Set theory as a foundation for mathematics


 Beatrice Washington
 2 years ago
 Views:
Transcription
1 Set theory as a foundation for mathematics Waffle Mathcamp 2011 In school we are taught about numbers, but we never learn what numbers really are. We learn rules of arithmetic, but we never learn why these rules are true. In this class we will see how we can construct numbers and rigorously prove the rules of arithmetic using only the notion of a set. What is a set? Informally, a set is a collection of objects. For example, there is a set {Waffle, Mathcamp, MCSP} which consists of me, Mathcamp, and the string MCSP. We could also have a set {0, 5, 2π, 734} consisting of the numbers 0, 5, 2π and 734. However, since we wish to define numbers in terms of sets, we cannot yet allow ourselves to talk about sets of numbers. Indeed, in order to build everything purely out of sets, we will require that all our sets be sets of sets. That is, every element of our sets is itself a set. This may sound circular if we can only consider sets of sets, how can we ever get started constructing any sets? However, one set that we can be sure exists is the empty set = {}, the set with no elements. Since has no elements, all of its elements are sets, so it is a set of sets. Once we have, we can then consider { }, {{ }}, {, { }}, and so on. In general, we will usually define sets using an expression like S = {x : P (x)}. This means that S is the set of all sets x such that the property P (x) is true. For example, assuming we already know what the integers Z are, we could define a set A = {x : x Z and x > 2} = {x Z : x > 2} of all integers greater than 2. We write x S to mean that x is an element of the set S. For example, for A as above, 3 A. Let X and Y be sets. Then we can construct the following sets: The union X Y = {x : x X or x Y }. The intersection X Y = {x : x X and x Y }. The difference X \ Y = {x : x X and x Y }. We say X is a subset of Y, or X Y, if every element of X is also an element of Y. I would like to emphasize that while we will be giving specific constructions of number systems as sets in these notes, there is nothing special about the exact constructions we use. There is more than one way (in fact, infinitely many ways) to exhibit sets that can serve as number systems, and what is important is not which one we choose but the fact that one does exist. For example, the natural numbers should not be defined as the set of Definition 1.6, but as any set equipped with an arithmetic structure which is isomorphic to that given in Section 2. Finally, it should be noted that for most of these notes we will be using what is called naive set theory, in which we are allowed to define any set we want to using an expression of the form {x : P (x)}. This approach to set theory is called naive because it doesn t actually work! At the end of the notes, we see why it doesn t work and why we must use more restrictive rules when constructing sets. Don t worry, though; all of the constructions made in these notes are in fact valid under these more restrictive rules (called the ZermeloFraenkel axioms for set theory). 1
2 1 The Natural Numbers The first, most basic mathematical object we will define using sets is the natural numbers (i.e., the nonnegative integers). We would like to define the set of natural numbers as N = {0, 1, 2, 3,...}, but there are a few difficulties with this. First, we need to define what 0, 1, 2, 3, and so on are as sets. Then we have to figure out what... means more precisely. Let s start at the very beginning: what is 0? We could take any set and call it 0, but perhaps the most natural choice is the empty set, the only set that has 0 elements. Thus we make a definition: Definition =. OK, we ve defined 0 as a set; how about 1? Just as we defined 0 to be the empty set, a set with 0 elements, perhaps we could define 1 to be a set with 1 element. Again, the choice is arbitrary, but we will choose: Definition = {0} = { }. That is, 1 is the set whose only element is 0. We can now similarly define all the natural numbers: Definition = {0, 1}, 3 = {0, 1, 2},..., n = {0, 1, 2,..., n 1},... That is, we onebyone define each natural number as the set of all the smaller natural numbers. This allows us to talk about any natural number individually if you hand me a number, I can (laboriously) write it down as a set built up from the empty set. For example, 4 = {0, 1, 2, 3} = {, {0}, {0, 1}, {0, 1, 2}} = {, { }, {, { }}, {, { }, {, { }}}}. However, it is still not clear how we can talk about all natural numbers at once. Indeed, Definition 1.3 is plagued by the same... that we worried about before. However, we can begin to make this more precise by seeing that there s an explicit pattern we can write down describing how to get from n to n + 1. In fact, since n = {0, 1,..., n 1} and n + 1 = {0, 1,..., n 1, n}, we can see that n + 1 = n {n}. We thus make the following definition: Definition 1.4. Let x be a set. Then the successor x + 1 of x is defined as x + 1 = x {x}. In this definition, we really only care about the case that x is a natural number, but we haven t yet defined what a natural number is! We can now state more precisely what we mean by N = {0, 1, 2, 3,...} : 1. 0 N 2. If k N, then k + 1 N as well. 3. All elements of N can be obtained using (1) and (2). This is still not completely satisfactory, since we have not written down an expression for N; we ve just listed some properties it should have. However, with a little more cleverness we can turn these properties into a definition. Definition 1.5. A set S is inductive if 0 S and whenever k S, k + 1 S as well. Definition 1.6. The set of natural numbers is N = {n : n S for every inductive set S}. 2
3 That is, an inductive set is a set that satisfies conditions (1) and (2) above. Since N only contains things that must be contained in any inductive set, it automatically satisfies (3). Let s check that N also satisfies (1) and (2). Proposition 1.7. The natural numbers are inductive. Proof. First, 0 N since by definition, 0 is in every inductive set. Now suppose k N and let S be an inductive set. Then k S, so k + 1 S since S is inductive. Since this holds for every inductive set S, it follows that k + 1 N. Thus N is inductive. You may have noticed that everything we ve been doing looks suspiciously like induction. If you hadn t noticed, hopefully the name inductive at least heightened your suspicions. In fact, our definition of the natural numbers is perfectly designed to make induction work. Theorem 1.8 (Proof by Induction). Let P (n) be a property that a natural number n can have. Suppose that: 1. P (0), and 2. For all k N, P (k) implies P (k + 1). Then P (n) holds for all natural numbers n. Proof. Let S = {k N : P (k)}. Then the hypotheses say exactly that S is an inductive set. By definition of N, then, S contains all of N. That is, P (n) holds for all n. This is the first major fact about numbers that we usually take for granted that we have been able to turn into a rigorously proved theorem using sets. We will soon use induction to prove several more. Looking back at the definition of N and the proof of Theorem 1.8, the two are really equivalent. Given this, there is a very important idea to keep in mind when attempting to solve problems about N: Principle 1.9. Because the natural numbers are defined to be such that induction works, essentially the only way to prove statements about them is by induction. For example, we can prove several things that ought be be true but aren t clear from our abstract definition of N. For instance, we intuitively know that N should only contain the sets 0 =, 1 = {0}, 2 = {0, 1}, 3 = {0, 1, 2}, and so on. But how do we know that other sets which we do not expect to see in the and so on are not in N? We prove it by induction! Here s an example of how this works. Proposition The set {1} is not a natural number. Proof. We prove by induction that for all n N, n {1}. First, clearly 0 = is not {1}. Now suppose k {1}; we wish to show that k + 1 = k {k} is not {1}. Since k k + 1 and the only element of {1} is 1, if k + 1 = {1} we must have k = 1. But then k + 1 = = {0, 1} = {1}. Thus it is impossible for k + 1 to be {1}. Hence {1} N. Here s a definition we ll use in the exercises and in the next section. Definition Let m and n be natural numbers. Then we write m < n if m n. Intuitively, this makes sense, because n = {0, 1, 2, 3,..., n 1} is exactly the set of smaller natural numbers. 3
