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Carry Minati 5 years, 6 months ago
Posted by Unique #Seraj Ali ??? 5 years, 6 months ago
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Royal Thakur 5 years, 6 months ago
Yogita Ingle 5 years, 6 months ago
If U is a universal set and A be any subset of U then the complement of A is the set of all members of the universal set U which are not the elements of A.
A′ = {x : x ∈ U and x ∉ A}
Alternatively it can be said that the difference of the universal set U and the subset A gives us the complement of set A.
Unique #Seraj Ali ??? 5 years, 6 months ago
Posted by Unique #Seraj Ali ??? 5 years, 6 months ago
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Royal Thakur 5 years, 6 months ago
Royal Thakur 5 years, 6 months ago
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Royal Thakur 5 years, 6 months ago
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Meghna Thapar 5 years, 6 months ago
In propositional logic and boolean algebra, De Morgan's laws are a pair of transformation rules that are both valid rules of inference. The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation. DeMorgan's first theorem states that two (or more) variables NOR´ed together is the same as the two variables inverted (Complement) and AND´ed, while the second theorem states that two (or more) variables NAND´ed together is the same as the two terms inverted (Complement) and OR´ed.
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Shilpi Kumari 5 years, 6 months ago
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Posted by Seraj Ali ??? #Srhmk 5 years, 6 months ago
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Yogita Ingle 5 years, 6 months ago
A pair of sets which does not have any common element are called disjoint sets. For example, set A={2,3} and set B={4,5} are disjoint sets. But set C={3,4,5} and {3,6,7} are not disjoint as both the sets C and D are having 3 as a common element.
Posted by Seraj Ali ??? #Srhmk 5 years, 6 months ago
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Gaurav Seth 5 years, 6 months ago
The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan’s laws.
For any two finite sets A and B;
(i) (A U B)' = A' ∩ B' (which is a De Morgan's law of union).
(ii) (A ∩ B)' = A' U B' (which is a De Morgan's law of intersection).
Proof of De Morgan’s law: (A U B)' = A' ∩ B'
Let P = (A U B)' and Q = A' ∩ B'
Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'
⇒ x ∉ (A U B)
⇒ x ∉ A and x ∉ B
⇒ x ∈ A' and x ∈ B'
⇒ x ∈ A' ∩ B'
⇒ x ∈ Q
Therefore, P ⊂ Q …………….. (i)
Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B'
⇒ y ∈ A' and y ∈ B'
⇒ y ∉ A and y ∉ B
⇒ y ∉ (A U B)
⇒ y ∈ (A U B)'
⇒ y ∈ P
Therefore, Q ⊂ P …………….. (ii)
Now combine (i) and (ii) we get; P = Q i.e. (A U B)' = A' ∩ B'
Proof of De Morgan’s law: (A ∩ B)' = A' U B'
Let M = (A ∩ B)' and N = A' U B'
Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'
⇒ x ∉ (A ∩ B)
⇒ x ∉ A or x ∉ B
⇒ x ∈ A' or x ∈ B'
⇒ x ∈ A' U B'
⇒ x ∈ N
Therefore, M ⊂ N …………….. (i)
Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'
⇒ y ∈ A' or y ∈ B'
⇒ y ∉ A or y ∉ B
⇒ y ∉ (A ∩ B)
⇒ y ∈ (A ∩ B)'
⇒ y ∈ M
Therefore, N ⊂ M …………….. (ii)
Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)' = A' U B'
Posted by Seraj Ali ??? #Srhmk 5 years, 6 months ago
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Gaurav Seth 5 years, 6 months ago
Since A is subset of B - Each element of the set A is in the set B
So therefore A union B = B
Posted by Pritam Saha 5 years, 6 months ago
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Khushboo Tiwari 5 years, 6 months ago
Posted by Shilpi Kumari 5 years, 6 months ago
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Shilpi Kumari 5 years, 6 months ago
Yogita Ingle 5 years, 6 months ago
Let x = 10's digit of a two digit number
Let y = units digit
Let z = the resulting equal digits
:Write an equation for each statement:
the sum of a two digit number is 10.
x + y = 10
or
y = (10-x)
if 18 be subtracted from it, the digits in the resulting number will be equal.
10x + y - 18 = 10z + z
10x + y = 11z + 18
Substitute (10-x) for y
10x + (10-x) = 11z + 18
10x - x = 11z + 18 - 10
9x = 11z + 8
The single digit integer values for x and z are limited to x=7 and z=5 in this equation:
x = 7, y = 3, z = 5
The number is 73
Check in the statement:
"if 18 be subtracted from it, the digits in the resulting number will be equal."
73 - 18 = 55
Posted by Gunjan Goyal 5 years, 6 months ago
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@S(Raj)Àĺì ???? 5 years, 6 months ago
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