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Math 140A - Sketch Notes Math 140A - Sketch Notes Neil Donaldson Fall 2019 1 Introduction The primary purpose of 140A/B is to organize the logical background of calculus. Calculus is excep- tionally useful, yet at its heart lies an ancient philosophical debate dating at least to Zeno’s paradoxes of the 5thC BC: how do we calculate with infinitesimally small quantities, and is it even legitimate to do so? Definite integration, for instance, requires the summation of the areas of vanishingly small rectangles. Is it legitimate to use such a method to ‘compute’ the area under a curve? Long after its pioneers1 had applied calculus to many important problems in the sciences, philosophers were ar- guing over its validity, and mathematicians were struggling to put calculus on a firm logical footing. Modern descriptions of calculus rely heavily on the notion of limit, though it took mathematicians over a century to realize the importance of this concept and how to formulate it clearly. The ‘modern’ e-d definition that we shall explore was developed in the early 1800’s, courtesy of Bolzano, Cauchy and Weierstrass (amongst others), all of whom feature prominently in 140A/B. The climax of 140B is a thorough description of the Riemann integral, which did not appear until the mid 1800’s. Elemen- tary Analysis is therefore a history of one of the great mathematical triumphs of the 16–1800’s. Most of the work on calculus will have to wait until 140B. In this course we primarily consider limits and continuity. We start, however, with something even more basic: the structure of the real numbers, without which the smooth, continuous changes of calculus cannot be described. While you will certainly be familiar with many examples of real numbers, they are, in general, significantly harder to describe than the integers or the rationals. It is perhaps surprising, though it is certainly a fact you’ve encountered before, that the set of irrational numbers has greater cardinality than the set of rationals.2 However, beyond specific examples such as p, e and simple radicals like p2, it is rather difficult to exhibit explicit examples of irrational numbers. The reason is simple: the integers are intuitive and you’ve been using them since infancy. The natural way to construct more numbers is to start with integers and build from there. 1.1 The Set N of Natural Numbers Here are one version of the standard axioms of the natural numbers, known as Peano’s Axioms. While reading them, think of the challenge mathematicians faced in getting here: what is the essential nature of the natural numbers, expressible without a prior understanding of number, and what dis- tinguishes the natural numbers from other sets? 1In particular Newton and Leibniz in the late 1600’s. 2Q is countable while R (and consequently) R Q is uncountable. n Axioms 1.1 (Peano). The natural numbers are a set N satisfying the following properties: 1. (Non-emptiness) N is non-empty. 2. (Successor function) There exists a function f : N N. This is usually denoted ‘+1’ so that we may ! write, n N, n + 1 N 8 2 2 3. (Initial element) f is not surjective. Otherwise said, there exists an element 1 range( f ) which is not 62 the successor of any element.3) 4. (Unique predecessor/order) f is injective. If m and n have the same successor, then m = n. 5. (Induction) Suppose that A N has the following properties: 1 A and n A = n + 1 A. ⊆ 2 2 ) 2 Then A = N. Axioms 1–4 should be obvious and can be summarized by saying that there exists a set N with an injective, non-surjective function f : N N. Axiom 5 is trickier, though it should remind you of ! proof by induction: it is essentially the same as the principle of mathematical induction and the notion of well-ordering. We won’t dwell on it, but you should review induction yourself. Here is an example of a simple induction argument written in the language of axiom 5. Example Prove that 7n 4n is divisible by 3 for all n N. − 2 Let A be the set of natural numbers for which 3 7n 4n. It is required to prove that A = N. j − • If n = 1, then 71 41 = 3, so 1 A. − 2 • Suppose that n A. Then 7n 4n = 3l for some l N. Hence 2 − 2 7n+1 4n+1 = 7 7n 4n+1 = 7(3l + 4n) 4n+1 = 3 7l + (7 4) 4n − · − − · − · which is divisible by 3. Hence n + 1 A. 2 Appealing to axiom 5, we see that A = N, hence result. Note