MATH 2400 LECTURE NOTES

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1 MATH 2400 LECTURE NOTES PETE L. CLARK 1. First Lecture 1.1. The Goal: Calculus Made Rigorous. The goal of this course is to cover the material of single variable calculus in a mathematically rigorous way. The latter phrase is important: in most calculus classes the emphasis is on techniques and applications; while theoretical eplanations may be given by the instructor e.g. it is usual to give some discussion of the meaning of a continuous function the student tests her understanding of the theory mostly or entirely through her ability to apply it to solve problems. This course is very different: not only will theorems and proofs be presented in class by me, but they will also be presented by you, the student, in homework and on eams. This course offers a strong foundation for a student s future study of mathematics, at the undergraduate level and beyond. As eamples, here are three of the fundamental results of calculus; they are called by me, at least the three Interval Theorems, because of their common feature: they all concern an arbitrary continuous function defined on a closed, bounded interval. Theorem 1. (Intermediate Value Theorem) Let f : [a, b] R be a continuous function defined on a closed, bounded interval. Suppose that f(a) < 0 and f(b) > 0. Then there eists c with a < c < b such that f(c) = 0. Theorem 2. (Etreme Value Theorem) Let f : [a, b] R be a continuous function defined on a closed, bonuded interval. Then f is bounded and assumes its maimum and minimum values. This means that there eist numbers m M such that a) For all [a, b], m f() M. b) There eists at least one [a, b] such that f() = m. c) There eists at least one [a, b] such that f() = M. Theorem 3. (Uniform Continuity and Integrability) Let f : [a, b] R be a continuous function defined on a closed, bounded interval. Then: a) f is uniformly continuous. 1 b) f is integrable: b f eists and is finite. a Ecept for the bit about uniform continuity, these three theorems are familiar results from the standard freshman calculus course. Their proofs, however, are not. Most freshman calculus tets like to give at least some proofs, so it is often the case that these three theorems are used to prove even more famous theorems 1 The definition of this is somewhat technical and will be given only later on in the course. Please don t worry about it for now. 1

2 2 PETE L. CLARK in the course, e.g. the Mean Value Theorem and the Fundamental Theorem of Calculus. Why then are the three interval theorems not proved in freshman calculus? Because their proofs depend upon fundamental properties of the real numbers that are not discussed in such courses. Thus one of the necessary tasks of the present course is to give a more penetrating account of the real numbers than you have seen before Numbers of Various Kinds. There are various kinds of numbers. most important to us: (1) Z + N Z Q R C. Let me remind you what these various numbers are. Here is a list of the ones which will be Z + = {1, 2, 3,..., n,...} is the set of positive integers (a.k.a. counting numbers ). On them we have defined the operations of addition + and multiplication. Moreover there is an identity element for the multiplication, namely 1. There is no additive identity. N = {0} Z + = {0, 1, 2, 3,..., n,...} is the set of natural numbers. Again we have defined operations of addition and multiplication, and now we have an additive identity, 0, as well as a multiplicative identity. Let me remark that Z + and N are clearly very similar: they differ only as to whether 0 is included or not. In analysis the subject we are beginning the study of here! the distinction between Z + and N is not very important, and in fact Spivak uses N to denote the positive integers. I am perhaps showing my stripes as an algebraically minded mathematician by making the distinction, but so be it. Recall that the operation of subtraction is nothing else than the inverse operation of addition: in other words, to say a b = c is to say that a = b + c. However the operation of subtraction is not everywhere defined on N: for instance, there is a natural number 5 3, but no natural number 3 5. Z = {..., 2, 1, 0, 1, 2,...} is the set of integers. This is formed out of the natural numbers N by formally allowing all subtractions: for instance, 17 is In the integers then every element n has an additive inverse n. However, the same cannot be said of multiplicative inverses. Recall that the operation of division is nothing else than the inverse operation of multiplication: to say a/b = c is to say that a = b c. However the operation of division is not everywhere defined on Z: thre is an integer 6/3, but no integer 6/4. Q = { a a, b Z, b 0}. b is the set of rational numbers. This is formed out of the integers Z by formally allowing all divisions by nonzero integers. Note that one subtlety here is that the same rational number has many different epressions as the quotient of two integers: for instance 6 4 = 3 2 = 3n 2n for any n Z+. So we just need to agree that a b = c d

