Practice for the Final Exam MAT 243 Discrete Math Structures (2025-2026 Spring C) Arizona State Univ

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.

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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 1 / 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 1 / 1 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 1 / 1 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.

Correct answer  Question 4 1 / 1 pts Select which one of the following statements is logically equivalent to (p∨q) → r . ( p→ r ) ∧ ( q→ r ) ( p→r ) ∨ ( q→ r ) ¬( p ∨ q ) → r (¬p∨¬q ) ∨ r None of these.

You can find the correct answer by doing a truth table, or, more efficiently, by using logical equivalences, the distributive and commutative properties, and De Morgan's law: (p∨q)→r ≡¬(p∨q)∨r ≡(¬p∧¬q)∨r ≡(¬p∨r)∧(¬q∨r) ≡(p→r)∧(q→r) Correct answer  Question 5 1 / 1 pts Consider the statement “Unless you stop smoking, you will get sick”. Select which one of the following statements are logically equivalent to that conditional above.

You will get sick if you keep smoking.

Continuing to smoke is sufficient to get sick.

Continuing to smoke is necessary to get sick.

If you get sick, that means you didn't stop smoking.

You will get sick only if you keep smoking.

Recall that "p only if q" does NOT mean "if q, then p". It means the converse: "If p, then q".

“Unless you stop smoking, you will get sick” means "If you don't stop smoking, then you will get sick" .

Equivalently, "If you keep smoking, you will get sick."

Correct answer 

Question 6 1 / 1 pts Evaluate 1010 OR 0011. Here, OR is the bitwise logical or, acting on bitstrings.

1,011

1,011 (with margin: 0) We apply the or bitwise, i.e. to each bit separately: 1010 OR 0011. = 1011.

Correct answer  Question 7 1 / 1 pts Determine the truth of the quantified statement ∀x ∃y (xy > x).

The domain of discourse is the set of real numbers.

True False

The statement is false because for x = 0, no such y can exist. With x = 0, xy = 0 regardless of what y is, and 0 is not greater than 0.

Correct answer  Question 8 2 / 2 pts True or False? There is a smallest real number k so that xlog(x) is O(xᵏ).

True False

xlog(x) is not O(x) because xlog(x)/x = log(x) goes to infinity.

On the other hand, xlog(x) is O(xᵏ) for any k > 1. This is because xlog(x) / xᵏ = log(x) / xᵏ⁻¹, which goes to zero as x goes to infinity. (Use L'Hospital's rule to justify this statement.)

It's a calculus fact that any positive power of x goes faster to infinity than log(x), so the denominator in log(x) / xᵏ⁻¹ (x has a positive power) "wins".

Correct answer  Question 9 2 / 2 pts Check all functions f(x) that are Ω(x²). f(x) = log(xˣ) f(x) = x³/2 + x² f(x) = ⌊x+2⌋ · ⌈x⌉ f(x)=2x1.9 +2x-1.9 f(x) = (x⁴+2x+3)/(x²-2) f(x) = (x⁵+2x+3)/(x²-2) f(x) = x² · log(x) f(x) = 2ˣ +x²

Remember that f(x) being Ω(x²) means that f grows "at least as fast" as x². This is the case if |f(x)| / x² is greater than or equal a positive constant, has a positive limit, or goes to infinity.

If the quotient goes to zero, then f(x) is not Ω(x²).

For f(x) = log(xˣ), f(x) / x² = log(x)/x goes to zero. For f(x) = x³/2 + x², f(x) / x² = x/2 + 1 goes to infinity. f(x) = ⌊x+2⌋ · ⌈x⌉ ≥ x · x , so f(x) / x² ≥ 1.

For f(x)=2x^1.9 +2x^(-1.9), f(x) / x² = 2x^(-0.1) +2x^(-3.9) goes to zero.

For f(x) = (x⁴+2x+3)/(x²-2), f(x) / x² = (x⁴+2x+3)/(x²(x²-2)) goes to 1, according to the horizontal asymptote rule for rational functions.

For f(x) = (x⁵+2x+3)/(x²-2), f(x) / x² is a linear function, which goes to infinity, plus a proper rational part, which goes to zero. Hence f(x) / x² goes to infinity.

For f(x) = x² · log(x), f(x) / x² = log(x) goes to infinity.

For f(x) = 2ˣ +x², f(x) / x² = 2ˣ / x² + 1 which goes to infinity. Correct answer  Question 10 2 / 2 pts Find the smallest integer k so that

is order of nᵏ.

2

2 (with margin: 0) As n gets large, the +1 in the denominator becomes negligible. Thus, f is approximately n + n² for large n, and therefore order of n².

To prove this exactly, we show that the limit of f(n)/n² is a positive number:

The last part of the equation is true because of the horizontal asymptote rule for rational functions.

Correct answer  Question 11 2 / 2 pts Enter the smallest integer n so that the following function is O(xⁿ). f(x)=x²(x³ + 1)+x⁵log(x).

6

6 (with margin: 0) The first term is order of x⁵ , hence also big-O of x⁵.

The second term is "between" x⁵ and x⁶ in order, hence big-O of x⁶ but not x⁵. That makes the first term negligible, and the sum big-O of x⁶ but not x⁵.

Correct answer  Question 12 2 / 2 pts Check all functions f(x) that are Θ(x²). f(x) = log(xˣ) f(x) = x³/2 + x² f(x) = ⌊x+2⌋ · ⌈x⌉ f(x)=2x1.9 +2x-1.9 f(x) = (x⁴+2x+3)/(x²-2) f(x) = (x⁵+2x+3)/(x²-2) f(x) = x² · log(x)

f(x) = 2ˣ +x²

f is Θ(x²) iff f is both O(x²) and Ω(x²). We get the answers to this problem by combining the answers of the two previous ones.

