chain of equivalants

• Oct 18th 2009, 08:05 PM
sprint10s
chain of equivalants
Im on my last problem and i i know the first one implies the 2nd, the 2nd implies the 3rd and the 3rd implies the first but i keep screwing up somewhere.

i : x2(squared) is odd
ii: 1-x is even
iii: x2 +1 is even
• Oct 18th 2009, 08:34 PM
tonio
Quote:

Originally Posted by sprint10s
Im on my last problem and i i know the first one implies the 2nd, the 2nd implies the 3rd and the 3rd implies the first but i keep screwing up somewhere.

i : x2(squared) is odd
ii: 1-x is even
iii: x2 +1 is even

Well, if you already "know" that i ==> ii --> iii ==> i then you're done since this is what you've got to do!
What's it that you think you've screwed?

Tonio
• Oct 19th 2009, 05:05 AM
oldguynewstudent
defintion of odd even
if x^2 is odd then x is odd; let x = 2k +1

then x^2 = (2k+1)(2k+1) = 4k^2 + 4k + 1

1 - (4k^2 + 4k + 1) = -4k^2 - 4k = 2(-2k^2 - 2k) which is of the form 2k which proves it's even

(4k^2 + 4k + 1) + 1 = 4k^2 + 4k + 2 = 2(2k^2 + 2k + 1) which again is of the form 2k which proves it's even.