(1) Which of the following are propositions? If it is a proposition, determine its truth value: A propositional function, but not a proposition.

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1 Math 231 Exam Practice Problem Solutions WARNING: This is not a sample test. Problems on the exams may or may not be similar to these problems. These problems are just intended to focus your study of the topics. Note: The final is cumulative. MIDTERM I: (1) Which of the following are propositions? If it is a proposition, determine its truth value: (a) Is Jack 20 years old? Not a proposition. Questions have no truth value and thusly are not propositions. (b) Do your homework! Not a proposition. Commands have no truth value and thusly are not propositions. (c) This statement is not a proposition. This is a false proposition. (d) x < 2. A propositional function, but not a proposition. (e) If 3 < 2 then 1 R. This is a proposition and its true (by default). (f) For all finite sets X and Y, it follows that X Y = X Y. This is a true proposition. (g) Does induction require the base case(s)? This is not a proposition (but the answer to the question is yes).

2 (h) x R, y R, y 4 = x. This is a false proposition (as there is no real even root of a negative). (i) x R, y R, y 4 > x. This is a true proposition (as one can always find y such that the fourth power is more than a given x). (j) y Q, x R, y > x This is a false proposition (for any rational there exists a real larger than that rational minus 10). (k) y Q, x R, y < x 2. This is a true proposition (use y = 1). (l) y Q, x R, y < x 3. This is a false proposition (as for any y there is an x such that x 3 y).

3 (2) Let U = {n Z 0 n 9}. Let A = {2, 3, 5, 7} and B = {2, 4, 6, 8}. (a) Find A B. {3, 5, 7} (b) Find A B. {0, 1, 9} (c) How many elements does the power set P(A) have? 16 (3) Assume p and q are true propositions, while r is a false propositions. Find the truth value of the following propositions. (a) (p q) (p (q r)). True (b) ((p q) r) (p q). False

4 (4) Assume P (x) and Q(x) are propositional functions on the domain D and r D. Suppose it is given that P (r) is true and Q(r) is false. For each quantified proposition, first determine if we have enough information to give it a true value. If there is enough information, report the truth value. (a) x, P (x). True since P (r) is true. (b) x, Q(x). There is not enough information as Q(x) could be true for some x D such that x r. (c) x, y, Q(y) P (x). True since for any x we can set y to r, making the implication true (by default). (d) x, y, P (x) Q(y). False since x and y both set to r makes the implication is false.

5 (5) Let the domain of discourse be the real numbers. Determine the truth value of the following quantified propositions. (a) x, (x < 2 x 2 < 4). False. Let x = 2. (b) x, (x < 2 x 2 < 4). True. For instance, pick x = 3. (6) Let the domain of discourse be the direct product of the real numbers with the real numbers. Determine the truth value of the following quantified propositions. (a) x, y, x 2 + y < 0. False. Let x be a real number. Set y = 1. Then x < 0 is not true. (b) x, y, x 2 + y < 0. True. Let x be a real number. Set y = x 2 1. Then x 2 + y = 1 < 0. (7) Negate the quantified proposition in Problem (6), Part (a). x, y, x 2 + y 0.

6 (8) Prove the following propositions or find a counterexample. (a) If m is an even integer and n is an odd integer then mn is even. Proof : Let m be an even integer and n be an odd integer. Then k Z such that m = 2k. Therefore, mn = (2k)n = 2(kn), which is even since kn Z (b) Let m, n be integers. If mn is an even integer then m and n are even integers. Counterexample: Let m = 1 and n = 2. Then mn = 2 is even, but not both m and n are even. (9) Use induction to prove the following proposition: For all integers n 4, n 2 2 n. Proof: First we prove the following claim: For all integers n 4, 2n n. We shall prove this by induction: Base Case: 2(4) + 1 = 9 16 = 2 4. Let n 4 be an integer and assume 2n+1 2 n. Then 2(n+1)+1 = 2n n +2 n = 2 n+1. So the claim has been established by induction. Now we prove the required inequality: Base case: 4 2 = = 2 4. Let n 4 be an integer and assume n 2 2 n. Then (n + 1) 2 = n 2 + 2n n + 2 n = 2 n+1 and so the proof is done by induction

