Discrete Math _ Exam Re-Take Flashcards
(13 cards)
Which one is a proposition?
2 × 2 = 5
Who is going to lunch?
x + 2 ÷ 2x
Get the key from the drawer.
- 2 × 2 = 5
You answered
x + 2 ÷ 2x
Assume p is true, q is false, and r is true.
What are the truth values of p ∧ q, q ∨ ¬r, and (p ∧ q) ∨ (q ∨ ¬r)?
F, F, F
F, F, T
T, T, F
F, T, F
- F, F, F
You answered
F, T, F
Given p is false and q is true.
Which expression has the same truth value as (p ∨ ¬q) → ¬p?
A.) q → p
B.) ¬(p → q)
C.) (p → q) ∧ ¬(q → p)
D.) ¬(p → q) ∧ (q → p)
(p → q) ∧ ¬(q → p)
You answered
¬(p → q)
Consider this truth table, which presents the values of predicate M(x,y)
, where x∈{1,2,3} and y∈{7,8,9}.
M 7 8 9
1 T T F
2 T F T
3 F T T
∀x M(x,x+6)
∃x ∀y M(x,y)
∀y ∃x M(x,y)
∃y M(10−y,y)
∀y ∃x M(x,y)
You answered
∃x ∀y M(x,y)
Given variables x,y,z
with values taken from the set D={−1,0,1}.
Which statement is true?
∀x(∃y(∀z(x+y+z=z)))
∃x(∀y(∀z(x+y+z=0)))
∀x(∀y(∃z(x+y+z=z)))
∃x(∃y(∀z(x+y+z=0)))
∀x(∃y(∀z(x+y+z=z)))
Your answered
∃x(∃y(∀z(x+y+z=0)))
Which statement has a free variable?
∀x (P(x)) ∧ R(x)
∀x (P(x) ∧ R(x))
∃x P(x) ∨ ∃y R(y)
∃x (P(x) ∨ R(x))
∀x (P(x)) ∧ R(x)
You answered
∃x (P(x) ∨ R(x))
Consider the theorem and its proof below.
Theorem: If the product of two positive real numbers is greater than 25, then at least one of the numbers is greater than 5.
Proof: For all positive real numbers a and b,
If a ≤ 5 and b ≤ 5, then ab ≤ 25.
If a ≤ 5, then ab ≤ 5b.
Since b ≤ 5, then ab ≤ 25.
Therefore, if ab > 25, then a > 5 or b > 5.
Is the proof valid, and which type of proof is used?
The proof is valid, and it is a direct proof.
The proof is not valid, and it is a direct proof.
The proof is valid, and it is proof by contrapositive.
The proof is not valid, and it is proof by contrapositive.
The proof is valid, and it is proof by contrapositive.
Your answer
The proof is valid, and it is a direct proof.
Consider the theorem and proof below:
Theorem: ¬(P∧ ¬Q) ≡ ¬P ∨ Q
Proof: Consider the truth table for the two statements. Since the truth table is correct and the truth values match, the two statements are equivalent.
P Q -P -Q PA-Q -(PA-Q) -PV Q
T T F F see screeshot
T F F
F T T
F F T
Proof by cases is used, but the proof is not valid.
Proof by cases is used, and the proof is valid.
Proof by contraposition is used, but the proof is not valid.
Proof by contradiction is used, and the proof is valid.
Proof by cases is used, but the proof is not valid.
You answered
Proof by cases is used, and the proof is valid.
Consider the proof below.
Theorem: Let n be an integer. If n2 is odd, then n is odd.
Proof: Let n be an integer. Assume n2 is odd and n is even. Then there is some integer k such that n=2k. Then n2=(2k)2 = 4k2, which is even. Therefore, n must be odd.
Is the proof valid, and which type of proof is used?
The proof is valid, and a direct proof is used.
The proof is not valid, and a direct proof is used.
The proof is valid, and proof by contradiction is used.
The proof is not valid, and a contradiction proof is used.
The proof is valid, and proof by contradiction is used.
You answered
The proof is not valid, and a contradiction proof is used.
Which logical statement is equivalent to 0?
1↑1
1↑0
0↑1
0↑0
1↑1
You answered
0↑0
Which expression is the Boolean equivalent of the CNF expression
(x+¯¯¯y)(¯¯¯x+y)
?
xy+¯¯¯x⋅¯¯¯y
x¯¯¯y+¯¯¯xy
¯¯¯x ⋅ ¯¯¯y
xy
xy+¯¯¯x⋅¯¯¯y
You answered
xy
Given this circuit:
Which description allows the lamp to glow?
P: True, Q: False
P: True, Q: True
R: False, P: False, Q: False
R: True, P: True, Q: True
P: True ,Q: False
Your Answer
R: True, P: True, Q: True
Given the circuit below:
Which input, given as (x, y, z), gives an output of 1?
Less
(0 ,0, 1)
(0, 1, 1)
(0 ,1, 0)
(1, 0, 0)
(0 ,1, 0)
You answered:
(0, 1, 1)
