JEE MAIN - Mathematics (2008 - No. 24)
Let R be the real line. Consider the following subsets of the plane $$R \times R$$ :
$$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$
$$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,
$$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$
$$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,
Which one of the following is true ?
Neither S nor T is an equivalence relation on R
Both S and T are equivalence relation on R
S is an equivalence relation on R but T is not
T is an equivalence relation on R but S is not
Explanation
Given $$S = \left\{ {\left( {x,y} \right):y = x + 1\,\,} \right.\,$$
and $$\,\,\,\left. {0 < x < 2} \right\}$$
As $$\,\,\,\,x \ne x + 1\,\,\,$$
for any $$\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$$
$$\therefore$$ $$S$$ is not reflexive.
Hence $$S$$ in not an equivalence relation.
Also $$\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$$ is an integer $$\left. {} \right\}$$
as $$x - x = 0$$ is an integer $$\forall x \in R$$
$$\therefore$$ $$T$$ is reflexive.
If $$x-y$$ is an integer then $$y-x$$ is also an integer
$$\therefore$$ $$T$$ 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.
$$\therefore$$ $$T$$ is transitive
Hence $$T$$ is an equivalence relation
and $$\,\,\,\left. {0 < x < 2} \right\}$$
As $$\,\,\,\,x \ne x + 1\,\,\,$$
for any $$\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$$
$$\therefore$$ $$S$$ is not reflexive.
Hence $$S$$ in not an equivalence relation.
Also $$\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$$ is an integer $$\left. {} \right\}$$
as $$x - x = 0$$ is an integer $$\forall x \in R$$
$$\therefore$$ $$T$$ is reflexive.
If $$x-y$$ is an integer then $$y-x$$ is also an integer
$$\therefore$$ $$T$$ 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.
$$\therefore$$ $$T$$ is transitive
Hence $$T$$ is an equivalence relation
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