Discrete Math Midterm 2 Practice Problems

2.1.19 What is the cardinality of each of these sets? a) {a} b) {{a}} c) {a, {a}} d) {a, {a}, {a, {a}}}

(a) 1 (b) 1 (c) 2 (d) 3

2.1.21 Find the power set of each of these sets, where a and b are distinct elements: {a} {a, b} {∅, {∅}}

(a) P({a}) = {∅, {a}} (b) P({a, b}) = {∅, {a}, {b}, {a, b}} (c) P({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}}

2.1.10 Determine whether these statements are true or false: a) ∅ ∈ {∅} b) ∅ ∈ {∅, {∅}} c) {∅} ∈ {∅} d) {∅} ∈ {{∅}} e) {∅} ⊂ {∅, {∅}} f) {{∅}} ⊂ {∅, {∅}} g) {{∅}} ⊂ {{∅}, {∅}}

(a) True (b) True (c) False (d) True (e) True, b/c ∅ ∈ {∅} and ∅ ∈ {∅, {∅}} but {∅} 6= {∅, {∅}}. (f) True, b/c {∅} ∈ {{∅}} and {∅} ∈ {∅, {∅}} but {{∅}} 6= {∅, {∅}}. (g) False, b/c {{∅}} = {{∅}, {∅}}, since repetition does not matter for sets.

2.1.32ad Let A = {a, b, c}, B = {x, y}, and C = {0, 1}. Find: a) A × B × C d) B × B × B

(a) {(a, x, 0),(a, x, 1),(b, x, 0),(b, x, 1),(c, x, 0),(c, x, 1), (a, y, 0),(a, y, 1),(b, y, 0),(b, y, 1),(c, y, 0),(c, y, 1)} (d) {(x, x, x),(x, x, y),(x, y, x),(x, y, y),(y, x, x),(y, x, y),(y, y, x),(y, y, y)}

2.3.26b Give an example of an increasing function from R to itself that is not one-to-one.

. Let f : R → R such that f(x) = 8675309 for each x ∈ R. Since f(x) = f(y) for every x, y ∈ R, f is not 1-1, yet f(x) ≤ f(y) whenever x < y, i.e., f is increasing (but clearly not strictly increasing :-).

2.3.24 Let f : R → R and let f (x) > 0 for all x ∈ R. Show that f (x) is strictly increasing if and only if the function g(x) = 1/f (x) is strictly decreasing.

. Let f : R → R such that f(x) > 0 for all x ∈ R. Then f is strictly increasing ⇐⇒ f(x) < f(y) whenever x < y ⇐⇒ 1/(f(y)) < 1/f(x) whenever x < y (b/c f(x) > 0) ⇐⇒ g(y) < g(x) whenever x < y ⇐⇒ g is strictly increasing.

2.4.36 Use the identity 1/(k(k + 1)) = 1/k - 1/(k + 1) and the fact that j = 1 => n Σ(a_j - a_(j - 1) = a_n - a_0, where a_0, a_1, . . . , a_n is a sequence of real numbers to compute k = 1 => n Σ (1/(k(k + 1))).

1 - 1/(n + 1)

4.1.26 List five integers that are congruent to 4 modulo 12.

4, 16, 28, 40, 52

2.2.14 Find the sets A and B if A − B = {1, 5, 7, 8}, B − A = {2, 10}, and A ∩ B = {3, 6, 9}.

A = (A−B)∪(A∩B) = {1, 3, 5, 6, 7, 8, 9} B = (B −A)∪(A∩B) = {2, 3, 6, 9, 10}

2.5.17 If A is an uncountable set and B is a countable set, must A − B be uncountable?

Claim: If A is an uncountable set and B is a countable set, then A−B be uncountable. Proof: Let A be an uncountable set, and let B be a countable set. Assume towards a contradiction that A − B is countable. By problem 16, A ∩ B is countable because A ∩ B ⊆ B and B is countable. Then A is the union of two countable sets, namely, A = (A − B) ∪ (A ∩ B). By Theorem 1 on page 174, the union of two countable sets is countable, thus, A is countable, which is a contradiction. Therefore, by contradiction, A − B must be uncountable. Theorem 1: f A and B are countable sets, then A ∪ B is also countable.

2.2.34 The symmetric difference of A and B, denoted by A ⊕ B, is the set containing those elements in either A or B, but not in both A and B. Draw a Venn diagram for the symmetric difference of the sets A and B.