4 1.1 Exercises Remark Many of the exercises provided in these notes have many parts, and may occasionally seem long and tedious. The purpose of these exercises is to give you practice working with the ideas on your own and understand them better, and also help convince you that you could rigorously prove any basic fact about numbers if you had to. If you ve already done a few exercises and it s getting easy but tedious, you should not feel like you have to do the rest. Just convince yourself that you could solve them if you had to, and beware that we will often use results from the exercises in later sections. On the other hand, a few of the exercises are more difficult, and if there s an exercise that you re finding really hard, come talk to me at TAU and I ll gladly help you through it! Exercise 1.1. Prove that the set {1, {1}} is not a natural number. (Hint: Imitate the proof of Proposition 1.10.) Exercise 1.2. Prove that if n N then either n = 0 or there is some m N such that n = m + 1. (Hint: Use induction.) Exercise 1.3. Show that m < n + 1 iff m n, where means either < or = (recall that m < n + 1 means m n + 1 by Definition 1.11). Exercise 1.4. Prove that the relation < is transitive: if k < m and m < n, then k < n. induction on n and Exercise 1.3). (Hint: Use The following is a slightly different form of induction that we will use occasionally. Exercise 1.5 (Strong Induction). Let Q(n) be a property that a natural number n can have. Suppose that whenever k N and Q(m) holds for all m < k, then Q(k) holds as well. Then prove that Q(n) holds for all natural numbers n. (Hint: Let P (k) be the statement that Q(m) holds for all m < k, and prove P (k) by ordinary induction on k.) Exercise 1.6. The relation < is antireflexive: n n for all natural numbers n. (Hint: Use strong induction.) 4
5 2 Counting We start with an important property of the relation < introduced at the end of last time. Theorem 2.1. The relation < is a (strict) total order on N. That is, it satisfies the following properties: 1. Transitivity: if k < m and m < n, then k < n. 2. Antireflexivity: n n for all natural numbers n. 3. Totality: for any m, n N, either m < n, m = n, or m > n. Proof. The first two properties are just Exercises 1.4 and 1.6. For totality, we will use induction on both m and n. Say that m and n are comparable if m < n, m = n, or m > n; we want to show any two natural numbers are comparable. First fix m. Let s prove by induction on n that if m < n, then m + 1 n. If n = 0, there is no m such that m < n, so this is vacuous. Now suppose that m < n implies m + 1 n, and suppose m < n + 1; we want to show m + 1 n + 1. By Exercise 1.3, this means that m n. If m = n, then m + 1 = n + 1 so m + 1 n + 1. If m < n, then by the induction hypothesis, m + 1 n < n + 1. Hence we have shown by induction that m < n implies m + 1 n. Also, note that if m n, then m + 1 > m n. This together with the previous result implies that if m and n are comparable, then m + 1 and n are comparable. Now we prove the totality by induction on m. First, let m = 0. We prove by induction on n that 0 n for all n. For n = 0, we have 0 = n. For the induction step, if 0 n, then 0 n < n + 1. Thus 0 is comparable to every n. Furthermore, we remarked above that if m is comparable to n, so is m + 1. Thus by induction on m, every m and n are comparable. OK, now we have the natural numbers and we know when one number is bigger than another. Let s see what else numbers are good for. The primary purpose of the natural numbers is to count things. What does it mean to count? Well, when we count that a set S has (say) 4 elements, we are really matching up its elements with the numbers 1, 2, 3, and 4. That is, we are establishing a onetoone correspondence, or bijection between S and the set {1, 2, 3, 4}. For us, it will be more convenient to start counting a 0, so instead we get a bijection with the set {0, 1, 2, 3}, which happens to be our set 4. What is a bijection more precisely? Well we could say we have a function f from S to 4 which sends an element of S to the number we re matching it with. This function has the property that for every n 4, there is exactly one s S such that f(s) = n. But what is a function? Traditionally, a function f : X Y between two sets is defined to be a set of ordered pairs (x, y) with x X and y Y satisfying certain properties. However, to make sense of this purely in terms of sets, we first have to define what an ordered pair is. There are many possible definitions; here is one. Definition 2.2. Let x and y be sets. Then the ordered pair (x, y) is the set {{x}, {x, y}}. The property that we really care about for ordered pairs is the following. Proposition 2.3. Let (x, y) and (a, b) be ordered pairs. Then (x, y) = (a, b) iff x = a and y = b. Proof. If (x, y) = (a, b) then x = a because x is uniquely defined by the property that x s for all s (x, y) = {{x}, {x, y}} and a is uniquely defined by the same property for (a, b). A similar argument shows that y = b (though we should consider separately the cases x = y and x y). Conversely, if x = a and y = b, then clearly (x, y) = (a, b). We can now define functions and bijections as usual. 5
6 Definition 2.4. Let X and Y be sets. Then a function from X to Y is a set f of ordered pairs (x, y) with x X and y Y such that for each x X, there is a unique y Y such that (x, y) f. We normally write f(x) = y instead of (x, y) f. We say f is an injection if for each y Y there is at most one x X such that f(x) = y, and f is a surjection if for each y Y there is at least one x X such that f(x) = y. We say f is a bijection if it is both an injection and surjection, i.e. for every y Y there is a unique x X such that f(x) = y. When there is a bijection between two sets, we can say they have the same size. If X is in bijection with Y and Y is in bijection with Z, X is in bijection with Z by composing the two bijections: if f : X Y and g : Y Z are bijections, then h(x) = g(f(x)) is a bijection from X to Z. Recall that we constructed the natural numbers so that for each n, we have n = {0, 1, 2, 3,..., n 1}. That is, n itself has n elements, informally speaking at least, so we can say a general set S has n elements if it is in bijection with n. Also, the number of elements in any finite set should be a natural number. This motivates the following definition. Definition 2.5. A set S is finite if it is in bijection with some n N. Here s another fact we may have thought we knew which we can now prove: the number of elements of a finite set is unique. In other words, counting works, and no matter what order you count a set in you get the same answer. Lemma 2.6. Let k be a natural number, n = k + 1, m < n, and let T = n \ {m} be the set of all elements of m except m. Then T is in bijection with k. Proof. Note that k = n \ {k}, as n = k + 1 = k {k}. Define f : k T as follows. If l < m, let f(l) = l. If l m, let f(l) = l + 1. It is not hard to check that this is a bijection. 1 Theorem 2.7 (Counting works). Let S be a finite set. Then there is only one natural number n that is in bijection with S. Proof. Suppose S is in bijection with both k and n and k n. Then by composing the bijections we obtain a bijection f : k n. By Theorem 2.1(d), we may assume k < n. We now prove by induction on n that there is no bijection from k to n for k < n. For n = 0, this statement is vacuous. Now suppose n = m + 1 and the statement is true for m. Suppose k < n and there is a bijection from n to k. Since n = m + 1 is nonempty, k is nonempty, so k 0. By Exercise 1.2, there is some l such that k = l + 1. Now restrict the bijection from n to k to the subset m n. We then get a bijection from m to the set T = k \ {f(m)}. By Lemma 2.6, T is also in bijection with l. But then m is in bijection with l < k m (by Exercise 1.3, k < n = m + 1 implies k m), contradicting the induction hypothesis. Hence there cannot be a bijection between k and n for k < n, so S cannot be in bijection with both k and n. Definition 2.8. Let S be a finite set. Then the cardinality S of S is the unique natural number that is in bijection with S. 2.1 Exercises Exercise 2.1 (Induction starting at n). Let P (k) be a property of natural numbers and n N. Suppose that P (n + 1) is true and for all n N, if P (k) is true then P (k + 1) is true. Show that P (k) is true for all k > n. (Hint: Prove by induction that either P (k) or k n. Exercise 2.2. Show that m < n iff m + 1 < n + 1. (Hint: Use Exercise 1.3.) 1 To see that this is a bijection, it is very important to know that k k so that n k. This follows from Exercise 2.1(c). 6
7 Exercise 2.3. Verify Lemma 2.6. That is, given m < n = k + 1, define f : k n \ {m} by f(l) = l if l < m and f(l) = l + 1 if m l < n. Show that for any j n \ {m}, there is a unique l < k such that f(l) = j. Exercise 2.4. Let n N. Show that there is no injective function f : n + 1 n. That is, if f : n + 1 n, then there are i, j n such that i j but f(i) = f(j). (Hint: Use induction on n and Lemma 2.6.) Exercise 2.5 (The Pigeonhole Principle). Suppose n < k are natural numbers. Show that there is no injective function f : k n. That is, if f : k n, then there are i, j n such that i j but f(i) = f(j). Exercise 2.6. When talking about cardinalities, it is customary to say that S T if there is an injection from S to T. Show that this agrees with our order on natural numbers: m n iff there is an injection from m to n. 7
