Practice for the Final Exam MAT 243: Discrete Math
Structures (2025-2026 Spring C) Arizona State
University
Practice for the Final Exam
Due May 6 at 11:59pm
Points 100
Questions 66
Available until May 7 at 11:59pm
Time Limit None
Allowed Attempts 3
Instructions
Take this exam to practice for the final exam. This practice exam earns a small amount of course credit.
Your final grade for this practice test will be the average of your three attempts, so be careful if you
attempt this again after already earning a satisfactory grade. Making another attempt but not completing
it, or submitting no answers, can drastically lower your grade for this test.
Furthermore, if you take additional attempts after the due date, take into account that the late
penalty will be applied. The responsibility for this is yours.
This quiz was locked May 7 at 11:59pm.
Attempt History
Attempt Time Score
LATEST Attempt 3 less than 1 minute 100 out of 100
Attempt 2 2 minutes 100 out of 100
Attempt 1 3,469 minutes 96 out of 100
Score for this attempt: 100 out of 100
Submitted May 6 at 11:59pm
This attempt took less than 1 minute.
Correct answer
Question 1
pts
Determine the truth of the quantified statement
,∀x ∃y (xy > x).
The domain of discourse is the set of positive real numbers.
True
, False
Suppose any positive real number x is given. Pick y = 2. Then xy = 2x. 2x > x since x > 0.
[Adding x to both sides of x > 0 produces x + x > x, or 2x > x.]
Correct answer
Question 2
pts
Match logically equivalent expressions.
p↔q
¬ ( (p → q) → ¬(q → p) )
p⊕q
(p ∨ q) ∧¬ (p ∧ q)
p∨q
¬p → q
p∧q
¬(¬p ∨¬q)
We know ¬p ∨ q ≡ p → q from the lecture. By substituting ¬p for p and using the double negation law, we
get p ∨ q ≡ ¬p → q.
By using De Morgan and the double negation law, we get p ∧ q ≡ ¬(¬p ∨¬ q).
We know from the lecture that p ⊕ q is true when exactly one of p,q is true. So we can define it as: p or
q, but (and) not both. This means p ⊕ q ≡ (p ∨ q) ∧¬ (p ∧ q).
- ---------------- Biconditional, Solution 1:
The biconditional is the negation of the exclusive or. By using the equivalence we just derived, p ↔ q ≡ ¬
((p ∨ q) ∧¬ (p ∧ q)).
We use De Morgan to simplify ¬ (p ∧ q):
p ↔ q ≡ ¬ ((p ∨ q) ∧ (¬p ∨ ¬q)).
, We now distribute the right side inside the parentheses. We will place unnecessary parentheses around
the conjunctions to emphasize the order of operations.
p ↔ q ≡ ¬ ( (p∧¬p) ∨ (q∧¬p) ∨ (p∧¬q) ∨ (q∧¬q) ).
We use the fact that p∧¬p and q∧¬q are contradictions, and then the identity law.
p ↔ q ≡ ¬ ( F ∨ (q∧¬p) ∨ (p∧¬q) ∨ F ) ≡ ¬ ( (q∧¬p) ∨ (p∧¬q) ).
We now use De Morgan on the two inside conjunctions and commute:
p ↔ q ≡ ¬ ( ¬(¬q∨p) C ¬(¬p∨q) ) ≡ ¬ ( ¬(¬p∨q) ∨ ¬(¬q∨p) )
Using the definition of conditional three times, we get
p ↔ q ≡ ¬ ( ¬(p → q) ∨ ¬(q→ p) ) ≡ ¬ ( (p → q) → ¬(q→ p) ).
- ---------------- Biconditional, Solution 2:
The biconditional is equivalent to (p → q) ∧ (q → p).
Using De Morgan, we get
(p → q) ∧ (q → p)≡¬(¬(p → q) ∨ ¬(q → p)).
Now, apply the definition of conditional to obtain
¬(¬(p → q) ∨ ¬(q → p))≡¬((p → q) → ¬(q → p)).
Correct answer
Question 3
pts
How many distinct ternary logical operators are there? A ternary logical operator is an operator that takes
three inputs, p, q and r, and for each truth configuration of the inputs, outputs true or false.
Recall that a logical operator is defined by its truth table.
256
256 (with margin: 0)
p can be true or false. Independently of p, q can be true or false. Independently of p and q, r can be true
or false. By the multiplication principle, there are 8 different configurations of truth values that a ternary
operator acts on (i.e. the truth table of such an operator has 8 rows, not counting the header).
Put differently, a ternary logical operator is a function from a domain of 8 elements to a domain of 2
elements. There are 2⁸ = 256 such functions, again by the multiplication principle.
