COS1501- Theoretical Computer Science
Xzero ≤ x < eight ⋂ x ∈ R, four ≤ x < 16 is the set:
a.Xfour ≤ x < 8
b.Xzero ≤ x ≤ 8
c. X ∈ Z, zero ≤ x ≤16
d.Four, five, 6, 7 - ANS-d.4, five, 6, 7
A right subset of a set constantly has much less elements than the set itself.
True or False - ANS-True
An element of a hard and fast can not be a fixed itself.
True or False - ANS-False
based on set A = 1, four, four, 1, five
Which one of the following is NOT a partition on A?
1. 1, 4, four, 1, five
2. 1, 4, 1, 5, 4
3. 4, 1, five, 1, four
four. 1, 4, 1, 5, 4 - ANS-2. 1, four, 1, 5, four
based on set A = 1, 4, four, 1, five
Which one of the following relations is NOT a valid relation on A?
1. (1, 4), (4, 1)
2. (1, 5), (four, four)
three. (four, 1, five), (1, 1), (four, 1)
four. (four, four) - ANS-2. (1, 5), (4, four)
based totally on set A = 1, 4, four, 1, 5
Which one of the following statements presents a right subset of A?
1. 1, four ,four
2. 1, 4
3. 1, 4, 4, 1, 5
4. 1, five - ANS-2. 1, 4
based totally on set A = 1, four, four, 1, five
Which one of the following statements presents one or greater elements of the set A?
,1. 1, four
2. 4
3. 1, 1, five
4. 4, 1, five - ANS-four. Four, 1, five
By the usage of logical equivalences and de Morgan's guidelines, we can show that the
statements¬p ⋁ q and (p ⋀ ¬q) →→ (¬p ⋁ q) are equivalent.
A.True
b.False - ANS-a.True
Commutative and Associative - ANS-Now we examine the regulations that we carried out:
((7 + 13) + 5) + 9
= nine + ((7 + 13) + 5) with the aid of commutativity
= nine + (5 + (7 + thirteen)) by means of commutativity
= nine + (5 + (13 + 7) ) by commutativity
= nine + ((five + 13) + 7) via associativity
= (9 + (5 + thirteen)) + 7 by associativity
= ((nine + 5) + thirteen) + 7 through associativity
Consider a definition of a setS = x ⋂ three < x < 7, x ∈ R.Which one of the following
alternatives pleasant describes S?
A.S = -2, -1, 0, 1, 2, 3, 4, 5, 6
b.S = x
c.S = x three < x < 7, x ∈ R
d.S = four, 5, 6 - ANS-d.S = 4, 5, 6
Consider the subsequent matrices:
Let A = zero 1 10 zero
zero 1 0 2 B = 2 0 and C = 0 10
3 0 4 zero' zero 3
4 zero
Which one of the following options regarding operations at the given matrices is FALSE?
Select one:
a.It is not possible to perform the operation A ∙ C.
B.The result of B ∙ A will bring about a four x 4 matrix.
C.The end result of A ∙ B is equal to matrix C.
D.The result of B ∙ C is the matrix - ANS-c.The end result of A ∙ B is identical to matrix C.
Consider the subsequent matrices:
1 2 1 2 three
Let A = [1 2], B = 3 3 and C = three 2 1
21
, Which one of the following alternatives concerning operations at the given matrices is FALSE?
Select one:
a.Calculating C + B will bring about a 2 x 2 matrix.
B.The result of B ∙ C is the matrix
765
12 12 12
567
c.The result of C ∙ B is the matrix
thirteen eleven
11 13
d.Performing the operation A ∙ C will result in a 1 x 3 matrix. - ANS-a.Calculating C + B will
result in a 2 x 2 matrix.
Consider the subsequent matrices:Which one of the following statements is FALSE?