In the "drawing" below, the universe is colored orange-yellow, the set A is the red square, the set B is the blue square, and their symmetric difference is the portion which is either red or blue. The corner square where A and B intersect is orangeyellow, because it is not part of the symmetric difference.

2.1.18 Find two sets A and B such that A ∈ B and A ⊆ B.

Let A = {a} and B = {a, {a}}. Then A ∈ B because A = {a} ∈ B, and A ⊆ B, because a is the only element of A, and a ∈ B.

2.5.15 Show that if A and B are sets, A is uncountable, and A ⊆ B, then B is uncountable.

Let A and B be sets such that A is uncountable, and A ⊆ B. Assume towards a contradiction that B is countable. Then by our result from Exercise 16, since A ⊆ B, the set A must also be countable, which is a contradiction. Therefore, by contradiction, it must be the case that B is uncountable. Exercise 16: Let A be a countable set, and let S ⊆ A. Since A is countable, we can write the elements of A as a list: a1, a2, a3, . . . . In order to write the elements of S as a list, simply take the list of the elements of A, and remove any a_k ∈ A − S. Then we are left with a list of the elements of S

2.5.16 Show that a subset of a countable set is also countable.

Let A be a countable set, and let S ⊆ A. Since A is countable, we can write the elements of A as a list: a1, a2, a3, . . . . In order to write the elements of S as a list, simply take the list of the elements of A, and remove any a_k ∈ A − S. Then we are left with a list of the elements of S.

4.1.5 Show that if a | b and b | a, where a and b are integers, then a = b or a = −b.

Let a, b ∈ Z be arbitrary. Suppose that a|b and b|a. (This means that a, b =/= 0.) Since a|b, then b = aj for some j ∈ Z. Since b|a, then a = bk for some k ∈ Z. Thus, a = bk = (aj) = a(jk). Since a =/= 0, then if we divide both sides of a = a(jk) by a we get 1 = jk. Since j, k ∈ Z and 1 = jk, then either j = k = 1 or j = k = −1. If j = k = 1, then a = b, and if j = k = −1, then a = −b.

4.1.4 Theorem 1 says that Let a, b, and c be integers, where a = 0. Then (i ) if a | b and a | c, then a | (b + c); (ii ) if a | b, then a | bc for all integers c; (iii ) if a | b and b | c, then a | c. Prove that part (iii ) of Theorem 1 is true.

Let a, b, c ∈ Z be arbitrary where a =/= 0. Suppose that a|b and b|c. Then b = aj and c = bk for some j, k ∈ Z. Thus, c = bk = (aj)k = a(jk). Since jk ∈ Z, then a|c.

4.1.34 Show that if a ≡ b (mod m) and c ≡ d (mod m), where a, b, c, d, and m are integers with m ≥ 2, then a − c ≡ b − d (mod m).

Let a, b, c, d, m ∈ Z be arbitrary with m ≥ 2. Suppose that a ≡ b (mod m) and c ≡ d (mod m). Then a = b + km and c = d + jm for some k, j ∈ Z. Considera − c = (b + km) − (d + jm) = b − d + (k − j)m. Since (k − j) ∈ Z, then a − c ≡ b − d (mod m).

4.1.36 Show that if a, b, c, and m are integers such that m ≥ 2, c > 0, and a ≡ b (mod m), then ac ≡ bc (mod mc).

Let a, b, c, m ∈ Z be arbitrary with m ≥ 2 and c > 0. Suppose that a ≡ b (mod m). Then a = b + km for some k ∈ Z. Consider ac = (b + km)c = bc + k(mc). Since k ∈ Z, then ac ≡ bc (mod mc).

4.1.35 Show that if n | m, where n and m are integers greater than 1, and if a ≡ b (mod m), where a and b are integers, then a ≡ b (mod n).

Let a, b, m, n ∈ Z be arbitrary with n, m ≥ 2. Suppose that n|m and a ≡ b (mod m). Since n|m, then m = kn for some k ∈ Z. Since a ≡ b (mod m), then a = b + jm for some j ∈ Z. Thus, a = b + j(kn) = b + (jk)n. Since jk ∈ Z, then a ≡ b (mod n).

2.3.39 Show that the function f (x) = ax + b from R to R is invertible, where a and b are constants, with a = 0, and find the inverse of f .

