Download CS 173 [A]: Discrete Structures, Fall 2012 Homework 2 Solutions

CS 173 [A]: Discrete Structures, Fall 2012
Homework 2 Solutions
Let’s make our life easier by proving a lemma about the divides relation. We’ll do it right
here at the start, because we’ll use it in three different places later.
Lemma 1: Let a, b, c ∈ Z. If a | b and a | c, then a | b + c.
Proof: Let a, b, c ∈ Z and suppose that a | b and a | c. Since a | b and a | c there are
integers k and j such that b = ak and c = aj. Then since b + c = ak + aj = a(k + j) and
k + j ∈ Z, it follows that a | b + c as claimed.
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Divisibility (8 points)
Recall the definition of divides: for integers a, b, a | b if and only if b = ak for some integer
k. Let a, b, c ∈ Z. Prove using the definition of divides that if a | b and a | c, then a | b2 +10c.
Solution: Let a, b, c be integers and suppose that a | b and a | c.
Since a | b, there is an integer k such that b = ak. So then b2 = a2 k 2 = a(ak 2 ). Since ak 2
(because a and k are integers), it follows that a | b2 .
Since a | c, there is an integer j such that c = aj. Then since 10c = a(10j). Since j is an
integer, so is 10j. So it follows that a | 10c.
By applying Lemma 1 to the results of the previous two paragraphs, it follows that
a | b2 + 10c as claimed.
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Congruence Modulo k (10 points)
Recall the definition of congruence mod k: for integers a, b, k where k ≥ 1, a ≡ b (mod k) if
and only if k | (a − b). Let a, b, c ∈ Z and k ∈ N. Use the definition of congruence mod k to
prove that if a ≡ b (mod k) and b ≡ c (mod k), then a ≡ c (mod k). Use the definition of
divides directly; do not use lemmas about divides from the book or lecture.
Solution: Let a, b, c be integers and let k be a natural number. Suppose that a ≡ b (mod k)
and b ≡ c (mod k).
a ≡ b (mod k) implies that k | (a − b).
b ≡ c (mod k) implies that k | (b − c).
By applying Lemma 1 to the two previous statements, it follows that k | [(a − b) + (b − c)].
So k | (a − c). And so a ≡ c (mod k).
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Divisibility and the GCD (15 points)
Part (a), 2 points:
Find integers m, n that are solutions to the equation ma + nb = 1 for the special case where
a = 5 and b = 13 (i.e. find integer solutions m and n to the equation 5m + 13n = 1).
Solution: Take m = −5 and n = 2. Then ma + nb = 5 · (−5) + 13 · 2 = −25 + 26 = 1.
(There are infinitely–many alternate solutions to the problem.)
Part (b), 10 points:
Let a, b, c ∈ Z such that all of a, b, c are non–zero. It is a fact that, if a, b ∈ Z are both
non–zero and gcd(a, b) = 1, then there are integers m, n such that ma + nb = 1. Use this
property of the GCD to prove that if gcd(a, b) = 1 and a | bc, then a | c.
Solution: Let a, b, c be non-zero integers, and suppose that gcd(a, b) = 1 and a | bc.
Since gcd(a, b) = 1, there are integers m and n such that ma + nb = 1 Multiplying this
equation by c, we get that mac + nbc = c (∗).
Since a | a, a | mac and since a | bc, a | nbc. So a divides each term in the sum on the
left–hand side of (∗), it follows by lemma 1 that a | c.
Part (c), 3 points:
Let a, b, c ∈ Z such that all of a, b, c are non–zero and gcd(a, b) > 1. Prove or give a counter
example to the claim that if a | bc, then a | c.
Counterexample: Take a = b = 3 and c = 2. Then bc = 6. So a | bc but a 6| c.
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Linear Congruences (8 points)
Definition: A linear congruence in the variable x is an equation of the form ax ≡ b (mod k)
where a, b, k ∈ Z, k ≥ 1. The solutions x to the congruence must be integers.
Part (a), 4 points:
Find all of the values x in the range 0 ≤ x < 15 satisfying the linear congruence 10x ≡ 5
(mod 15). Do all of these x also satisfy the linear congruence x ≡ 2 (mod 3)? Briefly justify
your answers.
Solution:
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Hint: Fill in the following table first (DO NOT submit this part of the homework):
x 10x remainder(10x, 15) remainder(10x, 15) = 5? (Yes / No)
0
0
0
N
1
10
10
N
2
20
5
Y
3
30
0
N
4
40
10
N
5
50
5
Y
6
60
0
N
7
70
10
N
8
80
5
Y
9
90
0
N
10 100
10
N
11 110
5
Y
12 120
0
N
13 130
10
N
14 140
5
Y
All solutions to the given linear congruence (unique modulo 15) are of the form x ≡
2, 5, 8, 11, 14 (mod 7). Each of these values for x also satisfies x ≡ 2 (mod 3) So our answer
is “yes.”
Part (b), 4 points:
Find all solutions x to the linear congruence 3x ≡ 2 (mod 7). Describe all possible values
of x in words and also using an equation of the form x ≡ b (mod 7) where b is an integer in
the range 0 ≤ b < 7. Briefly justify your answers.
Solution:
Hint: Fill in the following table first (DO NOT submit this part of the homework):
x 3x remainder(3x, 7) remainder(3x, 7) = 2 (Yes / No)
0 0
0
N
1 3
3
N
2 6
6
N
3 9
2
Y
4 12
5
N
5 15
1
N
6 18
4
N
3
This table shows one solution: 3. If we keep increasing x, this pattern of seven output values
will repeat. So any number of the form 3 + 7k (where k is an integer) is also a solution. So all
solutions to the given linear congruence (unique modulo 7) are of the form x ≡ 3 (mod 7).
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