Consider the following theorem. Theorem: The sum of any even integer and any odd integer is odd. Construct a proof for the theorem by selecting sentences from the following scrambled list and putting them in the correct order.Let m + n be any odd integer.By substitution and algebra, m + n = 2r + (2r + 1) = 2(2r) + 1.Suppose m is any even integer and n is any odd integer.Hence, m + n is two times an integer plus one. So by definition of odd, m + n is odd.Let t = r + s. Then t is an integer because it is a sum of integers.By substitution, m + n = 2t + 1.By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1.By definition of even and odd, there are integers r and s such that m = 2r and n = 2s + 1.So by definition of even, t is even.By definition of even and odd, there is an integer r such that m = 2r and n = 2r + 1.Let t = 2r. Then t is an integer because it is a product of integers.

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Description: Consider the following theorem. Theorem: The sum of any even integer and any odd integer is odd. Construct a proof for the theorem by selecting sentences from the following scrambled list and putting them in the correct order.Let m + n be any odd int

We have to prove the theorem by scrambling the given statements.

Answered: The sum of any even integer and any odd…

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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Consider the following theorem.

Theorem: The sum of any even integer and any odd integer is odd.

Construct a proof for the theorem by selecting sentences from the following scrambled list and putting them in the correct order.

Let m + n be any odd integer.
By substitution and algebra, m + n = 2r + (2r + 1) = 2(2r) + 1.
Suppose m is any even integer and n is any odd integer.
Hence, m + n is two times an integer plus one. So by definition of odd, m + n is odd.
Let t = r + s. Then t is an integer because it is a sum of integers.
By substitution, m + n = 2t + 1.
By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1.
By definition of even and odd, there are integers r and s such that m = 2r and n = 2s + 1.
So by definition of even, t is even.
By definition of even and odd, there is an integer r such that m = 2r and n = 2r + 1.
Let t = 2r. Then t is an integer because it is a product of integers.

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