1 Take-home exam and final exam study guide

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1 Math Introduction to Advanced Mathematics Fall Take-home exam and final exam study guide 1.1 Problems The following are some problems, some of which will appear on the final exam Number Theory Prove the following statements: Let a and b be integers and let p be a prime. If p ab then p a or p b. Let n be a natural number. If n > 1 then n is the product of prime numbers. For every prime p, p is not a rational number Functions, relations, etc. Suppose that f : X Y and g : Y Z are functions. Prove or disprove the following statements about the composite function g f : X Z: If f and g are both injective then g f is injective. If f and g are both surjective then g f is surjective. If f is injective and g is surjective then g f is surjective. If f is surjective and g is injective then g f is surjective. If f is surjective and g is injective then g f is injective. If f is a bijection and g is a bijection then g f is a bijection. If g f is injective then f is injective. If g f is injective then g is injective. If g f is surjective the f is surjective. If g f is surjective then g is surjective. If g f is a bijection then f and g are both bijections. 1.2 Dealing with mathematical statements Converse and contrapositive State the converse and contrapositive of the following statements (from the Number Theory worksheets): Propositions 6,7,8,10,16,18,22,25,43.

2 1.2.2 Negations Negate the following statements: For all integers a and b, if a is even and b is even, then a + b is even. For every real number ɛ > 0, there exists a natural number M such that for all natural numbers n, if n > M then a n L < ɛ. For every real number ɛ > 0 there exists a natural number M such that for all natural numbers n and m, if n > M and m > M, then a n a m < ɛ. For every real number ɛ > 0 there exists a δ > 0 such that for every real number x if 0 < x a < δ then f(x) L < ɛ. For every real number ɛ > 0 there exists a δ > 0 such that for every real number x if x a < δ then f(x) f(a) < ɛ. For every real number ɛ > 0 there exists a δ > 0 such that for all real numbers x and y if x y < δ then f(x) f(y) < ɛ Statements about real numbers Prove or disprove the following statements: For every y R there exists an x R such that x > y. For every y R there exists an x R such that y > x 2. For every x R there exists a y R such that y > x 2. For every m, n Z m n. For all y R there exists x R such that x + y = 0. There exists y R such that for all x R, x + y = 0. For all x R there exists y R such that xy = 0. There exists x R such that for all y R, xy = Induction Problems Definition 18. For a non-negative integer n, define the quantity n! (read as n factorial ) as follows: 0! = 1 and for all natural numbers n, n! = n.(n 1)!. Prove the following statements by induction: For every integer n 5, 2 n > n 2. For every integer n 4, n! 2 n. For every non-negative integer n and for every real number x 1, 1 + x + x x n = 1 xn+1 1 x.

3 For every integer n 1 and for any real numbers x 1, x 2,..., x n, n n x i x i. i=1 [For this last problem, you may assume that for any real numbers a and b, the triangle inequality holds: a + b a + b.] From now on, the material is in the form of slightly longer problems, some (or some parts of which) will appear on the final exam. 1.4 Induction, Functions, Equivalence Relations Definition 19. Suppose that X is a set and that is an equivalence relation on X. If x is an element of X then the equivalence class of x, denoted [x] is i=1 [x] = {y X x y}. Theorem 20. Suppose that X is a set and that is an equivalence relation on X. Suppose that x, y X. Then 1. If x y then x [y]. 2. If x y and z [x] then z [y]. 3. If there is a z so that z [x] but z is not in [y] then x y. (i.e. they are not related by ). 4. If x y then there are no element that are in both of [x] and [y]. 5. Every element z in X is in some equivalence class. 6. Every element z in X is in exactly one equivalence class. Remark 21. Two sets are disjoint is they have no elements in common. Theorem 20 says that equivalence classes are either disjoint or equal. A partition of a set Xis a collection of disjoint subsets X α so that every element of X is contained in some (and hence exactly one) of the X α. Theorem 20 says that if is an equivalence relation on X then the equivalence classes of partition X. Notation 22. The empty set (the set with no elements) is denoted. The intersection of two sets A and B is the set A B = {x x A and x B}. To say that sets A and B are disjoint exactly says that A B =. Notation 23. If X is a set, is an equivalence relation on X and C is an equivalence class of, then any element a of C is called a representative of C. Lemma 24. Let X be a set and let be an equivalence relation on X. If x is in X then x is a representative of [x]. Moreover, if y x then y is a representative of [x]. All the representatives of [x] are y so that y x.

