Homework 5 Solutions
Problem 1. Question 2.2 2 (a, f, g, j)
Solution
a.
A ∪ B = [2, 8)
f.
B − D = [2, 5]
g.
D − A = [8, ∞)
j.[5 points]
(A ∪ C) − (B ∩ D) = (1, 5] ∪ (6, 8)
Problem 2. Question 2.2 3(a, d)
Solution
a.[10 points]
A × B = {(1, a)(1, e)(1, k)(1, n)(1, r)(3, a)(3, e)(3, k)(3, n)(3, r)(5, a)(5, e)(5, k)(5, n)(5, r)}
B × A = {(a, 1)(e, 1)(k, 1)(n, 1)(r, 1)(a, 3)(e, 3)(k, 3)(n, 3)(r, 3)(a, 5)(e, 5)(k, 5)(n, 5)(r, 5)}
d.
A × B = {((2, 4)(4, 1))((3, 1)(2, 3))((2, 4)(2, 3))((3, 1)(4, 1))}
B × A = {((4, 1)(2, 4))((4, 1)(3, 1))((2, 3)(2, 4))((2, 3)(3, 1))}
Problem 3. Question 2.3 1 (d, k, n)
Solution
d.
S∞
Bn = N − {1}
Tn=1
∞
n=1 Bn = ∅
k.[5
S∞ points] 7
n=3 An = (0, 3 )
T∞
1
n=3 An = [ 3 , 2)
n.
S∞
Dn = (−∞, 1)
Tn=1
∞
n=1 Dn = (−1, 0]
1
Problem 4. Question 2.4 6 (b, d, e, h, k)
Solution
b. Prove 3 + 11 + 19 + ... + (8n − 5) = 4n2 − n, n ∈ N
Basic Step
8(1) − 5 = 3 and 4(1)2 − 1 = 3 so 1 ∈ S
Inductive Step
Assume 3 + 11 + 19 + ... + (8n − 5) = 4n2 − n
3 + 11 + 19 + ... + (8n − 5) + (8(n + 1) − 5) = 4n2 − n + (8(n + 1) − 5) = 4n2 − n + 8n + 3 =
4n2 + 7n + 3 = (4n − 3)(n + 1) = (n + 1)(4(n + 1) − 1) = 4(n + 1)2 − (k + 1) so n ∈ S implies
n+1∈S
d. [10 points] Prove 1 ∗ 1! + 2 ∗ 2! + 3 ∗ 3! + ... + n ∗ n! = (n + 1)! − 1, n ∈ N
Basic Step
1 ∗ 1! = 1 and (1 + 1)! − 1 = 2! − 1 = 1 so 1 ∈ S
Inductive Step
Assume 1 ∗ 1! + 2 ∗ 2! + 3 ∗ 3! + ... + n ∗ n! = (n + 1)! − 1
1 ∗ 1! + 2 ∗ 2! + 3 ∗ 3! + ... + n ∗ n! + (n + 1)(n + 1)! = (n + 1)! − 1 + (n + 1)(n + 1)! =
(n + 1)!(1 + n + 1) − 1 = (n + 1)!(n + 2) − 1 = (n + 2)! − 1 = ((n + 1) + 1)! − 1 so n ∈ S
implies n + 1 ∈ S
]2 , n ∈ N
e. Prove 13 + 23 + ... + n3 = [ n(n+1)
2
Basic Step
13 = 1 and [ 1(2)
]2 = 1 so 1 ∈ S
2
Inductive Step
Assume 13 + 23 + ... + n3 = [ n(n+1)
]2
2
2
2
3
]2 + (n + 1)2 = n (n+1)
+ 4(n+1)
=
13 + 23 + ... + n3 + (n + 1)2 = [ n(n+1)
2
4
4
(k+1)(k+2) 2
(k+1)2
2
(k + 2) = [
] so n ∈ S implies n + 1 ∈ S
4
2
(k+1)2
(k 2
4
+ 4k + 4) =
n
1
h. Prove 2!1 + 3!2 + ... + (n+1)!
= 1 − (n+1)!
,n ∈ N
Basic Step
1
= 12 and 1 − 2!1 = 12
2!
Inductive Step
n
1
Assume 2!1 + 3!2 + ... + (n+1)!
= 1 − (n+1)!
(n+1)
(n+1)
n
1
1
+ 3!2 + ... + (n+1)!
+ (n+2)!
= 1 − (n+1)!
+ (n+2)!
= 1 − n+2−n−1
= 1 − (n+2)!
so n ∈ S implies
(n+2)!
n+1∈S
1
2!