3 MATH 2400 LECTURE NOTES 3 iff ad = bc. Alternately, any nonzero rational number has a unique epression a b in lowest terms, i.e., with a and b not both divisible by any integer n > 1. 2 Thus with respect to the basic operations of + and present already on Z +, the rational numbers are ideally suited: we have an additive identity 0, every element has an additive inverse, we have a multiplicative identity 1, and every nonzero elmeent has a multiplicative inverse. 3 What then are the real numbers R? The geometric answer is that the real numbers correspond to points on the number line, but this does not make clear why there are such points other than the rational numbers. An answer that one learns in high school is that every real number has an infinite decimal epansion, not necessarily terminating or repeating, and conversely any integer followed by an infinite decimal epansion determines a real number. In fact this is perfectly correct: it gives a complete characterization of the real numbers, but it is not a cure-all: in order to pursue the implications of this definition and even to really understand it one needs tools that we will develop later in the course. Finally, the comple numbers C are epressions of the form a + bi where a and b are real numbers and i 2 = 1. They are etremely important in mathematics generally e.g. one needs them in order to solve polynomial equations but in this course they will play at most a peripheral role. Back to R: let us nail down the fact that there are real numbers which are not rational. One way to see this is as follows: show that the decimal epansion of every rational number is eventually periodic, and then ehibit a decimal epansion which is not eventually periodic, e.g. = where the number of 1 s after each 6 increases by 1 each time. But this real number has the ring of the abstract: it seems to have been constructed only to make trouble. The mathematicians of the ancient Pythagorean school discovered a much more real irrational real number. Theorem 4. (Pythagoras) The square root of 2 is not a rational number. Proof. The proof is the most famous (and surely one of the first) instances of a certain important kind of argument, namely a proof by contradiction. The strategy is simple: we assume that what we are trying to prove is false, and from that we reason until we reach an absurd conclusion. Therefore what we are trying to prove must in fact be true. Here goes: seeking a contradiction, we suppose that 2 is rational: this means that there eist integers a, b with b > 0, such that 2 = a b. Since the defining property of 2 is that its square is 2, there is really nothing to do but square both 2 Like most of the statements we have made recently, this requires proof! We remind the reader that we are not giving proofs here or even careful definitions; rather, we are reminding the reader of some of her mathematical past. 3 Trust me for now that we would not want 0 to have a multiplicative inverse: the eistence of a number such that 0 = 1 has undesirable consequences.

4 4 PETE L. CLARK sides to get clearing denominators, we get 2 = a2 b 2 ; 2b 2 = a 2. This shows that the integer a 2 is even, i.e., divisible by 2. It happens that for any integer a, if a 2 is even, then so is a: let us assume this for now; we can come back to it later. Thus we may write a = 2A with A Z. Substituting this into the equation, we get 2b 2 = a 2 = (2A) 2 = 4A 2, or b 2 = 2A 2. Thus b 2 is divisible by 2, so as above b = 2B for some B Z. Substitutin this into our last equation, we get or 4B 2 = (2B) 2 = b 2 = 2A 2, 2B 2 = A 2. Thus we are back where we started: assuming that 2b 2 = a 2, we found that both a and b were divisible by 2. This is suspect in the etreme, and we now have our choice of killing blow. One way to finish the argument