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. So by definition of even, t is even. By substitution and algebra, m + n = 2r + (2r + 1) = 2(2r) + 1. By substitution, m + n = 2t + 1. 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. Let t = r + s. Then t is an integer because it is a sum of integers. Let m + n be any odd integer. Suppose m is any even integer and n is any odd integer. By definition of even and odd, there are integers r and s such that m = 2r and n = 2s + 1. By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1. Hence, m + n is two times an integer plus one. So by definition of odd, m + n is odd. Proof: 1. Suppose m is any even integer and n is any odd integer. 2. By definition of even and odd, there are integers r and s such that m = 2r and n = 2s +1. 3. So by definition of even, t is even. 4. Let t = 2r. Then t is an integer because it is a product of integers. 5. Let m + n be any odd integer. 6. By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1. X
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. So by definition of even, t is even. By substitution and algebra, m + n = 2r + (2r + 1) = 2(2r) + 1. By substitution, m + n = 2t + 1. 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. Let t = r + s. Then t is an integer because it is a sum of integers. Let m + n be any odd integer. Suppose m is any even integer and n is any odd integer. By definition of even and odd, there are integers r and s such that m = 2r and n = 2s + 1. By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1. Hence, m + n is two times an integer plus one. So by definition of odd, m + n is odd. Proof: 1. Suppose m is any even integer and n is any odd integer. 2. By definition of even and odd, there are integers r and s such that m = 2r and n = 2s +1. 3. So by definition of even, t is even. 4. Let t = 2r. Then t is an integer because it is a product of integers. 5. Let m + n be any odd integer. 6. By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1. X
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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im not sure where i went wrong. if possible, what did i do wrong.
Transcribed Image Text: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.
So by definition of even, t is even.
By substitution and algebra, m + n = 2r + (2r + 1) = 2(2r) + 1.
By substitution, m + n = 2t + 1.
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.
Let t = r + s. Then t is an integer because it is a sum of integers.
Let m + n be any odd integer.
Suppose m is any even integer and n is any odd integer.
By definition of even and odd, there are integers r and s such that m = 2r and n = 2s + 1.
By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1.
Hence, m + n is two times an integer plus one. So by definition of odd, m + n is odd.
Proof:
1. Suppose m is any even integer and n is any odd integer.
2. By definition of even and odd, there are integers r ands such that m = 2r and n = 2s +1.
3. So by definition of even, t is even.
4. Let t = 2r. Then t is an integer because it is a product of integers.
5. Let m+n be any odd integer.
6. By substitution and algebra, m + n = 2r + (2s + 1) = 2(r + s) + 1.
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