Prove 1-n^2>0: 3n-2 is Even Integer

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The discussion centers on proving the inequality 1 - n² > 0, leading to the conclusion that 3n - 2 is an even integer. The proof presented indicates that if n = 0, then 3n - 6 equals -2, which is indeed an even integer. However, participants question the validity of the initial assumption that 1 - n² > 0 only implies n = 0, suggesting that the problem statement may be misinterpreted. The conversation highlights the need for clarity in mathematical proofs and the importance of understanding the implications of given conditions.

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bonfire09
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Let nεZ,Prove that 1-n^2>0, then 3n-2 is an even integer.

I proved it like this. I think its right but I am not able to word it correctly.

Since 1-n^2>0 therefore n=0. Then 3n-6=3(0)-2=-2. Since 0 is an integer, 3n-6 is even.

How can I learn to word this correctly because I am having some trouble with it?
 
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Are you sure you have the statement of the problem correct? It looks crazy. As you say 1 - n2 > 0 would imply n = 0, which makes the rest of it much too easy. Also, why would anyone ask you to prove 3n-2 is an even integer when n is already known to be an integer? Why not just ask you to show n is even?
 

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