Let f : R → R such that f(x) = ax+b, where a =/= 0 and b are constants. Let g : R → R satisfy g(x) = (x−b)/a. Notice that f(g(x)) = a((x − b)/a) + b = x, for each x ∈ R, and g(f(x)) = ((ax + b) − b)/a = x, for each x ∈ R. Thus, f is invertible and g = f^(−1) .

2.3.38 Let f (x) = ax + b and g(x) = cx + d, where a, b, c, and d are constants. Determine necessary and sufficient conditions on the constants a, b, c, and d so that f ◦ g = g ◦ f .

Let f(x) = ax + b and g(x) = cx + d, where a, b, c, d are constants. Suppose that f ◦ g = g ◦ f, then (f ◦ g)(x) = (g ◦ f)(x) f(g(x)) = g(f(x)) a(cx + d) + b = c(ax + b) + d (ac)x + (ad + b) = (ac)x + (bc + d). Since we have two polynomials which equal each other, their coefficients must equal each other, i.e., we must have that ac = ac and ad + b = bc + d.

2.3.34 If f and f ◦ g are one-to-one, does it follow that g is one-to-one? Justify your answer.

Let g : A → B and f : B → C such that f and f ◦ g are both 1-1. Assume towards a contradiction that g is not 1-1. Then there are a1, a2 ∈ A such that a1 6= a2, yet g(a1) = g(a2) = b. Thus, (f ◦ g)(a1) = f(g(a1)) = f(b) = c and (f ◦ g)(a2) = f(g(a2)) = f(b) = c, i.e., (f ◦ g)(a1) = (f ◦ g)(a2). But this is a contradiction since a1 6= a2 and f ◦ g is 1-1. Therefore, by contradiction, it must be the case that g is 1-1.

2.3.33b Suppose that g is a function from A to B and f is a function from B to C. Show that if both f and g are onto functions, then f ◦ g is also onto.

Let g : A → B and f : B → C such that f and g are both onto. Let c ∈ C be arbitrary. Since f is onto, f(b) = c for some b ∈ B. Since g is onto, b = g(a) for some a ∈ A. Thus, (f ◦ g)(a) = f(g(a)) = f(b) = c, i.e., f ◦ g : A → C is onto.

2.5.28 w that the set Z^+ × Z^+ is countable.

We could describe more verbally how to list all the elements of Z^+ × Z^+ as follows: 1: first list all elements (i, j) ∈ Z^+ × Z^+ such that i + j = 2: (1, 1) 2: then list all elements (i, j) ∈ Z^+ × Z^+ such that i + j = 3, where we list (a, b) before (c, d) if a < c: (1, 2),(2, 1) 3: then list all elements (i, j) ∈ Z^+ × Z^+ such that i + j = 4, where we list (a, b) before (c, d) if a < c: (1, 3),(2, 2),(3, 1)

2.1.22 Can you conclude that A = B if A and B are two sets with the same power set?

Yes. You can look at all of the singletons in P(A) and P(B). If P(A) and P(B) have exactly the same set of singletons, then A and B have exactly the same set of elements. Alternatively, let S ∈ P(A) be a maximal set in P(A) with respect to containment, i.e., there does not exist C ∈ P(A) such that S ⊂ C. You can check that this maximal set will be A itself. Do the same for B, then make an argument that if the maximal element of P(A) equals the maximal element of P(B), then A = B.

2.3.8 Find these values. a) FLR(1.1) b) CLG(1.1) c) FLR(−0.1) d) CLG(−0.1) e) CLG(2.99) f ) CLG(−2.99) g) FLR(1/2 + CLG(1/2)) h) CLG(FLR(1/2) + CLG(1/2) + 1/2)

a) 1 b) 2 c) -1 d) 0 e) 3 f) -2 g) 1 h) 2

4.1.21 Evaluate these quantities. a) 13 mod 3 b) −97 mod 11 c) 155 mod 19 d) −221 mod 23

a) 1 b) 2 c) 3 d) 9

4.1.20 Evaluate these quantities. a) −17 mod 2 b) 144 mod 7 c) −101 mod 13 d) 199 mod 19

a) 1 b) 4 c) 3 d) 9

2.4.30 What are the values of these sums, where S = {1, 3, 5, 7}? a) j ∈ SΣj b) j ∈ SΣj^2 c) j ∈ SΣ(1/j ) d) j ∈ SΣ1

a) 16 b) 84 c) 176/105 d) 4

4.1.9abc What are the quotient and remainder when a) 19 is divided by 7? b) −111 is divided by 11? c) 789 is divided by 23?

