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If $(3n+2)$ is odd then, prove $n$ is odd.

$$3n+2 = (2n+1)+(n+1)$$

We already have a fact that $2n+1$ is always odd. So, for $3n+2$ to be odd, $n+1$ should be even (For $x+y$ to be odd then either $x$ or $y$ should be odd not both)

As, $n+1$ is even, $n$ is always odd.

I should the solution to our teacher and he said the logic is wrong but denied to point our the specifics. Can you please help me with what I did wrong?

rlartiga
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Arun Poudel
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  • I personally like the proof you gave. It shows that you have a conceptual understanding of this problem. I can't see why your professor has a problem with this. – 1233dfv Apr 16 '14 at 19:41

3 Answers3

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This is perfectly correct as far as I can tell. Maybe you should submit a written response to your teacher, instead of an oral response maybe, which could get rid of the misunderstanding.

Sandeep Silwal
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Your proof is correct. Alternatively, similarly $\ n+(\overbrace{2n+2}^{\rm even})\,$ odd $\,\Rightarrow\,n\,$ odd.

Does "proof by direct method" have some technical meaning given by your teacher? If so, that might be why the proof was deemed wrong.

Bill Dubuque
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  • As far as I remember, by "Proof By direct method" he said we only use proven facts and idioms. – Arun Poudel Apr 16 '14 at 19:33
  • @Arun That's required of all proofs. Usually "direct" means not by way of contradiction, but that notion is difficult to make precise without serious study of mathematical logic. – Bill Dubuque Apr 16 '14 at 19:40
  • @Arun Proving the contrapositive is even easier (but might not be a "direct method") $$ {\rm even}\ n = 2k,\Rightarrow, 3n+2 = 3(2k)+2 = 2(3k+1)\ {\rm is\ even} $$ – Bill Dubuque Apr 16 '14 at 20:52
  • Yes, bill. Proving the contrapositive is easier. But the teacher's argument was, "It cannot be proved using direct method" – Arun Poudel Apr 16 '14 at 20:57
  • @Arun Well that gives a bit more of a clue as to what your teacher means by "direct method". Did they say anything more specific than "your logic is wrong"? – Bill Dubuque Apr 16 '14 at 21:00
  • Yah, he added that odd + even -> odd but the converse isn't true in every case.. But for this specific case that we are looking at(odd integers and separating them into two integers), the converse is always true, isn't it? :) – Arun Poudel Apr 16 '14 at 21:14
  • @Arun yes, the converse, as stated in your question, is true. – Bill Dubuque Apr 16 '14 at 21:41
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If $3n+2$ is odd, then $3n+2=2a+1$ where $a\in \mathbb{Z}$. Consider $n+(2n+2)=2a+1$ which imples that $n=2a-2n-1=2(a-n-1)+1$. Thus $n=2(a-n-2)+1$ where $(a-n+2)\in \mathbb{Z}$ and $n$ is odd.

1233dfv
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