ADVANCED CALCULUS 401 RECITATION NOTES, FALL 2014

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1 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 8/0: Proposition: is irrational. Proof. Assume toward a contradiction that is not irrational, i.e. rational. Then = p q, with p, q Z, q 0, such that gcd(p, q) = 1. Squaring both sides, we obtain = p q q = p. By definition of even, p is even, so p is even by lemma below. Hence p = k 1 for some k 1 Z, so by substitution, q = (k 1 ) = 4k 1 q = k 1. Thus, q is even, so q is even by same lemma, so q = k for some k Z. This shows that p and q share a factor of, so gcd(p, q). This contradicts the assumption that gcd(p, q) = 1. Hence, our original hypothesis was absurd. That is, must be irrational. Lemma (Jocelyn s Lemma): If p is even and p Z, then p is even. (Is this true if the hypothesis p Z is removed? No.) Proof. It s logically equivalent to prove that if p is odd, then p is odd. (Why? Read appendix middle of p. 315 for some insight.) Thus, assume p = k +1, k Z. Then p = (k + 1) = (k + 1)(k + 1) = 4k + 4k + 1 = (k + k) + 1. Since k + k Z, as the product and sum of integers is an integer, this shows that p is odd. Actually, the converse of Jocelyn s Lemma holds: Proposition: If p is even, then p is even. Proof. Since p is even, p = k for some k Z. Then p = (k) = 4k = (k ). Since k Z, this shows p is even. The previous proposition, combined with Jocelyn s lemma, give Theorem: p is even, p Z, if and only if p is even. The main reason we are combining the two above results into one is to point out that in many cases, we ll see if and only if statements such as this. In such statements, there are two implications, the forward and the converse. In our case, in the statement obtained by keeping if removing and only if, p is even is the hypothesis and p is even is the conclusion. In the statement obtained by keeping only if and removing if and, then p is even is the hypothesis and p is even is the conclusion. We separate the proof into two parts: the proof of the forward implication, and the proof of the converse. 1

2 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 Proof. ( ) Assume p Z and p is even. Then by Jocelyn s Lemma above, p is even. ( ) Assume p is even. Then by the previous proposition, p is even. Exercises: A.1.1: What is the negation of the statement either X is true, or Y is true, but not both? Solution: Rewrite statement as (X or Y ) and not (X and Y ). Then negation is (not X and not Y ) or both (X and Y ). A.1.: What is the negation of the statement X is true if and only if Y is true? Solution: Rewrite as X Y and Y X, so negation is (X Y ) or (Y X); in other words, (if X is true, Y is false) or (if Y is true, X is false). 8/7: Proposition : If a, b N such that a + b = 0, then a = 0 and b = 0. It s equivalent to prove the contrapositive: If a, b N such that a 0 or b 0, then a + b 0. Proof. (of contrapositive): First we prove the lemma: Lemma: If a is a positive natural number and b N, then a + b is positive. Proof of Lemma: We induct on b. If a = b, then a + b = a + 0 = 0 + a = a, which is positive by assumption. Suppose inductively that a + b is positive. Then a + (b + +) = (a + b) + +, which is nonzero by Axiom 3 of natural numbers (0 is not the successor of any natural number, and we know that a + b is a positive natural number, so (a + b) + + N as the successor of a natural number). Thus, a + (b + +) is positive, closing the induction. Suppose a, b N and a 0 or b 0. Then either a is positive or b is positive. Case 1: a is positive. Then a+b is positive by the preceding lemma, i.e., a+b 0. Case : b is positive. Then a + b = b + a is positive, i.e., a + b 0, by the same lemma. This concludes the proof.

3 ADVANCED CALCULUS 401 RECITATION NOTES, FALL Proposition (Distributive Law): For any a, b, c N, we have a(b + c) = ab + ac and (b + c)a = ba + ca. Proof. Let s induct on a. For a = 0, we have 0(b + c) = 0 by definition of multiplication (1), while 0a + 0c = (by definition of multiplication (1))= 0 (by definition of addition (1)), so the base case is proven. Assume inductively that a(b+c) = ab+ac. We want to show that (a++)(b+c) = (a + +)b + (a + +)c. Well, we have (a + +)(b + c) = a(b + c) + (b + c) by definition of multiplication () = (ab + ac) + (b + c) by inductive hypothesis = ab + b + ac + c by associativity and commutativity of addition = (a + +)b + (a + +)c by definition of multiplication (). This closes the proof of the first identity. The second identity follows from (b + c)a = a(b + c) by commutativity of multiplication = ab + ac by first identity = ba + ca by commutativity of multiplication. Recall the definition of for natural numbers: Definition Given n, m N, we say n m iff there exists a N such that n = m + a. Dominic and Tao both gave definitions of > for natural numbers. Definition (Dominic s): Given n, m N, we say n > m iff there exists a N, a > 0, such that n = m + a. Definition (Tao s): Given n, m N, we say n > m iff n m and n m. We show that Dominic s definition of > for natural numbers is equivalent to Tao s. Proof. ( ) Assume Dominic s definition. We want to show Tao s definition holds. ( T ) (We include the T in the subscript to remind ourselves we re showing Tao s definition holds.) Let n, m N such that n > m. Then by Dominic s definition, there exists a N, a > 0, such that n = m + a. Since a N, this shows n m. Moreover, n m; otherwise, if n = m, we would have n = m + a = m = m + 0 a = 0 by cancellation law, a contradiction.

4 4 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 ( T ) Assume n m and n m. Then by definition of, there exists a N such that n = m + a. Note furthermore that a > 0 because if a = 0, we have n = m + a = m + 0 = m, but we know n m. Thus, by Dominic s definition, n > m. ( ) Assume Tao s definition. We want to show Dominic s definition holds. ( D ) Suppose n > m. Then by Tao s definition, n m and n m. Thus, there exists a N such that n = m + a.we claim a > 0. If not, i.e. if a = 0, then n = m + a = m + 0 = m, contradicting n m. Hence, we have n = m + a with a N, a > 0. ( D ) Suppose there exists a N, a > 0, such that n = m + a. Then n m. Also, n m. Otherwise, n = m, so that n = m = m + a 0 = a, by cancellation, a contradiction. Thus, by Tao s definition, n > m. This concludes the proof. 9/3: Main comment on HW 1: For the two induction identities of HW 1 (Problems 1 and ), many of you appeared to assert the equality of both sides and then perform like operations to both sides of the equation until you ended up with one side clearly equalling the other. One problem with this is that we shouldn t say the conclusion is true before it s been shown. A second problem is that in certain situations, the like operations are not reversible. (In the case of those homework problems, the steps were reversible, i.e., the logic went both ways.) Eg. Suppose we have some mystery x, we know that this x satisfies x 5x+6 = 0, and I m trying to convince you that this implies x = 3. So I proceed as follows: Consider x = 3. Then x = 9 and 5x = 5(3) = 15, so x 5x + 6 = = 0 Since we ended up with a true statement by assumption, it must be the case that x indeed equals 3. Invalid argument! The problem is that although x = 3 implies x 5x + 6 = 0, the converse is not true; x 5x + 6 = 0 does not imply x = 3, as x = is also a solution to this equation. In summary: when proving an equation is true, start from one side of the desired equation and keep rewriting it till you reach the other, or rewrite both sides independently until you reach equal expressions, at which point we can say by transitivity of equality that the LHS equals the RHS. The same goes for proving any statement. Make sure not to assume the desired conclusion and then work until you reach a true fact. This would successfully prove the converse, but not necessarily the original statement. Remark about + +: Using what we currently know, we can officially discard the ++ notation. Note that for any n N, n + + = 1 + n (by Problem 3 of HW