a) 19 = 7(2) + 5, thus, 19 div 7 = 2 and 19 mod 7 = 5 b) −111 = 11(−11) + 10, thus, −111 div 11 = −11 and −111 mod 11 = 10 c) 789 = 23(34)+7, thus, 789 div 23 = 34 and 789 mod 23 = 7

2.4.32cd Find the value of each of these sums. c) j = 0 => 8Σ(2(3^(j)) + 3 * 2^(j)) d) j = 0 =. 8Σ(2^(j + 1) - 2^j)

a) 21215 b) 511

2.2.2 Suppose that A is the set of sophomores at your school and B is the set of students in discrete mathematics at your school. Express each of these sets in terms of A and B. a) the set of sophomores taking discrete mathematics in your school b) the set of sophomores at your school who are not taking discrete mathematics c) the set of students at your school who either are sophomores or are taking discrete mathematics d) the set of students at your school who either are not sophomores or are not taking discrete mathematics.

a) A ∩ B b) A − B c) A ∪ B d) A ∪ B

2.2.32 The symmetric difference of A and B, denoted by A ⊕ B, is the set containing those elements in either A or B, but not in both A and B. Find the symmetric difference of {1, 3, 5} and {1, 2, 3}.

a) B ⊆ A b) A ⊆ B c) A ∩ B = ∅ d) Doesn't imply anything about A and B e) A = B

2.5.1 Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and that set. a) the negative integers b) the even integers c) the integers less than 100 d) the real numbers between 0 and 1/2 e) the positive integers less than 1,000,000,000 f ) the integers that are multiples of 7

a) Countably infinite: f: Z^+ => {-1, -2, -3, . . . } where f(n) = -n. b) Countably infinite: f: Z^+ => 2Z where f(n) = (-1)^(n)2*FLR(n/2). c) Countably infinite: f: Z^+ => {99, 98, 97, 96, . . . } where f(n) = 100 - n. d) Uncountable. e) Finite. f) Countably infinite: f: Z^+ => 7Z where f(n) = (-1)^(n)7*FLR(n/2).

4.1.37 Find counterexamples to each of these statements about congruences. a) If ac ≡ bc (mod m), where a, b, c, and m are integers with m ≥ 2, then a ≡ b (mod m). b) If a ≡ b (mod m) and c ≡d (mod m), where a, b, c, d, and m are integers with c and d positive and m ≥ 2, then ac ≡ bd (mod m).

a) Let a = 1, b = 2, c = 2, m = 2. Then 1(2) ≡ 2(2) (mod 2), but 1 ≡/≡ 2 (mod 2)

2.3.20 Give an example of a function from N to N that is a) one-to-one but not onto. b) onto but not one-to-one. c) both onto and one-to-one (but different from the identity function). d) neither one-to-one nor onto.

a) Let f(n) = 2n. If f(n) = f(m), then 2n = 2m which implies m = n. Thus, f is 1-1. However, there is no preimage of 3 (or any odd natural number). Thus, f is not onto. b) Let f(n) = FLR(n/2). Since f(2) = 1 = f(3), then f is not 1-1. However, f is onto. Indeed, let k ∈ N be arbitrary. Since 2k ∈ N also, then f(2k) = FLR(2k/2) = k. c) Using the fact that N = {0, 1, 2, 3, . . .} we can define f as follows: f(n) = n + 2 if n is even. f(n) = n - 1 if n is odd. It can be easily checked that f is, in fact, its own inverse, i.e., f(f(n)) = n for each n ∈ N d) Let f(n) = 8675309 for each n ∈ N. Then f is neither 1-1 nor onto. For instance, f is not 1-1 because f(1) = f(2), and f not onto because there is no preimage of, say, 1.

