Abstract Math Problem
- Thread starter mrhoad1
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What is the definition of an even integer?
I know that an integer n is even if n = 2k. If the 4 is factored out as shown above to get 4(n^2 + 2n + 2), would I be able to let k = n^2 + 2n +2 to get 4k which would also be even?
Another approach I was thinking about after factoring is that n^2 = 4k^2 and 2n = 4k, which are also even, and then substitute these into the equation:
4(4k^2 + 4k +2) Could I then say that as a result the equation is even?
Thank you very much for the hints, I hope I am on the right track. I feel like I may be making this more complicated than it needs to be.
Please let me know if I am on the right track
Another approach I was thinking about after factoring is that n^2 = 4k^2 and 2n = 4k, which are also even, and then substitute these into the equation:
4(4k^2 + 4k +2) Could I then say that as a result the equation is even?
Thank you very much for the hints, I hope I am on the right track. I feel like I may be making this more complicated than it needs to be.
Please let me know if I am on the right track
Well EITHER hint will get you to a solution. I do think you are greatly overcomplicating it, perhaps in part because you are not reading Mark's hint carefully. You sort of got what he was driving at.I know that an integer n is even if n = 2k. If the 4 is factored out as shown above to get 4(n^2 + 2n + 2), But this is NOT what Mark said would I be able to let k = n^2 + 2n +2 to get 4k which would also be even?
Another approach I was thinking about after factoring is that n^2 = 4k^2 and 2n = 4k, which are also even, and then substitute these into the equation:
4(4k^2 + 4k +2) Could I then say that as a result the equation is even?
Thank you very much for the hints, I hope I am on the right track. I feel like I may be making this more complicated than it needs to be.
Please let me know if I am on the right track
But try this factoring \(\displaystyle 4n^2 + 4n + 8 = 2(2n^2 + 2n + 4).\)
How does that relate to the definition of even integers?
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My apologies, thank you for pointing that out to me, that helps out a lot.
After factoring: 2(2n^2 + 2n + 4) This relates to the definition of even integers n = 2k because when you multiply anything by 2 you will get an even integer.
Would it suffice to simply state this, or should I let k = (2n^2 + 2n + 4) and show it as 2k?
Thank you!
After factoring: 2(2n^2 + 2n + 4) This relates to the definition of even integers n = 2k because when you multiply anything by 2 you will get an even integer.
Would it suffice to simply state this, or should I let k = (2n^2 + 2n + 4) and show it as 2k?
Thank you!
I think a proof could go with either factoring. But being formal I'd probably mention closure under multiplication and addition as in:
\(\displaystyle n \in \mathbb Z \implies n^2 \in \mathbb Z \implies (n^2 + n + 2) \in \mathbb Z \implies\)
\(\displaystyle 4(n^2 + n + 2) \in \mathbb Z \implies 2\{2(n^2 + n + 2)\} = (4n^2 + 4n + 8)\ is\ an\ even\ integer\ by\ definition.\)
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