Given the following theorem/statement and its proof, MATCH the Method of Proof used to prove the given theorem/statement. PrompesSuomittea AnSwersProposition: Let n be an integer. If n+5is odd, then nisChoose a matchEven.Proof:Suppose that n+5is odd and n is also odd. By definition,then, there exists integers kand I so that n+5= 2k +1 andn= 21- 1. Hence, we have 2k +1=n+5= (2- 1 +5= 41 + 41 -1+5= 2(2 - 21+ 3).Therefore, 2k +1 s even.This is clearly impossible, and hence we cannot have that n 2+5 is odd and n is also odd.Therefore, if that n-5is odd, we must have n is even.Proposition: Let a, b be integers. If ab is even, then at leastChoose a matchone of a orbis even.Proof:Suppose that a and bare both odd. Then there are integerskand so thata= 2k+1 and b= 21- 1.Therefore, we have ab = (2k - 12+ 1) = 4kd+ 2k - 21 +1=2(2kl +k+)1, so ab is odd.Thus, If ab is cven, we must have at least one of a or b isEven.Proposition: Let a and b be integers. If a +bis even, then aChoose a matchand bare either both odd or bath even.Proot:Suppose that a and bare not both odd and not both even,so that one of a and bis odd, and the other is even. Withoutloss of generality, suppose that a is odd and bis even. Thenthere are integers kand I such that a = 2k +1 and b= 2LTherefore, a+b = (2k - 1) + 21 = 20+ )- 1, so a+bis odd.Hence, If a+bis even, then a and bare either both odd orboth even.Theorem: Ifa and b are consecutive integers, then the suma+bis odd.Choose a matchProof:Assume that oa and bare consecutive integers. Assume alsothat the sum a-bis not odd. Because the sum a-bis notodd, there exists no number k such that a+ b = 2k - 1.However, the integers a and b are consecutive, so we maywrite the sum abas 20 + 1. Thus, we have derived that a+bz k-1 for any integer kand also that a -b= 20 - 1. If wehold that a and b are consecutive then we know that thesum a +b must be odd.Theorem: The sum of an even integer and an odd integer isChoose a matchodd.Proof:Suppose a is an even integer andbis an odd integer.Then, by our definitions of even and odd numbers, we knowthat integers m and n exist so that a 2m and b= 2n + 1.Then a +b= (2m) + (2n + 1) = 2(m + n) +1= 2c +1 where c=m+nis an integer by the closure property of addition.We have shown that ab 2c1 for an integer c so a-bmust be odd.

Answered: Image /qna-images/answer/772de09c-78ec-42b5-811f-e65997a74623.jpg

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.