Question
Prove that $n^2-n$ divisible by 2 for every positive integer $n$
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Answer
To prove $n ^2- n$ divisible by 2 for every positive integer n .
We know that any positive integer is of the form $2 q$ or $2 q+1$, for some integer $q$. So, following cases arise:
Case I:
When $n =2 q$.
In this case, we have
$n^2-n=(2 q)^2-2 q=4 q^2-2 q=2 q(2 q-1)$
$\Rightarrow n^2-n=2 r \text { where } r=q(2 q-1)$
$\Rightarrow n^2-n \text { is divisible by } 2$
Case II:
When $n=2 q+1$
In this case, we have
$n^2-n=(2 q+1)^2-(2 q+1)$
$=(2 q+1)(2 q+1-1)=(2 q+1) 2 q$
$\Rightarrow n^2-n=2 r \text { where } r=q(2 q+1)$
$\Rightarrow n^2-n \text { is divisible by } 2$
Hence $n^2-n$ is divisible by 2 for every positive integer $n$.
We know that any positive integer is of the form $2 q$ or $2 q+1$, for some integer $q$. So, following cases arise:
Case I:
When $n =2 q$.
In this case, we have
$n^2-n=(2 q)^2-2 q=4 q^2-2 q=2 q(2 q-1)$
$\Rightarrow n^2-n=2 r \text { where } r=q(2 q-1)$
$\Rightarrow n^2-n \text { is divisible by } 2$
Case II:
When $n=2 q+1$
In this case, we have
$n^2-n=(2 q+1)^2-(2 q+1)$
$=(2 q+1)(2 q+1-1)=(2 q+1) 2 q$
$\Rightarrow n^2-n=2 r \text { where } r=q(2 q+1)$
$\Rightarrow n^2-n \text { is divisible by } 2$
Hence $n^2-n$ is divisible by 2 for every positive integer $n$.
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