Q
k. [10 points] Prove ni=1 (2i − 1) = (2n)!
,n ∈ N
n!2n
Basic Step
(2(1) − 1) = 1 and (2(1))!
= 1 so 1 ∈ S
1!21
InductiveQStep
Assume ni=1 (2i − 1) = (2n)!
n!2n
Qn+1
Qn
(2i
−
1)
=
(2i
−
1)(2(n + 1) − 1) = (2n)!
(2n + 1) =
i=1
i=1
n!2n
n ∈ S implies n + 1 ∈ S
2
(2n)!(2n+1)(2n+2)
n!(n+1)2n+1
=
(2(n+1))!
(n+1)!2( n+1)
so
Problem 5. Question 2.4 7(a, c, f, g, j)
Solution
a. Prove n3 + 5n + 6 is divisible by 3
Basic Step
13 + 5 + 6 = 12 and 12 is divisible by 3 so 1 ∈ S
Inductive Step
Assume n3 + 5n + 6 = 3k, k ∈ Z
(n+1)3 +5(n+1)+6 = (n2 +2n+1)(n+1)+5n+5+6 = n3 +2n2 +n+n2 +2n+1+5n+5+6 =
n3 + 3n2 + 8n + 6 + 6 = (n3 + 5n + 6) + (3n2 + 3n + 6) = 3k + 3(n2 + n + 2) = 3(k + l), l, k ∈ Z
so n ∈ S implies n + 1 ∈ S
c. Prove n3 − n is divisible by 6
Basic Step
13 − 1 = 0 and 0 is divisible by 6 so 1 ∈ S
Inductive Step
Assume n3 − n = 6k, k ∈ Z
(n+1)3 −(n+1) = n3 +2n2 +n+n2 +2n+1−n−1 = (n3 −n)+3n2 +3n = 6k +3(n(n+1)) =
6k + 3(2l) = 6(k + l), k, l ∈ Zso n ∈ S implies n + 1 ∈ S
f. [10 points] Prove 10n+1 + 3 ∗ 4n−1 + 5 is divisible by 9
Basic Step
102 + 3 + 5 = 108 and 108 is divisible by 9 so 1 ∈ S
Inductive Step
Assume 10n+1 + 3 ∗ 4n−1 + 5 = 9k, k ∈ Z
10(n+1)+1 + 3 ∗ 4(n+1)−1 + 5 = 10n+2 + 3 ∗ 4n + 5 = 10(10n+1 + 3 ∗ 4n−1 + 5) − 6 ∗ 3 ∗ 4n−1 − 45 =
10(9k) − 9(2 ∗ 4n−1 + 5) = 10(9k) − 9l, k, l ∈ Z so n ∈ S implies n + 1 ∈ S
g. Prove 8 divides 9n − 1
Basic Step
91 − 1 = 9 − 1 = 8 and 8 is divisible by 8 so 1 ∈ S
Inductive Step
Assume 9n − 1 = 8k, k ∈ Z
9n+1 − 1 = 9(9n ) − 1 = (8 + 1)9n − 1 = 8 ∗ 9n + 9n − 1 = 8(9n ) − 8k = 8l − 8kl, k ∈ Z so
n ∈ S implies n + 1 ∈ S
j. [10 points] Prove 4n+4 > (n + 4)4
Basic Step
45 = 1024 > 54 = 625 so 1 ∈ S
Inductive Step
Assume 4n+4 > (n + 4)4
4(n+1)+4 = 4n+5 = 4 ∗ 4n+4 > 4 ∗ (n + 4)4 = 4n4 + 64n3 + 256n2 + 1024n + 1024 >
n2 + 20n3 + 100n2 + 500n + 625 = (n + 5)4 so n ∈ S implies n + 1 ∈ S
Problem 6. Question 2.4 8 (b, d)
3
Solution
b. [10 points] Prove 2n > n2
Basic Step
25 = 32 > 52 = 25 so 1 ∈ S
Inductive Step
Assume 2n > n2
2n+1 = 2n ∗ 2 > 2(n2 ) = n2 + n2 + 2k + 1 − 2k − 1 = (k + 1)2 (k − 1)2 > (k + 1)2 so n ∈ S
implies n + 1 ∈ S
d. Prove n! > 3n
Basic Step
4! = 24 > 3(4) = 12 so 1 ∈ S
Inductive Step
Assume n! > 3n
(n + 1)! = n!(n + 1) > 3n(n + 1) > 3(n + 1) so n ∈ S implies n + 1 ∈ S
4