SOLUTION: If x is an integer, x is even if x=2p for some integer p, and x is odd if for no integer p such that x=2p. If ab is an odd number, show that a and b are both odd.

Algebra ->  Proofs -> SOLUTION: If x is an integer, x is even if x=2p for some integer p, and x is odd if for no integer p such that x=2p. If ab is an odd number, show that a and b are both odd.       Log On


   

Question 9579: If x is an integer, x is even if x=2p for some integer p, and x is odd if for no integer p such that x=2p. If ab is an odd number, show that a and b are both odd.
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
If a is an odd number, it has to be expressed as a = 2m + 1 for ANY integer m. The 2m forces it to be even, the +1 forces the expression to be odd. Similarly, b = 2n + 1 for some integer n. Now, we're going to prove that a*b is odd:

+ab+=+%282m+%2B+1%29%282n+%2B+1%29+

+ab+=+4mn+%2B2m+%2B+2n+%2B+1+ <---------- The product 4mn is even no matter what. So are 2m + 2n. The +1 throws off the "evenhood", so a*b is odd.