5 ADVANCED CALCULUS 401 RECITATION NOTES, FALL )= n + 1 by commutativity of addition, so instead of writing n + +, we can now write n + 1. Examples of functions: (1) Constant function: Consider f : N N defined by f(n) = 5. Then f is a function because for each n N, there exists exactly one m N such that f(n) = m. (In this case, for every n N, m = 5!) [What we wrote is a function because the RHS of the relation has only one value. If we had put a ± in front, say, it would not be a function!] Note that f is not one-to-one (injective): the output 5 does not have a unique input; for example, f(3) = f(14) = 5, although (In fact, every n N satisfies f(n) = 5 by definition!) Also, f is not onto (surjective). For example, consider m = 6 N. There exists no n N such that f(n) = 6. We can argue that f is not onto by choosing any m N other than 5. () Identity function: Consider g : N N given by g(n) = n. That is, g outputs whatever the input is. Then g is a function because for any n N, there exists exactly one m N such that g(n) = m. (In this case, m = n!) The identity function g is indeed 1-1, for if g(n 1 ) = g(n ) n 1 = n, using the definition of g. Also, g is onto: given any m N, we have that mn satisfies g(m) = m. Since g is 1-1 and onto, g is by definition bijective. (3) An injective but not onto function: (This example makes use of knowledge you have from previous courses but which we technically don t have yet in this course.) Consider the exponential function exp : R R defined by exp(x) = e x. This is a function because we just called it a function; actually, just kidding, it is so for the same reason as described above. It is injective because if e x1 = e x, then ln(e x1 ) = ln(e x ) since ln is the inverse of exp, so x 1 = x. It is not onto as, for example, the value 1 R has no a R such that e a = 1. Another example is the successor function f : N N we defined in class, given by f(n) = n + +. (4) An onto but not injective function: (This example, in using the concept of division, also is technically not legitimate at this point of our course. Nevertheless, it illustrates the concepts of 1-1 and onto well.) Consider the function h : N N given by { 0 if n is odd, h(n) = n if n is even. (Note that if n is even, n = k for some k N, so n = k = k N, so we are indeed mapping into N.) It helps to graph this by hand to see what s going on. We see that h is a function since every input n N has exactly one output: 0 if n is odd and n if n is even. Also, h is onto. Given any m N, it can be mapped to by h. Indeed, m N is even and

6 6 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 h(m) = m = m. However, h is not injective. Note that any two odd natural numbers give the same output 0, so for example, h(1) = h(3) = 0, but 1 3. Can you come up with your own examples? 9/3: Ordered n-tuples and n-fold Cartesian product: See Definition for the formal definition of what these are. We can denote the n-fold Cartesian product of sets X 1, X,..., X n by X 1 X X n or 1 i n X i. An element of the n-fold Cartesian product is an ordered n-tuple and is written (x 1, x,..., x n ), where x 1 X 1, x X,..., x n X n. (Note that the order matters; that is, the first component in an ordered n-tuple must be an element of the first set listed in the product,..., in general, the ith component must be an element of the ith set listed in the product. An ordered -tuple is also called an ordered pair. Two ordered n-tuples (x 1, x,..., x n ) and (x 1, x,..., x n) are equal iff x i = x i for all 1 i n. Example: Let X 1 = {1,, 3}, X = {3, 4, 5}, and X 3 = {4, 5, 6}. One element of the 3-fold Cartesian product X 1 X X 3 is (, 3, 6) because X 1, 3 X, and 6 X 3. However, (1,, 4) / X 1 X X 3 because / X (although 1 X 1 and 4 X 3 ). Note that (3, 4, 5) (3, 5, 4) because 4 5 and 5 4. Indeed, the tuples need only be different in one component to not be equal, so (3, 4, 5) (3, 5, 5). Cardinal Arithmetic (Prop ) (a) Let X be a finite set, and let x be an object which is not an element of X. Then X {x} is finite and #(X {x}) = #(X) + 1. Proof. To see that X {x} is finite, note first that X is finite by assumption. Let #(X) = n N. There there exists a bijection f : X {i N : 1 n}. Our intuition should tell us that X {x}, obtained by adding an element to X, has cardinality one more than X; that is, that #(X {x}) = n + 1. Define a bijection g : X {x} {i N : 1 n + 1} by g(y) = { f(y) if y X, n + 1 if y = x. Note that the domain of g is X {x} as the above rule tells us what g does to y in two cases: if y X, and if y = x. Since the range of f is {i N : 1 n} and the extra element x maps to n + 1, we are indeed mapping into g : X {x} {i N : 1 n + 1}. 1 1: Let g(y 1 ) = g(y ). We have a few cases to look at: Case 1: y 1, y X. Then g(y 1 ) = f(y 1 ) and g(y ) = f(y ), so f(y 1 ) = f(y ). Thus, since f is bijective, hence 1-1, we have y 1 = y.

7 ADVANCED CALCULUS 401 RECITATION NOTES, FALL Case : y 1 X, y = x. Then g(y 1 ) = f(y 1 ) and g(y ) = n + 1. Hence, f(y 1 ) = n + 1. But f(y 1 ) {i N : 1 n}, so f(y 1 ) n + 1. This is a contradiction, i.e., this case can t happen. Case 3: y 1 = x, y = x. Then by transitivity, y 1 = y. Case 4: y X, y 1 = x. Then by the same argument as in Case, exchanging the roles of y 1 and y, we obtain a contradiction. Hence g is 1-1. Onto: Let z {i N : 1 n + 1}. Case 1: z {i N : 1 n}. Then consider y X such that f(y) = z; we can do this because f is onto. Then g(y) = f(y) = z, using that y X. Case : z = n + 1. We have g(x) = n + 1. This shows that g is onto. Hence, g is a bijection, so X {x} is finite by definition, and #(X {x}) = n + 1 = #(X) + 1. (d) If X is a finite set, and f : X Y is a function, then f(x) is a finite set with #(f(x)) #(X). If in addition, f is one-to-one, then #(f(x)) = #(X). Sketch/idea of proof: Since X is a finite set, we have a bijection h : X {i N : 1 i n} for some n N. We want to construct a bijection g from f(x) to some finite subset of natural numbers. Observe the following diagram: X {i N : 1 i n} f f(x) h g We can label the elements of X as x 1, x,..., x n in correspondence with the bijection h : {i N : 1 i n}, i.e., x i := h 1 (i). To define g, the first inclination might be to say that g(f(x i )) = h(x i ) := i. The problem is, what if f(x i ) = f(x j ) for x i x j? (We re not given that f is 1-1.) This would create a problem because then g(f(x i )) would be both i and j, which are not equal, and possibly other values, meaning g is not a function. To get around this, in such a situation where f(x i ) can be represented in different ways, choose the smallest i for this element. (Alternatively, choose the largest.) Then one shows that g defined with this consideration in mind is therefore a function (has one, unambiguous output for each input) and is 1-1 because h is 1-1. It s not necessarily onto the set {i N : 1 i n}. It is, however, onto the image g(f(x)), which is a subset (why?) of {i N : 1 i n}. This is a clever little

