Direct Proof Discrete Math.Show that if n is an odd integer, then n 2 is...

Nathalie Fields

Nathalie Fields

Answered

2022-07-17

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.

Answer & Explanation

iljovskint

iljovskint

Expert

2022-07-18Added 18 answers

Step 1
An 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 2
In 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

enmobladatn

Expert

2022-07-19Added 6 answers

Step 1
By definition:
integer n is even iff there exists an integer k such that n = 2 k
integer n is odd iff there exists an integer k such that n = 2 k + 1
Step 2
So 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.

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