Prove that \[{{n}^{2}}-n\] is divisible by 2 for every positive integer n

Hint: First consider the case where n is even. Put, n = 2q, simplify and prove that \[\left( {{n}^{2}}-n \right)\] is divisible by 2. Now consider n as add, so put, n = 2q + 1. Simplify and prove that \[\left( {{n}^{2}}-n \right)\] is divisible by 2.

Complete step-by-step answer:
Given to us the expression, \[{{n}^{2}}-n\], which we need to prove that is divisible by 2. Let us first consider that n is an even positive integer. Let us first consider, n = 2q.
Now let us substitute, n = 2q in \[\left( {{n}^{2}}-n \right)\].
\[{{n}^{2}}-n={{\left( 2q \right)}^{2}}-\left( 2q \right)\]
\[{{n}^{2}}-n=4{{q}^{2}}-2q=2q\left( 2q-1 \right)\]
Thus let us put, \[r=q\left( 2q-1 \right)\].
So, \[{{n}^{2}}-n=2q\left( 2q-1 \right)=2r\].
i.e. \[{{n}^{2}}-n=2r\]
Thus we can say that, \[\left( {{n}^{2}}-n \right)\] is divisible by Q.
Now let us consider n to be an odd positive integer. Thus let us put, n = 2q + 1 in \[\left( {{n}^{2}}-n \right)\].
\[\therefore {{n}^{2}}-n={{\left( 2q+1 \right)}^{2}}-\left( 2q+1 \right)\]
\[\because \] We know, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
\[\begin{align}
  & \therefore {{n}^{2}}-n={{\left( 2q \right)}^{2}}+1+2\times 2q\times 1-2q-1 \\
 & {{n}^{2}}-n=4{{q}^{2}}+4q-2q=4{{q}^{2}}+2q \\
 & \therefore {{n}^{2}}-n=2q\left( 2q+1 \right) \\
\end{align}\]
Consider, \[r=q\left( 2q+1 \right)\].
\[\therefore {{n}^{2}}-n=2r\]
i.e. \[\left( {{n}^{2}}-n \right)\] is divisible by 2.
Thus we have proved that \[\left( {{n}^{2}}-n \right)\] is divisible by 2 for every positive integer n, whether n is even or odd.

Note: We can also prove it as, \[{{n}^{2}}-n=n\left( n-1 \right)\].
\[n\left( n-1 \right)\], if we consider n as even, then \[\left( n-1 \right)\] is odd. So the product is even i.e. it is divisible by 2.
If you consider \[\left( n-1 \right)\] as even then n is odd, still the product of odd and even numbers is even. So divisible by 2.