TITLE: Proof of $ x (x+1) $ is an even number by contradiction proof method
QUESTION [5 upvotes]: Prove that $ x(x+1) $ is always even by the method of contradiction.
I assumed $ x(x+1) = 2k + 1 $ as an odd integer where $ k $ is an integer
Add $1$ to both sides, we get $ x (x+1) + 1 = 2k + 1 + 1 $
$ x (x+1) + 1 = 2(k + 1) $
that is $2$ divides left hand side. How can I get a contradiction? Please guide.

REPLY [3 votes]: In the contradiction use this: If $ab$ be an odd number then $a$ and $b$ both are odd numbers.

REPLY [2 votes]: It seems like a strange question as the result is very easy to prove without using contradiction.  But anyway. . .
Here is an outline of a possible proof.  You didn't say what the numbers $x$ are so let's assume you want to prove it for positive integers.  Suppose that $x(x+1)$ is not always even: that is, it is odd for some $x$.  Choose the smallest such $x$.  Since $x(x+1)$ is odd, $x(x+1)-2x$ is odd minus even and therefore also odd, that is,
$$(x-1)x$$
is odd.  But this contradicts the minimality of $x$.
There are a couple of details still to be filled in, see how you go.