Q. Let $R$ be the real line. Consider the following subsets of the plane $R \times R$ $S = \{(x,\ y):y = x + 1$ and $o < x < 2\}$ $T = \{(x,\ y):x - y$ is an integer} Which of the following is true?
A) $T$ is an equivalent relation on $R$ but $S$ is not
B) Neither $S$ nor $T$ is an equivalence relation on $R$
C) Both $S$ and $T$ are equivalence relations on $R$
D) $S$ is an equivalence relations on $R$ and $T$ is not
Explanation
(a)
Since, $(1,\ 2) \in S$ but $(2,\ 1) \notin S$
$\therefore S$ is not symmetric.
Hence, $S$ is not an equivalent relation.
Given, $T = \left\{ (x,\ y):(x - y) \in I \right\}$
Now, $xTx \Rightarrow x - x = 0 \in I$, it is reflexive relation
Again, $xTy \Rightarrow (x - y) \in I$
$\Rightarrow y - x \in I \Rightarrow yTx$ it is symmetric relation.
Let $xTy$ and $yTz$
$\therefore x - y = I_{1}$ and $y - z = I_{2}$
Now, $x - z = (x - y) + (y - z) = I_{1} + I_{2} \in I$
$\Rightarrow x - z \in I$
$\Rightarrow \ \ xTz$
$\therefore T$ is transitive.
Hence, $T$ is an equivalent relation.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.