Exam (elaborations) TEST BANK FOR Principles of Mathematical Analysis By Walter Rudin (A Complete Solution Guide)

The Real and Complex Number Systems Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real. 1.1 Problems on rational numbers and fields Rudin Chapter 1 Exercise 1. Problem 1.1 Proof. Assume that r + x was rational. Then it follows from Definition 1.12(A1), (A4) and (A5) that x = (r + x) − x is also rational, a contradiction. Similarly, if rx was rational, then it follows from Definition 1.12(M1), M(4) and M(5) that x = rx x is also rational, a contradiction. This ends the proof of the problem.  Rudin Chapter 1 Exercise 2. Problem 1.2 Proof. Assume that √12 was rational so that √12 = m n , where m and n are co-prime integers. Then we have m2 = 12n2 and thus m is divisible by 3. Let m = 3k for some integer k. Then we have m2 = 9k2 and this shows that k2 = 4n2 3 , so n is divisible by 3. This contradicts the fact that m and n are co-prime, completing the proof of the problem.  Rudin Chapter 1 Exercise 3. Problem 1.3 Proof. Since x 6= 0, there exists 1 x ∈ F such that x · 1 x = 1. (a) Therefore, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = xz implies that y = z. 1 Chapter 1. The Real and Complex Number Systems 2 (b) Similarly, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = x implies that y = 1. (c) Similarly, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = 1 implies that y = 1 x. (d) Since 1 x ∈ F, we have 1 1 x ∈ F such that 1 x · 1 1 x = 1. Now we have 1 x · 1 1 x = 1 x ·x(= 1), then Proposition 1.15(a) implies that 1 1 x = x. This completes the proof of the problem.  1.2 Properties of supremums and infimums Rudin Chapter 1 Exercise 4. Problem 1.4 Proof. Since E ⊂ S, the definitions give α ≤ x and x ≤ β for all x ∈ E. Thus Definition 1.5(ii) implies that α ≤ β. This completes the proof of the problem.  Rudin Chapter 1 Exercise 5. Problem 1.5 Proof. Theorem 1.19 says that R is an ordered set with the least-upper-bound property. Since A is a non-empty subset of R and A is bounded below, inf A exists in R by Definition 1.10. Furthermore, −A is a non-empty subset of R. Let y be a lower bound of A, i.e. y ≤ x for all x ∈ A. Then we have −x ≤ −y for all x ∈ A. Thus −y is an upper bound of −A and sup(−A) exists in R by Definition 1.10. Let α = inf A and β = sup(−A). By definition, we have y ≤ β for all y ∈ −A, where y = −x for some x ∈ A. It implies that x = −y ≥ −β for all x ∈ A, so −β is a lower bound of A and then −β ≤ α. Similarly, we have α ≤ x for all x ∈ A so that −α ≥ −x for all x ∈ A. It implies that −α is an upper bound of −A, so β ≤ −α and then −β ≥ α. Hence we have α = −β, i.e. inf A = −sup(−A). This completes the proof of the problem.  1.3 An index law and the logarithm Rudin Chapter 1 Exercise 6. Problem 1.6 Proof. (a) Since bm 0 and n ∈ N, Theorem 1.21 implies that there exists one and only one real y such that yn = bm. Similarly, there exists one and only one real z such that zq = bp. We have ynq = (yn)q = (bm)q = bmq = bpn = (bp)n = (zq)n = zqn which implies that y = z, i.e., (bm) 1 n = (bp) 1 q . (b) Let br = b m n and bs = b p q . Without loss of generality, we may assume that n and q are positive. Then the corollary of Theorem 1.21 implies that br+s = b mq+np nq = (bmq+np) 1 nq = (bmq × bnp) 1 nq = (bmq) 1 nq × (bnp) 1 nq = b m n × b p q = br × bs. 3 1.3. An index law and the logarithm (c) By definition, B(r) = {bt | t ∈ Q, t ≤ r}, where r ∈ Q. It is clear that br ∈ B(r), so it is a nonempty subset of R. Since b 1, we have bt ≤ br for all t ≤ r so that br is an upper bound of B(r). Therefore, Theorem 1.19 and Definition 1.10 show that supB(r) exists in R. Now we show that br = supB(r). If 0 γ br, then γ is obviously not an upper bound of B(r) because br ∈ B(r). By Definition 1.8, we have br = supB(r). (d) By part (c), we