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(3) 1. a) Five people are taking part in a spelling competition. In how many ways can, first, second and third place be awarded? b) Bill has a choice of 4 pants and 6 shirts. In how many different ways can he dress? c) What is the number of 3-element subsets of a set with 7 elements?
(3) 2. Let A = { 1 , 3 , 5 }.
a) List all proper subsets of A. b) If U = {x ∈ N : 1 ≤ x ≤ 10 }, what is A? c) Suppose B is any subset of U such that A ⊆ B. What can you say about A ∩ B? What can you say about A ∪ B?
(4) 3. Draw Venn diagrams and show by hatching the following sets:
a) (A ∪ C) ∩ B b) (A − C) ∪ (B ∩ C)
(6) 4. For each expression, name the property from the given list and say whether it is a set property, a network property, or a logic property: associativity, commutativity, distributivity, identity, idempotent, de Morgan, closure, complement, property of 1 (or 0), tautology, contradiction.
a) ∼ (p ∨ q) ↔ (∼ p∧ ∼ q)
b) A + 1 = 1
c) A · (B + C) = A · B + A · C
d) A ∪ B ⊆ U
e) A ∪ A = A
f) (p ∨ t) ↔ t
(3) 5. Use a truth table to determine whether or not p ∧ (q ∨ r) is equivalent to p ∨ (q ∧ r)
(3) 6. Determine whether the given expression is a tautology, contradiction or neither?
(p →∼ q) ∧ (p ∨ q)
(5) 7. Use a truth table to determine whether the argument is valid or not.
H: If it snows, then it is winter. It is winter and it snows.
C: It is winter if and only if it snows.
(3) 8. Use a Venn diagram to determine the validity of the argument.
H: Some friends lie. Some friends are nice.
C: Friends who are nice do not lie.
(4) 9. If it is not 7 p.m. then I am not late for my bus.
a) Write the converse, the contrapositive, and the inverse of the implication above. b) Say which among the four statements are equivalent.
(5) 10.Find a Boolean table for each expression: a) (A + B)B b) (A + B) + AC
(2) 11.Draw a network to represent the given boolean expression: B(AB + CD) + AC
(5) 12.Simplify each expression, justifying each step using properties of Boolean algebra. a) AB + AB + B (2) b) A + B + C + A C (3) (6) 13.Classify each system below (Without solving) as dependent or independent, consistent with one solution only, consistent with infinitely solutions or inconsistent. Justify your answers. a) 3 x−^2 y^ =^3 y = 2 x + 1
b) (^) −^24 xx^ −+ 6^3 yy^ ==^ −^36 (2)
c) 4 x 2 x^ − −^2 yy^ ==^ −^52 (2)
(4) 14.Given the system 3 x − y = 7 x + 3y = 9 a) Estimate the solution by graphing. b) Solve the following system algebraically by substitution or elimination.
(12) 15.Given: A =
B =
[
]
C =
[
]
D =
[
]
I =
[
]
Find each of the following, if possible. If an operation is not possible, say why.
a) 3D − 2 C (2) b) CCT^ (2) c) B−^1 (2) d) D + I (1) e) IC (1) f) AD (3) g) DI (1)
(3) 16.Given A =
, explain why A−^1 does not exist.
(8) 17.Given A =
(^) find A−^1 using elementary row operations.
No, they are not equivalent
It is neither (it is equivalent to p ↔ q)
It is a valid argument (Show (p → q) ∧ (q ∧ p) → (p ↔ q) is a tautology)
Invalid, can have (F ∩ N ) ∩ L 6 = ∅
Converse: If I am not late for my bus then it is not 7 p.m. Contrapositive: If I am late for my bus then it is 7 p.m. Inverse: If it is 7 p.m. then I am late for my bus. The original is equivalent to contrapositive and the converse is equivalent to inverse.
(a) It is 1 when A = 1 and B = 0 and 0 otherwise. (b) It is 1 when A = B = 1, C = 0 or when A = 1, B = C = 0 or when A = B = 0 and C = 1 or when A = B = C = 0; it is 0 in all four other situations.
| −→ A −→ B −→ | | −→ B −→ | | −→ −→ | | −→ C −→ D −→ | | | | −→ | −→ A −→ C −→ |
- (a) It simplifies to A + B (b) It simplifies to 1.
- (a) Independent, consistent (a unique solution) (b) dependent, consistent (infinitely many solutions) (c) independent, inconsistent (no solution)
- Answer is (3, 2)
- (a)
[
]
(b)
[
]
(c)
[
]
(d) D + I is not possible; (e) IC = C
(f)
(^) (g) DI is not possible.
- A can not be reduced to I (a row of zeros appears in the left hand side of the [A|I] matrix.
17. A−^1 =
x y z
X = A−^1 B =
- x = 10, y = −3, z = − 1
- x = 3, y = 1, z = 2
- inconsistent
- P 1 is true as 1 = 1^2. Assume Pk is true then 1 + 3 + 5 + · · · + (2k − 1) = k^2 which implies (add 2k + 1 to both sides)
1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k^2 + 2k + 1 = (k + 1)^2
This shows that Pk+1 is true.
