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em: For every integer k, if k > 0 then k² + 2k + 1 is composite. Suppose k is any integer such that k > 0. If k² + 2k + 1 is composite, then k2 + 2k + 1 1

em: For every integer k, if k > 0 then k² + 2k + 1 is composite. Suppose k is any integer such that k > 0. If k² + 2k + 1 is composite, then k2 + 2k + 1 1

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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Theorem: For every integer k, if k > 0 then k² + 2k + 1 is composite. "Proof: Suppose k is any integer such that k > 0. If k² + 2k + 1 is composite, then k² + 2k + 1 = rs for some integers r and s such that 1<r < k² + 2k +1 and 1< s < k² + 2k + 1. Since k² + 2k +1 = rs and both r and s are strictly between 1 and k² + 2k + 1, then k² + 2k + 1 is not prime. Hence k2 + 2k + 1 is composite as was to be shown." In the above: it is correct because of boundedness O it is incorrect because it assumes what is to be proved it is incorrect because the definitions are incorrect it is correct because of deMorgan's law
Transcribed Image Text:Theorem: For every integer k, if k > 0 then k² + 2k + 1 is composite. "Proof: Suppose k is any integer such that k > 0. If k² + 2k + 1 is composite, then k² + 2k + 1 = rs for some integers r and s such that 1<r < k² + 2k +1 and 1< s < k² + 2k + 1. Since k² + 2k +1 = rs and both r and s are strictly between 1 and k² + 2k + 1, then k² + 2k + 1 is not prime. Hence k2 + 2k + 1 is composite as was to be shown." In the above: it is correct because of boundedness O it is incorrect because it assumes what is to be proved it is incorrect because the definitions are incorrect it is correct because of deMorgan's law
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