8 8 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 trick that is sometimes used in these kinds of arguments. A function always maps onto its image; in this case, note that any g(f(x i )) g(f(x)) can be obtained by plugging f(x i ) f(x) into the function g. Since g(f(x)) {i N : 1 i n}, then the elements of g(f(x)) can be put in bijection with the set {i N : 1 i m} for some m n. (To do this, send the smallest element of g(f(x)) to 1, the second smallest to, etc.) Therefore, since g is a bijection and this latter map is a bijection, we can map from f(x) to {i N : 1 i m} by composing these two bijections, and by Exercise of your homework this week, this composition is a bijection. So this will show that #(f(x)) = m n = #(X), so #f(x) #(X). Under the additional hypothesis that f is 1-1, we don t run into the potential hiccup encountered above. Try to figure out an efficient way to argue that we can actually get a bijection from f(x) to {i N : 1 i n}. (Ask me if you need a hint.) 9/11 (Class): HW 4 : Exercises 4.1., 4.1.5, 4..1, and Up until now we ve been working with the natural numbers. In this chapter we expand the boundaries by defining what integers and rational numbers are. We already have knowledge of how we want them to behave. Now we just need to define them and not proceed by undefined intuitive arguments. The Integers: Definition: An integer is an expression of the form a b, where a and b are natural numbers. For now, we can think of as a meaningless place-holder symbol, though it will turn out to be behave like a minus sign. Two integers a b and c d are called equal if and only if a+d = c+b. (Note that we have already defined addition of natural numbers, so we know what a + d and c + b mean.) Denote Z to be the set of all integers. Example: 14 Z and 5 17 Z, and 14 = 5 17 because + 17 = 19 = One should show that the definition of equality of integers satisfies the desired axioms of an equivalence relation. Tao shows transitivity in the book. I ll leave reflexivity to you. To see symmetry, note that if a b = c d, then a+d = c+b. To show, c d = a b, we need to show c+b = a+d. But this follows from a+d = c+b by symmetry of equality of natural numbers. Definition: The sum of two integers a b and c d is defined by (a b)+(c d) := (a + c) (b + d), and the product of these two integers is defined by (a b) (c d) := (ac + ) (ad + bc) Notice that the RHS s in each definition are in the form of an integer as we defined above, since a + c, b + d, ac +, ad + bc N, so we re just using our knowledge of algebra to rewrite the sum in a way consistent

9 ADVANCED CALCULUS 401 RECITATION NOTES, FALL with how we defined the way an integer is to be written. Example: We have (4 8) + (7 10) := (4 + 7) (8 + 10) = (11 18), and (6 3) ( 3) := (6() + 3(3)) (6(3) + 3()) = (1 + 9) (18 + 6) = 1 4. Lemma Addition and multiplication of integers are well-defined. That is, for addition, if we want to find (a b) + (c d) and we have other representations (a b) = (a b ) and (c d) = (c d ), then in fact (a b) + (c d) = (a b ) + (c d) and (a b) + (c d) = (a b) + (c d ). Therefore, as a result (why?), (a b) + (c d) = (a b ) + (c d ). Similarly, multiplication of integers does not depend on choice of representations of the integers. Proof. See book. Remark: What is a straightforward way to write a natural number n in integer form? We can write it as n 0 and observe that addition and multiplication of natural numbers according to the integer addition and multiplication rules behave like the natural number addition and multiplication rules: (n 0)+(m 0) = (n+m) (0+0) = (n+m) 0 and (n 0) (m 0) = (nm+0 0) (n 0+0 m) = (nm+0) (0 + 0) = nm 0. Also n 0 = m 0 if and only if n = m. (Use the definition of equality of integers to see why.) So natural numbers written and operating according to integer rules behave as they did before. Definition: The negation of an integer (a b) is (b a), denoted (a b). In past courses, this was alternatively called the additive inverse. That is, it s the element such that when we add it to (a b) we get zero. Check that (a b)+( (a b)) := (a b) + (b a) equals 0 0 using the definitions of addition of integers and equality of integers, and the commutative property of addition of natural numbers. Tao shows trichotomy for integers; that is, that any integer x is either equal to zero:= 0 0, n 0 for some positive n N, or 0 m for some positive m N. Proposition 4.1.6, the laws of algebra for integers, shows that the integers are a commutative ring (look up what a ring means if you re curious) the commutative referring to the fact that the multiplication is commutative. Let s prove one of the nine laws: Proposition: Let x, y, z Z. Then x(y + z) = xy + xz. Proof. Rewrite x = a b, y = c d, z = e f for some a, b, c, d, e, f N. Then we have

