Do not open this exam until you are told to begin. You will have 75 minutes for the exam.

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1 Math 2603 Midterm 1 Spring 2018 Your Name Student ID # Section Do not open this exam until you are told to begin. You will have 75 minutes for the exam. Check that you have a complete exam. There are 5 questions for a total of 45 points. You are allowed to have pen and paper on your desk. No notes of calculator are permitted. Show all your work. Unless explicitly stated otherwise in a particular question, if there is no work supporting your answer, you will not receive credit for the problem. If you need more space to answer a question, continue on the back of the page, and indicate that you have done so. Question Points Score Total: 45

2 Math 2603, Spring 2018 Midterm 1 Page 1 of 6 1. (5 points) For each statement below answer True or False. You don t have to justify your answer. (a) 3 < π and = 5 2. True False (b) If the statement A is True then the statement A B is always True. True False (c) The negation of the statement For every real number x, there exists an integer n such that n > x is There exists a real number x such that x n for every integer n. True False (d) Every poset has minimal elements but may not have a minimum element. True False (e) There are integers x and y such that 37x + 2y = 1. True False

3 Math 2603, Spring 2018 Midterm 1 Page 2 of 6 2. (12 points) Prove each of the following statements. (a) If n N, n 3 n is always a multiple of 3. Notice that n 3 n = n(n 2 1) = n(n 1)(n + 1). We look at three cases depending on the reminder in the division by three: Case1: n = 3k, n 3 n = 3k(3k 1)(3k + 1) which is a multiple of three. Case2: n = 3k + 1, n 3 n = (3k + 1)(3k)(3k + 2) which is a multiple of three. Case3: n = 3k + 2, n 3 n = (3k + 2)(3k + 1)(3k + 3) = 3(3k + 2)(3k + 1)(k + 1) which is also a multiple of three. (b) For a, b R \ {0}, define the relation a b if a b equivalence relation. is a rational number. Then is an We have to check three properties: Is reflexive: a = 1 is rational. a Is symmetric: If a = q is rational then b = 1 is also rational. b a q Is transitive: Given a and b rationals, their product is also rational, so a = a b is rational. b c c b c

4 Math 2603, Spring 2018 Midterm 1 Page 3 of 6 (c) If A, B and C are nonempty sets and A C = B C then A = B. Let s prove that A B. Pick a A, and c C. Then (a, c) A C so it is also an element of B C and, by definition, this means that a B. Analogously we can prove the reverse inclusion. (d) Let (A, ) be a totally ordered set with the following property: for every nonempty subset E of A, strictly contained in A, the poset (E, ) has a minimum element. Show that A has a minimum element. Let a A a fixed element. Consider the set E = A \ {a}. By hypothesis, there is a minimum element a E. Because (A, ) is totally ordered we can compare any two elements. We have two cases: Case1: a a. Then a is minimum in A. To see this, consider another element a A. Then a E and thus a a, by transitivity we conclude a a. since a is arbitrary, we conclude that a is minimum. Case2: a a. Since a is minimum in E = A\{a}, it will also be minimum in E {a} = A, as desired.

5 Math 2603, Spring 2018 Midterm 1 Page 4 of 6 3. (8 points) Consider the set A = {1, 2, 3,..., 12} and the relation a b if a divides b. You are told that (A, ) is a partially ordered set. (a) Draw the Hasse diagram of (A, ). (b) List all maximal and minimal elements. Minimal elements: 1 (it is also a minimum). Maximal elements: 7, 8, 9, 10, 11, 12. (c) Find the greatest lower bound of (6, 10) This is the g.c.d. so g.l.b.(6, 10) = 2.

6 Math 2603, Spring 2018 Midterm 1 Page 5 of 6 4. (8 points) (a) Let a and b be natural numbers greater than zero. Show that g.c.d.(a, b) = g.c.d.(a, a + b). Let d = g.c.d.(a, b) and d = g.c.d.(a, a + b). Since d/a and d/b we have d/a + b. So d is a common divisor of a and a + b and then d d. Similarly, Since d /a and d /(a + b) then d /(a+b) a = b and d is a common divisor of a and b hence d d. Both inequalities imply that d = d. (b) Use the result from part (a) to prove that g.c.d.(3k + 2, 5k + 3) = 1 for any natural number k. In the setting of part (a), put a = 3k + 2 and b = 2k + 1. Then g.c.d(3k + 2, 5k + 3) = g.c.d.(3k + 2, 2k + 1). For the right hand side apply part (a) again now with a = 2k + 1 and b = k + 1 to conclude that g.c.d.(3k + 2, 2k + 1) = g.c.d.(2k + 1, k + 1). We do this two more times to get: g.c.d.(2k + 1, k + 1) = g.c.d.(k + 1, k) = g.c.d.(k, 1) = 1.

7 Math 2603, Spring 2018 Midterm 1 Page 6 of 6 5. (12 points) (a) Let a, p N such that p is a prime number and p divides a 2. Show that p 2 divides a 2. Since p/a 2 = aa we have that p/a. This means there exists an integer q such that a = pq, then a 2 = (pq) 2 = p 2 q 2 which implies that p 2 /a 2. (b) Show that the only natural numbers m and n for which are m = n = 0. 6 m = 12 n. The left hand side can be decomposed uniquely as 2 m 3 m while the right hand side has prime factorization equal to 2 2n 3 n. If (m, n) is a solution these two factorizations have to be equal. In particular, we need the same power of 2 and the same power of 3, this is m = 2n Which implies n = 0 and m = 0. m = n