8 3 Addition, Multiplication, and Subtraction We now know how to count a set using the natural numbers. Now let s do arithmetic. First, what is addition? Well, if you have two sets, you can add their sizes by just mushing them together to form one big set. Here s a more formal way of putting that. Definition 3.1. Let S and T be sets. Then the disjoint union S T is the set {(0, s) : s S} {(1, t) : t T }. If n, m N, then their sum is n + m = n m. Wait a minute! We ve already defined n + 1 for natural numbers, and now we ve given a different definition. Do these two definitions agree? They do, since we can define a bijection from n {n} to n 1 by sending m to (0, m) for m < n and sending n to (1, 0). Some basic properties of addition are easy to prove using this definition. Proposition 3.2. Addition of natural numbers is commutative and associative: n + m = m + n and n + (m + k) = (n + m) + k. Also, for any n N, n + 0 = n. Proof. For commutativity, we define a bijection from n m to m n via (i, k) (1 i, k) (for i = 0, 1). Associativity is similar and left as an exercise. For the last property, note that the map m (0, m) is a bijection from n to n 0. The following property, called cancellation, is a bit harder, but not too bad. Lemma 3.3. If n N is nonzero, then there is a unique m N such that m + 1 = n. Proof. By Exercise 1.2, there exists such an m, so it suffices to show that it is unique. Suppose m + 1 = k + 1 = n. Then by Lemma 2.6, both m and k are in bijection with the set n \ {0}. By Theorem 2.7, this implies m = k. Proposition 3.4. If l + n = k + n, then l = k. Proof. We use induction on n. For n = 0, this follows from Proposition 3.2. Now suppose it is true for m and n = m + 1. Then if l + n = k + n, then l + m + 1 = k + m + 1, or (l + 1) + m = (k + 1) + m. By the induction hypothesis, l + 1 = k + 1, and by Lemma 3.3, l = k. The cancellation property allows us to define subtraction. If k + m = n, we write n m = k, and this is welldefined (assuming any such k exists) by cancellation. In the exercises you will show that n m exists iff n m. Now let s look at multiplication. Again, we look to what multiplication means in terms of sets. When we multiply m by n, we take a copy of n for each element of m. We could write this as having ordered pairs, the first coordinate of which tells us which copy we re in and the second of which tells us which element of n to take. This motivates the following definition. Definition 3.5. Let S and T be sets. Then their (Cartesian) product is the set S T of all ordered pairs (s, t) where s S and t T. If n, m N, then their product is nm = n m. Again, lots of basic properties are quite easy from this definition. Proposition 3.6. Multiplication of natural numbers is commutative and associative: n(mk) = (nm)k. Also, for any n N, n 0 = 0 and n 1 = n. nm = mn and Proof. For commutativity, define a bijection from n m to m n by (k, l) (l, k). Associativity is similar and left as an exercise. As a set, n 0 = n =. A bijection from n to n 1 is given by k (k, 0). Proposition 3.7. Multiplication distributes over addition: m(n + k) = mn + mk. Proof. Define a bijection from m (n k) to (m n) (m k) by (j, (i, l)) (i, (j, l)). 8
9 In the exercises you will prove some more basic properties of arithmetic. Now that we ve learned how to work with natural numbers as sets, let s try to understand more complicated numbers. Let s start with the integers. There are lots of ways we could go about adding negative numbers to the natural numbers to get all integers. For example, we could simply arbitrarily construct a set n for each positive natural number n (say, we could define n = (N, n)). This would work, but it would make defining arithmetic operations complicated because we would have to constantly split everything into cases depending on whether our numbers were positive or negative. The method we will use to construct the integers will be more natural in that it does not treat positive and negative numbers separately and in that the technique generalizes to many other settings (for example, we will use the same method later to construct the rational numbers). The idea is not to simply construct new sets n, but to construct sets m n for all natural numbers m and n. To construct a set that we could call m n is simple enough we can just use the ordered pair (m, n). However, a single integer has many representations as a difference m n. For example, 0 2, 1 3, 2 4, and so on should all represent the same integer. Which one should we actually pick to represent that integer? Instead of arbitrarily picking one of them, we will represent the integer 2 as the set of all ways of writing it as a difference. This is best captured using the notion of equivalence classes. Definition 3.8. A relation on a set S is a set R S S. We usually write x R y instead of (x, y) R. A relation is an equivalence relation if it satisfies the following properties: 1. Reflexivity: x x 2. Symmetry: If x y then y x 3. Transitivity: If x y and y z then x z If x S, then the equivalence class of x is [x] = {y S : y x}. We write S/ for the set of equivalence classes of elements of S. We assume the reader is either familiar with or can verify for themselves the following fact. If not, you can ask me about it at TAU. Proposition 3.9. Let S be a set with an equivalence relation. Then every x S is contained in exactly one equivalence class, and [x] = [y] iff x y. In our case, the set S should be the set N N of ordered pairs of natural numbers, and we should have (m, n) (k, l) if m n should equal k l. Unfortunately, if m < n, we haven t defined what m n is yet. What we really want is a condition that is equivalent to m n = k l but only uses operations we know are welldefined. One thing we can do is add n + l to both sides of the equation to obtain the condition m + l = k + n. Lemma The relation (m, n) (k, l) if m + l = k + n is an equivalence relation on N N. Proof. For reflexivity, we just use the tautology m + n = m + n. For symmetry, we use the fact that m + l = k + n implies k + n = m + l. For transitivity, suppose (m, n) (k, l) and (k, l) (i.j). Then adding the equalities m + l = k + n and k + j = l + i together gives m + l + k + j = k + n + l + i. Cancelling k + l gives m + j = n + i, or (m, n) (i, j). We can thus make the following definition. Definition The integers are the quotient Z = (N N)/, for defined as in Lemma We write [m, n] for the equivalence class of (m, n) N N. 9
10 3.1 Exercises Exercise 3.1. Prove that = 4 by explicitly giving a bijection from 2 2 to 4. Exercise 3.2. Construct bijections to prove the associativity of addition and multiplication (i.e., (m + n) + k = m + (n + k) and (mn)k = m(nk)). Exercise 3.3. This exercise explores how arithmetic interacts with the ordering on the natural numbers. (a): Show that for any m, l < k iff l + m < k + m. (Hint: Use Exercise 2.2 and induction.) (b): Show that n m exists (i.e., there exists k such that m + k = n) iff m n. (Hint: For the forward direction, apply (b) with l = 0. For the reverse direction, prove m n and let k = n \ m.) (c): Show that if m < n and l k, then m + l < n + k. (Hint: Write n + k = n + l + (k l).) (d): Show that if m > 0, then l < k iff lm < km. (Hint: Use distributivity, induction, and (d).) Exercise 3.4. Show that multipication (except by 0!) has cancellation: if k 0 and nk = mk, then n = m. It follows that division by natural numbers other than 0, when it can be done at all, is welldefined. (Hint: Use Exercise 3.3(e).) Exercise 3.5 is optional, but you should at least think about it if you haven t seen equivalence relations before. Exercise 3.5. (a): Prove Proposition 3.9. (b): A partition of a set X is a collection P of nonempty subsets of X such that if A, B P, then A B =, and such that the union P of all the sets in P is all of X. Define x y if x, y A for some A P. Show that is an equivalence relation and P = X/. The following exercise is optional (but interesting!). Exercise 3.6. If S and T are sets, let Fun(S, T ) be the set of all functions from S to T. For n, m N, define n m = Fun(m, n). Explain why this definition should be the same as the familiar notion of exponentiation and see what you can prove about it (e.g., properties like n m n k = n m+k and (n m ) k = n mk ). What is 0 0? 10
11 4 Integers and Rational Numbers But wait! The integer 0 should be the equivalence class of pairs (m, n) with m n = 0, i.e. the set {(0, 0), (1, 1), (2, 2),...}. This is very different than the set 0 = which we defined as the natural number 0. Similarly, for any n N the equivalence class [n, 0] should represent n as an integer, but it is very different from n as a set. Is this a problem? The answer is no, for a reason we ve been periodically mentioning, which is that the specific way we chose to define the natural numbers is not the definition of the natural numbers but just a definition. The reason we care about the natural numbers is the properties they have when we do arithmetic and count with them, not the specific sets we use to represent them. The only reason that we chose to define them the way we did is that it made it easier to prove those properties. However, any other set that looks like the natural numbers with elements called 0, 1, 2, and so on, and operations called addition and multiplication and an order relation that behave in exactly the same way is just ast good. This idea of looking the same is captured more precisely by the notion of an isomorphism, which is a bijection which preserves all of the relevant structure (addition, multiplication, order). 