is to observe that everything we have said above applies to A and B: thus we must also have A = 2A 1, B = 2B 2, and so forth. We can continue in this way factoring out as many powers of 2 from a and b as we wish. But the only integer which is arbitrarily divisible by 2 is 0, so our conclusion is a = b = 0, whereas we assumed b > 0: contradiction. Alternately, and perhaps more simply, any rational number may be written in lowest terms, so we could have assumed at the beginning that a b was written in this form and thus a and b are not both divisible by 2. Either way we get a contradiction, so 2 must not be a rational number Why do we not do calculus on Q? To paraphrase the title question, why do we want to use R to do calculus? there something stopping us from doing calculus over, say, Q? Is The answer to the second question is no: we can define limits, continuity, derivatives and so forth for functions f : Q Q eactly as is done for real functions. Moreover, the most routine results carry over with no change: it is still true, for instance, that sums and products of continuous functions are continuous. However almost all of the big theorems especially, the Interval Theorems become false over Q. For a, b Q, let [a, b] Q = { Q a b}. Eample: Consider the function f : [0, 2] Q Q given by f() = 1 if 2 < 2 and f() = 1 if 2 > 2. Note that we do not need to define f() at = ± 2, because by the result of the previous section these are not rational numbers. Then f is continuous in fact it is differentiable and has identically zero derivative. But

5 MATH 2400 LECTURE NOTES 5 f(0) = 1 < 0, f(2) = 1 > 0, and there is no c [0, 2] Q such that f(c) = 0. Thus the Intermediate Value Theorem fails over Q. Eample: Consider the function: f : [0, 2] Q Q given by f() = Again, this function is well-defined at all points of [0, 2] Q because 2 is not a rational number. It is also a continuous function. However it is not bounded above: by taking rational numbers which are arbitrarily close to 2, 2 2 becomes arbitrarily small and thus f() becomes arbitarily large. 4 In particular, f certainly does not attain a maimum value. Thus the Etreme Value Theorem fails over Q. Moreover, it can be shown (and will be later) that any function on a closed, bounded interval which is either uniformly continuous or integrable is bounded, so the above function f is neither uniformly continuous nor integrable. If you have had second semester freshman calculus, you should think about why the analogous function f : [0, 2] \ { 2} R is not improperly Riemann integrable: it builds up infinite area as we approach 2. The point of these eamples is in order to succeed in getting calculus off the ground, we need to make use of some fundamental property of the real numbers not possessed by (for intance) the rational numbers. This property, which can be epressed in various forms, is called completeness, and will play a major role in this course. 2. Second Lecture: Some Properties of Numbers 2.1. Aioms for a Field. In order to do mathematics in a rigorous way, one needs to identify a starting point. Virtually all mathematical theorems are of the form A = B. That is, assuming A, B must follow. For instance, in Euclidean geometry one lays down a set of aioms and reasons only from them. The aioms needed for calculus are a lot to swallow in one dose, so we will introduce them gradually. What we give here is essentially a codification of high school algebra, including inequalities. Specifically, we will give aioms that we want a number system to satisfy. At this point we will take it for granted that in our number system we have operations of addition, multiplication and an inequality relation <, and that there are distinguished numbers called 0 and 1. We require the following properties: (P0) 0 1. (P1) (Commutativity of +): For all numbers, y, + y = y +. (P2) (Associativity of +): For all numbers, y, z, ( + y) + z = + (y + z). (P3) (Identity for +): For all numbers, + 0 =. (P4) (Inverses for +): For all numbers, there eists y with + y = 0. (P5) (Commutativity of ): For all numbers, y, y = y. (P6) (Associativity of ): For all numbers, y, z ( y) z = (y z). (P7) (Identity for ): For all numbers, 1 =. (P8) (Inverses for ) For all numbers 0, there eists a number y with y = 1. (P9) (Distributivity of over +): For all numbers, y, z, (y +z) = ( y) +( z). 4 Once again, we will be much more precise about this sort of thing later on. This is just an overview.

