Question

Let X={a, e, i, o, u}and Y={a, g, o, f}. Verify that n(A-B) = n(A) - n(AB)

Answer 1

Answer:

Appropriate Question :-

Let A = {a,e,i,o,u} and B = {a,g,o,f} verify that n(A-B) = n[ (A) - n (A n B) ]

\large\underline{\sf{Solution-}}

Solution−

Given that,

\rm :\longmapsto\:A \: = \: \{a,e,i,o,u \}:⟼A={a,e,i,o,u}

and

\rm :\longmapsto\:B \: = \: \{a, \: g, \: o, \: f\}:⟼B={a,g,o,f}

So,

\rm :\longmapsto\:A - B = \{e, \: i, \: u\}:⟼A−B={e,i,u}

So,

\rm :\longmapsto\:n(A - B) = 3:⟼n(A−B)=3

Now,

\rm :\longmapsto\:A \: = \: \{a,e,i,o,u \}:⟼A={a,e,i,o,u}

and

\rm :\longmapsto\:B \: = \: \{a, \: g, \: o, \: f\}:⟼B={a,g,o,f}

\rm :\longmapsto\:A \: \cap \: B = \{a, \: o\}:⟼A∩B={a,o}

So,

\rm :\longmapsto\:A - (A \: \cap \: B) = \{e, \: i, \: u\}:⟼A−(A∩B)={e,i,u}

\rm :\longmapsto\:n[ \: A - (A \: \cap \: B) \: ]= 3:⟼n[A−(A∩B)]=3

Hence, We concluded that,

\rm :\longmapsto\: \boxed{ \bf{n(A - B) \: = \: n[ \: A - (A \: \cap \: B) \: ]}}:⟼

n(A−B)=n[A−(A∩B)]

Additional Information :-

1. Commutative Law :-

\boxed{ \tt{ \: A\cup B = B\cup A \: }}

A∪B=B∪A

\boxed{ \tt{ \: A\cap B = B\cap A \: }}

A∩B=B∩A

2. Associative Law :-

\boxed{ \tt{ \: (A\cup B)\cup C = A\cup (B\cup C) \: \: }}

(A∪B)∪C=A∪(B∪C)

\boxed{ \tt{ \: (A\cap B)\cap C = A\cap (B\cap C) \: \: }}

(A∩B)∩C=A∩(B∩C)

3. Distributive Law :-

\boxed{ \tt{ \: A\cup (B\cap C) = (A\cup B)\cap (A\cup C) \: }}

A∪(B∩C)=(A∪B)∩(A∪C)

\boxed{ \tt{ \: A\cap (B\cup C) = (A\cap B)\cup (A\cap C) \: }}

A∩(B∪C)=(A∩B)∪(A∩C)

4. Complement Law :-

\boxed{ \tt{ \: \phi' \: = \: U}}

ϕ

′

=U

\boxed{ \tt{ \: U' \: = \: \phi \: }}

U

′

=ϕ