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Define a set S of integers recursively as follows. Base: 0 ∈ S Recursion: If k ∈ S, then II(a). k + 3 ∈ S II(b). k − 3 ∈ S Restriction: Nothing is in S other than objects defined in I and II above. Use structural induction to prove that every integer in S is divisible by 3. Proof (by structural induction): Let property P(n) be the sentence, "n is divisible by 3." Continue the proof by selecting sentences from the following scrambled list and putting them in the correct order.

Define a set S of integers recursively as follows. Base: 0 ∈ S Recursion: If k ∈ S, then II(a). k + 3 ∈ S II(b). k − 3 ∈ S Restriction: Nothing is in S other than objects defined in I and II above. Use structural induction to prove that every integer in S is divisible by 3. Proof (by structural induction): Let property P(n) be the sentence, "n is divisible by 3." Continue the proof by selecting sentences from the following scrambled list and putting them in the correct order.

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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Define a set S of integers recursively as follows. Base: 0 ∈ S Recursion: If k ∈ S, then II(a). k + 3 ∈ S II(b). k − 3 ∈ S Restriction: Nothing is in S other than objects defined in I and II above. Use structural induction to prove that every integer in S is divisible by 3. Proof (by structural induction): Let property P(n) be the sentence, "n is divisible by 3." Continue the proof by selecting sentences from the following scrambled list and putting them in the correct order.

 

Proof (by structural induction): Let property P(n) be the sentence, "n is divisible by 3." Continue the proof by selecting sentences from the following scrambled list and putting them in the correct order. Let n be any integer in S and suppose that P(n) is true. In other words, suppose that n is divisible by 3. Since n is in S, applying rules II(a) and II(b) to n gives that n + 3 and n − 3 are in S, and since n is divisible by 3, both n + 3 and n - 3 are divisible by 3. Since P(n + 3) is true then we can conclude P(n) is true for any integer n. The only object in the base is 0 and 0 is divisible by 3. Suppose n is any integer in S such that P(n + 3) and P(n − 3) are true. Since II(a) and II(b) are the only rules in the recursion for S, every integer in S that is obtained from the base and recursion for S is divisible by 3. Because there are no integers in S other than those from the base and the recursion for S, we conclude that all the integers in S are divisible by 3. Show that P(n) is true for each integer n in the base for S: 1. Let n be any integer in S and suppose that P(n) is true. In other words, suppose that n is divisible by 3. Thus P(n) is true for each integer in the base for S. Show that for each integer n in S, if P(n) is true and if m is obtained from n by applying a rule from the recursion for S, then P(m) is true: ✓x 2. Suppose n is any integer in S such that P(n + 3) and P(n − 3) are true. 3. Since n is in S, applying rules Il(a) and II(b) to n gives that n + 3 and n - 3 are in S, and since n is divisible by 3, both n + 3 and n-3 are divisible by 3. ✓ 4. Hence both P(n + 3) and P(n − 3) are true. 5. Since P(n + 3) is true then we can conclude P(n) is true for any integer n. Conclusion: Since Il(a) and II(b) are the only rules in the recursion for S, every integer in S that is obtained from the base and recursion for S is divisible by 3. Need Help? Read It X
Transcribed Image Text:Proof (by structural induction): Let property P(n) be the sentence, "n is divisible by 3." Continue the proof by selecting sentences from the following scrambled list and putting them in the correct order. Let n be any integer in S and suppose that P(n) is true. In other words, suppose that n is divisible by 3. Since n is in S, applying rules II(a) and II(b) to n gives that n + 3 and n − 3 are in S, and since n is divisible by 3, both n + 3 and n - 3 are divisible by 3. Since P(n + 3) is true then we can conclude P(n) is true for any integer n. The only object in the base is 0 and 0 is divisible by 3. Suppose n is any integer in S such that P(n + 3) and P(n − 3) are true. Since II(a) and II(b) are the only rules in the recursion for S, every integer in S that is obtained from the base and recursion for S is divisible by 3. Because there are no integers in S other than those from the base and the recursion for S, we conclude that all the integers in S are divisible by 3. Show that P(n) is true for each integer n in the base for S: 1. Let n be any integer in S and suppose that P(n) is true. In other words, suppose that n is divisible by 3. Thus P(n) is true for each integer in the base for S. Show that for each integer n in S, if P(n) is true and if m is obtained from n by applying a rule from the recursion for S, then P(m) is true: ✓x 2. Suppose n is any integer in S such that P(n + 3) and P(n − 3) are true. 3. Since n is in S, applying rules Il(a) and II(b) to n gives that n + 3 and n - 3 are in S, and since n is divisible by 3, both n + 3 and n-3 are divisible by 3. ✓ 4. Hence both P(n + 3) and P(n − 3) are true. 5. Since P(n + 3) is true then we can conclude P(n) is true for any integer n. Conclusion: Since Il(a) and II(b) are the only rules in the recursion for S, every integer in S that is obtained from the base and recursion for S is divisible by 3. Need Help? Read It X
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