6 6 PETE L. CLARK Although it is not important for us now, the above aioms (P0) through (P9) are called the field aioms, and a structure which satisfies them is called a field. Eample: Both Q and R satisfy all of the above field aioms. (We take this as known information.) Eample: The comple numbers C satisfy all of the above field aioms. The only one which is not straightforward is the eistence of multiplicative inverses. For this: if z = + iy is a nonzero comple number i.e., the real numbers and y are not both zero then if w = iy 2 +y 2, zw = 1. Eample: Let F 2 = {0, 1} be a set consisting of two elements, 0 and 1. We define = 0, = = 1, = 0, 0 0 = 0 1 = 1 0 = 0, 1 1 = 1. Then F 2 satisfies all of the above field aioms. It is sometimes called the binary field. Proposition 5. In every system satisfying the field aioms, for every number we have 0 = 0. Proof. We have 0 = (0 + 0) = ( 0) + ( 0). Subtracting ( 0) from both sides gives 0 = 0. Proposition 6. In every system satisfying the field aioms: a) The only additive identity is 0. b) Every number has a unique additive inverse. If 1 denotes the additive inverse of 1, then the additive inverse of is ( 1). c) The only multiplicative identity is 1. b) Every nonzero number has a unique multiplicative inverse. Proof. a) Note that 0 is an additive identity by (P3). Suppose that z is another additive identity, and consider 0 + z. Since 0 is an additive identity, 0 + z = z. Since z is an additive identity, 0 + z = 0. Thus z = 0. b) Suppose y and z are both additive inverses to : + y = + z = 0. Adding y to both sides gives y = 0 + y = ( + y) + y = (y + ) + y = y + ( + y) = y + ( + z) = (y + ) + z = ( + y) + z = 0 + z = z, so y = z. Moreover, for any number, ( 1) + = (( 1) ) + (1 ) = ( 1 + 1) = 0 = 0. c),d) The proofs of these are the same as the proofs of parts a) and b) but with all instances of + replaced by and all instances of 0 replaced by 1. Proposition 7. In every system satisfying the field aioms, ( 1) 2 = 1. Proof. By Proposition 7b), ( 1) ( 1) is the additive inverse of 1, namely 1. Proposition 8. In every system satisfying the field aioms, if 0 and y 0 then y 0. Proof. Seeking a contradiction, suppose that y = 0. Since 0 it has a multiplicative inverse 1 and then by Proposition 5, 0 = 1 y = ( 1 )y = 1 y = y,

7 contradicting the assumption that y 0. MATH 2400 LECTURE NOTES 7 Note that a logically equivalent formulation of Proposition 8 is: in any system satisfying the field aioms, if y = 0 then 0 or y = Aioms for an ordered field. The remaining properties of numbers concern the inequality relation <. Instead of describing the relation < directly, it turns out to be simpler to talk about the properties of positive numbers. If we are given the inequality relation <, then we say that is positive if > 0, thus knowing < we know which numbers are positive. Conversely, suppose that we have identified a subset P of numbers as positive. Then we can define < y if y P. Now we want our set of positive numbers to satisfy the following properties. (P10) (Trichotomy) For all numbers, eactly one of the following holds: = 0, is positive, is positive. (P11) (Closure under +) For all positive numbers, y, + y P. (P12) (Closure under ) For all positive numbers, y, y P. A number system satisfying (P1) through (P12) is called an ordered field. Proposition 9. In every number system satisfying the ordered field aioms: a) 1 > 0 and 1 < 0. b) For every nonzero, 2 > 0. c) It follows that for all, 2 0. Proof. a) By (P0), 1 0. Thus by trichotomy, either 1 is positive and 1 is negative, or 1 is positive and 1 is negative. But by (P12) the product of two positive numbers is positive, so if 1 is positive and 1 is negative then 1 = ( 1) 2 is positive, a contradiction. So it must be that 1 is positive and 1 is negative. b) Since is nonzero, either > 0 or > 0. If > 0, then 2 = is the product of two positive numbers, hence positive. If < 0, then > 0 and then 2 = ( 1) 2 2 = ( ) ( ) is the product of two positive numbers, hence positive. c) Since 0 2 = 0, part c) follows immediately from part b). Eample: The binary numbers F 2 satisfy the field aioms (P0) through (P9), but are they an ordered field? Well, not on the face of it because we have not been given an inequality relation < satisfying (P10) through (P12). In fact we will now show that there is no such relation. Indeed, in any ordered field, since 1 > 0, also > 0, but in F = 0. In