2 We have not yet defined arithmetic and order on the integers, but we can start with the following result. Proposition 4.1. The function F (n) = [n, 0] is a bijection from N to {[n, 0] : n N} Z. Proof. This map is clearly surjective. For injectivity, suppose F (n) = F (m). Then [n, 0] = [m, 0] so (n, 0) (m, 0). But this just means that n + 0 = m + 0, so n = m. From now on, we will not normally distinguish notationally between the natural number n N and the integer F (n), and will pretend that N actually is a subset of Z. In Exercise 4.1, you will show that the definitions of arithmetic and order for integers agree with the definitions we already had for natural numbers. Let s now try to define addition of integers. Since (m n) + (k l) should equal (m + k) (n + l), we make the following definition. Definition 4.2. Let [m, n] and [k, l] be integers. Then their sum is [m, n] + [k, l] = [m + k, n + l]. But wait! The integer [m, n] is equal to the integer [m + 1, n + 1], so [m, n] + [k, l] should equal [m + 1, n + 1] + [k, l]. This is not obvious from the definition we have given. The property that addition should not depend on what representative (m, n) of the equivalence class [m, n] we choose is called addition being welldefined. Proposition 4.3. Addition of integers is welldefined: if [m, n] = [m, n ] and [k, l] = [k, l ], then [m + k, n + l] = [m + k, n + l ]. Proof. This is just algebra. We have m + n = n + m and k + l = k + l, and adding these two equations gives m + n + k + l = n + m + k + l, or Hence [m + k, n + l] = [m + k, n + l ]. (m + k) + (n + l ) = (n + l) + (m + k ). Note that addition of integers agrees with our earlier definition of addition of natural numbers: if we have integers [m, 0] and [n, 0] which should be the same as the natural numbers m and n, then [m, 0] + [n, 0] = [m + n, 0 + 0] = [m + n, 0] is the natural number m + n. Here are some basic properties of addition. Proposition 4.4. Addition of integers is commutative and associative. For any [m, n] Z, [m, n] + 0 = [m, n]. 2 In other words, an isomorphism a bijection f where m + n = k iff f(m) + f(n) = f(k), mn = k iff f(m)f(n) = f(k), and m < n iff f(m) < f(n) basically, everything works the same on either side of the bijection. 11
12 Proof. For commutativity, note that [m + k, n + l] = [k + m, l + n] by commutativity of addition of natural numbers. The other two similarly follow from the same properties for natural numbers (using that 0 = [0, 0]. Unlike the natural numbers, with the integers we can also always define subtraction. Definition 4.5. Let [m, n] Z. Then its negative is [m, n] = [n, m] (why does this definition make sense, and why is it welldefined?). We write [m, n] [k, l] = [m, n] + ( [k, l]). We intended for [m, n] to represent m n ; is it true that [m, n] = m n for this definition of subtraction? It is, because m n = [m, 0] + [0, n] = [m, n]. Note also that ( 1)[m, n] = [m, n], since [0, 1][m, n] = [m0 + n1, m1 + n0] = [n, m]. Proposition 4.6. For any integer [m, n], [m, n] [m, n] = 0. Proof. We have [m, n] [m, n] = [m, n] + [n, m] = [m + n, m + n] = 0. Remark 4.7. Propositions 4.4 and 4.6 together say that Z is an abelian group under addition, an abstract algebraic structure with addition and subtraction. This together with Proposition 4.11 says that Z is a commutative ring, an abstract algebraic structure with addition, subtraction, and multiplication. We can now show that Z indeed agrees with our intuitive idea of the integers. Proposition 4.8. Every integer is a natural number xor n for a unique nonzero natural number n. Proof. Let [k, l] be an integer. If k l, then k l N exists by Exercise 3.3(b), and it is easy to see that [k, l] = [k l, 0] is a natural number. If k < l, then [k, l] = [0, l k] = [l k, 0] and l > k so l k is a nonzero natural number. This n = l k is unique since if n = m, [0, n] = [0, m] so m = n. The proof of exclusivity of the two possibilities is similar and left to the reader. From now on, we will sometimes simply write n for an arbitrary integer. The results we have developed show that we can manipulate integers according to the ordinary rules of addition and subtraction. Now let s define multiplication. Since (m n)(k l) should equal (mk + nl) (ml + nk), we make the following definition. Definition 4.9. Let [m, n] and [k, l] be integers. Then their product is [m, n][k, l] = [mk + nl, ml + nk]. Proposition Multiplication of integers is welldefined: if [m, n] = [m, n ] and [k, l] = [k, l ], then [mk + nl, ml + nk] = [m k + n l, m l + n k ]. Proof. This is just algebra, but it s quite long and tedious. Basically, you make a bunch of formal manipulations starting from the equations m + n = n + m and k + l = k + l and eventually arrive at mk + nl + m l + n k = ml + nk + m k + n l. You are invited to work out the details if you enjoy messy algebra. Proposition Multiplication of integers is commutative and associative and distributes over addition. For any [m, n] Z, [m, n] 0 = 0 and [m, n] 1 = [m, n]. Proof. For commutativity, we have [m, n][k, l] = [mk + nl, ml + nk] = [km + ln, lm + kn] = [k, l][m, n]. The other assertions are similar but more complicated algebra and are left as exercises. The following facts will be important to us next section. Proposition Multiplication by integers other than 0 has cancellation: if mn = kn and n 0, then m = k. 12
13 Proof. If m, n, and k are all natural numbers, this is just Exercise 3.4. Suppose n N and m, k N. Then since mn = kn, m( n) = ( 1)mn = ( 1)kn = k( n), and we are done by the case where all the numbers are natural numbers. Suppose n, k N and m N. Then ( m)n N and mn = kn N, which implies mn = ( m)n = 0. Since n 0 and ( m) 0 = 0, Exercise 3.4 implies m = 0, and similarly we can show that k = 0. There are 5 other cases to consider, but they are all similar to the ones done above. There are a few other things to check to be sure that everything works as we expect, but their proofs are just as simple as all the proofs so far have been. Some of them are in the exercises, but others we leave without proof and trust that you can come up with a proof on the spot if we need to use them. To close this section, we define the order on the integers, whose properties are fleshed out in the exercises. Definition Let m, n Z. Then we say m n if n m N. 4.1 Exercises Exercise 4.1. Show that the map F of Proposition 4.1 is an isomorphism for all the structure we ve defined. That is, show that our definitions of addition, subtraction, multiplication, and order for integers all agree with the definitions in Section 2 for natural numbers. (The remark following Proposition 4.3 showed this for addition.) Exercise 4.2. Complete the proof of Proposition 4.4. Exercise 4.3. In this exercise we prove some basic properties of the ordering on the integers. (a): Show that n 0 iff n N. (b): Show that < is a strict total order, where < means and not = and strict total order means it satisfies the conditions of Theorem 2.1. (Hint: Use Proposition 4.8.) (c): Show that every part of Exercise 3.3 except (c) holds for integers, and that (c) fails. (d): Show that m n iff m n. You can skip the following exercise if we didn t get to talking about multiplication in class. Exercise 4.4. Finish the proof of Proposition
14 5 The Rational Numbers We will now construct the rational numbers from the integers in much the same way as we constructed the integers from the natural numbers. Just as integers were defined as differences of natural numbers, rational numbers will be quotients of integers. Just as we had m n = k l if m + l = k + n, we will have m/n = k/l if ml = kn. We must also not allow 0 as a denominator. Lemma 5.1. The relation (m, n) (k, l) if ml = kn is an equivalence relation on Z (Z \ {0}). Proof. The proof is exactly the same as that of Lemma 3.10 except we turn addition into multiplication everywhere. For example, for transitivity, suppose (m, n) (k, l) and (k, l) (i.j). Then multiplying the equalities ml = kn and kj = li together gives mlkj = knli. Cancelling kl (which is possible since k, l Z \ {0}) gives mj = ni, or (m, n) (i, j). Definition 5.2. The rational numbers are the quotient Q = (Z (Z \ {0})/, for defined as in Lemma 5.1. We write m/n for the equivalence class of (m, n) Z (Z \ {0}). Just as we could think of the natural numbers as sitting inside the integers, we can think of the integers as sitting inside the rational numbers. Proposition 5.3. The function G(n) = n/1 is a bijection from Z to {n/1 : n Z} Q. Proof. Again, the proof is exactly the same as Proposition 4.1, substituting multiplication for addition and 1 for 0. For example, for injectivity, suppose G(n) = G(m). Then (n, 1) (m, 1), so n 1 = m 1, so n = m. As before, we will not distinguish between n Z and G(n) Q. We will leave it as an exercise to see that this G preserves arithmetic and order as we define it below. Definition 5.4. Let m/n and k/l be rational. Then their sum is m/n + k/l = (ml + kn)/(nl). As for the integers, we must check that this definition is welldefined. Proposition 5.5. Addition of rational numbers is welldefined: if m/n = m /n and k/l = k /l, then (ml + kn)/(nl) = (m l + k n )/(n l ). Proof. We have (ml + kn)(n l ) = (mn ll + kl nn ) = (m nll + k lnn ) = (m l + k n )(nl). It is now straightforward to show the following. Proposition 5.6. Addition of rational numbers is commutative and associative. For any m/n Q, m/n + 0 = m/n and m/n + ( m)/n = 0. Proof. Exercise. Remark 5.7. Proposition 5.6 says that Q is an abelian group under addition. The treatment of multiplication is similar, and equally straightforward. Definition 5.8. Let m/n and k/l be rational. Then their product is (m/n)(k/l) = (mk)/(nl). 14