know that bx, by and bx+y are all well-defined in R. By definition, we have B(x) = {br | r ∈ Q, r ≤ x}, B(y) = {bs | s ∈ Q, s ≤ y}, B(x + y) = {bt | t ∈ Q, t ≤ x + y}. Before continuing the proof, we need to show several results: For every real x and y, we define B(x, y) = B(x) × B(y) = {br × bs | r, s ∈ Q, r ≤ x, s ≤ y}. Then we have bx × by = supB(x, y). Lemma 1.1 Proof of Lemma 1.1. By definition, bx and by are upper bounds of B(x) and B(y) respectively, so we have br ≤ bx and bs ≤ by for every br ∈ B(x) and bs ∈ B(y). Therefore, we have br × bs ≤ bx × by for every br × bs ∈ B(x, y). In other words, bx × by is an upper bound of B(x, y). Let 0 α bx × by. Then we have α bx by. We define the number p = 1 2 ( α bx + by). It is obvious from this definition that α bx p by. By α bx p, we have α p bx and so there exists br ∈ B(x) such that α p br. (1.1) Similarly, the inequality p by implies that there exists bs ∈ B(y) such that p bs. (1.2) Now inequalities (1.1) and (1.2) show that α br × bs for some br × bs ∈ B(x, y). Hence α is not an upper bound of B(x, y) and we have bx × by = supB(x, y), completing the proof of the lemma.  Let S be a set of positive real numbers and bounded above and S−1 = {x−1 | x ∈ S}. Then we have sup S = 1 inf S−1 . Lemma 1.2 Proof of Lemma 1.2. Suppose that α is an upper bound of S, i.e., 0 x ≤ α for all x ∈ S. Then we have 0 α−1 ≤ x−1 for all x−1 ∈ S−1. Hence the result follows directly from the definitions of the least upper bound and the greatest lower bound.  Chapter 1. The Real and Complex Number Systems 4 For every real x, we have b−x = 1 bx . Lemma 1.3 Proof of Lemma 1.3. We have two facts: – Fact 1. If b 1, then b 1 n 1 for every positive integer n. Otherwise, 0 b 1 n 1 implies that 0 b = (b 1 n )n 1n = 1 by Theorem 1.21, a contradiction. – Fact 2. If m and n are positive integers such that n m, then b 1 m b 1 n . Otherwise, it follows from Fact 1 that 1 b 1 m b 1 n and so it implies that b = (b 1 m )m (b 1 n )m (b 1 n )n = b, a contradiction. Let r and s be rational. Define A(x) = {bs | s ∈ Q, s ≥ x}. We next want to prove that supB(x) = inf A(x). In fact, it is clear that supB(x) ≤ inf A(x) by definitions. Suppose that D = inf A(x) − supB(x). Assume that D 0. By Fact 2 above, b 1 n − 1 is decreasing as n is increasing, so there exists a positive integer N such that bx(b 1 n − 1) D for all n ≥ N. By Theorem 1.20(b), we see that there exist r, s ∈ Q such that x − 1 2n s x and x r x + 1 2n. Thus we have r − s x + 1 2n − (x − 1 2n) = 1 n and this implies that D br − bs = bs(br−s − 1) ≤ bx(br−s − 1) D, a contradiction. Hence we must have supB(x) = inf A(x). Define B′(−x) = {b−t | t ∈ Q, t ≤ −x}. Let s = −t. Then we have b−t = bs and the inequality t ≤ −x is equivalent to s ≥ x. Thus we have B′(−x) = A(x). Now it follows from Lemma 1.2 and the above facts that b−x = supB(−x) = 1 inf B′(−x) = 1 inf A(x) = 1 supB(x) = 1 bx , completing the proof of the lemma.  We can continue our proof of the problem. Let br and bs be any elements of B(x) and B(y) respectively. Then we have r and s are rational numbers such that r ≤ x and s ≤ y. Since the sum t = r + s is also rational and t ≤ x + y, part (b) implies that br × bs = br+s ≤ bx+y. Thus bx+y is an upper bound of the set B(x, y) and Lemma 1.1 implies that bx × by ≤ bx+y. Assume that bxby bx+y. We deduce from Lemma 1.3 that by = (b−xbx)by = b−x(bxby) b−xbx+y ≤ b−x+(x+y) = by, a contradiction. Hence we have the desired result that bxby = bx+y. This completes the proof of the problem.  Rudin Chapter 1 Exercise 7. Problem 1.7 Proof. 5 1.4. Properties of the complex field (a) Since b 1, it is obvious that bn − 1 = (b − 1)(bn−1 + bn−2 + · · · + 1) ≥ (b − 1)(1 + 1 + · · · + 1) = n(b − 1) for any positive integer n. (b) It is obvious that b 1 n 1; otherwise, we have b = (b 1 n )n 1 which is impossible