Here you will learn what is cartesian product of sets and what is relation and inverse relation with example.<\/p>\n
Let’s begin –<\/p>\n
The cartesian product of two sets A, B is a non-void set of all ordered pair (a,b),<\/p>\n
where a \\(\\in\\) A and b \\(\\in\\) B. This is denoted by A \\(\\times\\) B.<\/p>\n
\n\\(\\therefore\\) A \\(\\times\\) B = {(a,b) \\(\\forall\\) a \\(\\in\\) A and b \\(\\in\\) B}<\/p>\n<\/blockquote>\n
e.g. A = {1,2}, B = {a,b}<\/p>\n
A \\(\\times\\) B = {(1,a), (1,b), (2,a), (2,b)}<\/p>\n
Note :<\/strong><\/p>\n
(i) A \\(\\times\\) B \\(\\ne\\) B \\(\\times\\) A (Non-commutative)<\/p>\n
(ii) n(A \\(\\times\\) B) = n(A)n(B) and n(P(A \\(\\times\\) B)) = \\(2^{n(A)n(B)}\\)<\/p>\n
(iii) A = \\(\\phi\\) and B = \\(\\phi\\) \\(\\iff\\) A \\(\\times\\) B = \\(\\phi\\)<\/p>\n
(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<\/p>\n
(v) A \\(\\times\\) (B \\(\\cup\\) C) = (A \\(\\times\\) B) \\(\\cup\\) (A \\(\\times\\) C)<\/p>\n
(vi) A \\(\\times\\) (B \\(\\cap\\) C) = (A \\(\\times\\) B) \\(\\cap\\) (A \\(\\times\\) C)<\/p>\n
(vii) A \\(\\times\\) (B – C) = (A \\(\\times\\) B) – (A \\(\\times\\) C)<\/p>\n
Relation<\/h2>\n
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<\/p>\n
\nR : {(a,b) | (a,b) \\(\\in\\) A \\(\\times\\) B and a R b}<\/p>\n<\/blockquote>\n
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}<\/p>\n
(i) ‘b’ is called image of ‘a’ under R.<\/p>\n
(ii) ‘a’ is called pre-image of ‘b’ under R.<\/p>\n
(iii) Domain of R : Collection of all elements of A which has a image in B.<\/p>\n
(iv) Range of R : Collection of all elements of B which has a pre-image in A.<\/p>\n
Note :<\/strong><\/p>\n
(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.<\/p>\n
(2) Elements of set A having image in B is not necessarily unique.<\/p>\n
(3) Basically relation is the number of subsets of A \\(\\times\\) B<\/p>\n
Number of relations = no. of ways of selecting a non-zero subset of A \\(\\times\\) B<\/p>\n
= \\(^{mn}C_1\\)+ \\(^{mn}C_2\\) + …….. + \\(^{mn}C_{mn}\\) = \\(2^{mn} – 1\\)<\/p>\n
Total number of relation = \\(2^{mn}\\)(including void relation)<\/p>\n\n\n
Example : <\/span> If A = {1, 3, 5, 7}, B = {2, 4, 6, 8}
\n\t\t Relation is aRb \\(\\implies\\) a > b, a \\(\\in\\) A, a \\(\\in\\) B<\/p>\nSolution : <\/span>R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}
\n\t\t Domain = {3, 5, 7}
\n\t\t Range = {2, 4, 6}
<\/p>\n\n\n\n