15 Proposition 5.9. Multiplication and reciprocals of rational numbers is well defined. Multiplication is commutative and associative and distributes over addition. For any m/n Q, (m/n) 0 = 0, (m/n) 1 = m/n. If m/n 0, we additionally have (m/n)(n/m) = 1. Proof. Exercise. Remark Proposition 5.9 together with Proposition 5.6 says that Q is a field, an abstract algebraic structure with addition, subtraction, multiplication, and division. Finally, we define the ordering on the rational numbers. Definition Let m/n Q. Then we write m/n > 0 if mn > 0 (Why does this definition make sense? Why is it welldefined?). For x, y Q, we write x > y if x y > 0. Proposition The relation < is a (strict) total order on Q. If x, y, z Q and x < y then x+z < y +z. If z > 0, we also have xz < yz, and if z < 0 we also have xz > yz. Proof. Exercise. Remark Proposition 5.12 together with the previous results says that Q is an ordered field, a field with a total order compatible with the field structure. 5.1 Exercises Except for Exercise 5.5, these exercises are all just checking basic properties of the rational numbers that we skipped in class. There are a lot of things to check but they are all pretty much straightforward, and it may get tedious and boring. If it is getting tedious and boring, don t feel like you have to do it all; what s important is that you understand how these proofs work, not that you do every single one of them. Exercise 5.1. Show that the map G of Proposition 5.3 is an isomorphism for all the structure we ve defined. That is, show that our definitions of addition, multiplication, and order for rationals all agree with the definitions in Section 4 for integers. Exercise 5.2. Prove Proposition 5.6. Exercise 5.3. Prove Proposition 5.9. Exercise 5.4. Prove Proposition Exercise 5.5. Prove that Q is an archimedean ordered field: for any x Q, there is an n N such that n > x. 15
16 6 The Real Numbers While the intuition behind constructing the integers and the rational numbers was fairly straightforward, the real numbers are another question entirely. It is not even necessarily clear what a real number is intuitively. In school, you ve probably seen real numbers presented as infinite decimals. Indeed, this is perfectly valid, and we could make it into a rigorous definition. However, this definition would be unrelated to our previous definition of the rationals and would unnaturally favor 10 as the base. Instead, we will look for a more natural definition. Passing from the natural numbers to the integers involved making addition betterbehaved (so that additive inverses exist and subtraction is possible). Passing from the integers to the rationals involved making multiplication betterbehaved (so that multiplicative inverses exist and division is possible). Passing from the rationals to the reals will involve making the order relation betterbehaved, though in a rather subtle way. A way to geometrically think of the irrational numbers is that they are holes in the rational number line. That is, they are places on the line whose location we can specify but where there is not any rational number. For example, we can define 2 as being right in between all the positive rationals whose square is less than 2 and all the positive rationals whose square is greater than 2. We could define π as being between 3, 3.1, 3.14, 3.141,... and 4, 3.2, 3.15, 3.142,.... In general, we can define any real number by saying what rational numbers are less than it and what rational numbers are greater than it. Alternatively, we could just give the rational numbers that are less than it, since the rational numbers greater than it are then all the rest. This motivates the following definition. Definition 6.1. A (left) cut in the rational numbers (or in any totally ordered set) is a subset L Q such that whenever x L and y < x, y L as well. We should think of a cut as the set of all rational numbers less than some real number. However, this definition isn t yet quite right a cut does not exactly correspond to a real number. The first problem is simple: according to Definition 6.1, both and all of Q are cuts. However, these geometrically would correpond to the points and, which we do not want to call real numbers. Thus we must exclude these two possibilities. Definition 6.2. A cut L Q is proper if L and L Q. The second problem with the notion of a cut is more subtle: more than one cut might correspond to the same number. For example, let L = {x Q : x < 2} and M = {x Q : x 2}. Then L and M both ought to represent the number 2. We could construct a similar example with 2 replaced by any rational number. To put it another way, we have to decide whether we want our cuts to be all rationals strictly less than a real number or a rationals less than or equal to a real number. We will (arbitrarily) choose strictly less than, so we will choose L rather than M to represent 2. How do we detect if a cut is of type L instead of type M? We could do it in an ad hoc way, but instead we will introduce the following definition which is very important to studying the real numbers. Definition 6.3. Let S Q (or any totally ordered set). An upper bound for S is an x Q such that x s for all s S. A least upper bound, or supremum of S is an upper bound that is less than every other upper bound. The supremum of S, if one exists, is unique. Indeed, if x and y are both suprema, then they are both upper bounds so x y and y x so x = y. We write sup S for the supremum of S if it exists. In the case of our cuts L and M above, it is easy to see that sup L = sup M = 2. The difference between L and M can be stated as the fact that M contains its supremum but L does not. On the other hand, cuts corresponding to irrational numbers do not have suprema, because what should be the supremum is the missing irrational number itself. Definition 6.4. A cut L Q is open if sup L, if it exists, is not contained in L. 16
17 Note that under this definition, a cut without a supremum is automatically open. It turns out that these two corrections to our definition of cuts are enough to get the real numbers. Definition 6.5. The real numbers R are the set of all proper open cuts in Q. Once again, we must see how our old Q embeds in this new number system. Proposition 6.6. The map H(x) = {y Q : y < x} is a bijection from Q to {L R : sup L exists}. Proof. Since sup H(x) = x, H clearly maps into {L R : sup L exists}. It is also clear that H is injective. Finally, if L R and sup L = x, then L H(x) since x is an upper bound for L and L is open. Furthermore, no y < x is an upper bound, so for all y < x there is a z L with y < z. Since L is a cut, this implies y L, for any y < x. That is, H(x) L, so H(x) = L. Hence H is surjective. As before, we should also check that this map H is compatible with arithmetic and order that we will define. As before, we leave this as an exercise. As before, we will not distinguish between x Q and H(x) R. Since the real numbers are defined in terms of the order relation on Q, it is easy to define the order relation on R. Indeed, it is intuitively not hard to see that one cut is to the left of another cut iff it is a subset. Definition 6.7. Let L, M R. Then we write L < M if L M. Proposition 6.8. The relation < is a (strict) total order on R. Proof. The only part that is not routine is totality: if L M, then L < M or M < L. Suppose L M, i.e. L M. Then there is some x L that is not in M. Since M is a cut, this also implies y M for all y > x. Thus M {y : y < x}. But since L is a cut and x L, every y < x is also in L. Thus M L, so M < L. Since the real numbers were defined in attempt to fill in the holes in the rational numbers, we would like to know whether there are any holes left in the real numbers. The following concept captures the idea of not having holes. Definition 6.9. A totally ordered set X is complete if whenever S X is nonempty and has an upper bound, then it has a supremum. It can be shown, in fact, that we get an equivalent notion if we restrict to the case that S is a cut. As we saw, the cuts which had suprema were those that corresponded to rational numbers, so completeness says that every cut already corresponds to some point in your set there are no holes to fill. Theorem The real numbers are complete. Proof. Let S R be nonempty and have an upper bound, and let L = S, the union of all the elements of S. Then L is nonempty since S is nonempty. If K R is an upper bound for S and x Q \ K, then x > y for all y K. Since every M S is contained in K, so is L. Thus x > y for all y L, so L is not all of R. Hence L is proper. Finally, if x = sup L, then x is an upper bound for L and hence for all M S, and it follows that x M for all M S so x L, so L is open. Hence L R. We now claim that L = sup S in R. It is clear that L M for all M S, so it is an upper bound. Furthermore, if L M for all M S, then L S = L by definition of the union, so L is the least upper bound. We now define arithmetic on R. Definition Let L and M be real numbers. Then their sum is L + M = {x + y : x L and y M}. The negative of a real number L is L = { x : x L and x sup L}. The absolute value of L is L if L 0 and L if L < 0 17
