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)
Relation
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}