10 10 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 x(y + z) = (a b)[(c d) + (e f)] = (a b)[(c + e) (d + f)] by definition of addition of integers = [a(c + e) + b(d + f)] [a(d + f) + b(c + e)] by definition of multiplication of integers = [ac + ae + + bf] [ad + af + bc + be] by distributive property for natural numbers, while xy + xz = (a b)(c d) + (a b)(e f) = [(ac + ) (ad + bc)] + [(ae + bf) (af + be)] by defn of mult of integers = [(ac + ) + (ae + bf)] [(ad + bc) + (af + be)] by defn of addition of integers = [ac + ae + + bf] [ad + af + bc + be] by assoc and comm of addition of natural numbers. By transitivity, we have that x(y + z) = xy + xz. Definition: Given two integers (a b), (c d), define subtraction by (a b) (b d) := (a b) + [ (c d)] = (a b) + (d c). One can show this subtraction behaves as we re used to. In particular, the meaningless placeholder is actually just subtraction: for natural numbers n, m, which as integers can be rewritten n 0 and m 0, we have (n 0) (m 0) (subtraction symbol in the middle):= (n 0) + (0 m) = (n + 0) (0 + m) = n m (old placeholder notation); that is, the original placeholder notation between natural numbers to form an integer is actually just subtraction, so we can discard the placeholder notation from now on. See more properties of integers in the book, such as the fact that the integers have no zero divisors and that there s a cancellation law. Also, define ordering in the same way as for natural numbers. Let s just prove one more fact for practice: Lemma (d) (Negation reverses order) Let a, b Z. a < b. If a > b, then Proof. Rewrite a = a 1 a and b = b 1 b for appropriate a 1, a, b 1, b N. Since a > b, then a b and a b by definition, so a = (a 1 a ) = b+m = (b 1 b )+(m 0) for some m N and (m 0) (0 0); otherwise, a = a 1 a = (b 1 b )+(0 0), which equals (b 1 +0) (b +0) = b 1 b = b by the definition of addition of integers, a contradiction. Hence, (a 1 a ) = (b 1 b ) + (m 0) for m N, m 0. We want to show a < b, or equivalently by definition, b > a, so we want to show (b 1 b ) = (a 1 a ) + (m 1 0) for some m 1 N, m 1 0; that is, we want to show that (b b 1 ) = (a a 1 ) + (m 1 0). Well, note that by adding b b 1 to both sides of (a 1 a ) = (b 1 b ) + (m 0), we get (b b 1 ) + (a 1 a ) = (b b 1 )+(b 1 b )+(m 0) (b b 1 )+(a 1 a ) = [(b +b 1 ) (b 1 +b )]+(m 0) = (0 0)+(m 0) = (0+m) (0+0) = m 0. Now, adding (a a 1 ) to both sides, we get (b b 1 )+(a 1 a )+(a a 1 ) = (m 0)+(a a 1 ) (b b 1 ) = (a a 1 )+(m 0)

11 ADVANCED CALCULUS 401 RECITATION NOTES, FALL by commutativity of addition in the integers. Thus, the m 1 we sought was actually m; we just showed b = a + m, m N, m 0, so a < b. Rational Numbers: Likewise, we define the rational numbers using our knowledge of how we want them to behave as a guide. Definition: A rational number is an expression of the form a//b, where a and b are integers and b 0. Two rational numbers a//b and c//d are called equal if and only if ad = cb. Denote the set of all rational numbers by Q. Example: 5//10 = 10//0 because ( 5)(0) := (0 5)(0 0) = [0(0) + 5(0)] [0(0)+5(0)] = (0+0) (0+100) = (0+0)+(0 100) = = 100, while 10(10) := (0 10)(10 0) = [0(10) +10(0)] [0(0) +10(10)] = (0+0) (0+100) = = 100. We can figure out from prior knowledge how we would like to define addition and multiplication in Q. Also, we know what we would like the negation (additive inverse) of a rational number to be. Definition: The sum of two rational numbers a//b and c//d is (a//b) + (c//d) := (ad + bc)//(); their product is (a//b) (c//d) := (ac)//(); and the negation of a//b is (a//b) := ( a)//b. Lemma 4..3 The sum, product, and negation operations on rational numbers are well-defined; if we add or multiply two rational numbers, then the results we get are equal to what we get by replacing either of the rational numbers by another equal rational number. Similarly, by replacing a rational number by another equal rational number and finding the negation of this equal rational number, we get the same value as the negation of the original. Proof. See book. Note that rational numbers a//1 and b//1 behave exactly like the integers: (a//1) + (b//1) := (a(1) + b(1))//(1 1) = (a + b)//1 (remember that integers which are natural numbers behave as we learned before, so 1 1 indeed is 1!); (a//1) (b//1) := (ab)//(1 1) = (ab)//1; (a//1) := ( a)//1; and a//1 = b//1 a(1) = b(1) a = b. In particular, the integer (and natural number) 0 can be written 0//1 in rational number form. Definition Let x = a//b be a nonzero rational, i.e., a, b 0. reciprocal of x to be x 1 := b//a. Define the Note that the notion of reciprocal is well-defined. Let x = a//b Q, a, b 0, and x also = a //b. Then by definition of equality, ab = a b. We want to show that b//a = b //a. Well, to show this, we need to show that ba = b a. But ba = ab (by commutativity of multiplication in integers)= ab = b a (by commutativity of

12 1 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 multiplication of integers again). The laws of algebra that hold for integers also hold for rationals. However, the rationals have one additional property that the integers don t have. If x Q, x 0//1, then xx 1 = x 1 x = 1. That is, every nonzero rational number has a multiplicative inverse. Check that this property holds using the definition of rational number multiplication. The rational numbers form a commutative ring just like the integers; even more, this additional property shows that they form a field. Definition: Define the quotient x/y of two rational numbers x and y, where y 0, by x/y := x y 1. Example: We have (7//6)/( 14//16) := (7//6) ( 14//16) 1 = (7//6) (16// 14) = (7)(16)//(6)( 14) = 11/ 84, which one can show using the definition of equality equals 4/3. Just as we could at the end of the day discard the meaningless placeholder symbol for integers and replace it by the standard subtraction symbol, we can also replace the // placeholder with the division bar / from the quotient definition above. Indeed, for two rationals a, b, b 0, rewrite a = a 1 //a and b = b 1 //b (so b 1 0). Indeed, we show that a rational number a//b (b 0 equals a/b. Note that a//b = (a 1 //a )//(b 1 //b ) and a/b := ab 1 := (a 1 //a ) (b //b 1 ) = (a 1 b )//(a b 1 ). To show this equals (a 1 //a )//(b 1 //b ), it s enough to show that (a 1 //a )(a b 1 ) = (a 1 b )(b 1 //b ). Well, (a 1 //a )(a b 1 ) := (a 1 //a )(a b 1 //1) = (a 1 a b 1 )//a = (a 1 b 1 )/1 (use definition of equality of rational numbers for this last step), while (a 1 b )(b 1 //b ) = (a 1 b //1)(b 1 //b ) = (a 1 b b 1 )//b = (a 1 b 1 )/1, so by transitivity the desired claim is true. Like the integers, the rational numbers can be ordered (see Prop 4..9). They form an ordered field. Why do we care? Well, the rational numbers behave nicely. Not every set of numbers can be ordered. (We haven t discussed them in this course, but for example, the complex numbers do not have this property.) 9/17: Proposition (Triangle Inequality for Absolute Value): Let x, y Q. Then x + y x + y. Proof. We prove the claim by splitting up into cases according to the trichotomy of rationals (Lemma 4..7): Recall that for z Q, z = { z if z 0, z if z < 0.

