I am to prove this statement:
For all integers n, n^2-n+3 is odd.
This is what I did, I just want to make sure its logical.
Suppose n is an integer such that n^2-n+3 is odd. this means there exists an integer, a, such that 2a+1 = n^2-n+3. It then follows that 2a = n^2-n+2 which is, by definition even. Since n^2-n+2 and n^2-n+3 are consecutive integers then it follows that n^2-n+3 is odd.
So it would be something like this?
Case 1: Let integer n be even. That means there exists some integer c such that 2c = n. n^2-n+3 = (2c)^2-2c+3 = 4c^2-2c+3 = 2(2c^2-c). Since 2c^2-c is an integer then n^2-n+3 is odd by definition.
Case 2: Let integer n be odd. That means there exists some integer c such that 2c+1 = n. n^2-n+3 = (2c+1)^2-(2c+1) +3 = 4c^2+4c+1-2c-1+3 = 4c^2+2c+3 = 2(2c^2+c) + 3. Since 2c^2+c is an integer then n^2-n+3 is odd by definition.