18 In this case we have no equivalence relation to worry about for welldefinedness, but we must check that L + M and L are indeed proper open cuts. This takes some work, but is too tedious to be worth going through here. Proposition Addition of real numbers is commutative and associative. For any L R, L + 0 = L and L + ( L) = 0. Proof. We prove that L+0 = L and leave the rest as an exercise. Note that here 0 = H(0) = {x Q : x < 0}. For any y L and x 0, x + y < y so x + y L, hence L + 0 L. Conversely, if x L, x sup L, so there is some z L such that z > x. Then y = x z 0, so x = z + y L + 0. Hence L L + 0, sol = L + 0. Remark Proposition 6.12 says that R is an abelian group under addition. Multiplication is rather more painful to define, and is most easily defined by splitting into cases depending on the signs of the factors Definition Let L 0 and M 0 be real. Then their product is LM = {xy : x L, y M, and x, y 0} {x : x < 0}. If L < 0 and M 0 or L 0 and M < 0, we define LM = L M, and if L < 0 and M < 0 we define LM = L M. The reciprocal of a real number L > 0 is L 1 = {x : xy < 1 for all y M such that y 0}. If L < 0, we define L 1 = L 1. Again, there is a fair amount of work to show that these products and reciprocals are themselves real numbers, and we omit this. Proposition Multiplication of real numbers is commutative and associative and distributes over addition. For any L R, L 0 = 0 and L 1 = L. If L 0, we additionally have LL 1 = 1. Proof. Exercise, though be warned that this is a fair deal of work. Remark Proposition 6.15 together with Proposition 6.12 says that R is a field. Finally, we record the fact that the ordering on R is compatible with the arithmetic. Proposition If K, L, M R and K < L then K + M < L + M. If M > 0, we also have KM < LM, and if M < 0 we also have KM > LM. Proof. Exercise. Remark Proposition 6.17 together with the previous results says that R is an ordered field. In fact, it can be shown that R is the only complete ordered field, up to isomorphism. This fact is very useful in showing that different methods of defining the real numbers are equivalent. 6.1 Exercises Exercise 6.1. Show that the map H of Proposition 6.6 is an isomorphism for all the structure we ve defined. That is, show that our definitions of addition, multiplication, and order for real numbers all agree with the definitions in Section 5 for rationals. The first three exercises have you fill in proofs that we skipped in class. I recommend that you do a few of these until you get the hang of how they go, but don t worry about doing all of them. Exercise 6.2. Complete the proof of Proposition Exercise 6.3. Prove Proposition Exercise 6.4. Prove Proposition Exercise 6.5. Prove that R is archimedean: for any L R, there is an n N such that n > L. Exercise 6.6. Define lower bounds and greatest lower bounds, or infima, in the same was as we defined upper bounds and suprema. Show that every nonempty subset of R with a lower bound has an infimum. 18
19 7 Problems with set theory Throughout these notes, we have worked with the assumption that a simple, intuitive notion of sets can and should be used as the basis to develop all of mathematics. There are various reasons why this assumption is not entirely correct, and in this section we will see the main problems with it. The first problem with the set theory we have been using thus far is that it is not selfconsistent! Let us be more precise. It turns out that everything we have done has used only the rules of logic plus the following two axioms about sets, the first of which is harmless but the second of which turns out to be too strong. 1. A set is determined by its elements 2. For any property P (x) of sets, there is a set {x : P (x)} of all sets having property P. Let s see how we can get a contradiction, called Russell s paradox, out of this. We use Axiom (2) with the property x x. Thus let S = {x : x x}. Now we ask, is S S? If S S, then by definition of S, S S. But if S S, then similarly S S! This is a contradiction! Here s another similar paradox, called Cantor s paradox. As you may have seen before, the real numbers are uncountable: they are not in bijection with the natural numbers. This is usually proved by the diagonal argument using the representation of real numbers as decimal (or binary) expansions. This diagonal argument can be extended to produce a set which is strictly larger than any given set. Definition 7.1. Let X be a set. Then the power set of X is P(X) = {S : S X}. Theorem 7.2 (Cantor). For any X, X is not in bijection with P(X). Proof. Suppose f : X P(X) is a bijection. Define S = {x X : x f(x)}. Then S P(X), so S = f(y) for some y X. But then y f(y) = S iff y S, a contradiction. Hence no such bijection f can exist. Now let s apply Cantor s theorem to the set of all sets. For example, we could define this set as V = {x : x = x}, since x = x is true for any x. By Cantor s theorem, V is not in bijection with P (V ). But V contains all sets, so it is clear that P (V ) = V. This is a contradiction. 3 It is thus clear that set theory as we have been doing it, also known as naive set theory, is inconsistent. If we want to be able to consistently use set theory, we must somehow modify Axiom (2). The standard solution that set theorists have accepted is to reject the idea that any property can define a set and conclude that Russell s Paradox and Cantor s Paradox are instead proofs that no set S = {x : x x} or V = {x : x = x} can exist! After all, assuming they did exist, we reached a contradiction, so therefore they must not exist. By rejecting Axiom (2), however, we must come up with new axioms that tell us how we can construct sets. The main idea behind these axioms, the standard formulation of which is known as the Zermelo Fraenkel Axioms (with Choice), or ZFC, is that instead of being about to create a set out of thin air, we must construct it out of set we already have. For example, if we know a set X exists, we are allowed to conclude that the power set P (X) exists. Or, in close analogy with Axiom (2), given a set X and any property P, the set {x X : P (x)} exists. If we knew the set V of all sets existed we could apply this with X = V to conclude Axiom (2), but V does not exist. While this proliferation of new axioms works, it certainly loses much of the aesthetic appeal of naive set theory. Instead of just using basic pure logic about how sets work to construct all of mathematics, we must now use some hodgepodge of arbitraryseeming rules about what sets we can make. This sentiment is perhaps worsened by the fact that one of these axioms is explicitly the fact that N itself exists we cannot, in fact, construct any infinite set without just directly assuming that one exists. Furthermore, a famous result of Gödel states that there is no perfect set of axioms. More precisely, for any consistent collection of axioms in which we can express arithmetic which we can write down, there is 3 You may have noticed that the proof of Cantor s theorem looks a lot like the argument in Russell s paradox. In fact, if you unwind the argument of Cantor s paradox to include the proof of Cantor s theorem in the case X = V, you will see that it is really the same as Russell s paradox. 19
MAT2400 Analysis I. A brief introduction to proofs, sets, and functions
MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take
More informationSets and functions. {x R : x > 0}.
Sets and functions 1 Sets The language of sets and functions pervades mathematics, and most of the important operations in mathematics turn out to be functions or to be expressible in terms of functions.
More informationCARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE
CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE With the notion of bijection at hand, it is easy to formalize the idea that two finite sets have the same number of elements: we just need to verify
More information13 Infinite Sets. 13.1 Injections, Surjections, and Bijections. mcsftl 2010/9/8 0:40 page 379 #385
mcsftl 2010/9/8 0:40 page 379 #385 13 Infinite Sets So you might be wondering how much is there to say about an infinite set other than, well, it has an infinite number of elements. Of course, an infinite
More informationIn mathematics you don t understand things. You just get used to them. (Attributed to John von Neumann)
Chapter 1 Sets and Functions We understand a set to be any collection M of certain distinct objects of our thought or intuition (called the elements of M) into a whole. (Georg Cantor, 1895) In mathematics
More informationMATH 321 EQUIVALENCE RELATIONS, WELLDEFINEDNESS, MODULAR ARITHMETIC, AND THE RATIONAL NUMBERS
MATH 321 EQUIVALENCE RELATIONS, WELLDEFINEDNESS, MODULAR ARITHMETIC, AND THE RATIONAL NUMBERS ALLAN YASHINSKI Abstract. We explore the notion of welldefinedness when defining functions whose domain is
More informationf(x) is a singleton set for all x A. If f is a function and f(x) = {y}, we normally write
Math 525 Chapter 1 Stuff If A and B are sets, then A B = {(x,y) x A, y B} denotes the product set. If S A B, then S is called a relation from A to B or a relation between A and B. If B = A, S A A is called
More informationSets and Cardinality Notes for C. F. Miller
Sets and Cardinality Notes for 620111 C. F. Miller Semester 1, 2000 Abstract These lecture notes were compiled in the Department of Mathematics and Statistics in the University of Melbourne for the use
More informationElementary Number Theory We begin with a bit of elementary number theory, which is concerned
CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,
More informationMath 4310 Handout  Quotient Vector Spaces
Math 4310 Handout  Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable
More informationTheorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q.
