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 explanations 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 exams. This course offers a strong foundation for a student’s future study of mathematics, at the undergraduate level and beyond. As examples, 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 exists c with a < c < b such that f (c) = 0. Theorem 2. (Extreme Value Theorem) Let f : [a, b] → R be a continuous function defined on a closed, bonuded interval. Then f is bounded and assumes its maximum and minimum values. This means that there exist numbers m ≤ M such that a) For all x ∈ [a, b], m ≤ f (x) ≤ M . b) There exists at least one x ∈ [a, b] such that f (x) = m. c) There exists at least one x ∈ [a, b] such that f (x) = 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 b) f is integrable: a f exists and is finite. Except 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 texts 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 1The 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

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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. 1.2. Numbers of Various Kinds. There are various kinds of “numbers”. Here is a list of the ones which will be most important to us: (1)

Z+ ⊂ N ⊂ Z ⊂ Q ⊂ R ⊂ C.

Let me remind you what these various numbers are. 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 0 − 17. 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. 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 expressions as the quotient of two intea c + gers: for instance 46 = 32 = 3n 2n for any n ∈ Z . So we just need to agree that b = d Q={

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iff ad = bc. Alternately, any nonzero rational number has a unique expression ab 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 expansion, not necessarily terminating or repeating, and conversely any integer followed by an infinite decimal expansion 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 complex numbers C are expressions of the form a + bi where a and b are real numbers and i2 = −1. They are extremely 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 expansion of every rational number is eventually periodic, and then exhibit a decimal expansion which is not eventually periodic, e.g. x = 0.16116111611116111116 . . . where the number of 1’s after each 6 increases by 1 each time. But this real number x 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 isa rational: this means that there exist integers a, b with b > 0, such that 2 = b . Since the defining √ property of 2 is that its square is 2, there is really nothing to do but square both 2Like 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. 3Trust me for now that we would not want 0 to have a multiplicative inverse: the existence of a “number” x such that 0 · x = 1 has undesirable consequences.

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sides to get 2=

a2 ; b2

clearing denominators, we get 2b2 = a2 . This shows that the integer a2 is even, i.e., divisible by 2. It happens that for any integer a, if a2 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 2b2 = a2 = (2A)2 = 4A2 , or b2 = 2A2 . Thus b2 is divisible by 2, so as above b = 2B for some B ∈ Z. Substitutin this into our last equation, we get 4B 2 = (2B)2 = b2 = 2A2 , or 2B 2 = A2 . Thus we are back where we started: assuming that 2b2 = a2 , we found that both a and b were divisible by 2. This is suspect in the extreme, 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 = 2A1 , B = 2B2 , 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 ab 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.  1.3. Why do we not do calculus on Q? To paraphrase the title question, why do we want to use R to do calculus? Is there something stopping us from doing calculus over, say, Q? The answer to the second question is no: we can define limits, continuity, derivatives and so forth for functions f : Q → Q exactly 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 = {x ∈ Q | a ≤ x ≤ b}. Example: Consider the function f : [0, 2]Q → Q given by f (x) = −1 if x2 < √2 and f (x) = 1 if x2 > 2. Note that we do not need to define f (x) at x = ± 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

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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. Example: Consider the function: f : [0, 2]Q → Q given by f (x) = x21−2 . 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, x2 − 2 becomes arbitrarily small and thus f (x) becomes arbitarily large.4 In particular, f certainly does not attain a maximum value. Thus the Extreme 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 examples 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 expressed in various forms, is called completeness, and will play a major role in this course. 2. Second Lecture: Some Properties of Numbers 2.1. Axioms 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 axioms and reasons only from them. The axioms 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 axioms 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 0. c) It follows that for all x, x2 ≥ 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 x is nonzero, either x > 0 or −x > 0. If x > 0, then x2 = x · x is the product of two positive numbers, hence positive. If x < 0, then −x > 0 and then x2 = (−1)2 x2 = (−x) · (−x) is the product of two positive numbers, hence positive. c) Since 02 = 0, part c) follows immediately from part b).  Example: The binary numbers F2 satisfy the field axioms (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 1 + 1 > 0, but in F2 1 + 1 = 0. In fancy language, F2 is a field which cannot be endowed with the structure of an ordered field. Example: The complex numbers C satisfy the field axioms (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 complex numbers we have an element i with i2 = −1. But the ordered field axioms imply both that −1 is negative and that any square is non-negative, contradiction. Proposition 10. For any x, y, z in a system satisfying the ordered field axioms: a) x < 0 ⇐⇒ 0 < −x. (We say “x is negative”.) b) The trichotomy property may be restated as: for any number x, exactly one of

