Prove that the product of any two consecutive positive integers is divisible by $2$.

(N/A) Let the two consecutive positive integers be $n$ and $n+1$,where $n$ is a positive integer.
Their product is $P = n(n+1) = n^2 + n$.
We can consider two cases for $n$:
Case $1$: If $n$ is even,then $n = 2k$ for some integer $k$. Thus,$P = 2k(2k+1) = 2(2k^2 + k)$,which is clearly divisible by $2$.
Case $2$: If $n$ is odd,then $n = 2k+1$ for some integer $k$. Thus,$P = (2k+1)(2k+1+1) = (2k+1)(2k+2) = 2(2k+1)(k+1)$,which is also divisible by $2$.
Since in both cases the product is divisible by $2$,the product of any two consecutive positive integers is always divisible by $2$.