Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Math 109, Winter 2017 Homework 2 Solutions A. Problems I 8. (i) First, suppose that x = −1 or x = 2. Then x + 1 = 0 or x − 2 = 0 and so x2 − x − 2 = (x + 1)(x − 2) = 0 by Proposition 4.4.1 in Eccles. Now, assume that x2 − x − 2 = 0. Suppose that x 6= −1, so x + 1 6= 0. Since 0 = x2 − x − 2 = (x + 1)(x − 2), we get that x − 2 = 0 by Proposition 4.4.1. Thus x = 2 as desired. (ii) First, assume that x < −1 or x > 2. Case 1. Suppose that x < −1. Then x + 1 < −1 + 1 = 0, and x − 2 < −1 − 2 = −3 < 0. Since x + 1 < 0 and x − 2 < 0, it follows that x2 − x − 2 = (x + 1)(x − 2) > 0 by Axiom 3.1.2.(iii) in Eccles. Case 2. Suppose that x > 2. Then x + 1 > 2 + 1 = 3 > 0, and x − 2 > 2 − 2 = 0. Since x + 1 > 0 and x − 2 > 0, it follows that x2 − x − 2 = (x + 1)(x − 2) > 0 by Axiom 3.1.2.(iii). Hence, in both cases we have that x2 − x − 2 > 0. Now, assume that x2 − x − 2 > 0. Suppose that x ≥ −1, so that x + 1 ≥ 0. Note, however, that x + 1 6= 0 since (x + 1)(x − 2) = x2 − x − 2 6= 0. Hence x + 1 > 0. Since (x + 1)(x − 2) > 0, we have that x − 2 > 0 by Axiom 3.1.2.(iii). Thus, x > 2 as desired. 9. By way of contradiction, suppose there is a largest integer, call it n. Since n + 1 is an integer, we have that n ≥ n + 1. Subtracting n from both sides of this inequality gives 0 ≥ 1, a contradiction. Hence, our assumption that there is a largest integer must be false, that is, there does not exist a largest integer. 11. By way of contradiction, suppose there is a smallest positive real number, call it x. Since x2 is an positive real number, we have that x ≤ x2 . Since x > 0, dividing both sides of the inequality by x gives 1 ≤ 21 , a contradiction. Hence, our assumption that there is a smallest positive real number must be false, that is, there does not exist a smallest positive real number. 12. We will use induction on n. For the base case, note that 41 + 5 = 9, which is divisible by 3. For the induction step, let k be a given positive integer and assume 3 divides 4k + 5. Then, by definition, there exists ` ∈ Z such that 3` = 4k + 5, so that 4k = 3` − 5. Then 4k+1 + 5 = 4(4k ) + 5 = 4(3` − 5) + 5 = 12` − 20 + 5 = 3(4` − 5). Since 4` − 5 is an integer, this shows that 3 divides 4k+1 + 5. Hence, 3 divides 4n + 5 for all positive integers n by induction. 1 13. We will use induction on n. For the base case, note that 4! = 24 > 16 = 24 . For the induction step, let k be an integer with k ≥ 4 such that k! > 2k . Then (k + 1)! = (k + 1)k! > (k + 1)2k k >2·2 k+1 =2 by the induction hypothesis since k ≥ 4 . Hence, (k + 1)! > 2k+1 . Thus, we have that n! > 2n for all positive integers n ≥ 4 by induction. B. By way of contradiction, suppose there exist positive real numbers a and b such that 2 1 1 . Note that a+b 2ab 2 2 1 1 = a+b = a + b , + a b ab so √ √ ab < 2ab . a+b (1) √ a + b < 2 ab, (2) ab < √ Since ab and a + b are both positive (because a and b are both positive), we can multiply both a+b sides of (1) by √ without changing the direction of the inequality to get ab √ √ √ 2 √ and therefore a + b − 2 ab < 0. However, we also have that a + b − √ 2 ab =2 ( a − b) ≥ 0 by Proposition 3.1.4 in Eccles, a contradiction. Thus, it follows that ab ≥ 1 + 1 for all positive a b real numbers a and b. C. (i) Let `, m, n be integers such that `|m and `|n. Then, by definition, there exist integers r and s such that m = `r and n = `s. Then 2m + 5n = 2(`r) + 5(`s) = `(2r + 5s). Since 2r + 5s is an integer, this shows that `|2m + 5n. (ii) Let `, m, n be integers such that `|m and `|n. Then, by definition, there exist integers r and s such that m = `r and n = `s. Then am + bn = a(`r) + b(`s) = `(ar + bs). Since ar + bs is an integer, this shows that `|am + bn. 2