7 (10) Use induction to prove the following proposition: For all integers, n 1, (2n 1) 2 = 4n3 n. 3 Proof: We begin with a base case: 1 2 = 1 and 4(1)3 1 3 Let n N and assume = 1. Then (2n 1) 2 = 4n3 n (2n 1) 2 + (2n + 1) 2 = 4n3 n 3 + (2n + 1) 2 n(2n 1)(2n + 1) = + (2n + 1) 2 3 [ ] n(2n 1) = (2n + 1) + (2n + 1) 3 [ 2n 2 ] n + 6n + 3 = (2n + 1) 3 [ 2n 2 ] + 5n + 3 = (2n + 1) 3 = = (2n + 1)(n + 1)(2n + 3) 3 (n + 1)[2(n + 1) 1][2(n + 1) + 1] 3 That completes the proof by induction = 4(n + 1)3 (n + 1). 3

8 (11) Use induction to prove the following proposition: n ( ) 2 n(n + 1) n N, i 3 =. 2 i=1 1 ( ) 2 (1)(1 + 1) Proof: We begin with the base case. i 3 = 1 3 = 1 and = 1. 2 Let n N. Assume that Then i=1 n ( ) 2 n(n + 1) i 3 =. 2 i=1 n+1 i 3 = i=1 ( n ) i 3 + (n + 1) 3 i=1 = ( ) 2 n(n + 1) + (n + 1) 3 2 = (n + 1) 2 [ n 2 4 = (n + 1) 2 ( n ) 2 ] + (n + 1) = ( ) 2 (n + 1)[(n + 1) + 1]. 2 That completes the proof by induction (12) Using induction prove that n N, 5 n 1 is divisible by 4. Proof: We begin with the base case = 4 is divisible by 4. Let n N. Assume that 5 n 1 is divisible by 4. So there exists k Z such that 5 n 1 = 4k. Consider that 5 n+1 1 = 5(5 n ) 1 = 4(5 n ) + 5 n 1 = 4(5 n + k). Since 5 n + k Z it follows that 5 n+1 1 is divisible by 4 thereby completing the proof by induction

9 (13) Prove the following proposition or find a counterexample: For all sets A, B and C, if A C = B C then A = B. Counterexample: Let A = {0}, B = {1}, and C =. Then A C = B C = but A B. (14) Prove the following proposition or find a counterexample: For all sets A, B and C, if C and A C = B C then A = B. Proof: Let A, B, C be sets and assume both C and A C = B C. If A is empty then = A C = B C, which means B must be empty since C is non-empty. By symmetry, if B is empty then so is A. Either way A = B. Assume neither A nor B is empty. Let x A and y C. Then (x, y) A C, but since A C = B C, it follows that (x, y) B C, which implies that x B. By symmetry if x B then x A. So A and B are subsets of each other and thus A = B

10 MIDTERM II: (15) True or False? (a) If f g is one-to-one, then g is one-to-one. True (can you prove it?) (b) If f g is onto, then g is onto. False (can you find a counterexample?) (c) The function f : N N given by f(n) = n 2 is injective (1-1). True (d) The function f : N N given by f(n) = n 2 is surjective (onto). False (e) The function f : N N given by f(n) = n 2 is bijective. False (f) There are 60 rearrangements of the word AP P LE. True

11 (16) How many 5-card deals from a standard deck of 52 contain at least one card from each of the four suits? A 5-card deal containing at least one card from each suit necessarily contains two cards of one suit and a single card each from the other three suits. So choose a suit (C(4, 1) = 4 choices), then two cards of that suit (C(13, 2) = 78 choices), then a card each from three remaining suits (C(13, 1) 3 = 2, 197 choices). So by the multiplication principle, there are......c(4, 1)C(13, 2)C(13, 1) 3 = 685, 464 such deals. (17) In how many ways can 3 boys and 5 girls stand in a line such that no two boys are standing next to each other? Arrange the girls. There are 5! = 120 ways to do this. That gives 6 potential positions for a boy (on either side of a girl). So there are P (6, 3) = 20 ways to arrange the boys into 3 of 6 of those positions. So by the multiplication principle, there are......(120)(20) = 2, 400 such arrangements.