fancy language, F 2 is a field which cannot be endowed with the structure of an ordered field. Eample: The comple numbers C satisfy the field aioms (P0) through (P9), but are they an ordered field? As above, we have not been given an inequality relation. Also as above we can show that there is no such relation. For in the comple numbers we have an element i with i 2 = 1. But the ordered field aioms imply both that 1 is negative and that any square is non-negative, contradiction. Proposition 10. For any, y, z in a system satisfying the ordered field aioms: a) < 0 0 <. (We say is negative.) b) The trichotomy property may be restated as: for any number, eactly one of

8 8 PETE L. CLARK the following holds: is positive, is zero, is negative. c) If is positive, 1 is positive. If is negative, 1 is negative. d) If is positive and y is negative, then y is negative. e) If and y are both negative, then y is positive. Proof. a) By definition, < 0 means 0 = is positive. Also 0 < means 0 = is positive. So there is nothing to show here. b) No further argument for this is needed; we just state it for future reference. c) Suppose is positive. Certainly 1 is not zero, so we need to rule out the possibility that it s negative. But if it were, then by part a) 1 would be positive and thus by (P12) 1 = 1 would be positive, contradicting XXX. If is negative then is positive so by what we just showed 1 = 1 is positive, and thus 1 1 = ( ) is negative. d) Suppose is positive and y is negative. In particular and y are not zero, so y 0. To show that y is negative, by part b) it is enough to rule out the possibility that y is positive. Suppose it is. Then, by part c), since is positive, 1 1 is positive, and thus y = y would be positive: contradiction. e) Suppose and y are both negative. Again, this implies y 0 and we need to rule out the possibility that y is negative. Suppose it is. Then y is positive, 1 1 is negative, so by part d) y = y is negative and thus y is positive: contradiction. Proposition 11. For all a, b, c, d in a system satisfying the ordered field aioms: a) If a < b and c < d, a + c < b + d. b) If a < b and c > 0, then ac < bc. c) If a < b and c < 0, then ac > bc. d) If 0 < a < b, then 0 < 1 b < 1 a. e) If a > 0 and b > 0, then a < b a 2 < b 2. Proof. a) Since a < b and c < d, b a is positive and d c is positive, and then by (P11) (b a) + (d c) = (b + d) (a + c) is positive, so b + d > a + c. b) Since a < b, b a is positive. Since c is positive, by (P12) bc ac = (b c)a is positive, so bc > ac. c) Left to you as an eercise. d) We have 1 a 1 1 b = (b a) ab. The hypotheses imply that b a and 1 ab are both positive, so by (P12) so is their product. e) Note that b 2 a 2 = (b + a)(b a). Since a and b are both positive, b + a is positive, and therefore b a is positive iff b 2 a 2 is positive Some further properties of Q and R. As we have mentioned before, the ordered field aioms (P0) through (P12) are just a list of some of the useful properties of Q and R. They are not a complete set of aioms for either Q or R in other words, there are other properties these fields have that cannot be logically deduced from these aioms alone. In fact this is already clear because Q and R each satisfy all the ordered field aioms but are essentially different structures: in Q the element 2 = is not the square of another element, but in R it is. Here we want to give some further familiar properties that do not hold for all ordered fields but both of which hold for R and one of which holds for Q. (We are still far away from the fundamental completeness

9 MATH 2400 LECTURE NOTES 9 aiom for R which is necessary to prove the Interval Theorems.) The first aiom is called the Archimedean property: it says that for any positive number, there is a positive integer n such that n. This clearly holds for R according to our description of real numbers as integers followed by infinite decimal epansions: a positive real number is of the form = n 0.a 1 a 2... a n... with n 0 Z + and a i {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, and thus is less than or equal to the integer n Since every positive real number is less than or equal to some integer, and every positive rational number is, in particular, a positive real number, then also every positive rational number is less than or equal to some integer. That is, Q also satisfies the Archimedean property. (Or, directly: any positive rational