Math 305 Fall 011 The Density of Q in R The following two theorems tell us what happens when we add and multiply by rational numbers. For the first one, we see that if we add or multiply two rational numbers
More informationCHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS
CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we construct the set of rational numbers Q using equivalence
More informationDiscrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 20
CS 70 Discrete Mathematics and Probability Theory Fall 009 Satish Rao, David Tse Note 0 Infinity and Countability Consider a function (or mapping) f that maps elements of a set A (called the domain of
More informationGod created the integers and the rest is the work of man. (Leopold Kronecker, in an afterdinner speech at a conference, Berlin, 1886)
Chapter 2 Numbers God created the integers and the rest is the work of man. (Leopold Kronecker, in an afterdinner speech at a conference, Berlin, 1886) God created the integers and the rest is the work
More informationIntroduction to mathematical arguments
Introduction to mathematical arguments (background handout for courses requiring proofs) by Michael Hutchings A mathematical proof is an argument which convinces other people that something is true. Math
More information2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try.
2. METHODS OF PROOF 69 2. Methods of Proof 2.1. Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. Trivial Proof: If we know q is true then
More informationEquivalence relations
Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence
More informationCHAPTER 5: MODULAR ARITHMETIC
CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called
More informationProblem Set. Problem Set #2. Math 5322, Fall December 3, 2001 ANSWERS
Problem Set Problem Set #2 Math 5322, Fall 2001 December 3, 2001 ANSWERS i Problem 1. [Problem 18, page 32] Let A P(X) be an algebra, A σ the collection of countable unions of sets in A, and A σδ the collection
More informationFinite and Infinite Sets
Chapter 9 Finite and Infinite Sets 9. Finite Sets Preview Activity (Equivalent Sets, Part ). Let A and B be sets and let f be a function from A to B..f W A! B/. Carefully complete each of the following
More informationIf f is a 11 correspondence between A and B then it has an inverse, and f 1 isa 11 correspondence between B and A.
Chapter 5 Cardinality of sets 51 11 Correspondences A 11 correspondence between sets A and B is another name for a function f : A B that is 11 and onto If f is a 11 correspondence between A and B,
More informationALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS. 1. Ideals in Commutative Rings In this section all groups and rings will be commutative.
ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS PETE L. CLARK 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. 1.1. Basic definitions and examples. Let R be a (commutative!)
More informationCHAPTER 3. Sequences. 1. Basic Properties
CHAPTER 3 Sequences We begin our study of analysis with sequences. There are several reasons for starting here. First, sequences are the simplest way to introduce limits, the central idea of calculus.
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More informationChapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.
Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to
More information1. R In this and the next section we are going to study the properties of sequences of real numbers.
+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real
More informationSo let us begin our quest to find the holy grail of real analysis.
1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers
More informationx if x 0, x if x < 0.
Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the
More informationChapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1
Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes
More information3. Mathematical Induction
3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)
More informationPART I. THE REAL NUMBERS
PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the representation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS
More informationProof: A logical argument establishing the truth of the theorem given the truth of the axioms and any previously proven theorems.
Math 232  Discrete Math 2.1 Direct Proofs and Counterexamples Notes Axiom: Proposition that is assumed to be true. Proof: A logical argument establishing the truth of the theorem given the truth of the
More informationBasic Category Theory for Models of Syntax (Preliminary Version)
Basic Category Theory for Models of Syntax (Preliminary Version) R. L. Crole ( ) Department of Mathematics and Computer Science, University of Leicester, Leicester, LE1 7RH, U.K. Abstract. These notes
More information0 ( x) 2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x 2.
SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then xy x + y. Proof. First we prove that if x is a real number, then x 0. The product of two positive
More informationINTRODUCTORY SET THEORY
M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H1088 Budapest, Múzeum krt. 68. CONTENTS 1. SETS Set, equal sets, subset,
More informationSets, Relations and Functions
Sets, Relations and Functions Eric Pacuit Department of Philosophy University of Maryland, College Park pacuit.org epacuit@umd.edu ugust 26, 2014 These notes provide a very brief background in discrete
More information2.1.1 Examples of Sets and their Elements
Chapter 2 Set Theory 2.1 Sets The most basic object in Mathematics is called a set. As rudimentary as it is, the exact, formal definition of a set is highly complex. For our purposes, we will simply define
More informationClassical Analysis I
Classical Analysis I 1 Sets, relations, functions A set is considered to be a collection of objects. The objects of a set A are called elements of A. If x is an element of a set A, we write x A, and if
More informationPOSITIVE INTEGERS, INTEGERS AND RATIONAL NUMBERS OBTAINED FROM THE AXIOMS OF THE REAL NUMBER SYSTEM
MAT 1011 TECHNICAL ENGLISH I 03.11.2016 Dokuz Eylül University Faculty of Science Department of Mathematics Instructor: Engin Mermut Course assistant: Zübeyir Türkoğlu web: http://kisi.deu.edu.tr/engin.mermut/
More informationFinite Sets. Theorem 5.1. Two nonempty finite sets have the same cardinality if and only if they are equivalent.
MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we
More informationReading 7 : Program Correctness
CS/Math 240: Introduction to Discrete Mathematics Fall 2015 Instructors: Beck Hasti, Gautam Prakriya Reading 7 : Program Correctness 7.1 Program Correctness Showing that a program is correct means that
More informationMathematics for Computer Science
Mathematics for Computer Science Lecture 2: Functions and equinumerous sets Areces, Blackburn and Figueira TALARIS team INRIA Nancy Grand Est Contact: patrick.blackburn@loria.fr Course website: http://www.loria.fr/~blackbur/courses/math
More informationCHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.
CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,
More informationCHAPTER 3 Numbers and Numeral Systems
CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural,
More informationDiscrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2
CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2 Proofs Intuitively, the concept of proof should already be familiar We all like to assert things, and few of us
More informationCONTENTS 1. Peter Kahn. Spring 2007
CONTENTS 1 MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 2 The Integers 1 2.1 The basic construction.......................... 1 2.2 Adding integers..............................
More information6. Metric spaces. In this section we review the basic facts about metric spaces. d : X X [0, )
6. Metric spaces In this section we review the basic facts about metric spaces. Definitions. A metric on a nonempty set X is a map with the following properties: d : X X [0, ) (i) If x, y X are points
More informationDedekind Cuts. Rich Schwartz. September 17, 2014
Dedekind Cuts Rich Schwartz September 17, 2014 1 Decimal Expansions How would you define a real number? It would seem that the easiest way is to say that a real number is a decimal expansion of the form
More informationStructure of Measurable Sets
Structure of Measurable Sets In these notes we discuss the structure of Lebesgue measurable subsets of R from several different points of view. Along the way, we will see several alternative characterizations
More informationIt is not immediately obvious that this should even give an integer. Since 1 < 1 5
Math 163  Introductory Seminar Lehigh University Spring 8 Notes on Fibonacci numbers, binomial coefficients and mathematical induction These are mostly notes from a previous class and thus include some
More informationIntroduction Russell s Paradox Basic Set Theory Operations on Sets. 6. Sets. Terence Sim
6. Sets Terence Sim 6.1. Introduction A set is a Many that allows itself to be thought of as a One. Georg Cantor Reading Section 6.1 6.3 of Epp. Section 3.1 3.4 of Campbell. Familiar concepts Sets can
More informationCourse 421: Algebraic Topology Section 1: Topological Spaces
Course 421: Algebraic Topology Section 1: Topological Spaces David R. Wilkins Copyright c David R. Wilkins 1988 2008 Contents 1 Topological Spaces 1 1.1 Continuity and Topological Spaces...............
More informationChapter 3. Cartesian Products and Relations. 3.1 Cartesian Products
Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing
More informationSet theory as a foundation for mathematics
V I I I : Set theory as a foundation for mathematics This material is basically supplementary, and it was not covered in the course. In the first section we discuss the basic axioms of set theory and the
More informationCartesian Products and Relations
Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special
More informationCHAPTER I THE REAL AND COMPLEX NUMBERS DEFINITION OF THE NUMBERS 1, i, AND 2
CHAPTER I THE REAL AND COMPLEX NUMBERS DEFINITION OF THE NUMBERS 1, i, AND 2 In order to mae precise sense out of the concepts we study in mathematical analysis, we must first come to terms with what the
More informationMODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.
MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on
More informationWRITING PROOFS. Christopher Heil Georgia Institute of Technology
WRITING PROOFS Christopher Heil Georgia Institute of Technology A theorem is just a statement of fact A proof of the theorem is a logical explanation of why the theorem is true Many theorems have this
More informationThis asserts two sets are equal iff they have the same elements, that is, a set is determined by its elements.
