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Solutions to Assignment 1, Math 220 1 Conjecture a formula for n X k=1 1 1 1 1 = + +···+ k(k + 1) 1·2 2·3 n · (n + 1) from the value of this sum for small integers n. Prove your conjecture using mathematical induction. Solution Let Sn = Pn 1 k=1 k(k+1) . Then 1 1 1 1 1 2 1 3 S 1 = , S2 = + = , S3 = + + = ,... 2 2 6 3 2 6 12 4 n and we conjecture that Sn = n+1 . We prove this via induction. Note that we have already shown this to be true for n = 1. Suppose Sm = m for m ∈ N. Then m+1 Sm+1 = m+1 X k=1 X 1 1 1 = + k(k + 1) k(k + 1) (m + 1)(m + 2) m k=1 and so Sm+1 = Sm + 1 . (m + 1)(m + 2) It follows from our induction hypothesis that Sm+1 = 1 m(m + 2) + 1 m+1 m + = = . m + 1 (m + 1)(m + 2) (m + 1)(m + 2) m+2 This completes the proof. 2 Use mathematical induction to prove Bernoulli’s inequality : If 1 + x > 0, then (1 + x)n ≥ 1 + nx, for all n ∈ N. Solution Note that if n = 1, (1 + x)n = 1 + nx. Suppose, then, that (1 + x)m ≥ 1 + mx, for some m ∈ N. Then (1 +x)m+1 = (1 +x)m · (1 +x) ≥ (1 +mx) · (1 +x) = 1 + (m+ 1)x+ mx2 . Since mx2 ≥ 0, this is ≥ 1 + (m + 1)x and so we reach the desired conclusion. 1 2 3 Show that any amount of postage that is an integer number greater than 11 cents can be formed using just 4-cent and 5-cent stamps. Solution We proceed by induction. Note that 12 = 3·4+0·5. Suppose, then, that we can write m = a · 4 + b · 5 for a and b nonnegative integers and m ≥ 12. If a ≥ 1, then m + 1 = (a − 1) · 4 + (b + 1) · 5 is a representation of m + 1 with nonnegative coefficients a − 1 and b + 1. If a = 0, since m ≥ 12, we have b ≥ 3 and so we can write m + 1 = 4 · 4 + (b − 3) · 5. This completes the induction proof. 4 Using our axioms for R, prove the the following assertions : a: If x ∈ R, then −(−x) = x. b: If x and y are real numbers of the same sign (x and y are either both positive or both negative), and x < y, then 1/y < 1/x. c: If x is a real number with x 6= 0 then x2 > 0. Solution Part (a). Note that Axiom A5 says that there is a unique number −(−x) such that −x + (−(−x)) = 0. Since, combining A2 and A5, −x + x = x + (−x) = 0, it follows that this unique number must in fact be x and so −(−x) = x. Part (b). Let us suppose first that x > 0 and y > 0. If y1 = x1 , then applying M5 twice (together with M2), we have 1= a contradiction. If 1 y 1 · y → x = y, x > x1 , then M5 and O4 imply that 1> 1 · y. x From M2, this gives 1 > y · x1 and, applying M5 and O4 again yields x > y, another contradiction. If x < 0 and y < 0, we argue as above, only replacing Axiom O4 with Theorem 11.1 (f). Part (c). If x < 0, then from Theorem 11.1 (f) (taking y = 0 and z = x), we have x2 = x · x > x · 0 = 0 (where we have used Theorem 11.1 (b)). If x > 0, then from O4 and Theorem 11.1 (b), x2 = x · x > x · 0 = 0. 3 5 Prove that there is no n ∈ N such that 0 < n < 1 (Use the WellOrdering Property of N). Solution The existence of an integer in (0, 1) contradicts the fact that N has a least element, namely 1.