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1 Exam topics 1. Basic structures: sets, lists, functions (a) Sets { }: write all elements, or define by condition (b) Set operations: A B, A B, A\B, A c (c) Lists ( ): Cartesian product A B (d) Functions f : A B defined by any input-output rule (e) Injective function: a 1, a 2 A: a 1 a 2 f(a 1 ) f(a 2 ) (f) Surjective function: b B, a A with f(a) = b (g) A, B have same cardinality: there is a bijection f : A B (h) A is countable: there is a bijection f : N A 2. Formal logic (a) Statements: definitely true or false (b) Conditional (open) statement P (x): true/false depends on variable x (c) Logical operations: and, or, not, implies (d) Truth tables and logical equivalence (e) Implication P Q equivalent to: contrapositive not(q) not(p ); independent from: converse Q P ; inverse not(p ) not(q) (f) Negate implication: not(p Q) is equivalent to: P and not(q) (g) Quantifiers: for all, there exists; (h) Negate quantifiers: not( x, P (x)) is equivalent to: x, not(p (x)) (i) Logical equivalences and set equations (j) Logic in mathematical language versus everyday language 3. Methods of proof (can be combined) (a) Direct proof (b) Proof by cases (c) Proof of the contrapositive (d) Proof by contradiction (e) Proof by induction (also complete induction) 4. Axioms of a Group (G, ) (All variables below mean elements of G.) (a) Closure: a b G. (b) Associativity: (a b) c = a (b c) (c) Identity: There is e with e a = a and a e = a for all a. (d) Inverses: For each a, there is some b with a b = e and b a = e. Extra axioms (e) Commutativity: a b = b a. (f) Distributivity of times over plus: a (b + c) = a b + a c and (b + c) a = b a + c a. 5. Divisibility of integers (All variables below mean integers.) (a) Divisibility: a b means b = ac for some c Z 1 11/20/13

2 (b) Properties of divisibility: a b, c = a mb+nc for all m, n a b and b c = a c. a b and b a = a = ±b. (c) Prime and composite Test: a is composite = a has prime factor p a. (d) Greatest common divisor gcd(a, b); relatively prime means gcd(a, b) = 1. (e) Division Lemma: a = qb + r with remainder 0 r < b. (f) Euclidean Algorithm computes remainders a > b > r 1 > > r k > 0. Computes gcd(a, b) = r k. Finds m, n with gcd(a, b) = ma + nb. (g) Consequences of gcd(a, b) = ma + nb Find integer solutions (x, y) to equation ax + by = c, if gcd(a, b) c. If e a and e b, then e gcd(a, b). Euclid s Lemma: If c ab and gcd(c, a) = 1, then c b. Prime Lemma: If p is prime with p ab, then p a or p b. For ā Z n, find multiplicative inverse b = ā 1, i.e ab 1 (mod n). (h) Fundamental Theorem of Arithmetic n > 1 is a product of primes uniquely, except for rearranging factors. There is a unique list of powers s 1, s 2, s 3,... 0 with: n = 2 s1 3 s2 5 s3 7 s4 11 s5. 6. Equivalence relation = on a set S (a) Defining properties: Reflexive: a a Symmetric: If a b, then b a. Transitive: If a b and b c, then a c (b) Equivalence class [a] = {b S b = a}. Following are logically the same: a = b a [b] [a] = [b], the same set 7. Clock arithmetic Z n (a) Modular equivalence: a b (mod n) means n a b. Class a = [a]. (b) Equivalence class a = [a]. Z n = {0, 1, 2,..., n 1} (c) Modular addition and multiplication satisfy all usual rules of algebra (d) Modular division: a 1 = b, where ab = 1, provided gcd(a, n) = 1. (e) In Z p with p prime, every a 0 has a 1 Z p. 8. Limits (a) Real number axioms: commutative group axioms for +, ; distribuitive law axioms of order <. (b) Completeness: If S R has upper bound, then lub(s) = sup(s) R. (c) Convergent sequence a n = l : ɛ>0, N, n N a n l < ɛ (d) Divergent sequence (a n ) : l, ɛ>0, N, n N with a n l ɛ. (e) Infinite it a n = : B, N, n N a n > B. (f) Thm: If (a n ) increasing bounded sequence, then (a n ) convergent. 2 11/20/13