3. Axioms of Set theory Before presenting the axioms of set theory, we first make a few basic comments about the relevant first order logic. We will give a somewhat more detailed discussion later, but
More informationThis chapter is all about cardinality of sets. At first this looks like a
CHAPTER Cardinality of Sets This chapter is all about cardinality of sets At first this looks like a very simple concept To find the cardinality of a set, just count its elements If A = { a, b, c, d },
More information1. Prove that the empty set is a subset of every set.
1. Prove that the empty set is a subset of every set. Basic Topology Written by MenGen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since
More information2. INEQUALITIES AND ABSOLUTE VALUES
2. INEQUALITIES AND ABSOLUTE VALUES 2.1. The Ordering of the Real Numbers In addition to the arithmetic structure of the real numbers there is the order structure. The real numbers can be represented by
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationCHAPTER 2. Inequalities
CHAPTER 2 Inequalities In this section we add the axioms describe the behavior of inequalities (the order axioms) to the list of axioms begun in Chapter 1. A thorough mastery of this section is essential
More informationREAL ANALYSIS I HOMEWORK 2
REAL ANALYSIS I HOMEWORK 2 CİHAN BAHRAN The questions are from Stein and Shakarchi s text, Chapter 1. 1. Prove that the Cantor set C constructed in the text is totally disconnected and perfect. In other
More informationSpring As discussed at the end of the last section, we begin our construction of the rational
MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 3 The Rational Numbers 1 3.1 The set Q................................. 1 3.2 Addition and multiplication of rational numbers............
More informationCourse 221: Analysis Academic year , First Semester
Course 221: Analysis Academic year 200708, First Semester David R. Wilkins Copyright c David R. Wilkins 1989 2007 Contents 1 Basic Theorems of Real Analysis 1 1.1 The Least Upper Bound Principle................
More informationStudents in their first advanced mathematics classes are often surprised
CHAPTER 8 Proofs Involving Sets Students in their first advanced mathematics classes are often surprised by the extensive role that sets play and by the fact that most of the proofs they encounter are
More information4. CLASSES OF RINGS 4.1. Classes of Rings class operator Aclosed Example 1: product Example 2:
4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets
More informationWe give a basic overview of the mathematical background required for this course.
1 Background We give a basic overview of the mathematical background required for this course. 1.1 Set Theory We introduce some concepts from naive set theory (as opposed to axiomatic set theory). The
More informationModule MA1S11 (Calculus) Michaelmas Term 2016 Section 3: Functions
Module MA1S11 (Calculus) Michaelmas Term 2016 Section 3: Functions D. R. Wilkins Copyright c David R. Wilkins 2016 Contents 3 Functions 43 3.1 Functions between Sets...................... 43 3.2 Injective
More informationPrime Numbers. Chapter Primes and Composites
Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are
More informationSets and Subsets. Countable and Uncountable
Sets and Subsets Countable and Uncountable Reading Appendix A Section A.6.8 Pages 788792 BIG IDEAS Themes 1. There exist functions that cannot be computed in Java or any other computer language. 2. There
More informationMathematical Induction
Chapter 2 Mathematical Induction 2.1 First Examples Suppose we want to find a simple formula for the sum of the first n odd numbers: 1 + 3 + 5 +... + (2n 1) = n (2k 1). How might we proceed? The most natural
More informationINTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS
INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28
More information2.3 Bounds of sets of real numbers
2.3 Bounds of sets of real numbers 2.3.1 Upper bounds of a set; the least upper bound (supremum) Consider S a set of real numbers. S is called bounded above if there is a number M so that any x S is less
More informationA set is a Many that allows itself to be thought of as a One. (Georg Cantor)
Chapter 4 Set Theory A set is a Many that allows itself to be thought of as a One. (Georg Cantor) In the previous chapters, we have often encountered sets, for example, prime numbers form a set, domains
More informationThis chapter describes set theory, a mathematical theory that underlies all of modern mathematics.
Appendix A Set Theory This chapter describes set theory, a mathematical theory that underlies all of modern mathematics. A.1 Basic Definitions Definition A.1.1. A set is an unordered collection of elements.
More informationChapter 1. Informal introdution to the axioms of ZF.
Chapter 1. Informal introdution to the axioms of ZF. 1.1. Extension. Our conception of sets comes from set of objects that we know well such as N, Q and R, and subsets we can form from these determined
More information5. ABSOLUTE EXTREMA. Definition, Existence & Calculation
5. ABSOLUTE EXTREMA Definition, Existence & Calculation We assume that the definition of function is known and proceed to define absolute minimum. We also assume that the student is familiar with the terms
More informationPOWER SETS AND RELATIONS
POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty
More informationIntegers and Rational Numbers
Chapter 6 Integers and Rational Numbers In this chapter, we will see constructions of the integers and the rational numbers; and we will see that our number system still has gaps (equations we can t solve)
More informationInduction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition
Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing
More informationvertex, 369 disjoint pairwise, 395 disjoint sets, 236 disjunction, 33, 36 distributive laws
Index absolute value, 135 141 additive identity, 254 additive inverse, 254 aleph, 466 algebra of sets, 245, 278 antisymmetric relation, 387 arcsine function, 349 arithmetic sequence, 208 arrow diagram,
More informationMATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:
MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,
More informationUndergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics
Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights
More informationA brief history of type theory (or, more honestly, implementing mathematical objects as sets)
A brief history of type theory (or, more honestly, implementing mathematical objects as sets) Randall Holmes, Boise State University November 8, 2012 [notes revised later; this is the day the talk was
More informationS(A) X α for all α Λ. Consequently, S(A) X, by the definition of intersection. Therefore, X is inductive.
MA 274: Exam 2 Study Guide (1) Know the precise definitions of the terms requested for your journal. (2) Review proofs by induction. (3) Be able to prove that something is or isn t an equivalence relation.
More information3.1 The Definition and Some Basic Properties. We identify the natural class of integral domains in which unique factorization of ideals is possible.
Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties We identify the natural class of integral domains in which unique factorization of ideals is possible. 3.1.1 Definition A Dedekind
More informationThe Rational Numbers. Peter J. Kahn
Math 3040: Spring 2009 The Rational Numbers Peter J. Kahn Contents 1. The Set Q 1 2. Addition and multiplication of rational numbers 4 2.1. Definitions and properties. 4 2.2. Comments 7 2.3. Connections
More informationFor the purposes of this course, the natural numbers are the positive integers. We denote by N, the set of natural numbers.
Lecture 1: Induction and the Natural numbers Math 1a is a somewhat unusual course. It is a proofbased treatment of Calculus, for all of you who have already demonstrated a strong grounding in Calculus
More information3. Recurrence Recursive Definitions. To construct a recursively defined function:
3. RECURRENCE 10 3. Recurrence 3.1. Recursive Definitions. To construct a recursively defined function: 1. Initial Condition(s) (or basis): Prescribe initial value(s) of the function.. Recursion: Use a
More informationCancelling an a on the left and a b on the right, we get b + a = a + b, which is what we want. Note the following identity.
15. Basic Properties of Rings We first prove some standard results about rings. Lemma 15.1. Let R be a ring and let a and b be elements of R. Then (1) a0 = 0a = 0. (2) a( b) = ( a)b = (ab). Proof. Let
More informationNotes on counting finite sets
Notes on counting finite sets Murray Eisenberg February 26, 2009 Contents 0 Introduction 2 1 What is a finite set? 2 2 Counting unions and cartesian products 4 2.1 Sum rules......................................
More information18.312: Algebraic Combinatorics Lionel Levine. Lecture 8
18.312: Algebraic Combinatorics Lionel Levine Lecture date: March 1, 2011 Lecture 8 Notes by: Christopher Policastro Remark: In the last lecture, someone asked whether all posets could be constructed from
More informationFoundations of Mathematics I Set Theory (only a draft)
Foundations of Mathematics I Set Theory (only a draft) Ali Nesin Mathematics Department Istanbul Bilgi University Kuştepe Şişli Istanbul Turkey anesin@bilgi.edu.tr February 12, 2004 2 Contents I Naive
More informationMath 140a  HW 1 Solutions
Math 140a  HW 1 Solutions Problem 1 (WR Ch 1 #1). If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Given that r is rational, we can write r = a b for some
More informationTopics in Number Theory
Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall
More informationSolution to Homework 2
Solution to Homework 2 Olena Bormashenko September 23, 2011 Section 1.4: 1(a)(b)(i)(k), 4, 5, 14; Section 1.5: 1(a)(b)(c)(d)(e)(n), 2(a)(c), 13, 16, 17, 18, 27 Section 1.4 1. Compute the following, if
More informationIt is time to prove some theorems. There are various strategies for doing
CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it
More information