2.2.16 Let A and B be sets. Show that a) (A ∩ B) ⊆ A. b) A ⊆ (A ∪ B). c) A − B ⊆ A. d) A ∩ (B − A) = ∅. e) A ∪ (B − A) = A ∪ B.

a) Let x ∈ A ∩ B. Then x ∈ A and x ∈ B. Thus, x ∈ A. Therefore, (A ∩ B) ⊆ A. b) Let x ∈ A. Then x ∈ A or x ∈ B. Thus, x ∈ A ∪ B. Therefore, A ⊆ A ∪ B. c) Let x ∈ A − B. Then x ∈ A and x 6∈ B. Thus, x ∈ A. Therefore, (A − B) ⊆ A. d) Assume towards a contradiction that x ∈ A∩(B−A). Then x ∈ A and x ∈ B−A. Since x ∈ B−A, then x ∈ B and x 6∈ A. Thus, x 6∈ A. But this contradicts the fact that x ∈ A. Therefore, by contradiction, it must be the case that A∩(B−A) = ∅. e) x ∈ A ∪ (B − A) ⇐⇒ (x ∈ A) ∨ (x ∈ B − A) Def. ⇐⇒ (x ∈ A) ∨ [(x ∈ B) ∧ ¬(x ∈ A)] Def. ⇐⇒ [(x ∈ A) ∨ (x ∈ B)] ∧ [(x ∈ A) ∨ ¬(x ∈ A)] Distrib. ⇐⇒ [(x ∈ A) ∨ (x ∈ B)] ∧ T Negation Law ⇐⇒ (x ∈ A) ∨ (x ∈ B) Identity Law ⇐⇒ (x ∈ A ∪ B) Def.

4.1.29 Decide whether each of these integers is congruent to 5 modulo 17. a) 80 b) 103 c) −29 d) −122

a) No b) No c) Yes d) No

2.1.46 This exercise presents Russell's paradox. Let S be the set that contains a set x if the set x does not belong to itself, so that S = {x | x ∉ x}. a) Show the assumption that S is a member of S leads to a contradiction. b) Show the assumption that S is not a member of S leads to a contradiction.

a) Suppose that S ∈ S. By definition, every x ∈ S satisfies x 6∈ x, thus, S 6∈ S. This contradicts the assumption that S ∈ S. b) Suppose that S 6∈ S. By its definition, S contains every set x such that x 6∈ x. Since S 6∈ S, then S ∈ S (by the definition of S). This, of course, is a contradiction.

2.3.11 Which functions are onto? a) f (a) = b, f (b) = a, f (c) = c, f (d) = d b) f (a) = b, f (b) = b, f (c) = d, f (d) = c c) f (a) = d, f (b) = b, f (c) = c, f (d) = d

a) Yes onto b) not onto b/c no preimage of a c) not onto b/c no preimage of a

2.1.5 Determine whether each of these pairs of sets are equal: a) {1, 3, 3, 3, 5, 5, 5, 5, 5}, {5, 3, 1} b) {{1}}, {1, {1}} c) ∅, {∅}.

a) Yes, they are equal. Repetition and order doesn't matter for sets. (b) No, 1 ∈ {1, {1}}, but 1 6∈ {{1}}. (c) No. The first set has no elements, while the second set has exactly one element. ∅ 6∈ ∅, but ∅ ∈ {∅}.

2.4.26abcd For each of these lists of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list.Assuming that your formula or rule is correct, determine the next three terms of the sequence. a) 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, . . . b) 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, . . . c) 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, . . . d) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, . . .

a) a_n = a_(n - 1) + (2n + 1) The next three terms are 123, 146, 171. b) a_n = 4n + 7, n ∈ N The next three terms are 47, 51, 55 c) The sequence is the positive integers written in binary. The next three terms are 1100, 1101, 1110. d) the sequence repeats the first number 1 time, the second number 3 times, the third number 5 times, the fourth number 7 times, . . . , the nth number 2n − 1 times. The numbers might be satisfying that the next number is derived from adding the previous two different numbers. Alternatively, it could be that for some reason the sequence starts with a 1, then switches to using the primes as the numbers in the sequence. The next three terms are: 8, 8, 8 if we use the "add the previous two different numbers rule" or: 7, 7, 7 if we use the "primes rule."

2.3.23 Determine whether each of these functions is a bijection from R to R. a) f (x) = 2x + 1 b) f (x) = x^2 + 1 c) f (x) = x^3 d) f (x) = (x^2 + 1)/(x^2 + 2)