Correct answer  Question 13 2 / 2 pts True or False? There is a smallest real number a so that 3ˣx is O(aˣ).

True False

Let us use the notations f(x) = 3ˣx and g(x) = 3ˣ. f(x) is not O(g(x)) because f(x)/g(x)=x goes to infinity as x goes to infinity.

On the other hand, f(x) is O(aˣ) for any real number a > 3. For example, f(x) is O(3.01ˣ). You can see that this is true by considering the quotient again: f(x)/g(x) = (3/a)ˣ·x. 3/a is less than 1, hence (3/a)ˣ goes to zero as x goes to infinity. Exponential decay beats polynomial growth (this is a calculus fact), so even (3/a)ˣ·x still goes to zero.

Since 3ˣx is O(aˣ) for all a > 3, but not for a = 3, there is no smallest real number a so that 3ˣx is O(aˣ).

Correct answer  Question 14 2 / 2 pts True or False? 5x² is O(x²).

True False

If the limit of |f(x)|/|g(x)| as x goes to infinity exists, then f(x) is O(g(x)).

In particular, if f(x)/g(x) is a constant, then f(x) is O(g(x)). Here, the quotient is a constant 5.

Correct answer  Question 15 2 / 2 pts Select the smallest k so that f(x)=x·(lnx)⁴+x² is big-O of O(xᵏ).

2

2 (with margin: 0) Both terms are big-O of x², hence their sum is too. Neither is big-O of x, hence their sum can't be big-O of x either.

Correct answer  Question 16 2 / 2 pts Find a big-O estimate of f(x)=(3x² + xlog(x²) + x²log(x²))·(πx³ + 2x + 4) + (x² + x⁴log(x)) such that in your estimate f(x) is O(g(x)) and g(x) is a simple function of the smallest order. g(x)=x⁵log(x) g(x)=x⁶ g(x)=x⁴log(x) g(x)=x⁵ None of these functions are big O estimates of f(x).

The big-O estimate of a sum is the big-O extimate of the largest function.

The big-O estimate of a product is the product of the big-O estimates.

Thus, 3x² + xlog(x²) + x²log(x²) is O(x² log(x)) and πx³ + 2x + 4 is O(x³ ).

Hence, (3x² + xlog(x²) + x²log(x²))·(πx³ + 2x + 4) is O(x⁵ log(x)).

We know that x² + x⁴log(x) is O( x⁴log(x)).

Using the sum rule, f(x)=(3x² + xlog(x²) + x²log(x²))·(πx³ + 2x + 4) + (x² + x⁴log(x)) is O(x⁵ log(x)). Correct answer  Question 17 2 / 2 pts Check all functions f(x) that are O(x²). f(x) = log(xˣ) f(x) = x³/2 + x² f(x) = ⌊x+2⌋ · ⌈x⌉ f(x)=2x1.9 +2x-1.9 f(x) = (x⁴+2x+3)/(x²-2)

f(x) = (x⁵+2x+3)/(x²-2) f(x) = x² · log(x) f(x) = 2ˣ +x²

f(x) = log(xˣ) = xlog(x) is O(x²). f(x) = x³/2 + x² is of higher degree than x² and therefore not O(x²). f(x) = ⌊x+2⌋ · ⌈x⌉ is approximately x².

More precisely: ⌊x+2⌋ ≤ x + 2 and ⌈x⌉ ≤ x + 1. Therefore, f(x) ≤ (x + 2)(x + 1). This means that f(x) is O(x² + 3x + 2). Since x² + 3x + 2 is O(x²), ⌊x+2⌋ · ⌈x⌉ is O(x²) as well. f(x)=2x^1.9 +2x^(-1.9) is O(x²) because each of the terms is. f(x) = (x⁴+2x+3)/(x²-2) < (x⁴+2x+3)/x² = x²+2/x+3/x² which is O(x²). f(x) = (x⁵+2x+3)/(x²-2) is a cubic function plus a proper rational part, and can't therefore be O(x²). f(x) = x² · log(x) can't be O(x²) because f(x)/x² = log(x) goes to infinity. f(x) = 2ˣ +x² can't be O(x²) because f(x)/x² = 2ˣ / x² + 1 goes to infinity.

Correct answer  Question 18 2 / 2 pts If f(x) is O(x³) and g(x) is O(x⁴), then what is the best we can say about f(x)·g(x)? Find the "lowest" big-O estimate that is guaranteed by the given information. f(x)·g(x) is O(x³). f(x)·g(x) is O(x⁴). f(x)·g(x) is O(x⁷). f(x)·g(x) is O(x¹²).

The example f(x) =x³ and g(x) = x⁴ shows that we can't conclude that f(x)·g(x) = x⁷ is O(x³) or O(x⁴). We can conclude f(x)·g(x) is O(x⁷) by the product rule.

This means that f(x)·g(x) is also O(x¹²), but the question asked for the "lowest" big-O estimate.

Correct answer  Question 19 2 / 2 pts True or False? x³ is O(3ˣ).

True False

Any polynomial function is big-O of any increasing (base > 1) exponential function.

Correct answer  Question 20 2 / 2 pts Find the lowest integer k so that 1⁵ + 2⁵ + 3⁵ + 4⁵ + 5⁵ + ... + n⁵ is O(nᵏ).

66(with margin: 0) Let us call 1⁵ + 2⁵ + 3⁵ + 4⁵ + 5⁵ + ... + n⁵ = S(n). Since each base is at most n, we have the inequality S(n) ≤ n⁵ + n⁵ + ... + n⁵ = n·n⁵ = n⁶. Therefore, S(n) is O(n⁶).