number may be written in the form a b with a, b Z+, and then a b a.) This Archimedean property is so natural and familiar (not to mention useful...) that the curious student may be well wonder: are there in fact systems of numbers satisfying the ordered field aioms but not the Archimedean property?!? The answer is yes, there are plenty of them, and it is in fact possible to construct a theory of calculus based upon them (in fact, such a theory is in many ways more faithful to the calculus of Newton and Leibniz than the theory which we are presenting here, which is a 19th century innovation). But we will not see such things in this course! The net property does provide a basic difference between Q and R. Theorem 12. Let be a real number and n Z +. a) If n is odd, there is a unique real number y such that y n =. We write y = n. b) If n is even and is positive, there is a unique positive real number y such that y n =. We write y = n. c) If n is even and is negative, then there is no real number y with y n =. The first two parts of Theorem 12 rely on the Intermediate Value Theorem so are not accessible to us at this time. (Thus we must guard aginst using the eistence of nth roots of real numbers in any of the theorems that lead up to the Intermediate Value Theorem. In fact we will not use such things in the proof of any theorem, but only as eamples.) As a supplement tto part b), note that if n is even and y is a positive real number such that y n =, then there is eactly one other real number with nth power equal to : y. (You might try to prove this as an eercise.) We can however prove part c) now, since this is true for elements of any system satisfying the ordered field aioms. Indeed, if n is even then n = 2k for an integer k, so if for some negative we have y n = then = y 2k = (y k ) 2, contradicting Proposition 9c). Here is a special case of Theorem 12 important enough to be recorded separately. 5 To be honest, when I first typed this I wrote that is necessarily less than n But actually this need not be true! Can you think of an eample? Beware: decimal epansions can be slightly tricky.

10 10 PETE L. CLARK Corollary 13. A real number is non-negative if and only if it is a square, i.e., if and only if there eists a real number y with y 2 =. Note that Corollary 13 does not hold in the number system Q, since 2 = is positive but is not the square of any rational number. Corollary 13 leads to a basic strategy for proving inequalities in the real numbers: for, y R, y (y ) = z 2 for some real number z. In the net section we will see some instances of this strategy in action Some Important Inequalities. For an element of an ordered field, we define the absolute value of to be if 0 and if < 0; it is denoted by. Thus 0 always and = ±. Proposition 14. For any number in an ordered field,. Proof. If 0 then =. If < 0 then < 0 < =, so <. Theorem 15. (Triangle Inequality) For all numbers, y, + y + y. Proof. Since is defined to be if 0 and if < 0, it is natural to break the proof into cases. Case 1:, y 0. Then + y = + y = + y. Case 2:, y < 0. Then + y < 0, so + y = ( + y) = y = + y. Case 3: 0, y < 0. Now unfortunately we do not know whether + y is non-negative or negative, so we must consider consider further cases. Case 3a: + y 0. Then + y = + y + y. Case 3b: + y < 0. Then + y = y + y = + y. Case 4: < 0, y 0. The argument is eactly the same as that in Case 3. In fact, we can guarantee it is the same: since the desired inequality is symmetric in and y meaning, if we interchange and y we do not change what we are trying to show we may reduce to Case 3 by interchanging and y. 6 The preceding argument is definitely the sort that one should be prepared to make when dealing with epressions involving absolute values. However, it is certainly not very much fun. Spivak gives an alternate proof of the Triangle Inequality which is more interesting and thematic. First, since both quantities + y and + y are non-negative, the inequality will hold iff it holds after squaring both sides (Proposition 11e). So it is enough to show ( + y ) 2 ( + y ) 2. now ( +y ) 2 = (+y) 2 = 2 +2y+y 2, whereas ( + y ) 2 = 2 +2 y + y 2 = 2 + 2y + y 2, so subtracting the left hand side from the right, it is equivalent to show that 0 ( 2 + 2y + y 2 ) ( 2 + 2y + y 2 ). But ( 2 + 2y + y 2 ( 2 + 2y + y 2 ) = 2y 2y 0 6 Such symmetry arguments can often by used to reduce the number of cases considered in a proof.

