Evaluate the proposed proof of the following result. Choose the most complete answer.Theorem: Let m,n €Z. If n is odd and m is odd, then nm is odd.Proof. Assume that n is odd and m is odd. Then there exists somekEZ where n = 2k + 1and m=2k + 1. Then nm = (2k + 1)(2k + 1) = 2(k² + 2k) + 1. Since k2 + 2k is an integer we know that nm is odd.O A.The theorem is correct, but the proof is incorrect.В.The theorem and proof are incorrect.C. The theorem and the proof are correct.D.The proof is correct but the theorem is incorrect.

we know that product of two odd integers is always odd so, given theorem is correct now, assuming…

Evaluate the proposed proof of the following result. Choose the most complete answer. Theorem: Let m,n €Z. If n is odd and m is odd, then nm is odd. Proof. Assume that n is odd and m is odd. Then there exists some k EZ where n=2k + 1 and m=2k + 1. Then nm = (2k + 1)(2k + 1) = 2(k² + 2k)+ 1. Since k² + 2k is an integer we know that nm is odd. А. O A. The theorem is correct, but the proof is incorrect. В. The theorem and proof are incorrect. C. The theorem and the proof are correct. D. The proof is correct but the theorem is incorrect.

Evaluate the proposed proof of the following result. Choose the most complete answer. Theorem: Let m,n €Z. If n is odd and m is odd, then nm is odd. Proof. Assume that n is odd and m is odd. Then there exists some k EZ where n=2k + 1 and m=2k + 1. Then nm = (2k + 1)(2k + 1) = 2(k² + 2k)+ 1. Since k² + 2k is an integer we know that nm is odd. А. O A. The theorem is correct, but the proof is incorrect. В. The theorem and proof are incorrect. C. The theorem and the proof are correct. D. The proof is correct but the theorem is incorrect.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.4: Mathematical Induction

Problem 41E

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