On the other hand, if n is even, then n/2 + 1 bases in the sum are at least n/2.

For example, S(8) = 1⁵ + 2⁵ + 3⁵ + 4⁵ + 5⁵ + 6⁵ + 7⁵ + 8⁵ , which has 5 bases that are at least 4.

Thus, for even n, S(n) > (n/2)⁵ + (n/2)⁵ + ... + (n/2)⁵, where we have more than n/2 terms (n/2)⁵ on the right side.

Therefore, S(n) > (n/2)⁶ = n⁶ / 64. This means that S(n) cannot be O(n⁵).

Another way to see that S(n) cannot be O(n⁵) is to use calculus. By making a Riemann sum diagram, you can see that S(n) is a left sum for the integral of f(x) = x⁵ from x=1 to x=n+1. Since f is decreasing, the left sum overestimates the integral. The value of the integral is 1/6 ( (n+1)⁶ - 1), an order n⁶ function.

Correct answer  Question 21 2 / 2 pts Given an increasing big-O order of the functions. This means that f1 is O(f2), f2 is O(f3), etc. f1 log(n)

f2

n

f3 n·log(n)

f4 n²

f5 n!

log(n) is O(n) because the ratio log(n)/n goes to zero. n is O(nlog(n)) because the ratio n/(nlog(n)) = 1/log(n) goes to zero. nlog(n) is O(n²) because the ratio nlog(n)/n² = log(n)/n goes to zero. n² is O(n!) because the ratio n²/n! goes to zero.

You can see that the last statement is true as follows: n²/n! = n/(n-1)! = n/((n-1)(n-2)!) = n/(n-1) · 1/(n-2)! . The factor n/(n-1) goes to 1, the factor 1/(n-2)! goes to zero.

Correct answer  Question 22 2 / 2 pts If f(x) is O(x³) and g(x) is O(x⁴), then what is the best we can say about f(x)+g(x)? Find the "lowest" big-O estimate that is guaranteed by the given information. f(x)+g(x) is O(x³). f(x)+g(x) is O(x⁴). f(x)+g(x) is O(x⁷). f(x)+g(x) is O(x¹²).

The example f(x) =x³ and g(x) = x⁴ shows that we can't conclude that f(x)+g(x) is O(x³).

We can conclude f(x)+g(x) is O(x⁴) by the sum rule.

This means that f(x)+g(x) is also O(x⁷) and O(x¹²), but the question asked for the "lowest" big-O estimate.

Correct answer

 Question 23 2 / 2 pts A device produces random 64-bit integers at a rate of one billion per second. After how many years of running is it unavoidable that the device produces an output for the second time? Round to the nearest number of years.

585

585 (with margin: 0) The device is capable of producing 2⁶⁴ many different outputs. The device produces 10⁹ outputs per second.

Therefore, it produces 10⁹ · 3600 · 24 · 365.25 outputs per year.

By the pigeonhole principle, it is unavoidable that the device has produced at least one output twice when it has produced 2⁶⁴ + 1 outputs.

It takes the device (2⁶⁴ + 1)/(10⁹ · 3600 · 24 · 365.25) years to produce 2⁶⁴ + 1 outputs. That is 584.54 years, or 585 years when rounded to the nearest year.

(Approximating the length of the year to 365 year does not change the rounded answer here.)

Correct answer  Question 24 2 / 2 pts How many 3 letter "words" can be created from the letters ABCDEFG when 1. repetition allowed 2. repetition is not allowed?

The words do not have to meaningful.

1.7³ 2. P(7,3) = 7·6·5 1. 3⁷ 2. P(7,3) = 7·6·5

1. 7+7+7 = 21 2. 7+6+5 = 18 1. 7³ 2. C(7,3) = 7·6·5/(3·2·1) None of these.

1. There are 7 choices for each of the three letters. There are 7 choices for the first letter and 7 choices for the second letter and 7 choices for the third letter. 2. There are 7 choices for the first letter and 6 choices for the second letter and 5 choices for the third letter.

Correct answer  Question 25 2 / 2 pts How many different four-digit code words can be constructed using the digits 0,1,2,3,4,5,6,7,8,9 with no repeated digits that start with either 0 or 9?

2·(9·8·7) 9·7·8 2·(9+8+7) 2·(9·9·9) None of these.

If the first digit is 0, then rest of the digits can be picked P(9,3) ways.

If the first digit is 9, then the rest of the digits can be picked P(9,3) ways.

Now we use the sum rule: the digits can be picked in P(9,3) + P(9,3) ways.

Correct answer  Question 26 2 / 2 pts What is the minimum number of people you need to guarantee that there are three among them who were born on the same day? There are 366 possible birthdays.

733

733 (with margin: 0) With 2·366 = 732 people, there could be exactly two people that were born on each birthday. With 733 people, that is impossible, because up to two people for each birthday can only add up to 732 people total.

Correct answer  Question 27 2 / 2 pts True or False? If there are 22 people in a room, then it is certain that at least 4 of them were born on the same day of the week.

True False

If there were at most 3 born on each day of the week, there would be at most 21 people, because there are only 21 days of the week.

Correct answer  Question 28 2 / 2 pts Four authors are going together on a speaking tour. In each city, they hold one event for one evening where each of them speaks for the same amount of time.

In consideration of the fact that the first and the last speaker of the evening have a special impact on the audience, they decide to rotate through all possible lineups, and to visit exactly as many cities as are needed to realize each lineup.

How many cities are they visiting?

