What is Cartesian Product of Sets – Definition and Example

Here you will learn what is cartesian product of sets and what is relation and inverse relation with example.

Let’s begin –

Cartesian Product of Sets

The cartesian product of two sets A, B is a non-void set of all ordered pair (a,b),

where a \(\in\) A and b \(\in\) B. This is denoted by A \(\times\) B.

\(\therefore\)   A \(\times\) B = {(a,b) \(\forall\) a \(\in\) A and b \(\in\) B}

e.g.   A = {1,2}, B = {a,b}

A \(\times\) B = {(1,a), (1,b), (2,a), (2,b)}

Note :

(i)  A \(\times\) B \(\ne\) B \(\times\) A    (Non-commutative)

(ii) n(A \(\times\) B) = n(A)n(B) and n(P(A \(\times\) B)) = \(2^{n(A)n(B)}\)

(iii) A = \(\phi\) and B = \(\phi\) \(\iff\) A \(\times\) B = \(\phi\)

(iv) If A and B are two non-empty sets having n elements in common, then (A \(\times\) B) and (B \(\times\) A) have \(n^2\) elements in common

(v) A \(\times\) (B \(\cup\) C) = (A \(\times\) B) \(\cup\) (A \(\times\) C)

(vi) A \(\times\) (B \(\cap\) C) = (A \(\times\) B) \(\cap\) (A \(\times\) C)

(vii) A \(\times\) (B – C) = (A \(\times\) B) – (A \(\times\) C)


Every non-zero subset of A \(\times\) B defined a relation from set A to set B.<br>If R is relation from A \(\rightarrow\) B

R : {(a,b) | (a,b) \(\in\) A \(\times\) B and a R b}

Let A and B be two non empty sets and R : A \(\rightarrow\) B be a relation such that R : {(a,b) | (a,b) \(\in\) R a \(\in\) A and b \(\in\) B}

(i) ‘b’ is called image of ‘a’ under R.

(ii) ‘a’ is called pre-image of ‘b’ under R.

(iii) Domain of R : Collection of all elements of A which has a image in B.

(iv) Range of R : Collection of all elements of B which has a pre-image in A.

Note :

(1) It is not necessary that each and every element of set A has a image in set B and each and every element of set B has preimage in set A.

(2) Elements of set A having image in B is not necessarily unique.

(3) Basically relation is the number of subsets of A \(\times\) B

Number of relations = no. of ways of selecting a non-zero subset of A \(\times\) B

= \(^{mn}C_1\)+ \(^{mn}C_2\) + …….. + \(^{mn}C_{mn}\) = \(2^{mn} – 1\)

Total number of relation = \(2^{mn}\)(including void relation)

Example : If A = {1, 3, 5, 7}, B = {2, 4, 6, 8}
Relation is aRb \(\implies\) a > b, a \(\in\) A, a \(\in\) B

Solution : R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}
Domain = {3, 5, 7}
Range = {2, 4, 6}

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