11 MATH 2400 LECTURE NOTES 11 by Proposition 14. So this gives a second proof of the Triangle Inequality. A similar argument can be used to establish the following variant. Proposition 16. (Reverse Triangle Inequality) For all numbers, y, y y. Proof. Again, since both quantities are non-negative, it is sufficient to prove the inequality after squaring both sides: ( y ) 2 = ( y ) 2 = 2 2 y + y 2 = 2 2y + y 2 2 2y + y 2 = ( y) 2 = ( y ) 2. Eercise: 7 Let, y be any numbers. a) Show that y y by writing = ( y) + y and applying the usual triangle inequality. b) Deduce from part a) that y y. Theorem 17. (Cauchy-Bunyakovsky-Schwarz Inequality, n = 2) a) For all numbers 1, 2, y 1, y 2, (2) ( 1 y y 2 ) 2 ( )(y y 2 2). b) Moreover equality holds in (2) iff y 1 = y 2 = 0 or there eists a number λ such that 1 = λy 1 and 2 = λy 2. Proof. By pure brute force, one can establish the following two squares identity: ( )(y y 2 2) = ( 1 y 2 2 y 1 ) 2 + ( 1 y y 2 ) 2. Now we need only rewrite it in the form ( )(y y 2 2) ( 1 y y 2 ) 2 = ( 1 y 2 2 y 1 ) 2 0, establishing part a). Moreover, equality holds iff 1 y 2 = 2 y 1. If in this equality y 1 and y 2 are both nonzero, we may divide by them to get 1 y 1 = 2 y 2 = λ. If y 1 = 0 and y 2 0 then we must have 1 = 0 and then we may take λ = 2 y 2. Similarly, if y 1 0 and y 2 = 0, then we must have 2 = 0 and then we may take λ = 1 y 1. Finally, if y 1 = y 2 = 0 then the equality 1 y 2 = 2 y 1 also holds. Theorem 18. (Cauchy-Bunyakovsky-Schwarz Inequality) For any n Z + and numbers 1,..., n, y 1,..., y n we have ( 1 y n y n ) 2 ( n)(y y 2 n). Proof. Again, the basic idea of showing that the right hand side minus the left hand side is equal to a sum of squares will work. Epanding out the right and left hand sides, we get n RHS = 2 i yj 2. LHS = 2 i yi 2 + i=1 i j n i y i j y j, i=1 2 i y 2 i + 2 i<j 7 All eercises appearing in these notes are informal and optional. However if you want to turn any eercise in and have me take a look at it, that can probably be arranged.

12 12 PETE L. CLARK so RHS LHS = i j 2 i y 2 j 2 i<j i y j j y i = i<j( i y j j y i ) 2 0. Theorem 19. (Arithmetic-Geometric Mean Inequality, n = 2) For all numbers 0 < a < b, we have ( ) 2 a + b a 2 < ab < < b 2. 2 Proof. First inequality: Since a > 0 and 0 < a < b, a a < a b. Second inequality: Epanding out the square and clearing denominators, it is equivalent to 4ab < a 2 + 2ab + ab 2, or to a 2 2b + b 2 > 0. But a 2 2ab + b 2 = (a b) 2, so since a b, (a b) 2 > 0. Third inequality: Since a+b 2 and b are both positive, it is equivalent to a+b 2 < b and thus to a + b < 2b. But indeed since a < b, a + b < b + b = 2b. Theorem 20. (Arithmetic-Geometric Mean Inequality) a) For all n Z + and any numbers 1,..., n, we have ( ) n n (3) 1 n. n b) Equality holds in (3) if and only if 1 =... = n. In order to prove this result we will use mathematical induction, which brings us to Chapter 2 of Spivak s book (and a different set of lecture notes).