12 (18) Suppose 8 fish are to distributed in 3 distinct tanks. In how many ways can this be done if... (a)...the fish are distinguishable? Each distinguishable fish can go into any of the 3 tanks. distribute the fish. So there are 3 8 = 6, 561 ways to (b)...the fish are indistinguishable? Selections of two slots to be s (separations symbols) among = 10 slots corresponds (bijectively) to distributions of the indistinguishable fish into 3 tanks. So there are C(10, 2) = 45 such distributions. (c)...the fish are indistinguishable and each tank must get a fish? Selections of two slots to be s (separations symbols) among 7 slots (F F F F F F F F ) corresponds (bijectively) to distributions of the indistinguishable fish into 3 tanks with at least one fish in each tank. So there are C(7, 2) = 21 such distributions. (19) How many non-negative integer solutions are there to x 1 + x 2 + x 3 = 20? Corresponds to counting how many choices of 2 slots there are among = 22 slots (two slots to be separation symbols). The other twenty slots are filled with +1 s. The the left of the first separation symbols is x 1, in between is x 2, and to the right of the second separation symbol is x 3. So there are C(22, 2) = 231 such solutions.

13 (20) Show that a nonempty set has the same number of subsets with an odd number of elements as it has with an even number of elements. Proof: Let n N and X be a set with n elements. n Using the binomial theorem, (x + y) n = C(n, k)x n k y k, with x = 1 and y = 1 we get: k=0 n C(n, k)( 1) k = C(n, 0) C(n, 1) + + ( 1) n C(n, n) = 0. k=0 By rearranging terms we get C(n, 0)+C(n, 2)+ +C(n, n 1) = C(n, 1)+C(n, 3)+ +C(n, n) when n is odd and C(n, 0) + C(n, 2) + + C(n, n) = C(n, 1) + C(n, 3) + + C(n, n 1) when n is even. Thus the number of subsets of X with an odd number of elements (sum of C(n, k) for odd k) is equal to the number of subsets of X with an even number of elements (sum of C(n, k) for even k) (21) Assume f : Z A given by f(x) = 2x 5 is onto. Determine the set A. A would be the set of odd integers. (22) Determine if g : R (0, 1] given by f(x) = 1 1+x 2 is an injection (1-1), a surjection (onto), and a bijection. f(1) = f( 1), so f is not 1 1 (so not a bijection). Let 0 < m 1. Set x = f(x) = m, so f is onto. 1 m 1. Then (23) Is the binary relation {(1, 1), (2, 3), (3, 2), (3, 3), (4, 4), (5, 6), (6, 6)} on N symmetric? No (24) Is the binary relation {(2, 2), (2, 3), (3, 3), (3, 4), (4, 4), (5, 6), (5, 7), (6, 7)} on N anti-symmetric? Yes

14 (25) Let f : N N be given by f(1) = 1 and for n 2, f(n) = n 1. Prove that f is a surjection (onto) but not a bijection. Proof : Let n N. Then f(n + 1) = n. Hence f is onto. Since f(1) = f(2) = 1, f is not 1 1 and therefore not a bijection (26) Let f : N N be given by f(n) = n + 1. Prove that f is an injection (1-1) but not a bijection. Proof : Let m, n N. Assume f(m) = f(n). Then m + 1 = n + 1 and so m = n. Thus f is 1 1. f is not onto since f(n) = n + 1 > 1 and 1 N. So f is not a bijection (27) Consider these definitions: A function f : R R is strictly increasing if x, y R, x > y f(x) > f(y) and strictly decreasing if x, y R, x > y f(x) < f(y). (a) Show that if f : R R is strictly increasing and x R, f(x) > 0 then g(x) = 1 f(x) decreasing. is strictly Proof : Let f : R R be strictly increasing. Define g : R R by g(x) = 1. Let x, y R. f(x) 1 Assume x > y. Then f(x) > f(y) > 0. Then f(x) < 1. This means g(x) < g(y). Hence g f(y) is strictly decreasing (b) Show that if f : R R is strictly increasing then g(x) = f(1 x) is strictly decreasing. Proof : Let f : R R be strictly increasing. Define g : R R by g(x) = f(1 x). Let x, y R. Assume x > y. Then 1 x < 1 y. Then f(1 x) < f(1 y). This means g(x) < g(y). Hence g is strictly decreasing