13 ADVANCED CALCULUS 401 RECITATION NOTES, FALL (I lumped the last two cases from our definition on Tuesday into one in order to make the arguments below quicker.) Case 1: x, y 0: Then by definition, x = a/b, y = c/d, for a, b, c, d Z, with b, d > 0, a, c 0. (Note that in the subcase x = 0 or y = 0, a = 0 or c = 0.) Thus, x + y = a/b + c/d = ad+bc. Since a, b, c, d 0, they each can be regarded as natural numbers, so ad, bc, N; also, ad + bc N, i.e., ad + bc 0, and > 0, by Lemma.3.3, since b, d > 0. Hence ad+bc 0 and so x + y = ad+bc. On the other hand, x + y = a/b + c/d, which equals a/b + c/d = ad+bc since, as already pointed out, b, d > 0, a, c 0. This shows x + y = x + y in this case, so x + y x + y. Case : x < 0 y. Then x = (a/b) = ( a)/b, y = c/d for a, b, c, d Z, with a, b, d > 0, c 0. We have x + y = ( a/b) + c/d = ( a)d+cb. We don t know whether ( a)d + cb = ad + cb is positive, zero, or negative, so we treat each of these subcases. Before we do this, note that x + y = ( a)/b + c/d = ( a/b) + c/d = a/b + c/d = ad+cb. Subcase.1: ad + cb 0. Then ad+cb 0 since also > 0, so x + y = ad+cb = ad+cb. We claim this is ad+cb, which is x + y. Indeed, we show ad+cb < ad+cb, which requires showing ad+cb ad+cb < 0. We have ad+cb ad+cb := ad+cb + [ ( ad+cb )] = ad+cb + [ (ad+cb) ] = ad+cb + ad cb = ad+cb ad cb = ad ad. Since a, d > 0, ad > 0, so ad < 0 = 0 by Lemma (d), so ad ad = ad is also negative, as the negation of a positive natural number. Hence, ad ad < 0, as desired. Subcase.: ad+cb < 0. Then ad+cb < 0, so x+y = ad+cb = ( ad+cb) = ad cb. We claim this is x + y ad+cb ad cb, so show = ad+cb or ad cb ad+cb < 0. If c = 0, then ad cb ad+cb and, so equality holds. If c > 0, then ad cb ad+cb = ad cb = ad + (ad+cb) = ad = ad bc ad cb = bc bc. Since b, c > 0, bc > 0, so bc < 0 and so bc bc < 0. Also, as before > 0, so this shows bc bc < 0, as desired. In both subcases, we found that x + y x + y. Case 3: y < 0 x. Then by Case, y + x y + x, so by commutativity of addition in Q, we have x + y x + y. Case 4: x, y < 0. Then x, y > 0, so by Case 1, x + ( y) = x + y, so 1(x + y) = 1 x + y (by homogeneity)= 1 x + y = x + y equals x + y = 1 x + 1 y = 1 x + 1 y = x + y. Alternate proof:

14 14 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 Proof. First we prove: Proposition 4.3.3(c): Let x, y Q. Then y x y if and only if y x. In particular, we have x x x. Proof: ( ) Assume y x y. Then if x 0, we have x = x y (by assumption), so y x. If x < 0, then x = x. We claim y x. Indeed, if not, then y < x, so y > ( x) = x, contradicting y x. Hence y x = x. ( ) Assume y x. If x 0, then y x = x 0, so x y; also, by Exercise 4..6, y 0 = 0, so y 0 x y x. Putting together the two results gives y x y. If x < 0, then y x = x, which is > x since x > 0 = 0 (by Exercise 4..6) so that x x > = 0 by Prop 4..9(d). Hence x y. Also, y x yields y x, so we have y x y. In particular, let y = x Q. x x x. Then since always x x, we always have Back to proof of triangle inequality: By the second statement of the proposition, we have x x x and y y y, so adding these inequalities together (you can do this by using Prop 4..9(d) twice while handling also the case of equality do you see how?) yields ( x + y ) x + y x + y. Now, by the proposition again, we have x + y x + y (observe that x + y takes the role of y from the original statement, while x+y takes the role of x), i.e. x+y x + y. 9/4: Proposition: The sequence a 1, a, a 3,... defined by a n := 1/n (i.e., the sequence 1, 1/, 1/3,... ) is a Cauchy sequence. Proof. (Tao gives this proof, and I provide no new mathematical ideas. Nevertheless, I didn t copy it word for word.) To show {a n } is Cauchy, we must show, by definition, that given ɛ > 0 (ɛ Q), there exists an N N \ {0} (here our sequence starts with n = 1) such that whenever n, m N, we have d(a n, a m ) ɛ. To get an idea of how to prove this, let s brainstorm how to find N by pretending we found N and doing some reverse engineering. We have d(a n, a m ) = 1 n 1 m can either equal 1 n 1 m (if the quantity inside the absolute values is zero or positive) or ( 1 n 1 m ) = 1 m 1 n (if the quantity inside the absolute values is negative). Note that if we ve found such an N, then since n, m N, for the first case, we have 1 n n 1 n N 1 N n 1 1 N N n 1 N 1 N 1 n, i.e., 1 n 1 N, so 1 n 1 m 1 n N. The second case is similar: we have 1 m 1 n 1 m N. That is, we showed that given an N as above, we have d(a n, a m ) 1 N. Now, if we can get this to be ɛ, then by transitivity, we would have d(a n, a m ) ɛ as desired. The challenge at this point is to find N such that 1 N ɛ. To find such N, it s equivalent to find N such that 1 ɛ N, i.e., N 1 ɛ, because then we can multiply

15 ADVANCED CALCULUS 401 RECITATION NOTES, FALL the inequality by ɛ 1 N and preserve order, as both ɛ and 1 N are positive. By Prop (latter statement, the Archimedean principle), indeed there exists an N natural number such that N > 1 ɛ Q, so N 1 ɛ. Moreover N is positive because if it were 0, then we would have 0 > 1 ɛ, contradicting that 1 ɛ is positive. Let s make sure we didn t make any mistakes in our reverse engineering. In fact, in writing a clean proof, you might start after this sentence ends; the brainstorming isn t typically included in the proof and would just be your scratch prep work. Let ɛ > 0 (ɛ Q). By the Archimedean Principle, we know N N such that N > 1 ɛ. Since 1 ɛ > 0, we have N > 0 by transitivity. Now, for n, m N, we have d(a n, a m ) = 1 n 1 m 1 N (fill in justification for this inequality as explained in brainstorming paragraphs). Since N > 1 ɛ, we have Nɛ > 1 ɛ > 1 N, i.e., 1 N < ɛ. Hence, by transitivity of the inequality, we have for all n, m N that d(a n, a m ) < ɛ. In particular, d(a n, a m ) ɛ. Hence, {a n } is Cauchy by definition. Definition: Let M 0 be rational. A sequence is bounded by M iff a i M for all i 1. Proposition Every finite sequence is bounded. Proof. In book. Note in the inductive step that once we get that for all 1 i n, a i M for some M 0 rational, we have a n+1 = 0 + a n+1 M + a n+1 using that addition preserves order, and also a i = a i + 0 M + a n+1 for all 1 i n for the same reason, since by definition of absolute value 0 a n+1. So our bound that works for all i {1,,..., n, n + 1} is M + a n+1. Proposition Every Cauchy sequence is bounded. Proof. In case this is on an upcoming homework, I won t give the proof in detail. However, the idea is that starting from some N N, the elements of a Cauchy sequence are all ɛ-close, for any ɛ > 0 rational, so one can show that this part of the sequence is bounded. We can consider the part of the sequence before such N as a finite sequence in itself since it has only the elements {a 1, a,..., a N 1 }. Thus, by previous proposition, this part of our sequence is bounded. Now show that since both parts of the sequence are bounded, the entire sequence is bounded. 10/8: Exercise (from Marcel Finan of Arkansas Tech): (a) Prove that if {a n } is a Cauchy sequence, then {a n} is also Cauchy. (b) Is the converse true? If yes, prove it. If not, provide a counterexample. Proof of (a): Let ɛ > 0. We want to find N such that n, m N implies a n a m ɛ. We know given ɛ 1 > 0, there exists N 1 such that whenever n, m N 1,

