- #1
hackedagainanda
- 52
- 11
Prove that for any arbitrary odd x, that x^2 is also odd.
By definition an odd number is an integer that can be written in the form of 2k + 1 for some integer k. This means that x = 2k + 1 where k is an integer
So let x^2 = (2k + 1)^2 we then get 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, This is where I get lost, I understand that k is an integer but how does it follow that 2k^2 + 2k is an integer?
By definition an odd number is an integer that can be written in the form of 2k + 1 for some integer k. This means that x = 2k + 1 where k is an integer
So let x^2 = (2k + 1)^2 we then get 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, This is where I get lost, I understand that k is an integer but how does it follow that 2k^2 + 2k is an integer?