Prompts Suomitted Answers Proposition: Let n be an integer. If n Sis odd, then nis Choose a match even. Proof Suppose that n+sis odd and nis also odd. By definition, then, there exists integers kand I so that n+5= 2k+1 and n= 21-1. Hence, we have 2k1=n+5 = (2- 17 -5 = 41-41-1+5 = 22 - 21- 31 Therefore, 2k+ 1seven. This is clearly impossible, and hence we cannot have that n +5is odd and nis also odd. Therefore, if that n²-sis odd, we must have nis even. Proposition: Let ab be integers. If ab is even, then at least Choose a match one of a orbis even. Proof Suppose that a and bare both odd. Then there are integers kand iso thata=2k+1 and b=21-1. Therefore, we have ab = (2k - 12- 1)= 4k+ 2k -2-1= 2zkl-k+-1, so ab is odd Thus, If ab is even, we must have at least one of a or bis Even. Proposition: Let a and b be integers. If a+bis even, then a Choose a match and bare either boch odd or bath even. Proof: Suppose that a and bare not both odd and nat both even, so that one of a and bis odd, and the other is even. Without lass of generality, suppose that a is odd andbis even. Then there are integers kand I such that a 2k 1 and b= 2L Therefore, a-b = (2k- 1)-21 = 2k +-1, so a+ bis odd. Hence, If a+ b is even, then a and bare either both odd or both even. Theorem: fa and bare consecutive integers, then the sum Choose a match a+bis odd. Proof Assume that a and bare consecutive integers. Assume also that the sum a-bis not odd. Because the suma-bis not odd, there exists no number k such that a+ b= 2k-1. However, the integers a and bare consecutive, so we may write the sum a -bas 20+ 1. Thus, we have derived that a bz k-1 for any integer kand also that a -b= 20-1. If we hoid that a andbare consecutive then we know that the sumab must be add. Theorem: The sum of an even integer and an odd integer is Choose a match odd. Proof: Suppose a is an even integer and bis an odd integer. Then, by our definitions of even and odd numbers, we know that integers mand n exist so that a 2m and b= 2n- 1. Then a+b= (2Zm) - (2n - 1)=2(m+n)+1=2c +1 where c= m-nis an integer by the closure property of addition. We have shown that a-b=2c-1 for an integer c so a-b must be odd.
Prompts Suomitted Answers Proposition: Let n be an integer. If n Sis odd, then nis Choose a match even. Proof Suppose that n+sis odd and nis also odd. By definition, then, there exists integers kand I so that n+5= 2k+1 and n= 21-1. Hence, we have 2k1=n+5 = (2- 17 -5 = 41-41-1+5 = 22 - 21- 31 Therefore, 2k+ 1seven. This is clearly impossible, and hence we cannot have that n +5is odd and nis also odd. Therefore, if that n²-sis odd, we must have nis even. Proposition: Let ab be integers. If ab is even, then at least Choose a match one of a orbis even. Proof Suppose that a and bare both odd. Then there are integers kand iso thata=2k+1 and b=21-1. Therefore, we have ab = (2k - 12- 1)= 4k+ 2k -2-1= 2zkl-k+-1, so ab is odd Thus, If ab is even, we must have at least one of a or bis Even. Proposition: Let a and b be integers. If a+bis even, then a Choose a match and bare either boch odd or bath even. Proof: Suppose that a and bare not both odd and nat both even, so that one of a and bis odd, and the other is even. Without lass of generality, suppose that a is odd andbis even. Then there are integers kand I such that a 2k 1 and b= 2L Therefore, a-b = (2k- 1)-21 = 2k +-1, so a+ bis odd. Hence, If a+ b is even, then a and bare either both odd or both even. Theorem: fa and bare consecutive integers, then the sum Choose a match a+bis odd. Proof Assume that a and bare consecutive integers. Assume also that the sum a-bis not odd. Because the suma-bis not odd, there exists no number k such that a+ b= 2k-1. However, the integers a and bare consecutive, so we may write the sum a -bas 20+ 1. Thus, we have derived that a bz k-1 for any integer kand also that a -b= 20-1. If we hoid that a andbare consecutive then we know that the sumab must be add. Theorem: The sum of an even integer and an odd integer is Choose a match odd. Proof: Suppose a is an even integer and bis an odd integer. Then, by our definitions of even and odd numbers, we know that integers mand n exist so that a 2m and b= 2n- 1. Then a+b= (2Zm) - (2n - 1)=2(m+n)+1=2c +1 where c= m-nis an integer by the closure property of addition. We have shown that a-b=2c-1 for an integer c so a-b must be odd.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
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