Let $R$ be the real line. Consider the following subsets of the plane $R \times R$. $S=\{(x, y): y=x+1$ and…
Let $R$ be the real line. Consider the following subsets of the plane $R \times R$.
$S=\{(x, y): y=x+1$ and $0 < x < 2\}, T=\{(x, y): x-y$ is an integer $\}$. Which one of the following is true?
neither $S$ nor $T$ is an equivalence relation on $R$
both $S$ and $T$ are equivalence relations on $R$
$S$ is an equivalence relation on $R$ but $T$ is not
$\mathrm{T}$ is an equivalence relation on $\mathrm{R}$ but $S$ is not
Solution
$
T=\{(x, y): x-y \in l\}
$
as $0 \in \mathrm{I} \mathrm{T}$ is a reflexive relation.
If $x-y \in 1 \Rightarrow y-x \in I$
$\therefore \mathrm{T}$ is symmetrical also
If $x-y=l_1$ and $y-z=l_2$
Then $x-z=(x-y)+(y-z)=l_1+I_2 \in I$
$\therefore \mathrm{T}$ is also transitive.
Hence $T$ is an equivalence relation.
Clearly $x \neq x+1 \Rightarrow(x, x) \notin S$
$\therefore S$ is not reflexive