Question

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.

• A. $R$ is an equivalence relation on 1.
• B. $R$ is not an equivalence relation on 1.
• C. $R$ is a symjetric relation on 1.
• D. $R$ is an identity relation on 1.

Answer 1

Answer: (A), (C)

The correct options are
A $R$ is an equivalence relation on 1.
C $R$ is a symjetric 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 $aRa\forall a\in I$
R is symmetric, $aRb$. Then by definition of $R$, $a-b=nk$ where $k\in I$.
Hence $b-a=(-k)n$ 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 $a-c=(a-b)+(b-c)=k_{1}n+k_{2}n=(k_1+k_2)n$