Discrete Math Problem

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Jkepler
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Discrete Math Problem

Post by Jkepler » Thu May 26, 2011 3:49 am

Prove the following about integers: x is an odd integer iff x2 + 2x + 1 is an even integer.
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hk
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Re: Discrete Math Problem

Post by hk » Thu May 26, 2011 7:54 am

Do you mean x2+2x+1?
If so I suspect that you copied the question from somewhere else.
Anyhow: try to rewrite x2+2x+1 (with brackets) and it will become pretty obvious IMHO.
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Jkepler
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Re: Discrete Math Problem

Post by Jkepler » Thu May 26, 2011 1:57 pm

Yes, correct on both counts. I meant that to be an exponent, and I had copied and pasted it from elsewhere.

Anyway, here's my solution:

Factoring tells us that x2 + 2x + 1 = (x + 1)2

So, x is odd iff x + 1 is even iff (x + 1)2 is even.

Would this be an acceptable proof?

Thanks again!
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thundre
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Re: Discrete Math Problem

Post by thundre » Thu May 26, 2011 3:23 pm

Jkepler wrote:Would this be an acceptable proof?
It depends how much rigor is expected in the class.

To prove "A iff B", you need to prove "B implies A" and "not B implies not A". You could do it in 2 separate sections, one starting with the statement x2 + 2x + 1 is even, the other starting with the opposite.
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Ellenion
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Re: Discrete Math Problem

Post by Ellenion » Sat Feb 24, 2018 1:48 am

Proof:
It can be seen that $x^2+2x+1$ is even, as given in the problem. Therefore, $x^2+2x$ is odd, since an even number minus $1$ is odd. Then, we know that any integer multiplied by $2$ will be even, and an odd number minus an even number will be odd. Therefore, $x^2$ is odd. Since squares of odd numbers are the only odd squares, then $x$ has to be odd.
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