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Unit 17: SETS AND FUNCTIONS
Explanation For Exercise 17.3
PROPERTIES OF UNION AND INTERSECTION
Give formal proof of the following fundamental properties of union and intersection of two or three sets.Commutative Property Of Union:
We know that for any two sets A and B,
AUB = BUA
This property is called commutative property of union.
Proof:
L.H.S = AUB
= {x | x ∈ A or x ∈ B} (By definition of union)
= {x | x ∈ B or x ∈ A} ∵ Order of elements in a set is not preserved
= BUA (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ AUB = BUA
Hence proved
Commutative Property Of Intersection:
We know that for any two sets A and B,
A∩B = B∩A
This property is called commutative property of intersection.
Proof:
L.H.S = A∩B
= {x | x ∈ A and x ∈ B} (By definition of intersection)
= {x | x ∈ B and x ∈ A} ∵ Order of elements in a set is not preserved
= B∩A (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ A∩B = B∩A
Hence proved
Associative Property Of Union:
We are already familiar with associative property of union which is as follows.
For any three sets A, B and C,
AU(BUC) = (AUB)UC
Proof:
L.H.S = AU(BUC)
= {x | x ∈ A or x ∈ BUC} (By definition of union)
= {x | x ∈ A or x ∈ B or x ∈ C}
= {x | x ∈ AUB or x ∈ C} (By definition of union)
= (AUB)UC (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ AU(BUC) = (AUB)UC
Hence proved
Associative Property Of Intersection:
We already know the associative property of intersection which states that:
For any three sets A, B and C,
A∩(B∩C) = (A∩B)∩C
Proof:
L.H.S = A∩(B∩C) = (A∩B)∩C
= {x | x ∈ A and x ∈ (B∩C)} (By definition of intersection)
= {x | x ∈ A and x ∈ B and x ∈ C}
= {x | x ∈ A∩B and x ∈ C} (By definition of intersection)
= (A∩B)∩C (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ A∩(B∩C) = (A∩B)∩C
Hence proved
Distributive Property Of Union Over Intersection:
We have already studied distributive property of union over intersection which is as follows.
For any three sets A, B and C,
AU(B∩C) = (AUB)∩(AUC)
Proof:
L.H.S = AU(B∩C)
= {x | x ∈ A or x ∈ (B∩C)} (By definition of union)
= {x | x ∈ A or (x ∈ B and x ∈ C)} (By definition of intersection)
= {x | (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)}
= {x | x ∈ AUB and x ∈ AUC} (By definition of union)
= (AUB)∩(AUC) (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ AU(B∩C)=(AUB)∩(AUC)
Hence proved
Distributive Property Of Intersection Over Union:
We are already familiar with distributive property of intersection over union which is as follow.
For any three sets A, B and C,
A∩(BUC)=(A∩B)U(A∩C)
Proof:
L.H.S = A∩(BUC)
= {x | x ∈ A and x ∈ (BUC)} (By definition of intersection)
= {x | x ∈ A and (x ∈ B or x ∈ C)} (By definition of union)
= {x | (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)}
= {x | x ∈ A∩B and x ∈ A∩C} (By definition of intersection)
= (A∩B)U(A∩C) (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ A∩(BUC)=(A∩B)U(A∩C)
Hence proved
De Morgan's Laws:
There are two De Morgan's Laws which are as under.
For any two sets A and B
(i) (AUB)' = A'∩B'
(ii) (A∩B)' = A'UB'
Proof:
(i) (AUB)' = A'∩B'
L.H.S = (AUB)'
= {x | x ∈ U and x ∉ AUB} (By definition of complement)
= {x | x ∈ U and (x ∉ A and x ∉ B)}
= {x | (x ∈ U and x ∉ A) and (x ∈ U and x ∉ B)}
= {x | x ∈ A' and x ∈ B'} (By definition of complement)
= A'∩B' (By definition of intersection)
= R.H.S
∵ L.H.S = R.H.S
∴ (AUB)' = A'∩B'
Hence proved.
(ii) (A∩B)' = A'UB'
Proof:
L.H.S = (AUB)'
= {x | x ∈ U and x ∉ A∩B} (By definition of complement)
= {x | x ∈ U and (x ∉ A or x ∉ B)}
= {x | (x ∈ U and x ∉ A) or (x ∈ U and x ∉ B)}
= {x | x ∈ A' and x ∈ B'} (By definition of complement)
= A'UB' (By definition of union)
= R.H.S
∵ L.H.S = R.H.S
∴ (A∩B)' = A'UB''
Hence proved.
VERIFY THE FUNDAMENTAL PROPERTIES OF GIVEN SETS:
Let us verify the fundamental properties with the help of the following examples.Example 1:
If A = {1,2,3,4,6,12} and B 4,6,8,9,10,12) then verify commutative property of union and intersection.
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