Question

Let $n$ be a fixed positive integer. Let a relation $R$ defined on $I$ (the set of all integers) as follows: $aRb$ iff $n/(a-b)$, that is, iff $a-b$ is divisible by $n$, then, the relation $R$ is

• A. Reflexive only
• B. Symmetric only
• C. Transitive only
• D. An equivalence relation

Answer 1

Answer: (D)

An equivalence relation

$R$ 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, let $aRb$.

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 have 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 $b-c=k_{2}n$, where $k_1,k_2\in I$.

It then follows that

$a-c=(a-b)+(b-c)=k_{1}n+k_{2}n=(k_1+k_2)n$

where $k_1+k_2\in I$