24

24 (with margin: 0) The lineups are ordered selections of 4 from 4 people without repetitions, aka permutations. The number of cities is 4! = 24.

Correct answer  Question 29 2 / 2 pts A true-false test contains 25 questions. In how many different ways can the 25-question test be answered?

2²⁵ 25² 50 C(15,2) = 25·24/2

None of these.

There are 2 choices for each of the 25 questions. The choices are independent from each other. Thus, there are 2²⁵ choices.

Correct answer  Question 30 2 / 2 pts There are 193 member states of the United Nations. If you have 195 diplomats from these states in a room, then 1. It is certain that at least 2 of them are from the same state.

2. It is certain that at least 3 of them are from the same state.

Both statements are true.

Both statements are false.

1. is true, 2. is false.

1. is false, 2. is true.

It is certain that at least 2 of them are from the same state, because if there was at most one diplomat from each state, then there would be at most 193 diplomats.

It is not certain that at least 3 of them are from the same state. For example, 191 states might be represented by exactly 1 diplomat, and 2 states might be represented by 2 diplomats each.

Correct answer  Question 31 2 / 2 pts There are 10 applied math and 13 pure math professors on the faculty in the mathematics department.

How many ways are there to select a committee of 7 members if the committee needs to have 4 applied math and 3 pure math members?

C(10,4) · C(13,3) = 10·9·8·7/4! · 13·12·11/3!

C(10,4) + C(13,3) = 10·9·8·7/4! + 13·12·11/3!

P(10,4) · P(13,3) = 10·9·8·7 · 13·12·11 P(10,4) + P(13,3) = 10·9·8·7 + 13·12·11 None of these.

We select 4 applied math people out of 10 (order of the selection does not matter). For each selection of the 4-member-applied math committee, we select pure math people out of the 13 (again the order of the selection does not matter).

Correct answer  Question 32 2 / 2 pts At a job fair, the organizers offer some free gifts to the participants. There are 4 different bags, 3 different water bottles and 5 different binders. Each participant can choose a bag and a water bottle and a binder. How many different choices does a participant have?

4·3·5 4+3+5 4!·3!·5!

4!+3!+5!

None of these.

A participant selects one item from each category. By the multiplication principle a participant can have 4·3·5 different choices.

Correct answer  Question 33 2 / 2 pts In a class of 3000 students, the professor gives a multiple choice quiz with 4 questions, possible answers A,B,C, D or E. Each student answers each question. Among the 3000 students, at least how many students will turn in the exact same answers for the quiz questions?

5

5 (with margin: 0) There are 5⁴=625 possible answers for the quiz. Considering 625 · 4 = 2600 625 · 5 = 3125, we realize that there must be 5 students who will turn in the exact same answer sheet. This is because if each answer sheet was only turned in by at most 4 students, there would at most be 625 · 4 = 2600 students total.

Correct answer  Question 34 1.6 / 1.6 pts Let S be a set of strings defined recursively as follows : 1. 0 ∈ S. 2. For all a ∈ S, 2a0 ∈ S and 2aa1 ∈ S.

(If a and b are strings then ab is the concatenation of a and b. For instance, if a = 0211 and b = 201 then ab = 0211201.)

Let S₀ be the initial population.

Let S₁ be all elements obtained by applying the recursion rule once to S₀.

Let S₂ be all elements obtained by applying the recursion rule once to S₁.

Check all elements of S ₀∪ S₁ ∪ S₂, and only those.

0 200 2001 22000 220010 22002001 2200120011 22001 220011 201 220000 2200020001 22200101 2200100

S₀ = {0} S₁ = {200,2001} S₂ = {22000, 220010, 22002001, 2200120011}.

Correct answer  Question 35 1.6 / 1.6 pts Match the linear, constant-coefficient, homogeneous recurrence relation represented by its characteristic polynomial, to the general solution of the recurrence. (r-1)(r-2) aₙ=p₁+p₂·2ⁿ

(r-1)(r-2)²

aₙ=p₁+(p₂+p₃n)·2ⁿ

(r-1)²(r-2)² aₙ=p₁+p₂n+(p₃+p₄n)·2ⁿ

r²+1

(r-1)(r-2): the roots are distinct, 1 and 2, and each have multiplicity 1. This means the general solution is aₙ=p₁·1ⁿ+p₂·2ⁿ = p₁+p₂·2ⁿ. (r-1)(r-2)²: the roots are distinct, 1 and 2, but now 2 has a multiplicity of 2. This means that the coefficient of the 2ⁿ term gets changed from a constant to a general linear function. The general solution is aₙ= p₁+ (p₂+p₃n)·2ⁿ. (r-1)²(r-2)²: now both roots have multiplicity 2. Both coefficients are general linear functions. The general solution is aₙ=(p₁+p₂n)·1ⁿ+(p₃+p₄n)·2ⁿ = p₁+p₂n+(p₃+p₄n)·2ⁿ. r²+1: this polynomial has imaginary roots, ±i, each with multiplicity 1. The general solution is aₙ=p₁(i)ⁿ+p₂(i)ⁿ.

Correct answer  Question 36 1.6 / 1.6 pts Find a closed-form representation of the following recursively defined sequence: aₙ = 9aₙ₋₂ for n ≥ 2, with initial conditions a₀ = 6 and a₁ = 12. aₙ =(-3)ⁿ +5·3ⁿ aₙ =3·(3)ⁿ +3 aₙ= 6 · 3ⁿ - 2n· 3ⁿ aₙ = 0.75 · 9ⁿ + 5.25 None of these.

The characteristic equation is r2 - 9 = 0, and its roots are r1 = -3 and r2 = 3.

The general solution is an =c1(-3)n +c23n.

