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) e S, then (a, b + 1) e S, (a + 1, b + 1) e S, and (a + 2, b+1) € S. Use structural induction to show that as 2b whenever (a, b) e S. Which statements are required to show the recursive step? (Check all that apply.) Check All That Apply Suppose (a, b) satisfies as 2b. If (a, b) e S and a ≤ 2b, then adding the inequality Os2 gives a ≤2(b + 1). If (a, b) e Sand a ≤ 2b, then adding the inequality 1s2 gives a +1≤2(b + 1). If (a, b) e S and as 2b, then adding the inequality 2 s 2 gives a + 2 ≤ 2(b + 1). If (a, b) e Sand as 2b, then adding the inequality O s1 gives a ≤ 2b +1.
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) e S, then (a, b + 1) e S, (a + 1, b + 1) e S, and (a + 2, b+1) € S. Use structural induction to show that as 2b whenever (a, b) e S. Which statements are required to show the recursive step? (Check all that apply.) Check All That Apply Suppose (a, b) satisfies as 2b. If (a, b) e S and a ≤ 2b, then adding the inequality Os2 gives a ≤2(b + 1). If (a, b) e Sand a ≤ 2b, then adding the inequality 1s2 gives a +1≤2(b + 1). If (a, b) e S and as 2b, then adding the inequality 2 s 2 gives a + 2 ≤ 2(b + 1). If (a, b) e Sand as 2b, then adding the inequality O s1 gives a ≤ 2b +1.
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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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 structural induction to show that a ≤ 2b whenever (a, b) = S.
Which statements are required to show the recursive step? (Check all that apply.)
Check All That Apply
Suppose (a, b) satisfies a ≤ 2b.
If (a, b) e Sand a ≤ 2b, then adding the inequality 0 ≤ 2 gives a ≤ 2(b + 1).
If (a, b) e Sand a ≤ 2b, then adding the inequality 1 ≤ 2 gives a +1 ≤2(b + 1).
If (a, b) e S and a ≤ 2b, then adding the inequality 2 ≤ 2 gives a + 2 ≤2(b + 1).
If (a, b) e Sand a ≤ 2b, then adding the inequality 0 ≤ 1 gives a ≤ 2b + 1.
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