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Select each step to complete a proof by contrapositive of the theorem below. Theorem: For every pair of integers x and y, if 5xy + 8 is even, then at least one of x or y must be even. 1.    Let x and y be integers. We will assume that it is not true that x or y is even and will show that 5xy + 8 is odd. 2) 2A) If it is not true that x or y is even, then x and y are both odd 2B) Therefore 5xy + 8 = 5(2k + 1)(2j+1) + 8 2C) x = 2k + 1 and y = 2j + 1 for some integers k and j 3) 3A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1 3B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer 3C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8 3D) x = 2k + 1 and y = 2j + 1 for some integers k and j   4) 4A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1 4B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer 4C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8 4D) x = 2k + 1 and y = 2j + 1 for some integers k and j   5) 5A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1 5B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer 5C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8 5D) x = 2k + 1 and y = 2j + 1 for some integers k and j

Select each step to complete a proof by contrapositive of the theorem below. Theorem: For every pair of integers x and y, if 5xy + 8 is even, then at least one of x or y must be even. 1.    Let x and y be integers. We will assume that it is not true that x or y is even and will show that 5xy + 8 is odd. 2) 2A) If it is not true that x or y is even, then x and y are both odd 2B) Therefore 5xy + 8 = 5(2k + 1)(2j+1) + 8 2C) x = 2k + 1 and y = 2j + 1 for some integers k and j 3) 3A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1 3B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer 3C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8 3D) x = 2k + 1 and y = 2j + 1 for some integers k and j   4) 4A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1 4B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer 4C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8 4D) x = 2k + 1 and y = 2j + 1 for some integers k and j   5) 5A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1 5B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer 5C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8 5D) x = 2k + 1 and y = 2j + 1 for some integers k and j

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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Select each step to complete a proof by contrapositive of the theorem below.
Theorem: For every pair of integers x and y, if 5xy + 8 is even, then at least one of x or y must be even.
1.    Let x and y be integers. We will assume that it is not true that x or y is even and will show that 5xy + 8 is odd.
2)
2A) If it is not true that x or y is even, then x and y are both odd
2B) Therefore 5xy + 8 = 5(2k + 1)(2j+1) + 8
2C) x = 2k + 1 and y = 2j + 1 for some integers k and j

3)
3A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1
3B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer
3C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8
3D) x = 2k + 1 and y = 2j + 1 for some integers k and j
 
4)
4A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1
4B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer
4C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8
4D) x = 2k + 1 and y = 2j + 1 for some integers k and j
 
5)
5A) 5(2k + 1)(2j + 1) + 8 = 20jk + 10j + 10k + 5 + 8 = 2(10jk + 5j + 5k + 6) + 1
5B) Since j and k are both integers, 10jk + 5j + 5k + 6 is also an integer
5C) Therefore 5xy + 8 = 5(2k + 1)(2j + 1) + 8
5D) x = 2k + 1 and y = 2j + 1 for some integers k and j
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