When you substitute n=0 and n=1 and the given initial values of the sequence, you can solve for c1 and c2 .

Correct answer  Question 37 1.6 / 1.6 pts Consider the following version of the Fibonacci sequence: f₀ = 1, f₁ = 1 and fₙ = fₙ₋₁ + fₙ₋₂ for all n ≥ 2.

Find f₄.

5

5 (with margin: 0) f₀ = 1 f₁ = 1 f₂ = f₁ + f₀ = 2 f₃ = f₂ + f₁ = 3 f₄ = f₃ + f₂ = 5.

Correct answer  Question 38 1.6 / 1.6 pts Match the following recursively defined sets to their descriptions.

0 ∈ S; for all x ∈ S, x+2 ∈ S. S is the set of non-negative e

1 ∈ S; for all x ∈ S, x+2 ∈ S and x-2 ∈ S. The set of odd integers.

1 ∈ S and 2 ∈ S; for all x ∈ S, x+3 ∈ S. The set of positive integers th

1 ∈ S; for all x ∈ S, 2x ∈ S. The set of powers of two, i.e.

0 ∈ S; for all x ∈ S, x+2 ∈ S:

Here we start with 0, and we get all the other elements by adding 2 to elements we already got. So we get 0, 2, 4, 6, 8, etc.

1 ∈ S; for all x ∈ S, x+2 ∈ S and x-2 ∈ S: Here we start with 1, and we get all the other elements by adding or subtracting 2 to/from elements we already got. Just with the addition, we get 1, 3, 5, 7, 9, etc, and with the subtraction, also -1, -3, -5, -7, etc. This way, we get all odd numbers. We can't get more than the odd numbers because the initial population 1 is odd, and the recursion rules always produce odd numbers from an odd number (that's structural induction reasoning).

1 ∈ S and 2 ∈ S; for all x ∈ S, x+3 ∈ S: We start with 1 and 2, and we find all the other elements by adding 3 to elements we already got. So we get 4 and 5, 7 and 8, etc. Thinking in terms of mod 3 clarifies what is going on: all those numbers have a mod 3 value of 1 or 2. We are getting all positive integers that are not divisible by 3.

1 ∈ S; for all x ∈ S, 2x ∈ S: We start with 1 and keep multiplying by 2. We get the numbers 1, 2, 4, 8, 16, etc. This means that we get the powers of 2 (with positive integer exponents).

Correct answer  Question 39 2 / 2 pts A and B are sets. A contains 12 elements, B contains 8. A and B have 5 elements in common.

How many elements are in A∪B ?

15

15 (with margin: 0) By the inclusion-exclusion principle, |A∪B| = |A| + |B| - |A ∩ B| = 12 + 8 - 5 = 15.

Correct answer  Question 40 2 / 2 pts The intersection of two sets contains 10 elements, and the union 45. What is the sum of the elements in the two sets?

55

55 (with margin: 0)

By the inclusion-exclusion principle, |A∪B| = |A| + |B| - |A ∩ B|.

We substitute the given information and get: 45 = |A| + |B| - 10. Thus|A| + |B|= 55.

Correct answer  Question 41 2 / 2 pts Suppose A and B are sets. No additional information is available. What can you say about |A∪B|? |A∪B| = |A| + |B| |A∪B| ≤ |A| + |B| |A∪B| ≥ |A| + |B| |A∪B| = |A| ∪ |B|

By the inclusion-exclusion principle, |A∪B| = |A| + |B| - |A ∩ B|.

Since |A ∩ B|≥ 0 (cardinalities of sets are always at least zero), we conclude that |A∪B| ≤ |A| + |B|.

Correct answer  Question 42 2 / 2 pts At College X, 130 students were surveyed about their electronic possessions.

90 had a cell phone.

50 had an iPod. 80 had a computer.

70 had a cell phone and a computer.

20 had a cell phone and an iPod. 16 had a computer and an iPod. 13 had a cell phone, an iPod, and a computer.

How many students had at least one of the devices?

127

127 (with margin: 0) We let P be the set of students with a cell phone, I the set with an iPod, and C the set with a computer.

Then the given information translates into the following cardinalities: |P| = 90 |I| = 50

|C| = 80 |P ∩ C| = 70 |P ∩ I | = 20 |I ∩ C| = 16 |P ∩ I ∩ C| = 13.

Using the inclusion-exclusion principle, we obtain |P ∪ I ∪ C| = |P| + |I| + |C| - |P ∩ C| - |P ∩ I | - |I ∩ C| + |P ∩ I ∩ C|. = 90 + 50 + 80 - 70 - 20 - 16 + 13 = 127.

Correct answer  Question 43 2 / 2 pts At a pet owner's meeting of 150 people, there are exactly 67 dog owners, and 56 cat owners. A sociologist wants to interview at least 10 people with unusual pets, which she defines as any pet that is neither cat nor dog. How many people does she need to select at random to interview, at least, to guarantee 10 people with unusual pets?

133

133 (with margin: 0) Let U be the universal set here, i.e. all pet owners at the meeting; A the set of dog owners at the meeting and B the set of cat owners at the meeting. Then |U| = 150; |A| = 67 and |B| = 56.

By the inclusion-exclusion principle,|A∪B| = |A| + |B| - |A ∩ B| ≤ |A| + |B| = 123.

The sociologist will have to interview 133 people to guarantee at least 10 pet owners who are not dog or cat owners.

Correct answer  Question 44 1 / 1 pts Consider the sequence aₙ = 2+5n.

It is arithmetic with common difference 2.

It is arithmetic with common difference 5.

It is geometric with common quotient 2.

It is geometric with common quotient 5.

It is neither arithmetic nor geometric.