16 16 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 a n a m ɛ 1. For n, m N 1, we have a n a m = (a n a m )(a n + a m ) = a n a m a n + a m a n a m ( a n + a m ) by triangle inequality = a n a m a n + a n a m a m ɛ 1 a n + ɛ 1 a m. We want this to equal, or be less than or equal to, ɛ. A potential issue, though, is that if either of the sequences {a n } or {a m } shoot off to infinity, then no ɛ 1 > 0 will be able to keep this expression in check for all n, m N. Then, we recall, as we ve encountered in Tuesday s lecture on a few occasions: Cauchy sequences are bounded! Thus, M 0 such that a n M for all n. Since this is true, choose M > 0 such that a n < M to avoid having to address below the possibility of M = 0. Continuing from above, we have a n a m ɛ 1 a n + ɛ 1 a m ɛ 1 M. If we can make ɛ 1 M equal ɛ, then this will show that for all n, m N 1, a n a m ɛ. Set ɛ 1 = ɛ M, which is > 0 since ɛ > 0 and M > 0. Thus, ɛ 1 M = ɛ, and by above, with N = N 1, we are done. Alternate proof of (a): Use that the product of two real numbers is a real number, by Prop (part of your homework this week). So since LIM n a n is a real number (because {a n } is a Cauchy sequence), then by this proposition (LIM n a n )(LIM n a n ) := LIM n (a n a n ) = LIM n a n is also a real number, so {a n} is Cauchy. Solution to (b): The converse is not true. For example, consider the sequence 1, 1, 1, 1,..., that is, the sequence defined by a n = ( 1) n. Note that {a n} is given by a n = [( 1) n ] = [( 1) ] n = 1 n = 1; that is, {a n} is the constant sequence 1, 1, 1,.... Then {a n} is Cauchy. (For any n, m, we have a n a m = 1 1 = 0 = 0, so for any ɛ > 0 we choose, any N will work to make a n a m ɛ. The same argument shows that any constant sequence is Cauchy.) However, {a n } is not Cauchy. To see this formally, suppose toward a contradiction that {a n } is Cauchy. Let ɛ = 1. Since we just assumed {a n } is Cauchy, we can find N such that for all n, m N, a n a m 1. But say N is even. We have N + 1, N N, and a N+1 a N = ( 1) N+1 ( 1) N = 1 1 = = > 1, i.e., a N+1 a N is not 1, a contradiction. The case N odd is argued similarly to get a contradiction.

17 ADVANCED CALCULUS 401 RECITATION NOTES, FALL We conclude that {a n } is not Cauchy. 10/15: Exercise Let x, y R. Then (1) x y + ɛ for all real numbers ɛ > 0 if and only if x y. and () x y ɛ for all real numbers ɛ > 0 if and only if x = y. Proof. (1): ( ) (By contradiction) Suppose x > y. Then x y > 0, so x y =LIM n a n for some sequence {a n } positively bounded away from 0; by definition, this means d > 0, d Q, such that n N, a n d. By a proposition from Tuesday s class, this implies that x y =LIM n a n d. We also know that x y ɛ for all ɛ > 0, so in particular, x y d, since d > 0 implies d > 0. Thus, x y d < d x y, so x y < x y by transitivity, contradicting that x y = x y. ( ) Let ɛ > 0 be real. We have that x y = y + 0 < y + ɛ x < y + ɛ using properties of order. Hence, x y + ɛ. (): ( ) By Exercise (part of your homework), we know that ɛ > 0, y ɛ x y + ɛ. Looking at the second of these inequalities, this tells us, by part (1), that x y. Also, by adding ɛ to both sides of the first inequality, we get that y x + ɛ for all ɛ > 0. Hence, again by part (1), we obtain y x. We have x y and y x, so by antisymmetry, x = y. ( ) Let x = y. Then given any ɛ > 0, we have x y = x x = 0 = 0, which is < ɛ by assumption, hence ɛ. Exercise Let (a n ) n=1 be a Cauchy sequence of rationals, and let x be a real number. Further suppose that a n x for all n 1. Then LIM n a n x. Similarly, if we suppose that a n x for all n 1, then LIM n a n x. Proof. We prove the first statement first. Proceed by contradiction. Suppose LIM n a n > x. Rewrite as x <LIM n a n. Then by Proposition (density of the rationals in R), there exists a q Q such that x < q <LIM n a n. Hence q LIM n a n :=LIM n q LIM n a n =LIM n (q a n ) < 0. Meanwhile, we have a n x < q for all n 1, so q a n 0 for all n 1. By Proposition (non-negative reals are closed), we have that LIM(q a n ) 0. (Notice that the point of introducing q into the proof is that we can then consider the Cauchy sequence {q a n } of rationals [since the difference of rationals is rational; also, Cauchy because it s a difference of the Cauchy sequences {q} and {a n }] in order to appeal to the stated proposition.) The two statements LIM n (q a n ) < 0 and LIM(q a n ) 0 yield a contradiction.

18 18 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 The other statement can be proven by a similar argument as above, reversing inequalities. Alternatively, it can be deduced from the first statement as follows. Since a n x for all n 1, we have a n x for all n 1. Hence, by the first statement, LIM( a n ) x. Thus, ( 1) LIMa n ( 1) x. Multiplying both sides of the inequality by 1 reverses the inequality by properties of order and so LIMa n x. Note: When Professor P proved the density of the rationals in the reals (Prop ), she actually used Exercise at the end of her proof. (See her on Tuesday s class to see how.) Therefore, our logic is circular if we rely on that proof, since in the proof of Exercise we used the density of the rationals. However, there is a way around this. We can prove the density of the rationals in a different way that doesn t rely on this exercise. A sketch of the proof that I believe Tao has in mind is as follows: Proposition : Given any two real numbers x < y, we can find a rational number q such that x < q < y. Proof. (I leave out a couple of details for you to try to fill in, but the main ideas are here. I m happy to discuss the omitted details with any of you if you d like.) Since x < y, we have y x > 0. By Exercise 5.4.4, there exists a positive integer N such that y x > 1/N. We claim that there must be a rational of the form m/n, where m Z, such that x < m/n < y. Suppose toward a contradiction there is not. Then choose the biggest rational m1 m1 N of this form such that N x. (This last step requires some argument.) Then we have m1+1 N y. [This is because otherwise we d have x < m1+1 m1+1 N < y and N is rational, against our assumption there is no rational between x and y.] Hence, we have m1 N x m1 m1+1 N x, and N y. Adding these two inequalities together gives 1/N y x. But y x > 1/N, so 1/N > 1/N, a contradiction. 10/: This week the homework hints are provided in this document. Exercise Let x, y > 0 be positive reals, and let n, m 1 be positive integers. (a) If y = x 1/n, then y n = x. Outline of proof: (By contradiction.) We use Tao s proof of Prop as a guide. Suppose y n < x or y n > x. Case 1: y n < x. Recall that by definition y = x 1/n := sup{z R : z 0 and z n x}. Let E = {z R : z 0 and z n x}. If we can find an ɛ > 0 such that (y + ɛ) n x, and this will contradict the fact that y is by definition an upper bound of E: we will have y + ɛ E (since y + ɛ 0 and

