Question
Prove that the product of two consecutive positive integers is divisible by 2
✓
Answer
Let $n-1$ and $n$ be two consecutive positive integer, then the product is $n(n-1)$
$n(n-1)=n^2-n$
We know that any positive integers is of the form $2 q$ or $2 q+1$ for same integer $q$
Case 1:
$\text { when } n=2 q$
$n^2-n=(2 q)^2-2 q$
$=4 q^2-2 q$
$=2 q(2 q-1)$
$=2[q(2 q-1)]$
$n^2-n=2 r$
$r=q(2 q-1)$
Hence $n ^2- n$. divisible by 2 for every positive integer.
Case 2:
$\text { when } n=2 q+1$
$n^2-n=(2 q+1)^2-(2 q+1)$
$=(2 q+1)[2 q+1-1]$
$=2 q(2 q+1)$
$n^2-n=2 r$
$r=q(2 q+1)$
$n^2-n$ divisible by 2 for every positive integer.
$n(n-1)=n^2-n$
We know that any positive integers is of the form $2 q$ or $2 q+1$ for same integer $q$
Case 1:
$\text { when } n=2 q$
$n^2-n=(2 q)^2-2 q$
$=4 q^2-2 q$
$=2 q(2 q-1)$
$=2[q(2 q-1)]$
$n^2-n=2 r$
$r=q(2 q-1)$
Hence $n ^2- n$. divisible by 2 for every positive integer.
Case 2:
$\text { when } n=2 q+1$
$n^2-n=(2 q+1)^2-(2 q+1)$
$=(2 q+1)[2 q+1-1]$
$=2 q(2 q+1)$
$n^2-n=2 r$
$r=q(2 q+1)$
$n^2-n$ divisible by 2 for every positive integer.
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