It is both arithmetic and geometric. The common quotient is 2, the common difference is 5.

It is both arithmetic and geometric. The common quotient is 5, the common difference is 2.

12/3/25, 1:21 AM

Practice for the Final Exam: MAT 243: Discrete Math Structures (2025 Spring C)

We can't determine whether it is arithmetic or geometric since the definition did not include the information whether it is 1-based or 0-based.

If you know that the general form of an arithmetic sequence is aₙ = a + nd, where d is the common difference, then you see that aₙ = 2 + 5n is an arithmetic sequence with common difference d = 5.

If you didn't know this, then you could confirm it algebraically. The sequence is arithmetic because aₙ₊₁ aₙ is always the same number: aₙ₊₁ - aₙ = (2 + 5(n + 1)) - (2 + 5n) = 5.

This shows not only that the sequence is arithmetic, but arithmetic with common difference 5.

You can see that aₙ = 2 + 5n is not geometric by looking at the first three terms: a₀ = 2 a₁ = 7 a₂ = 12.

The ratio a₁/a₀ is 7/2, but the ratio a₂/a₁ is 12/7, a different number.

Correct answer  Question 45 1 / 1 pts If f is the absolute value function f: ℝ → ℝ; f(x) = |x|, determine the pre-image f⁻¹([-2,1)).

(1,2] [0,1) (-1,1) [0,2]

f⁻¹([-2,1)) is by definition the set of x values that satisfy -2 ≤ |x| < 1.

Since any absolute values is at least zero, and therefore also at least -2, that inequality is equivalent to |x| < 1. This, in turn, is equivalent to -1 < x < 1.

Correct answer  Question 46 1 / 1 pts Pick all that apply. A constant function is always.. https://asu.instructure.com/courses/212193/quizzes/1609475?module_item_id=15542826

25/37

strictly increasing increasing strictly decreasing decreasing none of the other options.

Recall that f increasing means that for all a and b in the domain of f, a < b implies f(a) ≤ f(b). Written symbolically, f is increasing if and only if ∀a,b ( a < b → f(a) ≤ f(b) ).

For a constant function f, the conclusion of the conditional, f(a) ≤ f(b), is always true because f(a) = f(b) for all a, b in the domain. Thus the entire quantified statement is true. f strictly increasing means that for all a and b in the domain of f, a < b implies f(a) < f(b).

For a constant function, f(a) < f(b) is always false, and if there are at least two different numbers in the domain, then there is at least one case of a,b in the domain where a < b. Thus, there is at least one case of a,b in the domain where the conditional is false (true premise but false conclusion). Thus, the for all.. quantification of the conditional is also false.

We conclude that a constant function with a domain that has at least two elements is not strictly increasing. Therefore, it is not true that a constant function is always strictly increasing.

We can similarly justify the answers for decreasing and strictly decreasing.

Correct answer  Question 47 1 / 1 pts The function f: ℝ→ ℝ ; f(x) = x² + 1 is surjective but not injective injective but not surjective bijective neither surjective nor injective

The function is not injective because there are distinct inputs that have the same output, such as -1 and 1.

The function is not surjective because the codomain contains numbers less than 1, which are not produced as outputs.

Correct answer  Question 48 1 / 1 pts Solve the inequality 1 < ⌈ 2x+1⌉ < 6.

(0,2] (0,5/2] (-1/2,5/2] (1/2, 2]

1 < ⌈ 2x+1⌉ < 6 is equivalent to 1 < 2x+1 ≤ 5, which is equivalent to 0 < x ≤ 2.

Correct answer  Question 49 1 / 1 pts True or false? If a and b are real numbers and ⌊ ⌋ represents the floor function, then ⌊ab⌋ = ⌊a⌋ ⌊b⌋.

True False

An example quickly shows that multiplying first and then rounding down is not the same as rounding down and then multiplying: Let a = 1.5 b = 2.5 Then ab = 3.75 and ⌊a⌋ = 1 ⌊b⌋ = 2 ⌊ab⌋ = 3 but ⌊a⌋⌊b⌋ = 2.

Thus, ⌊a⌋⌊b⌋≠ ⌊ab⌋ when a = 1.5 b = 2.5.

Correct answer  Question 50 1 / 1 pts Use a summation formula we learned in class to compute the sum of the integers from 1000 to 5000.

Write the answer in un-simplified form. ½·5000·5001 - ½·999·1000 ½·5000·5001 - ½·1000·1001 5000·5001 - 999·1000 5000·5001 - 1000·1001

You find this sum as the difference of the sum of the integers from 1 to 5000 and the sum of the integers from 1 to 999 (NOT from 1 to 1000).

The sum of the integers from 1 to n is ½·n·(n+1).

Correct answer 

Question 51 1 / 1 pts Which methods may we use for correctly evaluating the sum 3⁴ + 3⁵ + .. + 3⁹ ?

We rewrite the sum as (3⁰ + 3¹ + 3² + 3³ + 3⁴ + 3⁵ + ... + 3⁹) - (3⁰ + 3¹ + 3² + 3³) and apply the geometric summation formula to each term. This produces the solution .

We may simplify this as follows:

The sum has only 6 terms with relatively low exponents. It is still practical to evaluate each term individually and add them: 3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹ = 81 + 243 + 729 + 2187 + 6561 + 19683 = 29484.

We can factor out the common power 3⁴ and thereby write the sum as 3⁴ (3⁰ + 3¹ + 3² + 3³ + 3⁴ + 3⁵). Then we apply the geometric summation formula to simplify this into .