19 ADVANCED CALCULUS 401 RECITATION NOTES, FALL (y+ɛ) n < x (y+ɛ) n x)) and y+ɛ > y. To ensure in the following steps that we can make (y + ɛ) n small enough, suppose up front that 0 < ɛ < 1. (You ll see why; remember our only stipulation is that ɛ must be > 0.) We have < y n + ny n 1 ɛ + (y + ɛ) n = y n + ny n 1 ɛ + ( ) n y n ɛ + + nyɛ + ɛ ( ) n y n ɛ + + nyɛ n 1 + ɛ n by binomial theorem since ɛ < 1 ɛ n < ɛ for all n (show this by induction) ( ) n = y n + ɛ(ny n 1 + y n + + ny + 1) Now, if we can make y n + ɛ(ny n 1 + ( n ) y n + + ny + 1) x, then this will show that (y + ɛ) n x by transitivity. Thus, find ɛ such that ɛ(ny n 1 + ( n ) y n + + ny + 1) x y n = d > 0 ɛ d ny n 1 +(. Note that c = n )y n + +ny+1 nyn 1 + ( n ) y n + + ny + 1 > 0, so show that can find ɛ d c. This step is left to you. Case : y n > x. We go on a similar quest for an ɛ, with 0 < ɛ < 1. This time we want (y ɛ) n > x. (Or x will work.) This will contradict that y is the least upper bound of E. (To see why y ɛ will be an upper bound for E, suppose that for some z E, z y ɛ, i.e., z > y ɛ. Then z n > (y ɛ) n > x by Prop 5.6.3, real version of (c), contradicting that z n x.) We have ( ) n 3 (y ɛ) n = y n ny n 1 ɛ + ( ) n y n ɛ + ny( ɛ) n 1 + ( ɛ) n y n ny n 1 ɛ y n 3 ɛ 3 remove all the positive terms with ɛ to get a smaller quantity ( ) n > y n ny n 1 ɛ y n 3 ɛ since ɛ < 1 ɛ > 1, so ɛ n > ɛ for all n 3 ( ) n = y n ɛ(ny n 1 + y n 3 + ) 3 We want to choose ɛ so that y n ɛ(ny n 1 + ( n 3) y n 3 + ) > x so that then (y ɛ) n > x. Try to proceed as in Case 1. (b) Conversely, suppose y n = x. Then y = x 1/n. Proof. Suppose y x 1/n. We have two cases: Case 1: y > x 1/n. Then y n > (x 1/n ) n by Prop (real version of (c)), and this equals x by part (a). We have y n > x, contradicting y n = x.

20 0 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 Case : y < x 1/n. Then y n < (x 1/n ) n = x, so y n < x, contradicting y n = x. (e) If x > 1, then x 1/k is a decreasing function of k. If x < 1, then x 1/k is an increasing function of k. If x = 1, then x 1/k = 1 for all k. Outline of proof (you should fill in the details): prove the last statement first. That is, show that 1 1/k = 1 for all positive integers k. Well, since 1 k = 1 (show this by induction), then by part (b) we have 1 = 1 1/k, so the last statement is proven. For x > 1, use part (c) to show that x 1/k > 1. (Hint: use result in first paragraph.) Now let k > k. We want to show that x 1/k < x 1/k. Write k = k + m for some positive natural number m. Write x 1/k = y and x 1/k = y. (This relabeling is not necessary but may help you see things better.) We want to show that y < y. Suppose y y. Use part (a) to show that (y ) k = x = y k. Thus, (y ) k+m = y k. Meanwhile, (y ) k y k by (how?). Use that y = x 1/k > 1 and again to show that (y ) k+m > y k. This contradicts that (y ) k+m = y k. For x < 1, go through the same argument, reversing inequalities.

21 ADVANCED CALCULUS 401 RECITATION NOTES, FALL Exercise Suppose (a n ) n=1 is a Cauchy sequence of rational numbers. Then (a n ) n=1 converges to LIM n a n, i.e., LIM n a n = lim n a n. Proof. Let LIM n a n = L and suppose lim n a n L, i.e., {a n } does not converge to L. Then there exists an ɛ 0 > 0 such that for all N > 0, have some n N N with a nn L > ɛ 0. But since {a n } is Cauchy, we know there exists an N such that n, m N, have a n a m ɛ0. We have an n N N such that a n N L > ɛ 0 and a n a nn ɛ for all n N. Thus, a nn L > ɛ 0 or a nn L < ɛ 0 and a n a nn ɛ0 for all n N. We want to show a n > L + ɛ0 or a n < L ɛ0 for al n N. In the case a nn L > ɛ 0, have a n a nn ɛ0 > L + ɛ. In the case a nn L < ɛ 0, have a n ɛ0 + a n N < L ɛ0. We have, by Exercise 5.4.8, that LIMa n L + ɛ0 > L or LIMa n L ɛ0 < L, contradicting LIM n a n = L. Theorem (Limit Laws): Let (a n ) n=m and (b n ) n=m be convergent sequences of real numbers, and let x, y be the real numbers x := lim n a n and y := lim n b n. (b) The sequence (a n b n ) n=m converges to xy; in other words, lim (a nb n ) = ( lim a n)( lim b n). n n n Proof. Let ɛ > 0. We want to find N so that n N, a n b n xy ɛ. We have a n b n xb n + xb n xy a n b n xy = a n b n xb n + xb n xy by triangle inequality = b n a n x + x b n y by factoring, then using homogeneity of absolute value Convergent sequences of reals are Cauchy by Prop 6.1.1, and Cauchy sequences of reals are bounded by Cor , so M > 0 such that b n M > 0 for all n. Choose N 1 so that a n x ɛ M for all n N 1 and, if x > 0, N such that b n y ɛ x for all n N. Then for n N = max(n 1, N ), b n a n x + x b n y M( ɛ M ) + x ɛ x = ɛ + ɛ = ɛ. If x = 0, then x b n y = 0, so b n a n x + x b n y = b n a n x M( ɛ M ) = ɛ ɛ, so N still works. Hence, (a n b m ) n=m converges to xy.

