Question

do it fast with explaination

Answer 1

E and F are any two sets.
Fourth option is not correct.
Let E={1,2,3,4}
and F={3,4,5,6,7,8}
EΠF={1,2,3,4}Π{3,4,5,6,7,8}
={3,4}
E(union)F={1,2,3,4}(union){3,4,5,6,7,8}
={1,2,3,4,5,6,7,8}
If U is a universal set and F is a subset of U,then the çomplement of F is the set which contains those elements of U ,which are not contained in F.
Therefore complement of F={1,2}
i)
E-F={1,2,3,4}-{3,4,5,6,7,8}
E-F={1,2}............................................1
E-(EΠF)={1,2,3,4}-{3,4}
E-(EΠF)={1,2}....................................2
By eq1 and eq2 we get,
E-F=E-(EΠF)
Hence above equation is true for any two sets E and F.
ii)
(EunionF)-F={1,2,3,4,5,6,7,8}-{3,4,5,6,7,8}
(EunionF)-F={1,2}.............................3
E-(EΠF)={1,2,3,4}-{3,4}
E-(EΠF)={1,2}.....................................4
By eq3 and eq4 we get,
(EunionF)-F=E-(EΠF)
Hence above equation is true for any two sets E and F.
iii)
E-(EΠF)={1,2}.....................................5
EΠFc={1,2,3,4}Π{1,2}
EΠFc={1,2}.........................................6
By eq5 and eq6 we get,
E-(EΠF)=EΠFc
Hence above equation is true for any two sets E and F.
iv)
(EunionF)-F={1,2}..............................7
(E-F)union(EΠF)={1,2}union{3,4}
(E-F)union(EΠF)={1,2,3,4)...............8
By eq7 and eq8 we get,
(EunionF)-F≠(E-F)union(EΠF)
Hence above equation is not true.

(E-F)union(EΠF)=E

Answer 2

Answer:

)

Hence above equation is not true.

option c is correct :(E-F)union(EΠF)=E