Using laws of exponentiation, we can just add up the exponents. The sum is equal to

You can't add the exponents in a geometric summation. Consider the example 2³ + 2⁴ . That's 8 + 16, which is 24. If you tried to evaluate this using the incorrect law of exponentiation that you can just get the exponents, you get 2⁷, which is 128.

Correct answer  Question 52 1 / 1 pts Given a positive integer n, evaluate

This telescoping sum is evaluated in the lecture.

Correct answer  Question 53 1 / 1 pts The function f: [0, ∞) → ℝ; f(x) = x² + 1 is surjective but not injective injective but not surjective bijective neither surjective nor injective

The function is injective because for non-negative numbers x and y, x² + 1 = y² + 1 implies x=y.

The function is not surjective because the range is [1, ∞), but the codomain is ℝ.

Correct answer  Question 54 1 / 1 pts Given the formula f(x) = x², pick all domain/codomain pairs A,B that would make f: A → B bijective.

A = {1}, B = {1} A = [-1,0], B = [0,1] A = [-1,0), B = (0,1] A = {-1, -1/2, 1/4, 1/8}, B = {1/64, 1/16, 1/4, 1}

All four choices result in a bijection.

Correct answer  Question 55 1 / 1 pts If f: [-2,2] → B; f(x) = x² is surjective, then B = [-4,4] [0,4] [4,4] The answer cannot be determined based on the given information.

f being surjective means that B must be the range of f.

When you square all the numbers in the interval [-2,2], you get exactly the numbers in the interval [0,4].

That means [0,4] is the range of f.

Correct answer  Question 56 1 / 1 pts Check the properties that the function f: [-2, 0) → [0,4], f(x) = x² has.

Injective Surjective Bijective Increasing Decreasing Strictly increasing Strictly decreasing

The function is not surjective because the range is (0,4], but the codomain is [0,4]. The 0 is included in the codomain but not in the range. f is strictly decreasing - the easiest way to verify this is by evaluating the derivative, f'(x) = 2x, which is negative on the domain of f.

Since f is strictly decreasing, it is also decreasing, and injective.

Correct answer  Question 57 1 / 1 pts Consider the sequence aₙ = 3·11ⁿ.

It is arithmetic with common difference 3.

It is arithmetic with common difference 11.

It is geometric with common quotient 3.

It is geometric with common quotient 11.

It is neither arithmetic nor geometric.

It is both arithmetic and geometric. The common quotient is 3, the common difference is 11.

You can see that aₙ = 3·11ⁿ is not arithmetic by computing the first three terms: a₀ =3 a₁ = 33 a₂ = 363.

The difference between the first two terms is 30, but the difference between the 2nd and 3rd term is 330.

The sequence is geometric because it has the general form of a geometric sequence: aₙ = a·qⁿ.

Correct answer  Question 58 1 / 1 pts What do you get when you index shift the sigma sum

so that k starts at 0?

Index shifting really means making a substitution for k and then renaming the new variable back to k.

To get the index started at 0, we need to introduce the new variable j = k - 5. k = 5 means j = 0. k = 9 means j = 4. k + 7 = (j + 5) + 7 = j + 12.

Thus, in terms of j, the sum is back to k.

Correct answer  Question 59 1 / 1 pts Check all correct ways of evaluating

We get the correct answer from there by renaming the j

.

Perhaps the most algebraically efficient evaluation is the one that starts with completing the square, to reduce the number of terms from three to one.

Then we perform an index shift to turn the k+3 into a k:

Finally, we break this sum up into two sums that start at k=1, so we can use summation formulas we know:

If we don't complete the square, then we have to deal with three separate terms. First we should split the sum into a difference, to prepare for the use of our summation formulas for the sum of squares and integers (which both start at index 1):

The first sigma sum can be split into three separate sigma sums (and the 6 can be factored out, i.e. taken in front of the sigma symbol). Each sigma sum can then be evaluated using summation formulas.

For the second sigma sum, this is not worth the formal effort since there are only four k values. There, we prefer to just evaluate the quadratic for all four k values.

.

Correct answer  Question 60 1 / 1 pts Pick all that apply. Any function f: {0} → {0} is.. increasing strictly increasing decreasing strictly decreasing

constant surjective injective

Functions defined on singletons (one-element sets) are always increasing, decreasing, strictly increasing and strictly decreasing. This may seem intuitively strange (how can you say a function has these properties when its graph consists of only one point?) but it makes sense based on the formal definition. f increasing means that for all a and b in the domain of f, a < b implies f(a) ≤ f(b). Written symbolically, f increasing is the case if and only if ∀a,b ( a < b → f(a) ≤ f(b) ).

The for all quantification produces only a single pair of values: a = b = 0. In that case, the premise of the conditional, a < b, is false. With a false premise, the conditional a < b → f(a) ≤ f(b) is true. Thus the entire quantified statement is true. f strictly increasing means that for all a and b in the domain of f with a < b, you have f(a) < f(b).

Symbolically written, f is strictly increasing if and only if ∀a,b ( a < b → f(a) < f(b) ).

This is true for the same reason that was just given for increasing.

You can similarly justify why f is decreasing and strictly decreasing. f is constant because its range only has one value. f is injective because each output only comes from one input. The sole output 0 comes from only one input, 0. f is surjective because its range is {0}, which is equal to the codomain.