a) f is a bijection. We illustrate this by presenting f^(−1)(x) = (x−1)/2. Since f(f^(−1)(x)) = 2((x−1)/2) + 1 = x and f^(−1)(f(x)) =((2x+1)−1)/2 = x for each x ∈ R, we can see that f is invertible, and thus a bijection. We could also show that f is 1-1 and f is onto separately. Suppose that f(x1) = f(x2), then 2x1 − 1 = 2x2 − 1, which implies, 2x1 = 2x2, thus x1 = x2. Therefore, f is 1-1. Let x ∈ R be arbitrary. Then (x−1)2 is the preimage of x, since f((x−1)/2) = 2(x−1)/2+ 1 = x. Therefore, f is onto. b) f is not 1-1. Since f(1) = 2 = f(−1), then f is not 1-1. Since f is not 1-1, then f is not a bijection. (It is also possible to show that f is not onto since f(x) ≥ 1 for each x ∈ R c) f is a bijection. We can illustrate this by presenting f^(−1)(x) = x^(1/3). Since f(f^(−1)(x)) = (x^(1/3))^3 = x and f^(−1)(f(x)) = x^(1/3)^3 = x for each x ∈ R, we can see that f is invertible, and thus a bijection. It is not that interesting to show that f is 1-1 and f is onto separately. d) f is not 1-1 because f(1) = 2/3 = f(−1). Since f is not 1-1, then f is not a bijection. (It is also possible to show that f is not onto, since f(x) ≥ 1/2 for each x ∈ R

2.3.2 Determine whether f is a function from Z to R if a) f (n) = ±n. b) f (n) = √(n^2 + 1). c) f (n) = 1/(n^2 − 4).

a) f is not a function because when n 6= 0, f(n) is not a unique value (it is two values ±n) b) f is a function c) f is not a function from Z to R, because f(2) and f(−2) do not exist (f does not assign a real number to 2 or −2)

2.3.14 Determine whether f : Z × Z → Z is onto if a) f (m, n) = 2m − n. b) f (m, n) = m^2 − n^2. c) f (m, n) = m + n + 1. d) f (m, n) = |m| − |n|. e) f (m, n) = m^2 − 4.

a) f is onto. Let k ∈ Z be arbitrary. f(0, −k) = 2(0) − (−k) = k b) f is not onto. There is no preimage of 2. Assume towards a contradiction that f(m, n) = 2. Then m2 − n 2 = 2, where m, n ∈ Z. Thus, (m − n)(m + n) = 2 where m, n ∈ Z. Since m − n ∈ Z and m + n ∈ Z, and the only factors of 2 are 1 and 2 or −1 and −2, then one of four cases must occur: Case 1: m − n = 1 and m + n = 2 2 Case 2: m − n = 2 and m + n = 1 Case 3: m − n = −1 and m + n = −2 Case 4: m − n = −2 and m + n = −1 In Cases 1 and 2, if we add both equations, we see that 2m = 3, or equivalently m = 3 2 6∈ Z. In Cases 3 and 4, by adding both equations we arrive at 2m = −3, which implies that m = − 3 2 6∈ Z. Thus, in each case we reach the contradiction that m 6∈ Z. Therefore, by contradiction, it must be the case that 2 has no preimage in Z × Z. c) f is onto. Let k ∈ Z be arbitrary. f(k, −1) = k + (−1) + 1 = k d) f is onto. Let k ∈ Z be arbitrary. If k ≥ 0, then f(k, 0) = |k| − |0| = |k| = k (because k ≥ 0). If k < 0, then f(0, k) = |0| − |k| = −|k| = −(−k) (because k < 0) = k. e) f is not onto. For any k < −4, k has no preimage. This is because m2 ≥ 0 for each m ∈ Z. Thus, f(m, n) ≥ −4 for each (m, n) ∈ Z × Z

2.4.37 Sum both sides of the sides k^2 - (k - 1)^2 = 2k - 1 from k =1 to k = n and use the fact that j = 1 => n Σ(a_j - a_(j - 1)) = a_n - a_0 to find a) a formula for k = 1 => n Σ(2k -1). b) a formula for k = 1 => n Σk

a) n^2 b) (n(n + 1))/2

2.3.10 Determine whether each of these functions from {a, b, c, d} to itself is one-to-one. a) f (a) = b, f (b) = a, f (c) = c, f (d) = d b) f (a) = b, f (b) = b, f (c) = d, f (d) = c c) f (a) = d, f (b) = b, f (c) = c, f (d) = d

a) yes; 1-1 b) not 1-1 b/c f(a) = b = f(b), but a =/= b c) not 1-1 b/c f(a) = d = f(d), but a =/= d

2.2.4 Let A = {a, b, c, d, e} and B = {a, b, c, d, e, f, g, h}. Find: a) A ∪ B. b) A ∩ B. c) A − B. d) B − A.

a) {a, b, c, d, e, f, g, h} b) {a, b, c, d, e} c) ∅ d) {f, g, h}

2.4.4d What are the terms a0, a1, a2, and a3 of the sequence {an}, where an equals 2^n + (−2)^n?

