Solutions to Exercises Chapter 1 1.1 If x = min were rational, then x 2 = m 2 /n2 would also be rational. The square of /2+V3 is 5+2V6.1f 5+2V6 = p/q E Q, then V6 = (p-5q)/lOq E Q, and this is not the case. 1.2 AnB = {1,3}, AU B = {1,2,3,4,5,7}, X\ A = {2,4,6,8}, X\ B = {5,6, 7,8}. X\(AnB) = (X\A)U(X\B) = {2,4,5,6, 7,8}, X\(AUB) = (X \ A) n (X \ B) = {6,8}. 1.3 437 = 19 x 23, 493 = 17 x 29. 1.4 a) True. Ifx = min E Q and x+y = p/q E Q, then y = (pn-qm)/qn E Q, a contradiction. b) False. Take x = 0. If we had insisted that x '" °this would have been true. c) False. Take x = ..;2, y = 1 - ..;2. d) False. Take x = y = ..;2. e) False. Take x =/2, y = V3. f) False. Again take x = ..;2, y = V3. 1.5 Take u = x + (l//2)(y - x). 1.6 Let N be a positive integer such that N > 1/(y - x). Then the difference between successive members of the sequence ... , -3/N, -2/N, -l/N, 0, l/N, 2/N, '" 238 Real Analysis is less than y - x, and so x < q < y for at least one rational number q=M/N. 1. 7 True for n = 1. Let n ~ 1 and suppose true for n. Then 12+...+ (n + 1)2 = (1/6)n(n + 1)(2n + 1) + (n + I? = (1/6)(n + 1)(2n2 + n + 6n + 6) = (1/6)(n + 1)[(n + 1) + 1][2(n + 1) + 1]. 1.8 True for n = 1. Let n ~ 1 and suppose true for n. Then 13+...+(n + 1)3 = (1/4)n2(n+l)2+(n+l)3 = (1/4)(n+l)2(n2+4n+4) = (1/4)(n+l)2[(n+ 1) + IF. 1.9 True for n = 1. Let n ~ 1 and suppose true for n. Then a + (a + d) + ... + (a + nd) = na + !n(n - l)d + a + nd = (n + l)a + Hn(n - 1) + 2n]d = (n + l)a + Hn + 1)[(n + 1) - l]d. 1.10 True for n = 1. Let n ~ 1 and suppose true for n. Then a+ar +... +arn = [a(l-rn)l/(I-r)+arn = [a-arn+arn-arn+l]/(I-r) = [a(l-r n +ll/(I­ r). 1.11 If n =1 then the right hand side is (x2 - 2x + 1)/(x - 1)2 = 1. So true for n = 1. Let n ~ 1 and suppose true for n. Then n n nxn+l - (n + l)x + 1 n 1 + 2x + ... + (n + l)x = (x _ 1)2 + (n + l)x nxn+l - (n + l)xn + 1 + (n + l)xn+2 - 2(n + l)xn+l + (n + l)xn =----'--~----'----:--'--~---'---'-----'-----=------ (x - 1)2 (n + l)xn+2 - (n + 2) xn+l + 1 = (x - 1)2 1.12 (n + 1)4 < 4n4 if and only if (n + 1)/n < .;2, that is, if and only if n(.;2 - 1) > 1, that is, if and only if n (being an integer) is at least 3. 4n > n4 if n = 5. Let n ~ 5, and suppose true for n. Then 4n +l = 4.4n > 4n4 > (n + 1)4. 1.13 The formula is true for n = 1. Let n ~ 1 and suppose true for n. Then qn+l = 3qn -1 = (3/2)(3n+ 1) -1 = (1/2)(3n +l +3 - 2) = (1/2)(3n +1 + 1). 1.14 The formula is true for n = 0 and n = 1. Suppose that it is true for all k < n. Then an = 4[2n- 2(n+ 1) - 2n- 3n] = 2n- 1 [2n+2 -n] = 2n- 1(n+2). 1.15 a) x2 + 4x + 5 = (x + 2)2 + 1 ~ 1 > O. b) x2 + 5xy + 7y2 = (x + (5/2)y)2 + (3/4)y2 ~ O. c) a2 + b2 + e2 + (l/a2) + (l/b2) + (l/e2) - 6 = [a - (l/a)F + [a - (l/bW + [a - (l/eW ~ O. Equality occurs if and only if a = l/a, b = l/b, e = l/e, that is, if and only if a, b, e E {-I, I}. 1.16 2a ~ a+b ~ 2b, and so 2ab/2b ~ 2ab/(a+b) ~ 2ab/2a; that is, a ~ H ~ b. Also, G2 - H 2 = ab - (4a2b2/(a + b)2) = [ab(a - b)2l/(a + b)2 ~ O. Solutions to Exercises 239 1.17 Trick question! A = {(x,y) : (x + 1)2 + (y - 2)2 + 1 = O} = 0, B = {(x,y): (x+2y+l)2+ y2+2=0}=0. 1.18 Ix - al < 6 ¢=} [x - a < 6 and -(x - a) < 6] ¢=} a - 6 < x < a + 6. 1.19 From Exercise 1.10, a+ar+'" arn = a/(I-r) -arn+1 /(I-r) < a/(I-r). 