that the two bulleted arguments are what you likely called the base case and the induction step in a previous course. What is different about the integers? It should be clear that the integers satisfy axioms 1, 2 and 4, but not 3 and 5. In particular: • 1 has a predecessor (namely zero). • A = N is a proper subset of Z satisfying axiom 5, but A = Z. 6 We can provide an explicit construction of the integers from the natural numbers however: simply extend the function f so that 0 is the unique predecessor of 1, 1 the unique predecessor of 2, etc. − 3It is purely a convention to denote the first natural number by 1: we could use 0, x, a, or any symbol you wish! In particular, you should not assume any additional properties of the element 1 beyond those described in the axioms. 2 1.2 The Set Q of Rational Numbers There are several ways to define the rational numbers from the integers. For example, we could consider the set of relatively prime ordered pairs Q = (p, q) : p Z, q N, gcd(p, q) = 1 Z N f 2 2 g ⊆ × Writing p instead of (p, q) and adopting the convention that lp = p for any non-zero l Z makes q lq q 2 things more familiar. It is easy to define the usual operations (+, , etc.) consistently with those for · the integers. An alternative, historical, approach involves equations: new sets of numbers can be ‘invented’ for the purpose of solving equations.4 The simplest situation involves considering linear equations qx p = 0 − where p, q Z and q = 0. Each such equation essentially describes a rational number! For example 2 6 27 13x + 27 = 0 ! x = −13 Extending this process further, we might consider higher degree polynomials. Definition 1.2. A number x is algebraic if it satisfies an equation of the form n n 1 anx + an 1x − + + a1x + a0 = 0 ( ) − ··· ∗ for some integers a0,..., an. (You should be alarmed by this! We have given up on constructing new numbers and instead are simply describing their properties. No matter, a construction of the real numbers will come later.) Examples 1. p2 is algebraic, since it satisfies the equation x2 2 = 0. − p 2. 5 p3 + 7 is also algebraic. This is slightly harder to check: clearly x5 7 = p3 = (x5 7)2 = 3 = x10 14x5 + 46 = 0 − ) − ) − It is far from clear, though is it true, that there exist non-algebraic (or transcendental numbers): e and p are classic examples. These satisfy no polynomial equation with integer coefficients, though demon- strating this is somewhat tricky and beyond this course. Deciding on the rationality or irrationality of an algebraic number is often an easy application of the next result. Theorem 1.3 (Rational Roots). Suppose that a ,..., a Z and that r Q satisfies ( ). If r = p in lowest 0 n 2 2 ∗ q terms, then p a and q a . j 0 j n Proof. Simply calculate n p p n n 1 n 1 n an + + a1 + a0 = 0 = an p + an 1 p − q + + a1 pq − + a0q = 0 q ··· q ) − ··· All terms except the last contain a factor of p, whence p a qn. Since gcd(p, q) = 1 it follows that j 0 p a . The result for q is almost identical. j 0 4Take a class on abstract algebra/ring theory for more of this approach. 3 Examples 1. p2 is irrational. Clearly p2 satisfies x2 2 = 0. If p2 = p were rational in lowest terms, then − q the rational roots theorem says that p 2 and q 1 = p2 1, 2 j j ) 2 f± ± g This is clearly a contradiction since none of the numbers 1, 2 satisfy x2 2 = 0. ± ± − You should compare this argument to the standard proof of the irrationality of p2 as seen in a previous course. Note how easy it is to extend this argument to p3, p29, p3 2, p5 8 etc. 2. (p3 1)1/3 is irrational. It satisfies (x3 + 1)2 = 3, from which − x6 + 2x3 2 = 0 − By the Theorem, if x = p we’d necessarily have p 2 and q 1, whence x = 1, 2, none of q j j ± ± which satisfies (x3 + 1)2 = 3. 1/2 3. 4+p3 is irrational. It satisfies 5x2 4 = p3, from which 5 − 25x4 40x2 + 13 = 0 − whence, if x = p we’d have p 13 and q 25. There are twelve possibilities in all: q j j 1 13 1 13 x = 1, 13, , , , ± ± ±5 ± 5 ±25 ±25 It is tedious to check all cases, but none satisfy the required polynomial. In fact it is much easier with this example to bypass the Theorem entirely: note that if x Q 2 then p3 = 5x2 4 Q immediately gives the desired irrationality! − 2 The rational roots theorem can also be use to help find rational solutions to high-degree polynomials and aid in factorization.
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