3 1. (a) Use a multiplication table to find all values a Z 7 for which the equation x 2 = a has a solution x Z 7. For each such a, list all of the solutions x. (b) Find all solutions x Z 7 to the equation x 2 + 2x + 6 = Use quantifiers to express what it means for a sequence (x n ) n N to diverge. You cannot use the terms not or converge. 3. Suppose A, B R are bounded and non-empty. Show that sup(a B) = max {sup(a), sup(b)}. 4. Let A, B be sets, and suppose there is a surjection f : A B. Prove that there is an injection g : B A. 5. Use the formal definition of it to prove the following. (a) (b) (c) (d) n n 3 4 = 0 4n 5 2n + 7 = 2 n 3 3n n + 5 = + n n = 6. For each of the following, determine if defines an equivalence relation on the set S. If it does, prove it and describe the equivalence classes. If it does not, explain why. (a) S = R R. For (a, b) and (c, d) S, define (a, b) (c, d) if 3a + 5b = 3c + 5d. (b) S = R. For a, b S, a b if a < b. (c) S = Z. For a, b S, a b if a b. (d) S = R R. For (a, b) and (c, d) S, define (a, b) (c, d) if a = c and b = d. Here x is the smallest integer greater than or equal to x. 7. Consider Z n. (a) Under what conditions on n does every nonzero element have a multiplicative inverse? How about an additive inverse? (b) Does every nonzero element have a multiplicative inverse in Z 21? (c) Does 5 have a multiplicative inverse in Z 21? Explain why or why not. If it does, find 5 1. (d) Solve the equation 5x 14 = 19 in Z /20/13

4 8. Let A = {a, b, c} and B = {a, x}. List all elements of (a) A B (b) A B (c) A\B (d) A B (e) Power set of A 9. Let S(n) = {(x, y) R R max{x, y} = n}. Prove that S(3) S(5) is the empty set. 10. Let f : N N, given by f(n) = n 4. (a) Prove that f is surjective (b) Prove that f is not injective 11. Let f : A B and g : B A be functions satisfying f(g(x)) = x for all x B. Prove that f is surjective. 12. Describe a concrete bijection from N to N {1, 2, 3}. Briefly tell why it is injective and surjective. 13. Make a truth table for not (A B) = A B. Find a shorter logically equivalent expression. 14. Find the negations of the following statements: (a) (A B) (B C) (b) A = (B C) (c) x y (P (x) (not Q(y))) 4 11/20/13

5 1. (a) Use a multiplication table to find all values a Z 7 for which the equation x 2 = a has a solution x Z 7. For each such a, list all of the solutions x. Solution. We only need to look at the diagonal of the multiplication table for Z 7. Then the equation x 2 = a has a solution x Z 7 if and only if a { 0, 1, 2, 4 }. When a = 0, the only solution is x = 0. When a = 1, the solutions are x = 1 and x = 6. When a = 2, the solutions are x = 3 and x = 4. When a = 4, the solutions are x = 2 and x = 5. (b) Find all solutions x Z 7 to the equation x 2 + 2x + 6 = 0. Solution. Adding 2 to both sides, the given equation is equivalent to x 2 + 2x + 1 = 2. We can factor the left-hand side to get (x + 1) 2 = 2. It follows from part (a) that x + 1 = 3 or x + 1 = 4, and hence x = 2 or x = Use quantifiers to express what it means for a sequence (x n ) n N to diverge. You cannot use the terms not or converge. Solution. A sequence (x n ) n N diverges if for every L R there is some ɛ > 0 such that for all N N there is some natural number n N for which x n L ɛ. In terms of quantifiers this is L R ɛ > 0 N N n N, x n L ɛ. 3. Suppose A, B R are bounded and non-empty. Show that sup(a B) = max {sup(a), sup(b)}. Solution. First note that since A and B are both bounded and non-empty, the same is true of A B and so sup(a B) R exists. It follows immediately from Beck Proposition 8.50 that sup(a B) max {sup(a), sup(b)}, since A and B are both subsets of A B. We need to prove the reverse inequality. For sake of contradiction, suppose sup(a B) > max {sup(a), sup(b)}. Then neither sup(a), nor sup(b) can be an upper bound for A B. So there is some x A B with x > sup(a) and x > sup(b). But x A B implies that x A or x B, so this cannot happen. 5 11/20/13

