8. Find Icm(143, 227 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):
8. Find Icm(143, 227 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):
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.7: Problem Solving: Consecutive Integers
Problem 14OE
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Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
Topic Video
Question
9
![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 foliewing:
(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 + b?) = 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 |.
(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 Icm(a, b):
(a) gcd(a, b)
(b) If k > 0, then Icm(ka, kb) = k lcm(a, b).
(c) If m is any common multiple of a and b, then lcm(a, b)|m.
[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
= lcm(a, b) if and only if a = ±b.
%3D
%3D
nod(ardla, 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%2F608621f2-91f9-445b-ad55-2861da4192da%2Fuz5pvom_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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 foliewing:
(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 + b?) = 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 |.
(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 Icm(a, b):
(a) gcd(a, b)
(b) If k > 0, then Icm(ka, kb) = k lcm(a, b).
(c) If m is any common multiple of a and b, then lcm(a, b)|m.
[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
= lcm(a, b) if and only if a = ±b.
%3D
%3D
nod(ardla, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b)
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