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Union and intersections - class-IX
Let $n$ be a fixed positive integer. Define a relation $R$ on $I$ (the set of all integers) as follows: a R b iff $n|(a-b)$ i.e., iff (a-b) is divisible by n. Show that $R$ is an equivalence relation on 1.
If $A\subseteq B$
$\therefore A\cap B=A$
R is reflexive since for any integer $a$ we have $a-a=0$ and $0$ is divisible by $n$.
Hence $aRa\quad \forall a\in I$
R is symmetric, $aRb$. Then by definition of $R$, $a-b=nk$ where $k\in I$.
Hence $b-a=\left( -k \right) n$ where $-k\in I$ and so $bRa$.
Thus we shown that $aRb\Rightarrow bRa$
R is transitive, let $aRb$ and $bRc$. then by definition of $R$, we have
$a-b={ k } _{ 1 }n$ and $b-a={ nk } _{ 2 }$
where ${ k } _{ 1 },{ k } _{ 2 }\in I$
It follow that $a-c=\left( a-b \right) +\left( b-c \right) ={ k } _{ 1 }n+{ k } _{ 2 }n=\left( { k } _{ 1 }+{ k } _{ 2 } \right) n$