Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 2 Solutions Please write neatly, and in complete sentences when possible.

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1 Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 2 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 1.4.2, 1.4.4, 1.4.9, , , 1.5.1, 1.5.7, Solution (a) Given two rational numbers a, b Q, there exist integers p 1, q 1, p 2, and q 2 Z so that q 1, q 2 0, and a = p 1 q 1 and b = p 2 q 2. In this case a + b = p 1q 2 + p 2 q 1 and ab = p 1p 2. q 1 q 2 q 1 q 2 Since q 1, q 2 0, we must have q 1 q 2 0. Thus both a + b and ab are represented by the ratio of two integers, where the denominator is nonzero. Thus both a + b, ab Q. (b) Suppose a Q, t I, and a+t Q. In this case, t = (a+t)+( a) is a sum of rational numbers ( a), a + t Q. By part (a), this implies t Q, contradicting the assumption that t I. Suppose now that a is non-zero, so that a = p/q for p, q 0. In this case, 1/a = q/p Q. Suppose that at Q. Then t = at (1/a) is the product of rational numbers at, 1/t Q. By part (a) again, this implies that t Q, a contradiction. (c) No, I is not closed under addition or multiplication: Consider s = 2, t = 2, and u = 1/ 2. These are all irrational: We proved in class that s I.By part (b), t, u I as well, since s + t = 0 Q and su = 1 Q. On the other hand, this also implies that I is not closed under addition and multiplication. Solution Let S indicate the set {1/n n N}, and let a = inf S, the greatest lower bound for the set S. Note that 0 is a lower bound for S, since 1/n 0 for all n N, so that by definition we have a 0. On the other hand, suppose a > 0. In this case, by part (ii) of Theorem (the Archimidean property of R), there is an n N so that 1/n < a. This violates the assumption that a is a 1

2 2 lower bound for S. Solution (a) Suppose that A B, so that there is a 1-1 and onto map f : A B. This implies that there is an inverse map g : B A. Namely, for each b B, since f is onto, there is an element a A so that f(a) = b, and we define g(b) = a. Since f is a function, we cannot have f(a) = b and f(a ) = b for a a, so that g is 1-1. Since f is 1-1, g is onto: Suppose b = f(a). We must have g(b) = a, since a is the only element of A with b = f(a). Thus every element a A is in the image of g. Thus g forms a 1-1 correspondence, and B A. (b) Suppose that A B and B C, so that there are 1-1 and onto maps g : A B and f : B C. Consider f g : A C. Suppose that f g(a) = f g(a ). In this case f(g(a)) = f(g(a )), so that g(a) = g(a ) since f is 1-1. Since g is also 1-1, we have a = a, and we may conclude that f g is also 1-1. Suppose c C. Since f is onto, there is an b B with f(b) = c. Since g is onto, there is an a A with g(a) = b. Thus f g(a) = f(b) = c, and f g is onto. Thus f g is 1-1 and onto, and we conclude that A C. Solution (a) Let f : (0, 1) S be defined by f(x) = (.5, x). If f(a) = f(b) then (.5, a) = (.5, b), so that a = b, and we conclude f is 1-1. (b) Each element x (0, 1) has a decimal expansion. (Technically, as the book points out, x may have more than one decimal expansion. At the moment this is immaterial just choose one for each x (0, 1)). We define a function g : S (0, 1) as follows: Given a point (x, y) S, with decimal expansions x =.a 1 a 2 a 3... y =.b 1 b 2 b 3..., and let g(x, y) be the real number with decimal expansion.a 1 b 1 a 2 b 2 a 3 b Then g is 1-1: Suppose that x, y, w, z (0, 1) satisfy g(x, y) = g(w, z) =.e 1 e 2 e 3 e 4 e 5 e In this case, the decimal expansions of x and w are both given by.e 1 e 3 e 5..., and likewise the decimal expansions of y and z are both