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the following holds: x is positive, x is zero, x is negative. c) If x is positive, x1 is positive. If x is negative, x1 is negative. d) If x is positive and y is negative, then xy is negative. e) If x and y are both negative, then xy is positive. Proof. a) By definition, x < 0 means 0 − x = −x is positive. Also 0 < −x means −x − 0 = −x 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 x is positive. Certainly x1 is not zero, so we need to rule out the possibility that it’s negative. But if it were, then by part a) −1 x would be positive and thus by (P12) x · −1 = −1 would be positive, contradicting XXX. If x is x 1 negative then −x is positive so by what we just showed −x = −1 is positive, and x 1 −1 thus x = −( x ) is negative. d) Suppose x is positive and y is negative. In particular x and y are not zero, so xy ̸= 0. To show that xy is negative, by part b) it is enough to rule out the possibility that xy is positive. Suppose it is. Then, by part c), since x is positive, 1 1 x is positive, and thus y = xy · x would be positive: contradiction. e) Suppose x and y are both negative. Again, this implies xy ̸= 0 and we need to rule out the possibility that xy is negative. Suppose it is. Then −xy is positive, 1 1 x is negative, so by part d) −y = −xy · x is negative and thus y is positive: contradiction.  Proposition 11. For all a, b, c, d in a system satisfying the ordered field axioms: 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 < 1b < a1 . e) If a > 0 and b > 0, then a < b ⇐⇒ a2 < b2 . 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 exercise. 1 1 d) We have a1 − 1b = (b − a) · ab . The hypotheses imply that b − a and ab are both positive, so by (P12) so is their product. e) Note that b2 − a2 = (b + a)(b − a). Since a and b are both positive, b + a is positive, and therefore b − a is positive iff b2 − a2 is positive.  2.3. Some further properties of Q and R. As we have mentioned before, the ordered field axioms (P0) through (P12) are just a list of some of the useful properties of Q and R. They are not a “complete set of axioms” for either Q or R – in other words, there are other properties these fields have that cannot be logically deduced from these axioms alone. In fact this is already clear because Q and R each satisfy all the ordered field axioms but are essentially different structures: in Q the element 2 = 1 + 1 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

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axiom for R which is necessary to prove the Interval Theorems.) The first axiom is called the Archimedean property: it says that for any positive number x, there is a positive integer n such that x ≤ n. This clearly holds for R according to our description of real numbers as integers followed by infinite decimal expansions: a positive real number x is of the form x = n0 .a1 a2 . . . an . . . with n0 ∈ Z and ai ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, and thus x is less than or equal to the integer n0 + 1.5 +

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 ab with a, b ∈ Z+ , and then ab ≤ 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 axioms 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 next property does provide a basic difference between Q and R. Theorem 12. Let x be a real number and n ∈ Z+ . √ a) If n is odd, there is a unique real number y such that y n = x. We write y = n x. b) If n is even and x is√positive, there is a unique positive real number y such that y n = x. We write y = n x. c) If n is even and x is negative, then there is no real number y with y n = x. 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 existence 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 examples.) As a supplement tto part b), note that if n is even and y is a positive real number such that y n = x, then there is exactly one other real number with nth power equal to x: −y. (You might try to prove this as an exercise.) We can however prove part c) now, since this is true for elements of any system satisfying the ordered field axioms. Indeed, if n is even then n = 2k for an integer k, so if for some negative x we have y n = x then x = y 2k = (y k )2 , contradicting Proposition 9c). Here is a special case of Theorem 12 important enough to be recorded separately. 5To be honest, when I first typed this I wrote that x is necessarily less than n + 1. But 0 actually this need not be true! Can you think of an example? Beware: decimal expansions can be slightly tricky.

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Corollary 13. A real number x is non-negative if and only if it is a square, i.e., if and only if there exists a real number y with y 2 = x. Note that Corollary 13 does not hold in the number system Q, since 2 = 1 + 1 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 x, y ∈ R, x ≤ y ⇐⇒ (y − x) = z 2 for some real number z. In the next section we will see some instances of this strategy in action. 2.4. Some Important Inequalities. For an element x of an ordered field, we define the absolute value of x to be x if x ≥ 0 and −x if x < 0; it is denoted by |x|. Thus |x| ≥ 0 always and x = ±|x|. Proposition 14. For any number x in an ordered field, x ≤ |x|. Proof. If x ≥ 0 then x = |x|. If x < 0 then x < 0 < −x = |x|, so x < |x|.



