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Basis step: (0,0) e S Recursive step: If (a, b) = S, then (a, b + 1) = S, (a + 1, b + 1) = S, and (a + 2, b + 1) € S. Jse strong induction on the number of applications of the recursive step of the definition to s D) E S. Let P(n) be the proposition that a ≤ 2b whenever (a, b) e S is obtained by n applications of the

Basis step: (0,0) e S Recursive step: If (a, b) = S, then (a, b + 1) = S, (a + 1, b + 1) = S, and (a + 2, b + 1) € S. Jse strong induction on the number of applications of the recursive step of the definition to s D) E S. Let P(n) be the proposition that a ≤ 2b whenever (a, b) e S is obtained by n applications of the

Advanced Engineering Mathematics
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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Let S be the subset of the set of ordered pairs of integers defined recursively by Basis step: (0,0) = S Recursive step: If (a, b) = S, then (a, b + 1) = S, (a + 1, b + 1) = S, and (a + 2, b + 1) = S. Use strong induction on the number of applications of the recursive step of the definition to show that a ≤ 2b whenever (a, b) E S. Let P(n) be the proposition that a ≤ 2b whenever (a, b) e S is obtained by n applications of the recursive step. Click and drag the steps given in the right to the corresponding step names given in the left to show the inductive steps to prove that P(n) is true. This pair was obtained as (a, b + 1) or (a + 1, b + 1) or (a + 2, b + 1) from some element (a, b), which was itself obtained in k or fewer steps. Consider an ordered pair in S obtained by k + 1 applications of the recursive step. This pair was obtained as (a, b + 1) or (a + 1, b + 1) or (a + 2, b + 1) from some element (a, b), which was itself obtained in k steps. By the inductive hypothesis, a ≤ 2b. Add the inequality 0 ≤ 2 to get a ≤ 2(b + 1), which implies that a + 1 ≤ 2(b + 1) and a + 2 ≤ 2(b + 1). By the inductive hypothesis, a ≤ 2b. For the case that the new pair is (a, b + 1), add the inequality 0 ≤ 2 to a ≤ 2b to get a ≤ 2(b + 1). For (a + 1, b + 1), add 1 ≤ 2 to a ≤ 2b to get a + 1 ≤2(b + 1). For (a + 2, b + 1), add 2 ≤2 to a ≤ 2b to get a + 2 ≤2(b + 1).
Transcribed Image Text:Let S be the subset of the set of ordered pairs of integers defined recursively by Basis step: (0,0) = S Recursive step: If (a, b) = S, then (a, b + 1) = S, (a + 1, b + 1) = S, and (a + 2, b + 1) = S. Use strong induction on the number of applications of the recursive step of the definition to show that a ≤ 2b whenever (a, b) E S. Let P(n) be the proposition that a ≤ 2b whenever (a, b) e S is obtained by n applications of the recursive step. Click and drag the steps given in the right to the corresponding step names given in the left to show the inductive steps to prove that P(n) is true. This pair was obtained as (a, b + 1) or (a + 1, b + 1) or (a + 2, b + 1) from some element (a, b), which was itself obtained in k or fewer steps. Consider an ordered pair in S obtained by k + 1 applications of the recursive step. This pair was obtained as (a, b + 1) or (a + 1, b + 1) or (a + 2, b + 1) from some element (a, b), which was itself obtained in k steps. By the inductive hypothesis, a ≤ 2b. Add the inequality 0 ≤ 2 to get a ≤ 2(b + 1), which implies that a + 1 ≤ 2(b + 1) and a + 2 ≤ 2(b + 1). By the inductive hypothesis, a ≤ 2b. For the case that the new pair is (a, b + 1), add the inequality 0 ≤ 2 to a ≤ 2b to get a ≤ 2(b + 1). For (a + 1, b + 1), add 1 ≤ 2 to a ≤ 2b to get a + 1 ≤2(b + 1). For (a + 2, b + 1), add 2 ≤2 to a ≤ 2b to get a + 2 ≤2(b + 1).
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