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

Solution:

To Prove: that the product of two consecutive integers is divisible by 2.

Proof: Let n − 1 and n be two consecutive positive integers. 

Then their product is $n(n-1)=n^{2}-n$

We know that every positive integer is of the form 2q or 2q + 1 for some integer q.

So let n = 2q 

So, $n^{2}-n=(2 q)^{2}-(2 q)$

$\Rightarrow n^{2}-n=(2 q)^{2}-2 q$

$\Rightarrow n^{2}-n=4 q^{2}-2 q$

$\Rightarrow n^{2}-n=2 q(2 q-1)$

$\Rightarrow n^{2}-n=2 r$ (where $r=q(2 q-1)$ )

$\Rightarrow n^{2}-n$ is even and divisble by 2

Let n = 2q + 1 

So, $n^{2}-n=(2 q+1)^{2}-(2 q+1)$

$\Rightarrow n^{2}-n=(2 q+1)((2 q+1)-1)$

$\Rightarrow n^{2}-n=(2 q+1)(2 q)$

$\Rightarrow n^{2}-n=2 r(r=q(2 q+1))$

$\Rightarrow n^{2}-n$ is even and divisble by 2

Hence it is proved that that the product of two consecutive integers is divisible by 2