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I am attempting to learn about mathematical proofs on my own and this is where I've started. I think I can prove this by induction. Something like:

$n = 2k+1$ is odd by definition

$n = 2k+1 + 2$ (this is where I'm stuck, how do I show that this is odd?)

$n = 2(k+1) + 1$ (if I can show that it's odd, I can do the same here and prove my conjecture by induction, right?)

Thanks for any assistance

Michael Albanese
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hereslookinateuclid
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    Though not related to the post, +1 for your username. Welcome to the site! – Rick Decker Aug 25 '13 at 21:07
  • ... I fixed an arithmetic error in your third $n=$ line. The correction (along with walcher's answer) should make it obvious that you were on the right track. – Rick Decker Aug 25 '13 at 21:15
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    Induction is not needed and should not be used, since the small step needed for the induction argument is essentially the same as the non-induction proof. – André Nicolas Aug 25 '13 at 22:57

2 Answers2

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An integer $n$ is odd if and only if it is not divisible by $2$ or again if and only if it is of the form $2k+1$ for some integer $k$. If $n=2k+1$ is odd, then $n+2=2k+1+2=2k+2+1=2(k+1)+1$ is obviously odd as well.

walcher
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Odd numbers have a remainder of $1$ when divided by $2$, thus

$$n+2 = 2k+3 \equiv 3 \equiv 1 \pmod 2 $$

Argon
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