Question

Prove that $\left( {{n}^{2}}-n \right)$ is divisible by 2 for every positive integer n?

Answer 1

Hint: We start solving the problem by recalling the definition of positive integers. We then assume n is an even positive number and substitute the general form of it in place of n to solve the required result. Similarly, we prove the result by assuming n as an odd positive number and proving the result as required.

Complete step-by-step answer:
According to the problem, we need to prove that $\left( {{n}^{2}}-n \right)$ is divisible by 2 for every positive integer n.
We know that the positive integers are always greater than 0. So, we have $n>0$.
Let us prove the result by assuming n as an even positive integer. We know that the general form of the positive even integers is $2r$, $\left( r\ge 1 \right)$. Let us substitute $n=2r$ in $\left( {{n}^{2}}-n \right)$.
So, we get ${{n}^{2}}-n={{\left( 2r \right)}^{2}}-\left( 2r \right)$.
$\Rightarrow {{n}^{2}}-n=4{{r}^{2}}-2r$.
$\Rightarrow {{n}^{2}}-n=2r\left( 2r-1 \right)$.
Let us assume $r\left( r-1 \right)=d$, to represent it as a factor.
$\Rightarrow {{n}^{2}}-n=2d$ ---(1).
From equation (1), we can see that $\left( {{n}^{2}}-n \right)$ is clearly divisible by 2 for n being an even positive integer.
Now, let us prove the result by assuming n as an odd positive number. We know that the general form of the positive odd numbers is $\left( 2r-1 \right)$, $\left( r>1 \right)$. Let us substitute $n=2r-1$ in $\left( {{n}^{2}}-n \right)$.
So, we get ${{n}^{2}}-n={{\left( 2r-1 \right)}^{2}}-\left( 2r-1 \right)$.
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
$\Rightarrow {{n}^{2}}-n=4{{r}^{2}}-4r+1-2r+1$.
$\Rightarrow {{n}^{2}}-n=4{{r}^{2}}-6r+2$.
$\Rightarrow {{n}^{2}}-n=2\left( 2{{r}^{2}}-3r+1 \right)$.
Let us assume $2{{r}^{2}}-3r+1=e$, to represent it as a factor.
$\Rightarrow {{n}^{2}}-n=2e$ ---(2).
From equation (2), we can see that $\left( {{n}^{2}}-n \right)$ is clearly divisible by 2 for n being an odd positive integer.
From equations (1) and (2), we have proved $\left( {{n}^{2}}-n \right)$ is divisible by 2 for all positive integers n.

Note: Alternatively, we can solve the problem by using mathematical induction technique. Let us recall the steps to do mathematical induction:
We first check whether the given statement holds true for $n=1$.
We then assume the statement holds true for $n=k$ and try to prove it for $n=k+1$.
So, we have given the statement $\left( {{n}^{2}}-n \right)$ is divisible by 2 for every positive integer n.
Let us check the value of ${{n}^{2}}-n$ for $n=1$.
$\Rightarrow {{1}^{2}}-1=1-1$.
$\Rightarrow {{1}^{2}}-1=0$.
We know that 0 is divisible by 2.
Now, we assume $\left( {{n}^{2}}-n \right)$ is divisible by 2 for a positive integer $n=k$.
So, we have ${{k}^{2}}-k$ is divisible by 2.
So, let us assume ${{k}^{2}}-k=2q$.
Let us add $\left( -2k+2 \right)$ on both sides.
$\Rightarrow {{k}^{2}}-k-2k+2=2q-2k+2$.
$\Rightarrow {{k}^{2}}-2k+1-k+1=2q-2k+2$.
$\Rightarrow {{\left( k-1 \right)}^{2}}-\left( k-1 \right)=2\left( q-k+1 \right)$.
We can see that $\left( {{n}^{2}}-n \right)$ is divisible by 2 for a positive integer $n=k+1$.