Direct Proof Discrete Math.Show that if n is an odd integer, then n 2 is odd.Proof : Assume that n is an odd integer. This implies that there is some integer k such that n = 2 k + 1. Then n 2 = ( 2 k + 1 ) 2 = 4 k 2 + 4 k + 1 = 2 ( 2 k 2 + 2 k ) + 1. Thus, n 2 is odd.Why does the solution assume n to be 2 k + 1?How do you know n 2 is odd based on 2 ( 2 k 2 + 2 k ) + 1?I don't see how 2 k + 1 for n and 2(2k2+2k)+1 for n 2 means an odd integer. Some clarification would be helpful.

Question

Direct Proof Discrete Math.Show that if n is an odd integer, then       n    2   is odd.Proof : Assume that n is an odd integer. This implies that there is some integer k such that   n  =  2  k  +  1. Then       n          2        =  (  2  k  +  1      )          2        =  4      k          2        +  4  k  +  1  =  2  (  2      k          2        +  2  k  )  +  1. Thus,       n          2       is odd.Why does the solution assume n to be   2  k  +  1?How do you know       n          2       is odd based on   2  (  2      k          2        +  2  k  )  +  1?I don&#039;t see how   2  k  +  1 for n and 2(2k2+2k)+1 for       n          2       means an odd integer. Some clarification would be helpful.

iljovskint · Accepted Answer

Step 1An integer m is odd if and only if it can be written as the sum of an even integer and 1, if and only if there exists an integer q such that   m  =  2  q  +  1..Step 2In the proof, assumption that n is odd implies the existence of an integer k such that   n  =  2  k  +  1 (that is the only if part of the lemma). As for       n    2    =  2  (  2      k    2    +  2  k  )  +  1, letting   q  =  2      k    2    +  2  k we have       n    2    =  2  q  +  1  ,, from which it follows that n2 is odd (that is the if part of the lemma).

enmobladatn · Accepted Answer

Step 1By definition:integer n is even iff there exists an integer k such that   n  =  2  kinteger n is odd iff there exists an integer k such that   n  =  2  k  +  1Step 2So in the first step we use the definition to go from n is odd to   n  =  2  k  +  1, and in the other step we again use the definition to go from       n    2    =  2  (  2      k    2    +  2  k  )  +  1 to       n    2   is odd.