Let A = [ 4 -1 ], B = -1 0 C = 2 2 -three
-four three four 0 -1
-2 zero,
a.The result of B ∙∙ C is
b.The end result of A ∙∙ C is
c.B ∙∙ C = C ∙∙ B
d.The operation (C ∙∙ B) ∙∙ A isn't viable. - ANS-c.B ∙∙ C = C ∙∙ B
Consider the following proposition:"For any predicates P(x) and Q(x) over a website D, the
negation of the statement ∃x ∈ D, P(x) ∧ Q(x)"is the statement "∀x ∈ D, P(x) →→
¬Q(x)".We can use this fact to write the negation of the subsequent assertion:"There exist
integers a and d such that a and d are bad and a/d = 1 + d/a".Which one of the options provides
the negation of this assertion?
A.There exist integers a and d such that a and d are high quality and a/d = 1 + d/a.
B.For all integers a and d, if a and d are advantageous then a/d ≠≠ 1 + d/a.
C.For all integers a and d, if a and d are negative then a/d ≠≠ 1 + d/a.
D.For all integers a and d, a and d are effective and a/d ≠≠ 1 + d/a. - ANS-c.For all integers a
and d, if a and d are terrible then a/d ≠≠ 1 + d/a.
Consider the following quantified announcement:∀x ∈ Z [(x2 ≥ 0) ∨ (x2 + 2x - 8＞0)]Which
one of the options provides a real announcement regarding the given statement or its negation?
A.The negation ∃x ∈ Z [(x2 < 0) ∨ (x2 + 2x - 8 ≤ 0)] is not actual.
B.X = - three could be a counterexample to prove that the negation isn't always actual.
C.X = - 6 would be a counterexample to prove that the statement isn't always authentic.
D.The negation ∃x ∈ Z [(x2 < 0) ∧ (x2 + 2x - 8 ≤ 0)] is true. - ANS-b.X = - 3 would be a
counterexample to prove that the negation is not true.
Consider the following relation on set B = 1, b, 1, b, 1, b:
P = (1, b), (b, 1, b), (1, b, 1), (b, 1), (1, 1).
Xzero ≤ x < eight ⋂ x ∈ R, four ≤ x < 16 is the set:
a.Xfour ≤ x < 8
b.Xzero ≤ x ≤ 8
c. X ∈ Z, zero ≤ x ≤16
d.Four, five, 6, 7 - ANS-d.4, five, 6, 7
A right subset of a set constantly has much less elements than the set itself.
True or False - ANS-True
An element of a hard and fast can not be a fixed itself.
True or False - ANS-False
based on set A = 1, four, four, 1, five
Which one of the following is NOT a partition on A?
1. 1, 4, four, 1, five
2. 1, 4, 1, 5, 4
3. 4, 1, five, 1, four
four. 1, 4, 1, 5, 4 - ANS-2. 1, four, 1, 5, four
based on set A = 1, 4, four, 1, five
Which one of the following relations is NOT a valid relation on A?
1. (1, 4), (4, 1)
2. (1, 5), (four, four)
three. (four, 1, five), (1, 1), (four, 1)
four. (four, four) - ANS-2. (1, 5), (4, four)
based totally on set A = 1, 4, four, 1, 5
Which one of the following statements presents a right subset of A?
1. 1, four ,four
2. 1, 4
3. 1, 4, 4, 1, 5
4. 1, five - ANS-2. 1, 4
based totally on set A = 1, four, four, 1, five
Which one of the following statements presents one or greater elements of the set A?
,1. 1, four
2. 4
3. 1, 1, five
4. 4, 1, five - ANS-four. Four, 1, five
By the usage of logical equivalences and de Morgan's guidelines, we can show that the
statements¬p ⋁ q and (p ⋀ ¬q) →→ (¬p ⋁ q) are equivalent.
