I think the mistake with the original proof is, it didn't take into account, "What if n is odd?" It assumed n = 2m (which, in the end, is true), but didn't take care of the "possibility" that n is odd.
Hello All,
As the instructions say, I need to evaluate everything that is wrong with this...let me share what I figured the correct proof to be. Any help would be great!!
=> n is even if n^{2} is even
<= n^{2}is even id n is even
1st direction:
n=2k by definition of an even integer
(2k)^{2 }is even because even integers are closed under multiplication (I feel that I'm lacking something on this direction...seems too easy)
2nd direction:
Contrapositive says: n is off if n^{2 }is odd.
n is odd=2k+1 by definition of an odd integer
n^{2 }= (2k+1)^{2}
= 4k^{2 }+4k+1
= 2(2k^{2 }+2K)+1 This is odd by definition of an odd integer. This is true, by way of contrapositive. Thus, the we have proven both directions....this is TRUE.
Does it look good? What are the mistakes in the original, since that is the actual question?
Thanks a lot!!