,
2),(3, 1),(3, 2)}

=
{(2,
1),(3, 1)} ... (4)

From (3) and (4), *A* × (*B* ∩*C* ) = (*A*×*B* ) ∩ (*A*×*C*) is verified.

The above two verified
properties are called distributive property of cartesian product over union and
intersection respectively. In fact, for any three sets *A*, *B*, *C
*we have

(i) *A* × (*B* ∪*C*) = (*A* ×*B*) ∪ (*A* ×*C*) (ii) *A* × (*B* ∩*C* ) = (A×B) ∩ (A×*C*).

If A, B,
C are three non-empty sets then the cartesian
product of three sets is the set of
all possible ordered triplets given by

A× B ×C= {(a,b,c) for
all a ∈
A,b ∈
B,c ∈
C }

Let *A *= {0,1}, *B
*= {0,1}, *C *= {0,1}

A×B = {0,1}×{0,1}
= {(0, 0),(0,1),(1, 0),(1,1)}

Representing A×B in the xy - plane we get a picture shown in Fig. 1.5.

(*A*×*B*)×*C*=
{(0, 0),(0,1),(1, 0),(1,1)} ×{0,1}

= {(0, 0, 0),(0,
0,1),(0,1, 0),(0,1,1),(1, 0, 0),(1, 0,1)(1,1, 0),(1,1,1)}

Representing *A×B ×C* in the *xyz* - plane we get a picture as shown in Fig. 1.6

Thus, A×B
represent vertices of a square in two dimensions and A×B ×C represent
vertices of a cube in three dimensions.

**NOTES**

In general,
cartesian product of two non-empty sets provides a shape in two dimensions and
cartesian product of three non-empty sets provide an object in three
dimensions.

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