15 (28) Consider the binary relation R on R given by (x, y) R if x 2 = y 2. Determine if this relation is reflexive, symmetric, transitive, and anti-symmetric. This relation is not anti-symmetric, just consider that ( 1, 1), (1, 1) R. But it is reflexive, symmetric, and transitive. (29) Consider the binary relation R on N given by (m, n) R if m divides n. Determine if this relation is reflexive, symmetric, transitive, and anti-symmetric. This relation is not symmetric (for example, 2 divides 6 but not vice versa). But it is reflexive, anti-symmetric, and transitive. (30) Consider these definitions: A function f : R R is even if x R, f( x) = f(x) and odd if x R, f( x) = f(x). Suppose f : R R and g : R R are odd functions. Show that h(x) = (f g)(x) is odd. Show that k(x) = f(x)g(x) is even. Proof : Let f : R R and g : R R be odd. Define h : R R by h(x) = (f g)(x) = f(g(x)) and define k : R R by k(x) = f(x)g(x). Let x R. Then h( x) = f(g( x)) = f( g(x)) = f(g(x)) = h(x) and k( x) = f( x)g( x) = ( f(x))( g(x)) = f(x)g(x) = k(x). Hence h(x) is odd and k(x) is even (31) Let A, B be nonempty sets. Let f : A B and S B. The inverse image of S under f is f 1 (S) = {x A f(x) S}. Prove that f 1 (S) = f 1 (S). Proof : Let A, B be nonempty sets. Let f : A B be a function and S B. If f 1 (S) is the empty set then f 1 (S) f 1 (S). If f 1 (S) is the empty set then f 1 (S) f 1 (S). So assume f 1 (S) and f 1 (S). Let x f 1 (S). Then f(x) S. So f(x) S which means x f 1 (S). Hence x f 1 (S) and f 1 (S) f 1 (S). Let x f 1 (S). Then x f 1 (S). So f(x) S which means f(x) S. Hence x f 1 (S) and f 1 (S) f 1 (S) Thus f 1 (S) = f 1 (S)

16 FINAL: (32) True or False? (a) A connected graph on n vertices must have at least n 1 edges. True (b) The sum of all the vertex degrees in a graph is even. True (c) Every Euler cycle is a Hamiltonian cycle. False (d) There exists a graph whose vertices have the following degrees: 2, 2, 1, 1, 1, 1. True (33) Your friend says that his favorite (simple) graph has the following degrees: 1, 1, 1, 2, 3, 4, 4, 5. Does such a graph exist? If so, draw it. If not, justify. Such a graph cannot exist as there are an odd number of odd vertices, resulting in an odd sum of vertex degrees.

17 (34) Draw a graph such that... (a)...there is a Hamiltonian cycle but no Euler cycle.

18 (b) There is an Euler cycle but no Hamiltonian cycle. (35) Draw a simple undirected graph with the following degrees or explain why such a graph does not exist: 1, 1, 2, 3, 3, 6. No such graph exists as the maximum degree vertex for a simple graph with six vertices is 5.

19 (36) Draw a graph with the following adjacency matrix:

20 (37) Let G = (V, E) be a simple non-empty graph. Show that the relation R on V such that (u, v) R if and only if there is an path between u and v is symmetric, and transitive. Proof : Let G = (V, E) be a simple non-empty graph and R be the relation on V such that (u, v) R if and only if there is an path between u V and v V. Let u, v V. Assume (u, v) R. Then there is a path between u and v. By reversing the edges there is a path between v and u. Hence (v, u) R and R is symmetric. Now let u, v, w V. Assume (u, v), (v, w) R. Then there is a path between u and v as well as a path between v and w. By concatenating these paths we get a path between u and w. Hence (u, w) R and R is transitive (38) Show that in a simple graph with at least two vertices there must be (at least) two vertices of the same degree. Proof : Let G be a simple graph with n 2 vertices. To yield a contradiction assume all the vertex degrees are different. Then the degrees must be 0, 1, 2,..., n 1. Then there is a vertex v of degree n 1. Hence v is connected to all other vertices. This means all vertices have degree at least 1, which is a contradiction

21 (39) Apply Dijkstra s algorithm to find the length of the shortest path between vertices u, v in the graph. Is the path unique?

22 (40) Apply Kruskal s algorithm to find a minimal spanning tree in the graph.!! " #$%&!$ '( ) % * + #, -.! '( +!$!!!$ / ",!.!!0! & %$!.! &! #1,!, -