Download Math 109, Winter 2017 Homework 2 Solutions

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.
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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.
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