3 given by.e 2 e 4 e We conclude that x = w and y = z, so that (x, y) = (w, z), and g is 1-1. The ambiguity in the decimal expansions will cause this map not to be onto: Suppose we ve chosen 1/2 = and x =.e 1 e 2 e 3..., so that ( ) 1 g 2, x =.5e 1 0e 2 0e Note that in this case, the real number.4e 1 9e 2 9e is not in the image of g: If it was, then we would find that the pre-image under g of this point is ( , x). But the latter is the same point as (1/2, x), and for this point we ve already decided that its image is.5e 1 0e 2 0e e 1 9e 2 9e Thus g is not onto. The same argument will work if we had chosen the decimal expansion 1/2 = Solution (a) The polynomials x 2 2, and x 3 2 have 2 and 3 2, respectively, as roots, so that each is an algebraic number. Note that ( 2 + 3) 2 = so that 1 ( ( 2 + ) 3) 2 5 = 6. 2 Thus is a root of the polynomial 1 ( x 2 5 ) 2 1 ( 6 = x 4 10x ), 4 4 and is also algebraic. (b) Following the instructions in the problem, we let A n indicate the algebraic numbers obtained as roots of polynomials of degree n, with integer coefficients. Each polynomial of degree n can be written as a n x n + a n 1 x n a 1 x + a 0, for integers a 0, a 1,..., a n, with a n 0. Let P m indicate the set of polynomials of degree n as above that further satisfy a n + a n a 1 + a 0 m. 3

4 4 Let B m A n indicate the algebraic numbers inside of A n that can be obtained as roots of polynomials in P m. We claim that each set B m is finite: By recording the coefficients of each polynomial, we obtain a bijection from P m to { m, m + 1,..., m 1, m} n, a finite set. Note that there is a natural map φ : B m P m which associates to each algebraic number x B m a polynomial φ(x) in P m having x as a root. Since each polynomial p P m has a finite number of roots, the size of the set φ 1 (p) is finite. Since B m = φ 1 (P m ), we find that B m is finite. Note that we have A n = m=1 B m. By Theorem , part (ii), we may conclude that A n is countable. Solution By definition of cardinality, it suffices to find a 1-1 and onto map from (0, 1) to R. Among many possibilities, the following works: Let f : (0, 1) R be the function defined by ( ) 1 f(x) = log x 1. Note first that x 1 provides a 1-1 correspondence of the interval x (0, 1) with the interval (1, ), and x x 1 provides a 1-1 correspondence of (1, ) with (0, ). Finally, x log x provides a 1-1 correspondence of (0, ) with R. We conclude that f is a 1-1 correspondence of (0, 1) with R. Solution We examine the maps arising from Exercise 1.5.6: (a) Here are two 1-1 maps from A = {a, b, c} to P (A): Let f 1, f 2 : A P (A) be given by f 1 (a) = {a}, f 1 (b) = {b}, f 1 (c) = {c}, and f 2 (a) = {b, c}, f 2 (b) = {a, c}, f 2 (c) = {a, b}. In this case the set B arising in Cantor s construction for f 1 is give by B =, while for f 2 we obtain B = {a, b, c}. In either case, we note that the set B is not in the image of f 1 or f 2. (b) Given B = {1, 2, 3, 4}, an example of a 1-1 map g : B P (B) is given by g(1) = {1, 2}, g(2) = {1, 3}, g(3) = {2, 3}, g(4) = {2}. In this case, Cantor s set B is given by {2, 4}. Evidently this set is not in the

5 image of g. Solution (a) Suppose that a B. In this case, since f(a ) = B by assumption, a is an element of A that is mapped by f to a set containing a. By the definition of the set B, we must have a / B, a contradiction. (b) Suppose that a / B. In this case, since f(a ) = B by assumption, a is an element of A that is mapped by f to a set not containing a. By definition of the set B, we must have a B, a contradiction.

Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible.

Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 4.2.1, 4.2.3, 4.2.6, 4.2.8,