Result Proof For each integer n ≥ 8, there are nonnegative integers a and b such that n = 3a + 5b. We proceed by induction. Since 8 = 3·1+5.1, the statement is true for n = 8. Assume for each integer i with 8 ≤ i ≤ k, where k ≥ 8 is an arbitrary integer, that there are non- negative integers s and † such that i = 3s + 5t. Consider the integer k + 1. We show that there are nonnegative integers x and y such that k + 1 = 3x + 5y. Since 9 = 3·3+5.0 and 10 = 3.0+5 ∙ 2, this is true if k + 1 = 9 and k + 1 = 10. Hence, we may assume that k + 1 ≥ 11. Thus, 8 ≤ (k + 1) − 3 < k. By the induction hypothesis, there are non- negative integers a and b such that (k+ 1) − 3 = 3a + 5b and so k + 1 = 3(a+1)+5b. Letting x = a + 1 and y = b, we have the desired conclusion. By the Strong Principle of Mathematical Induction, for every integer n ≥ 8, there are nonnegative integers a and b such that n = 3a + 5b.
Result Proof For each integer n ≥ 8, there are nonnegative integers a and b such that n = 3a + 5b. We proceed by induction. Since 8 = 3·1+5.1, the statement is true for n = 8. Assume for each integer i with 8 ≤ i ≤ k, where k ≥ 8 is an arbitrary integer, that there are non- negative integers s and † such that i = 3s + 5t. Consider the integer k + 1. We show that there are nonnegative integers x and y such that k + 1 = 3x + 5y. Since 9 = 3·3+5.0 and 10 = 3.0+5 ∙ 2, this is true if k + 1 = 9 and k + 1 = 10. Hence, we may assume that k + 1 ≥ 11. Thus, 8 ≤ (k + 1) − 3 < k. By the induction hypothesis, there are non- negative integers a and b such that (k+ 1) − 3 = 3a + 5b and so k + 1 = 3(a+1)+5b. Letting x = a + 1 and y = b, we have the desired conclusion. By the Strong Principle of Mathematical Induction, for every integer n ≥ 8, there are nonnegative integers a and b such that n = 3a + 5b.
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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Question
Please show the following proof in more detail, I dont get any part of it.
Transcribed Image Text:Result
For each integer n ≥ 8, there are nonnegative integers a and b such that n = 3a + 5b.
Proof We proceed by induction. Since 8 = 3·1+5.1, the statement is true for n = 8. Assume
for each integer i with 8 ≤ i ≤ k, where k ≥ 8 is an arbitrary integer, that there are non-
negative integers s and t such that i = 3s + 5t. Consider the integer k + 1. We show that
there are nonnegative integers x and y such that k + 1 = 3x + 5y. Since 9 = 3·3+5.0
and 10 = 3.0 +5.2, this is true if k + 1 = 9 and k + 1 = 10. Hence, we may assume
that k + 1 ≥ 11. Thus, 8 ≤ (k+ 1) − 3 < k. By the induction hypothesis, there are non-
negative integers a and b such that
(k+ 1)-3 = 3a + 5b and so k + 1 = 3(a + 1) + 5b.
Letting x = a + 1 and y = b, we have the desired conclusion.
By the Strong Principle of Mathematical Induction, for every integer n ≥ 8, there
are nonnegative integers a and b such that n = 3a + 5b.
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