if A={1,2,3} ,B={2,3,4,5} ,and C={2,4,6,8} verify that A∪B=(A-B)∪B
Answers
Step-by-step explanation:
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8}
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5}
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.So, A−C={1,2}
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.So, A−C={1,2} Intersection of two sets has the common elements of both the sets.
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.So, A−C={1,2} Intersection of two sets has the common elements of both the sets. ⇒(A−B)∩(A−C)={1} ..... (2)
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.So, A−C={1,2} Intersection of two sets has the common elements of both the sets. ⇒(A−B)∩(A−C)={1} ..... (2)From (1) and (2), we have
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.So, A−C={1,2} Intersection of two sets has the common elements of both the sets. ⇒(A−B)∩(A−C)={1} ..... (2)From (1) and (2), we haveA−(B∪C)=(A−B)∩(A−C)
Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}For the LHS:Union of two sets will have the elements of both sets.So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C)So, A−(B∪C)={1} ..... (1)For the RHS:A−B will have elements of A which are not in B.So, A−B={1,3,5} A−C will have elements of A which are not in C.So, A−C={1,2} Intersection of two sets has the common elements of both the sets. ⇒(A−B)∩(A−C)={1} ..... (2)From (1) and (2), we haveA−(B∪C)=(A−B)∩(A−C)Hence, the given expression is true.