Exercise 1.3
1. If A∩B = ɸ, prove
that:
(a) A $ \subseteq
{\rm{\bar B}}$
Solution:
Let x be an element of A.
Then x ԑ A à x ∉ B
[A ∩ B = ɸ]
àx ԑ ${\rm{\bar B}}$
So, A $ \subseteq {\rm{\bar B}}$
(b) B ∩
${\rm{\bar A}}$
Solution:
B ∩ ${\rm{\bar A}}$ = {x:x ԑ B and x ԑ ${\rm{\bar A}}$}
= {x:x ԑ B and x∉ A} = {x:x ԑ B} [x∉
A]
(c) A U ${\rm{\bar
B}}$
Solution:
A U ${\rm{\bar B}}$ = {x:x ԑ A or x ԑ ${\rm{\bar B}}$}
= {x:x ԑ ${\rm{\bar
B}}$}
[A∩B = ɸ]
= ${\rm{\bar B}}$
2) If A ⊆ B,Prove that:
(a) ${\rm{\bar B}}
\subseteq $${\rm{\bar A}}
Solution:
Let x be an element of ${\rm{\bar B}}$.
Then, x ԑ ${\rm{\bar B}}$à x ∉ B.
= x∉ A [A ⊆
B]
àx ԑ ${\rm{\bar A}}$.
So, ${\rm{\bar B}} \subseteq $${\rm{\bar A}}$
(b) A U B=B
Solution:
A U B = {x:x ԑ A or x ԑ B}
={x:x ԑ
B} [A ⊆
B]
= B
(c) A ∩ B =A
Solution:
A ∩ B = {x:x ԑ A and x ԑ B}
={x:x ԑ
A} [A⊆
B]
= A
3. Prove that:
(a) B – A = B ∩
${\rm{\bar A}}$
Solution:
B – A = {x:x ԑ B and x∉ A}
= {x:x ԑ B and x ԑ ${\rm{\bar A}}$} = {x:x ԑ B ∩ ${\rm{\bar
A}}$}
So, B – A = B ∩ ${\rm{\bar A}}$
(b) A – ${\rm{\bar
B}}$=A∩B
Solution:
A – ${\rm{\bar B}}$ = {x:x ԑ A and x∉${\rm{\bar
B}}$}
= {x:x ԑ A and x ԑ B} = {x:x ԑ A∩B} = A∩B
So, B – A = B ∩ ${\rm{\bar A}}$
(c)A-B=${\rm{\bar
B}}$ – ${\rm{\bar A}}$
Solution:
A – B = {x:x ԑ A and x∉ B}
= {x:x∉${\rm{\bar A}}$ and x ԑ ${\rm{\bar
B}}$} = {x:x ԑ{ ${\rm{\bar B}}$ – ${\rm{\bar A}}$} = ${\rm{\bar B}}$ –
${\rm{\bar A}}$
4) If A,B and C are
the subsets of the universal set U, Prove that:
(a) AU(B∩C) = (A U B)
∩ (A U C)
Solution:
AU(B∩C) = {x:x ԑ A or x ԑ (B∩C)}
= {x:x ԑ A or (x ԑ B and x ԑ C)}
= {x:(x ԑ A or x ԑ B) and (x ԑ A or, x ԑ C)}
= {x:x ԑ (A U B) and x ԑ (A U C)}
= (A U B) ∩ (A U C)
(b) A∩(BUC) = (A ∩
B)U(A∩C)
A∩(BUC) = {x:x ԑ A and x ԑ (BUC)}
= {x:x ԑ A and (x ԑ B or x ԑ C)}
= {x:(x ԑ A and x ԑ B) or (x ԑ A and x ԑ C)}
= {x:x ԑ (A ∩ B) or x ԑ (A ∩ C)}
= (A ∩ B)U(A∩C)
5) If A,B and C are
the subsets of the universal set U, Prove that:
a) (AUB)’=A’∩B’
Solution:
(AUB)’ = {x:x∉ (A U B)}
= {x:x∉ A and x∉ B} = {x:x ԑ
${\rm{\bar A}}$ and x ԑ ${\rm{\bar B}}$}
= {x:x ԑ (${\rm{\bar A}}$ ∩ ${\rm{\bar B}}$)} = ${\rm{\bar
A}}$ ∩ ${\rm{\bar B}}$
b) (A∩B)’=A’UB’
Solution:
(A∩B)’ = {x:x∉ (A ∩
B)}
= {x:x∉ A or x∉ B} = {x:x ԑ
${\rm{\bar A}}$ or x ԑ ${\rm{\bar B}}$}
= {x:x ԑ (${\rm{\bar A}}$ U ${\rm{\bar B}}$)} = ${\rm{\bar
A}}$ U ${\rm{\bar B}}$
c) A – (B U C)= (A –
B) ∩ (A – C)=(A-B)-C
Solution:
A – (B U C) = {x;x ԑ A and x∉ (B U C)}
= {x:x ԑ A and (x ∉ B and x ∉ C)}
= x:(x ԑ A and x∉ B) and (x ԑ A and x ∉
C)
= {x:x ԑ (A – B) and x ԑ (A – C)}
So, A – (B U C) = (A – B) ∩ (A – C)
Again, A – (B U C) = {x:x ԑ A and x ∉ (B U C)}
= {x:x ԑ A and (x∉ B or x∉ C)}
= {x:(x ԑ A and x ∉B)and x ∉ C}
= {x:x ԑ (A – B) and x ∉ C}
= {x:x ԑ ((A – B) – C)} = (A – B) – C.
d) A – (B∩C)= (A – B)
U (A – C)
Solution:
A – (B∩C) = {x:x ԑ A and x ∉ (B ∩C)}
= {x:x ԑ A and (x∉ B and x ∉ C)}
= {x:(x ԑ A and x∉ B) or (x ԑ A and x ∉
C)}
= {x:x ԑ (A – B) or x ԑ (A – C)}
= (A – B) U (A – C)