Result For every integer n ≥ 7, there exist positive integers a and b such that n = 2a +3b. 2q or n = 2q + 1 Proof Let n be an integer such that n ≥ 7. Since n is even or odd, either n = for some integer q. We consider these two cases. Case 1. n = 2q. Since n ≥ 7, it follows that q≥ 4. Thus, n = 2q = 2(q −3)+6=2(q −3) +3.2. Since q≥ 4, it follows that q - 3 € N. Case 2. n = 2q + 1. Since n ≥ 7, it follows that q ≥ 3. Thus, n = 2q+1 = 2(q − 1) +2+1 = 2(q − 1) + 3.1. Since q3, it follows that q- 1 € N.
Result For every integer n ≥ 7, there exist positive integers a and b such that n = 2a +3b. 2q or n = 2q + 1 Proof Let n be an integer such that n ≥ 7. Since n is even or odd, either n = for some integer q. We consider these two cases. Case 1. n = 2q. Since n ≥ 7, it follows that q≥ 4. Thus, n = 2q = 2(q −3)+6=2(q −3) +3.2. Since q≥ 4, it follows that q - 3 € N. Case 2. n = 2q + 1. Since n ≥ 7, it follows that q ≥ 3. Thus, n = 2q+1 = 2(q − 1) +2+1 = 2(q − 1) + 3.1. Since q3, it follows that q- 1 € N.
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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Please explain the following example in more detail, I especially dont understand the yellow underlined parts.
Transcribed Image Text:Result
For every integer n ≥ 7, there exist positive integers a and b such that n = 2a + 3b.
2q or n = 2q + 1
Proof Let n be an integer such that n ≥ 7. Since n is even or odd, either n =
for some integer q. We consider these two cases.
Case 1. n = 2q. Since n ≥ 7, it follows that q ≥ 4. Thus,
n = 2q = 2(q −3) +6=2(q − 3) +3.2.
Since q≥ 4, it follows that q- 3 € N.
Case 2. n = 2q + 1. Since n ≥ 7, it follows that q ≥ 3. Thus,
n = 2q+ 1 = 2(q − 1) + 2 + 1 = 2(q − 1) + 3 · 1.
Since q3, it follows that q - 1 € N.
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