6 4. Let A, B be finite sets, and suppose there is a surjection f : A B. Prove that there is an injection g : B A such that f g : B B is the identity function. Solution. Let b B, we want to define g(b) A. Since f is surjective, it follows that f 1 (b) is a non-empty set. Let a f 1 (b) be any element of this set, and declare g(b) = a. We obviously have that f g is the identity, since f(g(b)) = f(a) = b for all b B. To show g is injective, suppose there are b, b B with g(b) = g(g ). Let a = g(b) denote this common value. Then by construction of g, we have a f 1 (b) and a f 1 (b ). Applying f to a therefore gives f(a) = b and f(a) = b, and so b = b. 5. Use the formal definition of it to prove the following. (a) n n 3 4 = 0 Solution. Let ε > 0 be given, aribitrary. Define N = max{3, 2 + 2}. Let n N, n N be arbitrary. Then, ε n n = n n 3 4 n2 + 3n 2 2n 3 4n 2 (since n 3) (We increased the numerator and decreased the denominator, = 2 n 2 2 N 2 keeping in mind 2n 3 2n 2 > 0, as n > 2) (Factor and cancel out common terms) (n N) ε (N 2 ε + 2) Thus, ε > 0, N such that n > N with n N, n n < ε. Thus, indeed n n 3 4 = 0. (b) 4n 5 2n + 7 = /20/13

7 Solution. Let ε > 0 be given, aribitrary. Define N = 1. Let n N, n N be arbitrary. Then, ε 4n 5 2n = < 19 2n 19 2n + 7 (since n > 0) (We decreased the denominator, < 1 n 1 N (n N) = ε (N = 1 ε ) Thus, ε > 0, N such that n > N with n N, 4n 5 2n n 5 < ε. Thus, indeed 2n + 7 = 0. (c) n 3 3n n + 5 = + Solution. Let M > 0 be given, aribitrary. Define N = 6M + 3. Let n N, n N be arbitrary. Then, n 3 3n n + 5 n3 3n n + 5n = n3 3n 6n = n2 3 6 N (since n 1) (n N) = M (N = 6M + 3) Thus, M > 0, N such that n > N with n N, n3 3n n + 5 n 3 3n M. Thus, indeed n + 5 = +. (d) n n = 7 11/20/13

8 Solution. Let M < 0 be given, aribitrary. Define N = 7 M. Let n N, n N be arbitrary. Then, n n < n2 7 n n2 7n n (since n > 7, thus n 2 7 > 0) (since the denominator is negative and n > 7, decreasing the numerator, while still keeping it positive ) = 7 n 7 N (n N) = M (N = 7 M) Thus, M < 0, N such that n > N with n N, n2 7 1 n n 2 7 M. Thus, indeed 1 n =. 6. For each of the following, determine if defines an equivalence relation on the set S. If it does, prove it and describe the equivalence classes. If it does not, explain why. (a) S = R R. For (a, b) and (c, d) S, define (a, b) (c, d) if 3a + 5b = 3c + 5d. Solution. The relation as defined above is indeed an equivalence relation, since it satisfies reflexivity, symmetry and transitivity, as shown below. Reflexivity: Let (a, b) S. Then 3a + 5b = 3a + 5b, and therefore (a, b) (a, b). Symmetry: Let (a, b), (c, d) S such that (a, b) (c, d). Then 3a + 5b = 3c + 5d. This is equivalent to 3c + 5d = 3a + 5b, which implies (c, d) (a, b). Transitivity: Let (a, b), (c, d), (e, f) S, such that (a, b) (c, d) and (c, d) (e, f). Then 3a + 5b = 3c + 5d and 3c + 5d = 3e + 5f. By transitivity of equality for real numbers we have 3a + 5b = 3e + 5f, and therefore (a, b) (e, f). The equivalence classes are the lines 3x + 5y = c, i.e. each equivalence class is a line with slope 3 5 and the different equivalence classes have different y intercepts (given by c 5 ). (b) S = R. For a, b S, a b if a < b. 8 11/20/13