Structures (2025-2026 Spring C) Arizona State
University
Practice for the Final Exam
Due May 6 at 11:59pm
Points 100
Questions 66
Available until May 7 at 11:59pm
Time Limit None
Allowed Attempts 3
Instructions
Take this exam to practice for the final exam. This practice exam earns a small amount of course credit.
Your final grade for this practice test will be the average of your three attempts, so be careful if you
attempt this again after already earning a satisfactory grade. Making another attempt but not completing
it, or submitting no answers, can drastically lower your grade for this test.
Furthermore, if you take additional attempts after the due date, take into account that the late
penalty will be applied. The responsibility for this is yours.
This quiz was locked May 7 at 11:59pm.
Attempt History
Attempt Time Score
LATEST Attempt 3 less than 1 minute 100 out of 100
Attempt 2 2 minutes 100 out of 100
Attempt 1 3,469 minutes 96 out of 100
Score for this attempt: 100 out of 100
Submitted May 6 at 11:59pm
This attempt took less than 1 minute.
Correct answer
Question 1
pts
Determine the truth of the quantified statement
,∀x ∃y (xy > x).
The domain of discourse is the set of positive real numbers.
True
, False
Suppose any positive real number x is given. Pick y = 2. Then xy = 2x. 2x > x since x > 0.
[Adding x to both sides of x > 0 produces x + x > x, or 2x > x.]
Correct answer
Question 2
pts
Match logically equivalent expressions.
p↔q
¬ ( (p → q) → ¬(q → p) )
p⊕q
(p ∨ q) ∧¬ (p ∧ q)
p∨q
¬p → q
p∧q
¬(¬p ∨¬q)
We know ¬p ∨ q ≡ p → q from the lecture. By substituting ¬p for p and using the double negation law, we
get p ∨ q ≡ ¬p → q.
By using De Morgan and the double negation law, we get p ∧ q ≡ ¬(¬p ∨¬ q).
We know from the lecture that p ⊕ q is true when exactly one of p,q is true. So we can define it as: p or
q, but (and) not both. This means p ⊕ q ≡ (p ∨ q) ∧¬ (p ∧ q).
- ---------------- Biconditional, Solution 1:
The biconditional is the negation of the exclusive or. By using the equivalence we just derived, p ↔ q ≡ ¬
((p ∨ q) ∧¬ (p ∧ q)).
We use De Morgan to simplify ¬ (p ∧ q):
p ↔ q ≡ ¬ ((p ∨ q) ∧ (¬p ∨ ¬q)).
, We now distribute the right side inside the parentheses. We will place unnecessary parentheses around
the conjunctions to emphasize the order of operations.
p ↔ q ≡ ¬ ( (p∧¬p) ∨ (q∧¬p) ∨ (p∧¬q) ∨ (q∧¬q) ).
We use the fact that p∧¬p and q∧¬q are contradictions, and then the identity law.
p ↔ q ≡ ¬ ( F ∨ (q∧¬p) ∨ (p∧¬q) ∨ F ) ≡ ¬ ( (q∧¬p) ∨ (p∧¬q) ).
We now use De Morgan on the two inside conjunctions and commute:
p ↔ q ≡ ¬ ( ¬(¬q∨p) C ¬(¬p∨q) ) ≡ ¬ ( ¬(¬p∨q) ∨ ¬(¬q∨p) )
Using the definition of conditional three times, we get
p ↔ q ≡ ¬ ( ¬(p → q) ∨ ¬(q→ p) ) ≡ ¬ ( (p → q) → ¬(q→ p) ).
- ---------------- Biconditional, Solution 2:
The biconditional is equivalent to (p → q) ∧ (q → p).
Using De Morgan, we get
(p → q) ∧ (q → p)≡¬(¬(p → q) ∨ ¬(q → p)).
Now, apply the definition of conditional to obtain
¬(¬(p → q) ∨ ¬(q → p))≡¬((p → q) → ¬(q → p)).
Correct answer
Question 3
pts
How many distinct ternary logical operators are there? A ternary logical operator is an operator that takes
three inputs, p, q and r, and for each truth configuration of the inputs, outputs true or false.
Recall that a logical operator is defined by its truth table.
256
256 (with margin: 0)
p can be true or false. Independently of p, q can be true or false. Independently of p and q, r can be true
or false. By the multiplication principle, there are 8 different configurations of truth values that a ternary
operator acts on (i.e. the truth table of such an operator has 8 rows, not counting the header).
Put differently, a ternary logical operator is a function from a domain of 8 elements to a domain of 2
elements. There are 2⁸ = 256 such functions, again by the multiplication principle.
