(i) A - B = A ∩ B'
A - B means everything in A expect for anything in A ∩ B.
Hence, the required term is proved.
(iii) A - B = A ⇔ A ∩ B - ϕ
In order to prove that A-B = A ⇔ A⋂B = Φ,
we will first prove that A-B = A ⇒ A⋂B = Φ and then we'll prove
that A⋂B = Φ ⇒ A-B = A
First part : Proving that A-B = A ⇒ A⋂B = Φ :-
Let x be an arbitrary element of A
So, x ∈ A
Since A = A-B, so x ∈ A-B
So, x ∈ A and x ∉ B
This means that for an arbitrary element of A, it is not an element of B
So, A and B are both disjoint sets
So, they have no element in common
So, A⋂B = Φ
So, A-B = A ⇒ A⋂B = Φ
Second part : Proving that A⋂B = Φ ⇒ A-B = A :-
A-B = A-(A⋂B)
Since A⋂B = Φ, so, A-B = A-Φ = A
So, A⋂B = Φ ⇒ A-B = A
So, A-B = A ⇔ A⋂B = Φ
Hence, proved
(v) (A−B)∩B = ϕ
The relative complement or set difference of sets A and B, denoted A–B, is the set of all elements in
A that are not in B. In set-builder notation,
A – B = {x ∈ U : x ∈ A \ and \ x ∉ B} = A ∩ B'.
So, using set laws, we get
(A−B)∩B = (A∩B′)∩B = A∩(B′∩B) = A∩ϕ = ϕ
as ϕ = empty set, so there is nothing common in between A and ϕ
(vi) (A-B) ∪ (B-A) = (A∪B) - (A ∪ B)
( A - B ) U ( B - A ) gives us what is called as the symmetric difference of the sets A and B. Actually, this is also equal to the quantity (A ∪ B)- (A ∩ B)
For example, let us take the sets A and B as:-
A = {1,2,3,4,5} and B = {3,4,5,6}
Then, A-B= {1,2} and B-A = {6}
And (A-B) U (B-A) = {1,2,6}
Also,
A ∪ B = {1,2,3,4,5,6} and
A ∩ B = {3,4,5}
And (A∪B)-(A ∩ B) = {1,2,6}, which is same as (A-B) ∪ (B-A)…
So, in order to find (A-B) U (B-A) i.e. the symmetric difference of the sets A and B, it is sufficient to find (AUB)-(A ∩ B)
(vii) A - (B∪C) = (A-B) ∩ (A-C)
A − (B∪C)
= A ∩ (B∪C)′
∵ (A−B) = A∩B′
= A∩(B′∩C′)
∵ (B∪C)′ = (B′∩C′)
= (A∩B′) ∩ (A∩C′)
= (A−B) ∩ (A−C)
(viii) A - (B∩C) = (A-B) ∪ (A-C)
A-(B ⋂ C) = (A-B) ⋃ (A-C)
According to definition,
A-B={x| x ∈A and x∉B}
A-C={x |x ∈A and x∉ C}
the
(A-B) ⋃ (A-C) = {x|x∈A and x∉(B and C)
Let X = A and Y = (B ⋂ C)
the
X-Y = {x | x∈X and x∉Y}
x∉Y
x∉(B ⋂ C)
x∉(B and C)
A-(B ⋂ C) or X-Y = {x|x∈A and x∉(B and C)
Therefore,
A-(B ⋂ C) = (A-B)⋃(A-C)
Hence the proof
