Solution : Let A be a set. Then, the relation \(I_A\) = {(a, a) : a \(\in\) A} on A is called the identity relation on A. In other words, then the relation \(I_A\) on A is called the identity relation if every element of A is related to itself only. Example : If A …
Solution : Let A be a set. Then, A \(\times\) A \(\subseteq\) A \(\times\) A and so it is a relation on A. This relation is called the universal relation on A. In other words, a relation R on a set is called universal relation, if each element of A is related to every element …
Solution : Let A be a set. Then, \(\phi\) \(\subseteq\) A \(\times\) A and so it is a relation on A. This relation is called the void or empty relation on set A. In other words, a relation R on a set A is called void or empty relation, if no element of A is …
Solution : Let A be a finite set containing n elements. Let 0 \(\le\) r \(\le\) n. Consider those subsets of A that have r elements each. We know that the number of ways in which r elements can be chosen out of n elements is \(^nC_r\). Therefore, the number of subsets of A having …
Prove that the total number of subsets of a finite set containing n elements is \(2^n\). Read More »
Solution : Let A be any set and \(\phi\) be the empty set. In order to show that \(\phi\) \(\subseteq\) A, we must show that every element of \(\phi\) is an element of A also. But, \(\phi\) contains no element. So, every element of \(\phi\) is in A. Hence, \(\phi\) is the subset of A.
Solution : A set consisting of a single element is called a singleton set. Example : The set {5} is a singleton set. Example : The set {x : x \(\in\) N and \(x^2\) = 9} is a singleton set equal to {3}. Note : The cardinal number of a singleton set is 1.
Solution : The cardinality of a set is the number of elements in a set. For example : Let A be a set : A = {1, 2, 4, 6} Set A contains 4 elements. Therefore, Cardinality of set is 4.
Solution : Given, n(A) = 2 and n(B) = 4 \(\therefore\) n(A\(\times\)B) = 8 The number of subsets of (A\(times\)B) having 3 or more elements = \(^8C_3 + {^8C_4} + ….. + {^8C_8}\) = \(2^8 – {^8C_0} – {^8C_1} – {^8C_2}\) = 256 – 1 – 8 – 28 = 219 [\(\because\) \(2^n\) = …
Solution : Clearly P(A) = Power set of A = set of all subsets of A = {\(\phi\), {x}, {y}, {x,y}} Similar Questions Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is If aN = {ax : x \(\in\) …
Solution : 6N = {6, 12, 18, 24, 30, …..} 8N = {8, 16, 24, 32, ….} \(\therefore\) 6N \(\cap\) 8N = {24, 48, …..} = 24N Similar Questions If A = {x,y}, then the power set of A is If A = {2, 4} and B = {3, 4, 5} then (A \(\cap\) B) …
If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Read More »