A.True
b.False - ANS-a.True
Commutative and Associative - ANS-Now we examine the regulations that we carried out:
((7 + 13) + 5) + 9
= nine + ((7 + 13) + 5) with the aid of commutativity
= nine + (5 + (7 + thirteen)) by means of commutativity
= nine + (5 + (13 + 7) ) by commutativity
= nine + ((five + 13) + 7) via associativity
= (9 + (5 + thirteen)) + 7 by associativity
= ((nine + 5) + thirteen) + 7 through associativity
Consider a definition of a setS = x ⋂ three < x < 7, x ∈ R.Which one of the following
alternatives pleasant describes S?
A.S = -2, -1, 0, 1, 2, 3, 4, 5, 6
b.S = x
c.S = x three < x < 7, x ∈ R
d.S = four, 5, 6 - ANS-d.S = 4, 5, 6
Consider the subsequent matrices:
Let A = zero 1 10 zero
zero 1 0 2 B = 2 0 and C = 0 10
3 0 4 zero' zero 3
4 zero
Which one of the following options regarding operations at the given matrices is FALSE?
Select one:
a.It is not possible to perform the operation A ∙ C.
B.The result of B ∙ A will bring about a four x 4 matrix.
C.The end result of A ∙ B is equal to matrix C.
D.The result of B ∙ C is the matrix - ANS-c.The end result of A ∙ B is identical to matrix C.
Consider the subsequent matrices:
1 2 1 2 three
Let A = [1 2], B = 3 3 and C = three 2 1
21
, Which one of the following alternatives concerning operations at the given matrices is FALSE?
Select one:
a.Calculating C + B will bring about a 2 x 2 matrix.
B.The result of B ∙ C is the matrix
765
12 12 12
567
c.The result of C ∙ B is the matrix
thirteen eleven
11 13
d.Performing the operation A ∙ C will result in a 1 x 3 matrix. - ANS-a.Calculating C + B will
result in a 2 x 2 matrix.
Consider the subsequent matrices:Which one of the following statements is FALSE?
Let A = [ 4 -1 ], B = -1 0 C = 2 2 -three
-four three four 0 -1
-2 zero,
a.The result of B ∙∙ C is
b.The end result of A ∙∙ C is
c.B ∙∙ C = C ∙∙ B
d.The operation (C ∙∙ B) ∙∙ A isn't viable. - ANS-c.B ∙∙ C = C ∙∙ B
Consider the following proposition:"For any predicates P(x) and Q(x) over a website D, the
negation of the statement ∃x ∈ D, P(x) ∧ Q(x)"is the statement "∀x ∈ D, P(x) →→
¬Q(x)".We can use this fact to write the negation of the subsequent assertion:"There exist
integers a and d such that a and d are bad and a/d = 1 + d/a".Which one of the options provides
the negation of this assertion?
A.There exist integers a and d such that a and d are high quality and a/d = 1 + d/a.
B.For all integers a and d, if a and d are advantageous then a/d ≠≠ 1 + d/a.
C.For all integers a and d, if a and d are negative then a/d ≠≠ 1 + d/a.
D.For all integers a and d, a and d are effective and a/d ≠≠ 1 + d/a. - ANS-c.For all integers a
and d, if a and d are terrible then a/d ≠≠ 1 + d/a.
Consider the following quantified announcement:∀x ∈ Z [(x2 ≥ 0) ∨ (x2 + 2x - 8＞0)]Which
one of the options provides a real announcement regarding the given statement or its negation?
A.The negation ∃x ∈ Z [(x2 < 0) ∨ (x2 + 2x - 8 ≤ 0)] is not actual.
B.X = - three could be a counterexample to prove that the negation isn't always actual.
C.X = - 6 would be a counterexample to prove that the statement isn't always authentic.
D.The negation ∃x ∈ Z [(x2 < 0) ∧ (x2 + 2x - 8 ≤ 0)] is true. - ANS-b.X = - 3 would be a
counterexample to prove that the negation is not true.
Consider the following relation on set B = 1, b, 1, b, 1, b:
P = (1, b), (b, 1, b), (1, b, 1), (b, 1), (1, 1).