4 1.5 Equivalence classes mod n Lemma 25. Let n be a natural number. Define a relation M n on by saying (a, b) M n if a b mod n. Then M n is an equivalence relation. Remark 26. Rather than writing (a, b) M n it makes sense to keep writing a b mod n. We ll call the equivalence classes of this relation equivalence classes of integers mod n, or simply classes mod n. Proposition 27. The equivalence classes mod n are exactly [0], [1],..., [n 1], and all of these are disjoint. Definition 28. Let n be a natural number and let Y n = {[0], [1],..., [n 1]} be the equivalence classes mod n. Define the addition function A : Y Y Y by A([a], [b]) = [a + b]. Proposition 29. The function A is well-defined. A function being well-defined is a centrally important notion in mathematics. On the face of it, there are lots of elements of [a], and if c is such an element then we have (by Proposition 20) that [a] = [c]. Thus we need to prove, in particular, that in these circumstances [c+b] = [a + b]. If a function is not well-defined then it is not really a function, since this means that the definition doesn t really make sense. The point is that the function A is defined by making some choices. The choice we made were of representatives of equivalence classes, and we have to check that if we made any of our choices differently we d get the same answer. This is in contrast to the following (non-)example. Non-example 30. If we took Y as in Definition 28 and tried to define a function N : Y Y Z by N([a], [b]) = a + b then it would not be well-defined. This is because if we took different representatives, we d get different answers. Make sure you believe this by giving some examples of this. Definition 31. Let n be a natural number and let Y n be as in Definition 28. Define the multiplication function M : Y n Y n Y n by M([a], [b]) = [ab]. Proposition 32. The function M is well-defined. 1.6 The rational numbers In this section, we walk through a construction of the rational numbers from the integers. Definition 33. Define a relation on Z (Z {0}) by setting (a, b) (c, d) if ad bc = 0. Proposition 34. The relation defined above is an equivalence relation.

5 Definition 35. The set Q is the set of equivalence classes in Z (Z {0}) under the relation defined above. We define addition and multiplication functions on Q as follows. Definition 36. Define addition and multiplication on Q as follows: [(a, b)] + [(c, d)] = [(ad + bc, bd)] and [(a, b)]. [(c, d)] = [(ac, bd)] Proposition 37. The addition and multiplication functions on Q are well-defined. Lemma 38. If a is an integer and b and c are in Z {0} then [(a, b)] = [(ac, bc)]. Lemma 39. If a is an integer and b is in Z {0} then [(a, b)] + [(0, 1)] = [(a, b)] and [(a, b)]. [(1, 1)] = [(a, b)]. Lemma 40. If a and c are integers and b, d Z {0} then [(a, b)] + [(c, d)] = [(c, d)] + [(a, b)]. Question 41. If the set Q above is supposed to be the rational numbers, what does the class [(a, b)] represent? 1.7 Cauchy sequences In this problem, we ll consider sequences of rational numbers. You are allowed to consider rational numbers as ordinary fractions in this problem. Definition 42. Let X be a set. A sequence (of elements) from X is a function a : N X. We ll often denote the element a(n) of the sequence by a n, and write (a n ) or (a n ) n N for the sequence. Definition 43. A sequence (a n ) of rational numbers is a Cauchy sequence if for every ɛ > 0 (a rational number) there is a natural number M so that for every i and j natural numbers so that i > M and j > M we have a i a j < ɛ. Proposition 44. Let (a n ) be a Cauchy sequence of rational numbers. Then there exists a rational number B so that for every natural number i we have B < a i < B.

6 Definition 45. Let X be the set of Cauchy sequences of rational numbers. Define a relation on X as follows: Suppose that (a n ) and (b n ) are Cauchy sequences of rational numbers. We say (a n ) (b n ) if for every ɛ > 0 a rational number, there exists a natural number M so that for every i > M we have a i b i < ɛ. Theorem 46. The relation on Cauchy sequences of rational numbers is an equivalence relation. Remark 47. It is possible to define the real numbers to be the set of equivalence classes of Cauchy sequences of rational numbers. A real number x = k.x 1 x 2 x 3 x 4 where k is a natural number and x i are the digits of the decimal expansion. Then define a n = k.x 1 x 2 x n and prove that (a n ) is a Cauchy sequence. [Note that k.x 1 x 2 x n is a rational number!] Any other Cauchy sequence equivalent to (a n ) should be a sequence of numbers converging to x. Extra-credit Problem 48. If the real numbers are defined as in the above remark, what is 1? What is 0? How is addition defined? What about multiplication? 1.8 Multiplication by a in the integers mod p [Here Y n is the set of equivalence classes mod n from Definition 28.] Definition 49. Suppose that p is a prime number, and suppose that 0 < a < p is an integer. Define a function m a : Y p Y p by m a ([i]) = [ai]. Lemma 50. The function m a is well-defined. Proposition 51. The function m a is injective. Proposition 52. The function m a is surjective. Theorem 53. If p is a prime and a is an integer such that 0 < a < p then there exists a unique integer x such that 0 < x < p and ax 1 mod p. Definition 54. Given a prime p and an integer 0 < a < p we will call the unique integer x in the conclusion of the above theorem the inverse of a mod p. Proposition 55. Let p be a prime. Then the inverse of 1 mod p is 1 and the inverse of p 1 mod p is p 1. Theorem 56. Let p be a prime and let a be an integer so that 1 < a < p 1. Then the inverse of a mod p is not equal to a. Theorem 57. Let p be a prime. Then 1.2..(p 1) 1 mod p.