Question
Let $$R$$ be the real line. Consider the following subsets of the plane $$R \times R:$$
$$\eqalign{
& S = \left\{ {\left( {x,y} \right):y = x + 1\,{\text{and }}0 < x < 2} \right\} \cr
& T = \left\{ {\left( {x,y} \right):x - y\,\,{\text{is an integer}}} \right\}, \cr} $$
Which one of the following is true?
A.
Neither $$S$$ nor $$T$$ is an equivalence relation on $$R$$
B.
Both $$S$$ and $$T$$ are equivalence relation on $$R$$
C.
$$S$$ is an equivalence relation on $$R$$ but $$T$$ is not
D.
$$T$$ is an equivalence relation on $$R$$ but $$S$$ is not
Answer :
$$T$$ is an equivalence relation on $$R$$ but $$S$$ is not
Solution :
$$\eqalign{
& {\text{Given }}S = \left\{ {\left( {x,y} \right):y = x + 1\,{\text{and }}0 < x < 2} \right\} \cr
& \because \,\,x \ne x + 1\,\,{\text{for any }}x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S \cr
& \therefore \,\,S\,\,{\text{is not reflexive}}{\text{.}} \cr} $$
Hence $$S$$ in not an equivalence relation.
$$\eqalign{
& {\text{Also }}T = \left\{ {\left( {x,y} \right):x - y\,\,{\text{is an integer}}} \right\} \cr
& \because \,\,x - x = 0\,\,{\text{is an integer }}\forall \,x \in R \cr
& \therefore \,\,T\,\,{\text{is reflexive}}{\text{.}} \cr} $$
If $$x - y$$ is an integer then $$y - x$$ is also an integer
∴ $$T$$ is symmetric.
If $$x - y$$ is an integer and $$y - z$$ is an integer then
$$(x - y) + (y - z) = x - z$$ is also an integer.
∴ $$T$$ is transitive
Hence $$T$$ is an equivalence relation.