Correct answer  Question 61 1 / 1 pts Let f:R→R;f(x)=x². Evaluate the two sets f((-1,1)) and f⁻¹((0,1]). f((-1,1))=[0,1); f⁻¹((0,1])=(-1,1) f((-1,1))=[0,1]; f⁻¹((0,1])=(-1,0) U (0,1) f((-1,1))=[0,1); f⁻¹((0,1])=(0,1) f((-1,1))=[0,1); f⁻¹((0,1])=(-1,0) U (0,1) f((-1,1))=[0,1); f⁻¹((0,1])=[-1,0) U (0,1] f((-1,1))=(0,1); f⁻¹((0,1])=(0,1) f((-1,1))=[0,1]; f⁻¹((0,1])=(-1,1)

Let f : A → B. Then

1. The image of a subset X of A is the set of outputs of the inputs in X.

Formally, f(X)={f(x)|x in X}.

2. The preimage of a subset Y of B is the set of inputs in A whose outputs lie in Y.

Formally, f⁻¹(Y)={x| f(x) in Y}.

Correct answer  Question 62 1 / 1 pts Determine which one of the following students answered the following problem correctly: Is the function f: [0, 1] → ℝ; f(x) = x² injective? Is it surjective? Prove both of your answers based on the definitions of injective and surjective.

Star: the function is neither injective nor surjective. It is not injective because f(-1)=f(1).

It is not surjective because the negative numbers are missing from its range.

Sun: the function is injective. Suppose f(a)=f(b) for some a and b in [0,1]. By definition of f, that means a² = b² which implies a² - b² =0. Factoring the right hand side of the equation, we obtain (a-b)(a+b)=0. Thus, either a=b or a=-b. Since a and b are in [0, 1], a=-b is only possible when a=b=0. Thus, we have proved that" if f(a)=f(b), then a = b" which proves that f is injective.

The function is not surjective. Let a ∈ [0,1]. Then a² ≥ 0. Hence y values less than 0, which occur in the codomain, do not occur as outputs of f.

Sky: the function is injective because it is strictly increasing.

It is not surjective because the range is [0,1], which is a proper subset of the codomain.

Blaze: the function is injective. Suppose f(a)=f(b). By definition of f, that means a² = b². By applying the square root function to both sides, we get a = b.

The function is not surjective. Let a ∈ [0,1]. Then a² ≥ 0. Hence y values less than 0 do not occur as outputs of f.

Shadow Moon: the function is injective. Suppose a=b. Then f(a)=f(b). Thus, the function satisfies the definition of injectivity.

The function is not surjective. Let a ∈ [0,1]. Then a² ≥ 0. Hence y values less than 0, which occur in the codomain, do not occur as outputs of f.

Flow: the function is injective because its derivative f(x)=2x is positive for all x > 0. The mean value theorem we know from calculus 1 then implies that the function is injective.

The function is not surjective. Using calculus again, the absolute maximum of the function is 1 and the absolute minimum is 0. We can now apply the intermediate value theorem from calculus since f is continuous. Therefore, the range of f is [0,1], which is not the stated codomain.

Star's apparent counter-example to injectivity ignores the fact that the domain is [0,1]. There is no f(-1).

The explanation of why f is not surjective is correct, but it does not constitute proof. Star would have had to show that elements of the codomain are missing from the range, not just plausibly claim it.

Similarly, Sky's argument is not proof founded on the definitions. It "delegates" the properties to be proved to other, similarly unproved properties.

Blaze's attempted proof of injectivity is question-begging (assuming the conclusion). Applying the square root function implicitly assumes that g : [0, 1] → [0, 1]; g(x) = x² is bijective , because otherwise, it would have no inverse. So the attempted proof implicitly argues that f is injective because g is bijective, and thus also injective.

How do we know g is injective? Because of the exact same algebra that we would need to carry out (which is shown in Sun's proof) to show that f is injective. Assuming that g is injective is therefore equivalent to assuming that f is injective, i.e. assuming the conclusion.

Shadow Moon's argument for injectivity is based on misunderstanding of that property. That the same input results in the same output is a property of any function.

Blaze, Sun and Shadow Moon all give the same correct proof of surjectivity.

Flow's argument is not incorrect, but it appeals to powerful (and actually fairly difficult to prove) calculus facts. It is not the elementary proof that was asked for (elementary meaning based on the definitions of injective and surjective.)

Correct answer  Question 63 1 / 1 pts True or false? The sum of the squares of the first n positive integers is n(n+1)(2n+1)/6.

True False

This is the summation formula for squares we covered in class.

Correct answer 

Question 64 1 / 1 pts If f is the ceiling function from ℝ to ℝ, what is f((1/2, 3/2))? {1,2} (1,2) [1,2] the empty set

The ceiling of a real number is always an integer. Thus, the ceiling-image of any set of real numbers is always a set of integers.

The real numbers x that satisfy 1/2 < x ≤ 1 have a ceiling of 1.

The real numbers x that satisfy 1 < x < 3/2 have a ceiling of 2.

Correct answer  Question 65 1 / 1 pts Solve the inequality 1 ≤ ⌊2x+1⌋ ≤ 6.

[0, 3) [0,3] [0, 5/2) [0, 5/2]

1 ≤ ⌊2x+1⌋ ≤ 6 is equivalent to 1 ≤ 2x+1 < 7, which is equivalent to 0 ≤ x < 3.

Correct answer  Question 66 1 / 1 pts Check all true statements.

You can always modify a non-injective function f: A→B to become injective, by replacing A by a suitable proper subset of A. You can always modify a non-injective function f: A→B to become injective, by replacing B by the range of f.

By redefining the codomain of a function to make it equal to its range, you can always force the function to become surjective.

By modifying the codomain of a function to be equal to the domain, you make the function bijective.

Failure to be injective means that a function has "too many" inputs. You can always artificially create injectivity by restricting the function to a subset of its domain where outputs are unique.

Failure to be surjective means that a function has "too many" potential outputs. You can always artificially create surjectivity by changing the codomain to the range.

Quiz Score: 100 out of 100