22 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 (c) For any real number c, the sequence (ca n ) n=m converges to cx; in other words, lim (ca n) = c( lim a n). n n Proof. We use part (b). By part (b), we have lim n (ca n ) = (lim n c)(lim n a n ). But one can show that lim n c = c, since c c = 0 < ɛ for any ɛ > 0. Hence, lim n (ca n ) = c(lim n a n ). (d) The sequence (a n b n ) n=m converges to x y; in other words, lim (a n b n ) = lim a n lim b n. n n n Proof. We use part (a) (part of your homework). We have lim n (a n b n ) = lim n (a n + ( b n )) = lim n a n + lim n ( b n ). (One can show that since (b n ) n=m is convergent, so too is ( b n ) n=m.) But lim n ( b n ) = lim n (( 1) b n ), so this equals ( 1) lim n b n by part (c). We thus have lim n (a n b n ) = lim n a n + ( 1) lim n b n = lim n a n lim n b n. (f) Suppose that y 0 and that b n 0 for all n m. Then the sequence (a n /b n ) n=m converges to x/y; in other words, Proof. We have a n lim = lim n a n. n b n lim n b n 10/9: lim n a n = ( lim lim n b a n)( lim b n) 1 n n n = ( lim a n)( lim n n b 1 n ) by part (e) (part of your homework) = lim (a nb 1 n ) by part (b) n a n = lim. n b n Once again, this week the homework hints are provided in this document. Hints:

23 ADVANCED CALCULUS 401 RECITATION NOTES, FALL : To show sup(a n ) n=m sup(b n ) n=m, suppose toward a contradiction that sup(a n ) n=m > sup(b n ) n=m, and try to contradict the fact that sup(a n ) n=m is the least upper bound of (a n ) n=m. Use a similar idea to show that inf(a n ) n=m inf(b n ) n=m. Try to show the last two inequalities in the problem by using the first two : Apply lim sup and lim inf to the terms in the given inequality a n b n c n to get another couple strings of inequalities, using previous exercise. Now try to use Prop 6.4.1(f) : Come up with a sequence whose first, fourth, seventh,... terms are all 0, whose second, fifth, eighth,... terms are increasing in value, and whose third, sixth, ninth,... terms are decreasing in value. (You can choose c 0 instead of 0, but this might make some of the steps of the proof a little more annoying.) For example, we can define 0 if n = 3k, k N, f(n) = k if n = 3k + 1, k N, k if n = 3k +, k N. (This is a a little different from how we defined it in recitation, but it may be a little easier to work with.) Based on this, show that the three limits points are 0,, and.you can use the equivalent definition of limit point, that x is a limit point of (a n ) n=1 iff for every ɛ > 0 and every N 1, there exists an n N such that a n x ɛ. No matter what N you choose, you ll always be able to find an n past that N whose term is ɛ close to 0 for any ɛ > 0. (Why?) And refer to the definition of and being limit points in Exercise to show that they are also limit points. To show there is no other limit point, assume there is some M that isn t 0,, or, then choose ɛ and N wisely so that for no n N will any term of your sequence be ɛ-close to M. To show M is not ɛ-close to the terms of the part your sequence that is increasing and the part of the sequence that is decreasing will require some playing with the algebra. You ll probably have to define your sequence explicitly as a piecewise function in order to justify fully. 6.5.: Tao provides a pretty helpful hint for this exercise, and I ll just comment on how to use the squeeze test to prove for the case 1 < x < 0. We have x n <= x n <= x n for all n (this is the generalization for reals of Prop 4.3.3(c)) x n x n x n for all n by (cf (d)) Since 1 < x < 0, we have 0 < x < 1. Now try to apply the squeeze theorem given this inequality. (You ll need to use the result of another case in this problem,

24 4 ADVANCED CALCULUS 401 RECITATION NOTES, FALL 014 as well as one of the limit laws.) Exercise 1: Consider the sequence (a n ) n=1 given by a n = 1/n. Find the supremum and infimum of (a n ) n=1, using the definitions of supremum and infimum to justify your answers. Solution: We claim that sup(a n ) n=1 = 1 and inf(a n ) n=1 = 0. (Convince yourself by listing off some terms that these are reasonable guesses.) Indeed, 1 is an upper bound of (a n ) n=1 = ( 1 n ) n=1 because n 1 implies 1 n n 1 1 n 1 1 n. It is the least upper bound because given any upper bound M of ( 1 n ) n=1, we must have M 1 ( 1 n ) 1 n=1 by definition of upper bound. Now, n > 0 for all n 1 from the definition of > for rational numbers, as numerators are denominators are both positive for all terms; hence 0 is a lower bound of ( 1 n ) n=1. 0 is the greatest lower bound because we can t have a c > 0 that is a lower bound of ( 1 n ) n=1 since we can always find an N 1 natural number such that 1 N < c by Archimedean principle, meaning c 1 N. Therefore, any lower bound c of ( 1 n ) n=1 satisfies c 0. Thus, 0 is the greatest lower bound. Exercise : Show that the sequence {b n } defined by b n = n, n 1 is (a) convergent and (b) Cauchy using the ɛ, N definitions of convergent and Cauchy. Proof. (a) To show ( n ) n=1 is convergent, we need to make the correct guess at 1 the value to which it converges. List off some terms:, 1 4, 1 8, 1 16, and by noticing that the terms are getting smaller and smaller yet are all positive, we guess that lim b n = 0. We confirm this using the ɛ, N definition. Let ɛ > 0. We seek N 1 such that for all n N, n 0 ɛ n ɛ. Note that for all n 1, n := 1 = ( 1 n )n > 0 by Proposition 4.3.1(b), since 1 > 0. Thus, we want N 1 such that n N, ɛ. We re tempted to cite some Archimedean Principle, but n such principles that we discussed didn t involve the natural number in the exponent. To get around this, we claim that 1 1 n n for all n 1. To prove this, we ll prove that n n for all n 1, from which our claim will follow by multiplying by 1 n on both sides of the inequality. Indeed, 1 = 1. Assume n n. Then n n+1 n +1 n + n = ( n ) = n+1, so n+1 n+1. (We used that 1 n for all n 1, which you can also prove by induction.) Thus, the claim follows. Now, by Archimedean Principle, choose N a natural number such that 1 N ɛ. Then for all n N, we have 1 n 1 1 N, and so 1 n n 1 N ɛ. Thus, n N, 1 ɛ, as n desired, showing ( n ) n=1 is convergent to 0. (b) We show ( n ) n=1 is Cauchy. Let ɛ > 0. We seek N so that n, m N, n m ɛ. For any such n, m pair, suppose WLOG that n m. (The