4. Assuming that gcd(a, b) = 1, prove the following: (a) gcd(a +b, a – b) = 1 or 2. %3D %3D
4. Assuming that gcd(a, b) = 1, prove the following: (a) gcd(a +b, a – b) = 1 or 2. %3D %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Topic Video
Question
4a,8,9
![nples:
ers
or and lea
gcd(39, 42, 54) = 3
and
gcd(49, 210, 350) = 7
The reader is cautioned that it is possible for three integers to be relatively prime as
a triple (in other words, gcd(a, b, c) = 1), yet not relatively prime in pairs; this is
brought out by the integers 6, 10, and 15.
PROBLEMS 2.4
1. Find gcd(143, 227), gcd(306, 657), and gcd(272, 1479).
2. Use the Euclidean Algorithm to obtain integers x and y satisfying the following:
(a) gcd(56, 72) = 56x + 72y.
(b) gcd(24, 138) = 24x + 138y.
(c) gcd(119, 272) = 119x + 272y.
(d) gcd(1769, 2378) = 1769x + 2378y.
3. Prove that if d is a common divisor of a and b, then d = gcd(a, b) if and only if
gcd(a/d, b/d) = 1.
[Hint: Use Theorem 2.7.]
4. Assuming that gcd(a, b) = 1, prove the folowing:
(a) gcd(a +b, a – b) = 1 or 2.
[Hint: Let d = gcd(a + b, a – b) and show that d|2a, d| 2b, and thus that
d< gcd(2a, 2b) = 2 gcd(a, b).]
(b) gcd(2a + b, a + 2b) = 1 or 3.
(c) gcd(a +b, a² + b²) = 1 or 2.
[Hint: a? + b² = (a + b)(a – b) + 2b².]
(d) gcd(a +b, a² – ab + b²) = 1 or 3.
[Hint: a? - ab +b² = (a + b)² – 3ab.]
5. For n > 1, and positive integers a, b, show the following:
(a) If gcd(a, b) = 1, then gcd(a", b")= 1.
[Hint: See Problem 20(a), Section 2.2.]
(b) The relation a" | b" implies that a | b.
[Hint: Put d = gcd(a, b) and write a = rd,b = sd, where gcd(r, s) = 1. By part (a),
gcd(r", s") = 1. Show that r = 1, whence a = d.]
6. Prove that if gcd(a, b) = 1, then gcd(a +b, ab) = 1.
7. For nonzero integers a and b, verify that the following conditions are equivalent:
(a) a |b.
(b) gcd(a, b) =|a l.
(c) lcm(a, b) = |b|.
8. Find lcm(143, 227), lcm(306, 657), and lcm(272, 1479).
9. Prove that the greatest common divisor of two positive integers divides their least common
multiple.
10. Given nonzero integers a and b, establish the following facts concerning lcm(a, b):
(a) gcd(a, b) = lcm(a, b) if and only if a = ±b.
(b) If k > 0, then lcm(ka, kb) = k lcm(a, b).
(c) If m is any common multiple of a and b, then lcm(a, b) | m.
and s. If
common
1 b; say,
atisfying
rdance
y if
he
st
n.
[Hint: Put t = lcm(a, b) and use the Division Algorithm to write m = qt +r, where
0 <r <t. Show that r is a common multiple of a and b.]
11. Let a, b, c be integers, no two of which are zero, and d = gcd(a, b, c). Show that
d = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39c12f77-fd5e-41ae-b73a-621f9f2bab30%2Fbe8d205e-ee9e-4592-9f27-76657034317e%2Fhf7e53f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:nples:
ers
or and lea
gcd(39, 42, 54) = 3
and
gcd(49, 210, 350) = 7
The reader is cautioned that it is possible for three integers to be relatively prime as
a triple (in other words, gcd(a, b, c) = 1), yet not relatively prime in pairs; this is
brought out by the integers 6, 10, and 15.
PROBLEMS 2.4
1. Find gcd(143, 227), gcd(306, 657), and gcd(272, 1479).
2. Use the Euclidean Algorithm to obtain integers x and y satisfying the following:
(a) gcd(56, 72) = 56x + 72y.
(b) gcd(24, 138) = 24x + 138y.
(c) gcd(119, 272) = 119x + 272y.
(d) gcd(1769, 2378) = 1769x + 2378y.
3. Prove that if d is a common divisor of a and b, then d = gcd(a, b) if and only if
gcd(a/d, b/d) = 1.
[Hint: Use Theorem 2.7.]
4. Assuming that gcd(a, b) = 1, prove the folowing:
(a) gcd(a +b, a – b) = 1 or 2.
[Hint: Let d = gcd(a + b, a – b) and show that d|2a, d| 2b, and thus that
d< gcd(2a, 2b) = 2 gcd(a, b).]
(b) gcd(2a + b, a + 2b) = 1 or 3.
(c) gcd(a +b, a² + b²) = 1 or 2.
[Hint: a? + b² = (a + b)(a – b) + 2b².]
(d) gcd(a +b, a² – ab + b²) = 1 or 3.
[Hint: a? - ab +b² = (a + b)² – 3ab.]
5. For n > 1, and positive integers a, b, show the following:
(a) If gcd(a, b) = 1, then gcd(a", b")= 1.
[Hint: See Problem 20(a), Section 2.2.]
(b) The relation a" | b" implies that a | b.
[Hint: Put d = gcd(a, b) and write a = rd,b = sd, where gcd(r, s) = 1. By part (a),
gcd(r", s") = 1. Show that r = 1, whence a = d.]
6. Prove that if gcd(a, b) = 1, then gcd(a +b, ab) = 1.
7. For nonzero integers a and b, verify that the following conditions are equivalent:
(a) a |b.
(b) gcd(a, b) =|a l.
(c) lcm(a, b) = |b|.
8. Find lcm(143, 227), lcm(306, 657), and lcm(272, 1479).
9. Prove that the greatest common divisor of two positive integers divides their least common
multiple.
10. Given nonzero integers a and b, establish the following facts concerning lcm(a, b):
(a) gcd(a, b) = lcm(a, b) if and only if a = ±b.
(b) If k > 0, then lcm(ka, kb) = k lcm(a, b).
(c) If m is any common multiple of a and b, then lcm(a, b) | m.
and s. If
common
1 b; say,
atisfying
rdance
y if
he
st
n.
[Hint: Put t = lcm(a, b) and use the Division Algorithm to write m = qt +r, where
0 <r <t. Show that r is a common multiple of a and b.]
11. Let a, b, c be integers, no two of which are zero, and d = gcd(a, b, c). Show that
d = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b)
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