Math 201-B
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Homework #4 (solutions)
1. Exercise 1 from Chapter 8 of the textbook BP.
Solution: Let A = {12n : n ∈ Z}, B = {2n : n ∈ Z}, C = {3n : n ∈ Z}. We want to show
that A ⊆ A ∩ B, that is
A⊆B
and
A ⊆ C.
Let x ∈ A. Then x = 12n for some n ∈ N. Let k = 6n and m = 4n. Then
x = 2k
and
x = 3m,
and hence x ∈ B ∩ C. The proof is completed.
2. Exercise 20 from Chapter 8 of the textbook BP.
Solution: Let A = {9n : n ∈ Q} and B = {3n : n ∈ Q}. We want to show that A = B, that
is
A⊆B
and
B ⊆ A.
First, let x ∈ A. Then x = 9n for some n ∈ Q. Let k = 2n. Then k ∈ Q and
x = (32 )n = 3k ,
and hence x ∈ B. Thus A ⊆ B.
Similarly, pick any x ∈ B. Then x = 3n for some n ∈ Q. Let k = n/2. Then k ∈ Q and
n
x = 91/2 = 9k ,
and hence x ∈ A. Thus B ⊆ A. The proof is completed.
3. Exercise 28 from Chapter 8 of the textbook BP.
Solution: We want to show that any n ∈ Z can be represented in the form
n = 12a + 25b
with suitable a, b ∈ Z. Toward this end observe that any n ∈ Z can be written in the form
n = 12k + r,
k ∈ Z, r ∈ {0, 1, 2, . . . , 11},
1
where r is the remainder after division of n by 12. To get from this the desired result one
can set, for instance,
b=r
a = k − 2b,
and
so that a + 2b = k, b = r, and hence 12a + 25b = 12(a + 2b) + b = 12k + r = n.
4-6. Exercises 1, 5, 11 from Chapter 9 of the textbook BP.
Solution: Please see solutions at the end of the textbook BP.
7. Exercise 30 from Chapter 9 of the textbook BP.
Solution: False. Indeed,
1 = 42a + 7b = 7(6a + b)
for some a, b ∈ Z, would imply that 1 divides 7.
8-11. Exercises 3, 7, 11, 21 from Chapter 10 of the textbook BP.
Solution: Please see solutions at the end of the textbook BP.
12. Exercise 22 from Chapter 10 of the textbook BP.
Solution: For n ∈ N, let
1 1 1
1
1 1−
1−
1−
··· 1 − n
Sn = 1 −
2
4
8
16
2
and
an =
1
1
+ n+1 .
4 2
We want to show by induction that Sn ≥ an ∀ n ∈ N.
Base of induction: For n = 1, we have Sn = 1 − 12 = 12 and an = 14 + 41 = 12 . Thus S1 = a1 .
Step of induction: Assume Sn ≥ an (induction hypothesis). In order to establish that
Sn+1 ≥ an+1 it siffices to show that
Sn+1
an+1
≥
,
Sn
an
which is equivalent to
1−
1
2n+1
≥
1
4
1
4
+
+
1
2n+2
1
2n+1
.
One can rewrite (1) as follows:
1 1
1 1
1
1 − n+1
+ n+1 ≥ + n+2 .
2
4 2
4 2
2
(1)
Open the brackets on the left-hand side of teh above equation to obtain
1
1
1
1
1
1
+ n+1 − n+3 − 2n+2 ≥ + n+2 .
4 2
2
2
4 2
This is equivalent to
1
2n+1
−
1
2n+3
−
1
22n+2
−
1
2n+2
≥0
which holds true if and only if (multiply both the sides by 22n+2 )
2n+3 − 2n−1 − 1 − 2n ≥ 0.
Since 2n+3 = 2 × 2n+2 = 8 × 2n−1 + 4 × 2n , the latter inequality holds true and so does (1).
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