18.100C Real Analysis: Problem Set 1 Solution

Problem Set 1 Solutions, 18.100C, Fall 2012 September 12, 2012 1 Let m and n be positive integers with no common factor. Prove that if m/n is rational, then m and n are both perfect squares, that is to say there exist integers p and q such that m = p2 and n = q 2 . (This is proved in Proposition 9 of Book X of Euclids Elements). Assume m/n is rational. Then there exist positive integers M and N with no common factor such that m/n = M/N and so mN 2 = nM 2 . Claim: M 2 divides m and N 2 divides n. Assume the claim for now. Then m = M 2 m/ and n = N 2 n/ for some m/ and n/ . Substituting we obtain M 2 m/ N 2 = N 2 n/ M 2 which gives m/ = n/ . m/ = n/ divides m and n so m/ = n/ = 1 and we have shown m and n are perfect squares. Proof of claim: We show that M 2 divides m; the argument that N 2 di­ vides n is identical. Write M as a product of primes p1 · · · pr and note that no pi divides N . Assume inductively that p21 · · · p2t divides m. Then p2t+1 M2 p21 · · · p2t m N2 p21 · · · p2t Since pt+1 does not divide N 2 we see m p2t+1 , which gives p21 · · · p2t+1 m. 2 p1 · · · pt2 The inductive hypothesis holds when t = 0; the empty product is 1. Thus, by induction p21 · · · pr2 = M 2 divides m. 1 2 Problem 8 from Page 22: Prove that no order can be defined in the complex field that turns it into an ordered field. Suppose that such an order exists. We know that in any ordered field squares are greater than or equal to zero. Since i2 = −1, this means that 0 ≤ −1. Thus 1 = 0 + 1 ≤ −1 + 1 = 0 ≤ 1, which implies 0 = 1, a contradiction. 3 Problem 9 from Page 22: suppose z = a + bi and w = c + di. Define z < w if a < c, and also if a = c but b < d. Prove that this turns the set of all com­ plex numbers into an ordered set. Does this set have the least-upper-bound property? Suppose z = a + bi and w = c + di are distinct complex numbers. If a = c, then either a > c or a < c, and hence either z < w or z > w. If a = c, then b = d, since otherwise z = w. In this case, either b < d or b > d, and then z > w or z < w. So in either case either z < w or z > w. Now suppose x = a + bi, y = c + di, and z = e + f i are complex num­ bers with x < y and y < z. We need to show that x < z. We have a ≤ c, since otherwise x > y, and similarly c ≤ e, and so a ≤ e. If a < e, then x < z. If a = e, then in fact a = c = e. In this case, since x < y, we must then have b < d and similarly d < f , so b < f and again x < z. In either case x < z. So this relation makes C into an ordered set. This order does not have the least-upper-bound property. To see this, consider the purely imaginary line, i.e. the set L = {a + bi ∈ C|a = 0} This set is bounded above by 1. Let z = a + bi be any upper bound for L. Then a > 0; indeed, if a ≤ 0, then the complex number w = (b + 1)i satisfies z < w and w ∈ L, so z cannot be an upper bound for L. Now consider the complex number y = a + (b − 1)i. Clearly y < z. However, since a > 0 y 2 is also an upper bound for L. This shows that for every upper bound on L there exists a strictly smaller upper bound, and so C cannot have the least-upper-bound property with this ordering. 4 Let R be the set of real numbers and suppose f : R → R is a function such that for all real numbers x and y the following two equations hold (1)f (x + y) = f (x) + f (y) (2)f (xy) = f (x)f (y) Claim: f (x) = 0 for all x or f (x) = x for all x. a Setting x = 1 and y = 0 in equation (1) gives f (1) = f (1) + f (0) so that f (0) = 0. Setting x = y = 1 in equation (2) gives f (1) = f (1)2 . Thus f (1) is equal to 0 or 1 as we may see by solving the equation x − x2 = x(1 − x) = 0. Remark: If I did not have to answer all parts of the question fully and I just wanted to prove the claim as quickly as possible I would proceed by noting that f (1) = 0 gives f (x) = f (1)f (x) = 0 for all x ∈ R. From this moment on I could then assume f (1) = 1. b By a) we have f (0) = 0 = 0f (1). Also, f (x) + f (−x) = f (x − x) = f (0) = 0 so that f (−x) = −f (x) for all x ∈ R. In particular, f (−1) = −f (1). Let n ∈ Z and assume that f (n) = nf (1). Then f (n + 1) = f (n) + f (1) = nf (1) + f (1) = (n + 1)f (1) 3 and f (n − 1) = f (n) + f (−1) = nf (1) − f (1) = (n − 1)f (1). By induction f (n) = nf (1) for all n ∈ Z. For n, m ∈ Z, m 6= 0 f n mf (1) = f m and so f n m =f n f (m) = f (n) = nf (1) m n n f (1) = f (1). m m Thus f (q) = qf (1) for all q ∈ Q and by a) either f (q) = 0 for all q ∈ Q or f (q) = q for all q ∈ Q. c Suppose x ≥ 0. Then there exists a y ∈ R such that y 2 = x and f (x) = f (y 2 ) = f (y)2 ≥ 0. Thus x ≥ y ⇒ x−y ≥ 0 ⇒ f (x)−f (y) = f (x)+f (−y) = f (x−y) ≥ 0 ⇒ f (x) ≥ f (y). d Suppose f (1) = 0. Given any x ∈ R we can find p, q ∈ Q such that p ≤ x ≤ q. Then 0 = f (p) ≤ f (x) ≤ f (q) = 0 ⇒ f (x) = 0. Alternatively, we proceed as remarked in a). Suppose f (1) = 1. Let x ∈ R and n ∈ N. Then there exist p, q ∈ Q such that 1 1 x− ≤p≤x≤q ≤x+ n n and so 1 1 x − ≤ p = f (p) ≤ f (x) ≤ f (q) = q ≤ x + . n n 4 So for all x ∈ R and all n ∈ N we have x− 1 1 ≤ f (x) ≤ x + n n which gives f (x) = x for all x ∈ R. 5