Let p be a fixed prime number. Use strong induction to prove the following: Proposition P (n) for n a positive integer: There exists a non-negative integer s and a positive integer m that is not divisible by p such that n = p°m. (a) Verify the Basis Step is True: Confirm that P (1) is true: (b) State the Inductive Hypothesis: For any positive integer k state that P (1), P (2), .. ., P (k) are true. Note: You need to state this using words and math symbols, for example "For any integer g such that 1 <= q <= k , there exists a non-negative integer s and...". Write out the whole sentence using words and math symbols. (c) Prove the Conclusion of the Inductive Step: For any positive integer k prove P (k + 1) is true assuming that P (1), P (2), ..., P (k) are true, by filling in the steps below: Case 1: k + 1 is not divisible by p (Note: If k + 1 is not divisible by p then there isn't much to do! This step is easy, and you don't need to use the inductive hypothesis here.) Case 2: k + 1 is divisible by p (Note: Let (k + 1) = pq where q is a positive integer. Then 1 <= q <= k. Apply %3D the inductive hypothesis to q. Write all of this out below and finish the argument to prove P(k+1).)

Let p be a fixed prime number. Use strong induction to prove the following: Proposition P (n) for n a positive integer: There exists a non-negative integer s and a positive integer m that is not divisible by p such that n = p°m. (a) Verify the Basis Step is True: Confirm that P (1) is true: (b) State the Inductive Hypothesis: For any positive integer k state that P (1), P (2), .. ., P (k) are true. Note: You need to state this using words and math symbols, for example "For any integer g such that 1 <= q <= k , there exists a non-negative integer s and...". Write out the whole sentence using words and math symbols. (c) Prove the Conclusion of the Inductive Step: For any positive integer k prove P (k + 1) is true assuming that P (1), P (2), ..., P (k) are true, by filling in the steps below: Case 1: k + 1 is not divisible by p (Note: If k + 1 is not divisible by p then there isn't much to do! This step is easy, and you don't need to use the inductive hypothesis here.) Case 2: k + 1 is divisible by p (Note: Let (k + 1) = pq where q is a positive integer. Then 1 <= q <= k. Apply %3D the inductive hypothesis to q. Write all of this out below and finish the argument to prove P(k+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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