Question

prove that n square - n is divisible by 2 for every positive integer​

Answer 1

$\large\underline{\underline{\mathfrak{\sf{\red{Correct\:Question-}}}}}$

Prove that $\sf{n^2-n}$ is divisible by 2 for every positive integer n.

$\large\underline{\underline{\mathfrak{\sf{\red{Explanation-}}}}}$

We know that,

$\boxed{\sf{\blue{a=bq+r}}}$

Here a = n, b = 2 and r = 0 or 1 [ According to the question ]

$\bold\pink{Case-1}$

Let's suppose n be an even positive integer.

$\huge\boxed{\sf{\orange{n=2q}}}$

Now, we have to check whether $\sf{n^2-n}$ is divisible by 2 or not.

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

$\implies$ $\sf{n^2-n=4q^2-2q}$

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

$\implies$ $\sf{n^2-n=2m}$

Where $\sf{m=q(2q-1)}$

$\therefore$ In case 1, $\sf{n^2-n}$ is divisible by 2.

$\bold\pink{Case-2}$

Let's suppose n be an odd positive integer.

$\huge\boxed{\sf{\orange{n=2q+1}}}$

Now, we have to check whether $\sf{n^2-n}$ is divisible by 2 or not.

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

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

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

$\implies$ $\sf{n^2-n=2m}$

Where $\sf{m=q(2q-1)}$

$\therefore$ In case 2, $\sf{n^2-n}$ is divisible by 2.

From both the cases, we concluded that $\bold{\sf{n^2-n}}$ is divisible by 2 for every integer n.

Hence proved!

______________________

#AnswerWithQuality

#BAL

Answer 2

Answer:

n² - n is divisible by 2 for every integer n

Proved.

Step-by-step explanation:

Let's Check out that in these situations n² - n is divisible by 2 or not.

$\bigstar\;\bf{\underline{\underline{Situation-1:}}}$

Let n be an even positive integer.

When n = 2q.

In this situation,

$\sf\\ \\\implies n^2 -n = (2q)^2-2q = 4q^2 - 2q = 2q (2q - 1 )\\ \\ \implies n^2 - n = 2r\quad[\textsf Where\;r = q (2q - 1)]$

Here,

n² - n is divisible by 2.

$\rule{300}{1.5}$

$\bigstar\;\bf{\underline{\underline{Situation-2:}}}$

When n = 2q + 1

In this situation,

$\sf \\ \\\implies n^2 -n = (2q + 1)^2 - (2q + 1)= (2q +1) ( 2q+1 -1)= 2q (2q + 1)\\ \\ \implies n^2 - n = 2r\quad[\textsf Where\;r = q (2q + 1)}$

Here,

n² - n is divisible by 2.

Hence, Proved.

_______________________

Remember :

$\bf \implies a=bq+r$

$\rule{300}{1.5}$

#AnswerWithQuality.

#BAL! :)