9 Solution. The relation defined by a b if a < b is not an equivalence relation, since it does not satisfy reflexivity. Namely, a a, since a a. (c) S = Z. For a, b S, a b if a b. Solution. The relation defined by a b if a b is not an equivalence relation, since it does not satisfy symmetry. Namely, a b does not necessarily imply b a. For example, 2 8, but 8 2. (d) S = R R. For (a, b) and (c, d) S, define (a, b) (c, d) if a = c and b = d. Here x is the smallest integer greater than or equal to x. Solution. The relation as defined above is indeed an equivalence relation, since it satisfies reflexivity, symmetry and transitivity, as shown below. Reflexivity: Let (a, b) S. Then a = a and b = b, and therefore (a, b) (a, b). Symmetry: Let (a, b), (c, d) S such that (a, b) (c, d). Then a = c and b = d. This is equivalent to c = a and d = b, which implies (c, d) (a, b). Transitivity: Let (a, b), (c, d), (e, f) S, such that (a, b) (c, d) and (c, d) (e, f). Then a = c and b = d, as well as c = e and d = f. By transitivity of equality for real numbers we have a = e and b = f, and therefore (a, b) (e, f). The equivalence classes are squares in the plane R 2 with sides parallel to the coordinate axes, in particular, they are sets of the form (i, i + 1] (j, j + 1] (Cartesian product of intervals), where the ordered pair (i, j) Z Consider Z n. (a) Under what conditions on n does every nonzero element have a multiplicative inverse? How about an additive inverse? Solution. Every nonzero element in Z n has a multiplicative inverse if n is prime. Indeed, if n is prime, gcd(n, m) = 1 m Z such that 0 < m < n, and therefore by Bezout s Lemma there exist integers x, y such that nx + my = 1, thus my 1 mod n, i.e. m ȳ = 1, which implies that m 1 = ȳ. Every element in Z n does have an additive inverse n N. (b) Does every nonzero element have a multiplicative inverse in Z 21? Solution. No, one can check that 3 and 7 do not have multiplicative inverses in Z 21. (c) Does 5 have a multiplicative inverse in Z 21? Explain why or why not. If it does, find /20/13

10 Solution. One can express gcd(5, 21) in the form 5x + 21y for some integers x, y by applying the Euclidean Algorithm, 21 = , therefore 5 ( 4) = 1, thus 5 1 = 17. (Note that the equivalence classes 4 = 17.) (d) Solve the equation 5x 14 = 19 in Z 21. Solution. The equation 5x 14 = 19 is equivalent to 5x = 12, which, using that 5 1 = 17 yields x = = Let A = {a, b, c} and B = {a, x}. List all elements of (a) A B (b) A B (c) A\B (d) A B (e) Power set of A Solution. A B = {a, b, c, x}, A B = {a}, A\B = {b, c}, A B = {(a, a), (a, x), (b, a), (b, x), (c, a), (c, x)}, Power set of A is {, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}}. 9. Let S(n) = {(x, y) R R max{x, y} = n}. Prove that S(3) S(5) is the empty set. Solution. Assume the contrary: let (a, b) S(3) S(5). Then max{a, b} = 3 and max{a, b} = 5, but 3 5, hence our assumption leads to a contradiction. Therefore the intersection is empty. 10. Let f : N N, given by f(n) = n 4. (a) Prove that f is surjective (b) Prove that f is not injective Solution. Given y N, f(y + 4) = y = y since y 0, which shows that f is surjective. f(1) = 3 = f(7), hence f is not injective. 11. Let f : A B and g : B A be functions satisfying f(g(x)) = x for all x B. Prove that f is surjective /20/13

11 Solution. First attempt: Assume f is not surjective. Then there is a b B such that there are no a A with f(a) = b. Let c = g(b), then f(c) = f(g(b)) = b by assumption, hence we found a c A with f(c) = b which contradicts with the assumption. Second attempt: Given b B, let a = g(b), and compute f(a) = f(g(b)) = b by assumption. Since b was arbitrary, this shows that f is surjective. 12. Describe a concrete bijection from N to N {1, 2, 3}. Briefly tell why it is injective and surjective. Solution. By division lemma, given n, there is a unique q and r with 0 r < 3 with n = 3 q + r, we could define f(n) by f(n) = (q, r + 1). Then f has the inverse function given by g(a, b) = 3 a + b. Check that f(g(a, b)) = (a, b) and g(f(n)) = n. 13. Make a truth table for not (A B) = A B. Find a shorter logically equivalent expression. Solution. Consider all possibilities for simultaneous truth values for A and B: A B not (A B) A B not (A B) = not A T T F T T T F F F T F T F F T F F T F F We see that the only time the expression is False is when both A and B are False, hence this expression is logically equivalent to A B. 14. Find the negations of the following statements: (a) (A B) (B C) (b) A = (B C) (c) x y (P (x) (not Q(y))) Solution. not((a B) (B C)) not(a B) not(b C) (nota notb) (notb notc) not(a = (B C)) not(nota (B C)) A not(b C) A (notb notc) not( x y (P (x) (not Q(y)))) x y (notp (x)) Q(y) 11 11/20/13

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