Question

Prove that ${n^2} - n$ is divisible by 2 for any positive number n.

Answer 1

Hint- In order to solve the problem separately consider n to be odd and even. Use the general form of even and odd number in place on n to prove the given problem.

Complete step-by-step answer:

To prove that: ${n^2} - n$ is divisible by 2 for any positive number n.
Case (I): let n be an even positive integer.
As we know that the general form of even numbers is 2q.
So let $n = 2q$
Now let us substitute the value of n in the given function.
In this case we have:
$ \Rightarrow {n^2} - n = {\left( {2q} \right)^2} - \left( {2q} \right)$
Now let us simplify the equation, we get:
$ \Rightarrow {n^2} - n = 4{q^2} - 2q$
Now let us take 2 common from the RHS.
$
   \Rightarrow {n^2} - n = 2q\left( {2q - 1} \right) \\
   \Rightarrow {n^2} - n = 2r,{\text{ where }}r = q\left( {2q - 1} \right) \\
 $
So, in this case as ${n^2} - n$ is a multiple of 2 so ${n^2} - n$ is divisible by 2.
Case (II): let n be an odd positive integer.
As we know, the general form of an odd number is 2q+1.
So let $n = 2q + 1$
Now let us substitute the value of n in the given function.
In this case we have:
$ \Rightarrow {n^2} - n = {\left( {2q + 1} \right)^2} - \left( {2q + 1} \right)$
Now let us simplify the equation by taking 2q+1 common from RHS, we get:
$
   \Rightarrow {n^2} - n = \left( {2q + 1} \right)\left( {\left( {2q + 1} \right) - \left( 1 \right)} \right) \\
   \Rightarrow {n^2} - n = \left( {2q + 1} \right)\left( {2q} \right) \\
 $
Now let us take 2 common from the RHS.
$
   \Rightarrow {n^2} - n = 2q\left( {2q + 1} \right) \\
   \Rightarrow {n^2} - n = 2r,{\text{ where }}r = q\left( {2q + 1} \right) \\
 $
So, in this case also as ${n^2} - n$ is a multiple of 2 so ${n^2} - n$ is divisible by 2.
Hence, ${n^2} - n$ is divisible by 2 for every positive number n.

Note- In order to solve such problems students must remember the general form of odd and even numbers. Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. Any number that can be exactly divided by 2 is called an even number. Even numbers always end up with the last digit as 0, 2, 4, 6 or 8.