Theorem 15. (Triangle Inequality) For all numbers x, y, |x + y| ≤ |x| + |y|. Proof. Since |x| is defined to be x if x ≥ 0 and −x if x < 0, it is natural to break the proof into cases. Case 1: x, y ≥ 0. Then |x + y| = x + y = |x| + |y|. Case 2: x, y < 0. Then x + y < 0, so |x + y| = −(x + y) = −x − y = |x| + |y|. Case 3: x ≥ 0, y < 0. Now unfortunately we do not know whether |x + y| is non-negative or negative, so we must consider consider further cases. Case 3a: x + y ≥ 0. Then |x + y| = x + y ≤ |x| + |y|. Case 3b: x + y < 0. Then |x + y| = −x − y ≤ | − x| + | − y| = |x| + |y|. Case 4: x < 0, y ≥ 0. The argument is exactly the same as that in Case 3. In fact, we can guarantee it is the same: since the desired inequality is symmetric in x and y – meaning, if we interchange x and y we do not change what we are trying to show – we may reduce to Case 3 by interchanging x and y.6  The preceding argument is definitely the sort that one should be prepared to make when dealing with expressions 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 |x + y| and |x| + |y| are non-negative, the inequality will hold iff it holds after squaring both sides (Proposition 11e). So it is enough to show (|x + y|)2 ≤ (|x| + |y|)2 . now (|x+y|)2 = (x+y)2 = x2 +2xy +y 2 , whereas (|x|+|y|)2 = |x|2 +2|x||y|+|y|2 = x2 + |2xy| + y 2 , so subtracting the left hand side from the right, it is equivalent to show that 0 ≤ (x2 + |2xy| + y 2 ) − (x2 + 2xy + y 2 ). But (x2 + |2xy| + y 2 − (x2 + 2xy + y 2 ) = |2xy| − 2xy ≥ 0 6Such symmetry arguments can often by used to reduce the number of cases considered in a proof.

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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 x, y, ||x| − |y|| ≤ |x − y|. Proof. Again, since both quantities are non-negative, it is sufficient to prove the inequality after squaring both sides: (||x| − |y||)2 = (|x| − |y|)2 = |x|2 − 2|x||y| + |y|2 = x2 − |2xy| + y 2 ≤ x2 − 2xy + y 2 = (x − y)2 = (|x − y|)2 .  Exercise:7 Let x, y be any numbers. a) Show that |x| − |y| ≤ |x − y| by writing x = (x − y) + y and applying the usual triangle inequality. b) Deduce from part a) that ||x| − |y|| ≤ |x − y|. Theorem 17. (Cauchy-Bunyakovsky-Schwarz Inequality, n = 2) a) For all numbers x1 , x2 , y1 , y2 , (2)

(x1 y1 + x2 y2 )2 ≤ (x21 + x22 )(y12 + y22 ).

b) Moreover equality holds in (2) iff y1 = y2 = 0 or there exists a number λ such that x1 = λy1 and x2 = λy2 . Proof. By pure brute force, one can establish the following two squares identity: (x21 + x22 )(y12 + y22 ) = (x1 y2 − x2 y1 )2 + (x1 y1 + x2 y2 )2 . Now we need only rewrite it in the form (x21 + x22 )(y12 + y22 ) − (x1 y1 + x2 y2 )2 = (x1 y2 − x2 y1 )2 ≥ 0, establishing part a). Moreover, equality holds iff x1 y2 = x2 y1 . If in this equality y1 and y2 are both nonzero, we may divide by them to get xy11 = xy22 = λ. If y1 = 0 and y2 ̸= 0 then we must have x1 = 0 and then we may take λ = xy22 . Similarly, if y1 ̸= 0 and y2 = 0, then we must have x2 = 0 and then we may take λ = xy11 . Finally, if y1 = y2 = 0 then the equality x1 y2 = x2 y1 also holds.  Theorem 18. (Cauchy-Bunyakovsky-Schwarz Inequality) For any n ∈ Z+ and numbers x1 , . . . , xn , y1 , . . . , yn we have (x1 y1 + . . . + xn yn )2 ≤ (x21 + . . . + x2n )(y12 + . . . + yn2 ). 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. Expanding out the right and left hand sides, we get n ∑ ∑ RHS = x2i yi2 + x2i yj2 . i=1

LHS =

n ∑ i=1

x2i yi2 + 2

i̸=j

∑

xi yi xj yj ,

i 0. a+b Third inequality: Since a+b 2 and b are both positive, it is equivalent to 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 x1 , . . . , xn , we have ( )n x1 + . . . + xn (3) x1 · · · xn ≤ . n b) Equality holds in (3) if and only if x1 = . . . = xn . 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).