1.20 Put a = 1 and r = IX2/Xll in the previous inequality, to obtain 1 + Ixdxli +.. ·IX2/Xlln-l < 1/(I-lx2/xll). The left hand side is not greater than n times its smallest term, so not greater than nlx2/xlln-l. Hence nIX2/Xlln-l < 1/(I-lx2/xll = IXll/(lxd -lx21). 1.21 (3x+2)/(x+l) <1 ¢=} (3x+2)/(x+l)-1<0 ¢=} (3x+2-x­ 1)/(x + 1) < 0 ¢=} (2x + 1)/(x + 1) < 0 ¢=} x E (-1, -1/2]. 1.22 If x ~ y, then Ix - yl = y - x, so (I/2)(x + y + Ix - yl) = y = max {x,y}, (I/2)(x + y -Ix - yl) = x = min {x, y}. The case x ~ y is similar. 1.23 lab - cdl = Ib(a - c) + c(b - d)1 ~ Iblla - cl + Icllb - dl. So with the given information lab - cdl ~ Iblla - cl + Icllb - dl < K(€/2K) + L(€/2L) = €. 1.24 Since labl > k2, 1(I/a) - (l/b)1 = la - bl/labl ~ la - bl/k2. Chapter 2 2.1 The following table gives the answer: 0.0001 100,000,000 2.2 If k = 0 then the sequence is (1,1,1, ...), with limit 1. If k > 0, then k k II/n - 01 < € whenever n > (I/€)l/k. Hence (I/n ) ~ O. If k = -l < 0, k l l then I/n = n' > M for any given M > 0, provided n > M / • (Note that, for any positive a in IR and any positive integer k, we can define al / k as sup {x E IR : x k < a}.) 2.3 n/(n2+ 1) < 0.0001 if and only if n2-10,000n +1> 0, that is, if and only if n > 5,000 + y'25,OOO,OOO - 1, that is, if and only if n > 9,999. 2 2.4 n + 2n ~ 9999 ¢=} n ~ -1 + VI + 9999 = 99. 2.5 an ~ -00 if and only if, for every K > 0 there exists a positive integer N with the property that an < - K for every n > N. 2.6 The formula is correct for n = 1 and n = 2. Suppose that it is true for l 2 all k < n. Then an = (1/2)[2 + 4(-(I/2)t- + 2 + 4(-(I/2)t- ] = 240 Real Analysis (1/2)[4 - 8(-(1/2))n + 16(-(1/2)tl = 2 + 4(-(1/2)t. It is then clear that (an) --* 2. 2.7 Suppose that f3 > B. There exists N such that Ibn - f31 < f3 - B for all n > N. Thus f3 - bn < f3 - B and so bn > B for all n > N. This is a contradiction to the definition of B. 2.8 If lanl ~ A, then -A ~ an ~ A. So "bounded" implies "bounded above and below". Conversely, suppose that A ~ an ~ B. If 0 ~ A ~ B then lanl = an ~ B. If A ~ B ~ 0 then lanl = -an ~ -A. If A ~ 0 ~ B then lanl ~ max {IAI, IBI}. SO in all cases (an) is bounded. 2.9 We know that for every € > 0 there exists a positive integer N such that Ian - al < € for every n > N. It follows that there exists a positive integer N', namely N' = N - 1, such that Ibn - al < f for every n > N'. Thus (bn ) --* a. 2.10 Since (an) --* a, for every f > 0 there exists N such that Ian - al < f for all n > N. There exists an integer M with the property that bM = aK, with K ~ N. Then, for all m > M, bm = ak for some k > N, and so Ibm - al < f. 2.11 Suppose, for a contradiction, that L < O. Taking f = ILI/2 = -L/2, we know that there exists a positive integer N such that Ian - LI < ILI/2 for all n > N. This implies in particular that an < L/2 < 0 -a contradiction. 2.12 Suppose first that L > O. Since an is positive for all n, Iva;- JII = (Ian ­ LD/(~ + JI) < (l/JI)lan - LI· Choosing N so that Ian - LI < (JI)f for all n > N, we see that Iva;- JII < f for all n > N. So (va;) --* JI. Suppose now that L = O. If, for a given f, we choose N so that lanl < €2 for every n > N, then Iva;I < €, and so (va;) --* O. 2.13 Since (an - xn) is a positive sequence with limit ~ - a, we deduce from Exercise 2.10 that ~ - a > O. Similarly f3 - ~ ~ O. 2.14 By Theorems 2.1 and 2.8, lim max {an,bn} = max {a,f3} , lim min {an,bn} =min{a,f3}. n-too n-too Let an = (_l)n, bn = (_l)n+1. Then max {an, bn} = 1, min {an, bn} = -1.