a_0 = 2 a_1 = 0 a_2 = 8 a_3 = 0

2.4.2d What is the term a8 of the sequence {an} if an equals -(−2)^n?

a_8 = -256

2.4.33bc Compute each of these double sums. b) i = 0 => 2 Σ j = 0 => 3 Σ(2i + 3j) c) i = 1 => 3 Σ j = 0 => 2 (i)

b) 78 c) 18

2.4.35 Show that j = 1 => n Σ(a_j - a_(j - 1)) = a_n - a_0, where a_0, a_1, . . . , a_n is a sequence of real numbers. This type of sum is called telescoping.

j = 1 => n Σ(a_j - a_(j - 1)) = (a_1 - a_0) + (a_2 - a_1) + (a_3 - a-2) + . . . + (a_(n - 1) - a_(n-2)) + (a_n - a_(n - 1) = -a_0 + a_n = a_n - a_0.

2.1.35 How many different elements does A × B have if A has m elements and B has n elements?

mn

2.2.24 Let A, B, and C be sets. Show that (A − B) − C = (A − C) − (B − C).

x ∈ (A − B) − C ⇐⇒ (x ∈ A − B) ∧ ¬(x ∈ C) Def. ⇐⇒ [(x ∈ A) ∧ ¬(x ∈ B)] ∧ ¬(x ∈ C) Def. ⇐⇒ [(x ∈ A) ∧ ¬(x ∈ C)] ∧ ¬(x ∈ B) Comm. & Assoc. ⇐⇒ (x ∈ A − C) ∧ ¬(x ∈ B) Def. ⇐⇒ [(x ∈ A − C) ∧ ¬(x ∈ B)] ∨ F Identity ⇐⇒ [(x ∈ A − C) ∧ ¬(x ∈ B)] ∨ [(x ∈ A) ∧ F] Domination ⇐⇒ [(x ∈ A − C) ∧ ¬(x ∈ B)] ∨ [(x ∈ A) ∧ [¬(x ∈ C) ∧ (x ∈ C)]] Negation ⇐⇒ [(x ∈ A − C) ∧ ¬(x ∈ B)] ∨ [[(x ∈ A) ∧ ¬(x ∈ C)] ∧ (x ∈ C)] Assoc. ⇐⇒ [(x ∈ A − C) ∧ ¬(x ∈ B)] ∨ [(x ∈ A − C) ∧ (x ∈ C)] Def. ⇐⇒ (x ∈ A − C) ∧ [¬(x ∈ B) ∨ (x ∈ C)] Distrib. ⇐⇒ (x ∈ A − C) ∧ ¬[(x ∈ B) ∧ ¬(x ∈ C)] DeMorgan & Dbl Neg. ⇐⇒ (x ∈ A − C) ∧ ¬(x ∈ B − C) Def. ⇐⇒ x ∈ (A − C) − (B − C) Def

2.2.19a Show that if A and B are sets, then A − B = A ∩ B.

x ∈ A − B ⇐⇒ (x ∈ A) ∧ ¬(x ∈ B) Def. ⇐⇒ (x ∈ A) ∧ (x ∈ B) Def. ⇐⇒ (x ∈ A ∩ B) Def.

2.2.13 Prove the second absorption law from Table 1 by showing that if A and B are sets, then A ∩ (A ∪ B) = A.

x ∈ A ∩ (A ∪ B) ⇐⇒ (x ∈ A) ∧ (x ∈ A ∪ B) Def. ⇐⇒ (x ∈ A) ∧ [(x ∈ A) ∨ (x ∈ B)] Def. ⇐⇒ (x ∈ A) Logical Absorption

2.2.35 The symmetric difference of A and B, denoted by A ⊕ B, is the set containing those elements in either A or B, but not in both A and B. Show that A ⊕ B = (A ∪ B) − (A ∩ B).

x ∈ A ⊕ B ⇐⇒ [(x ∈ A) ∨ (x ∈ B)] ∧ ¬[(x ∈ A) ∧ (x ∈ B)] Def. of Symmetric Difference ⇐⇒ (x ∈ A ∪ B) ∧ ¬(x ∈ A ∩ B) Def's of Union & Intersection ⇐⇒ [x ∈ (A ∪ B) − (A ∩ B)] Def. of Difference of Sets