Let R be the real line.

Question

Let R be the real line. Consider the following subsets of the plane $$R \times R$$ :

$$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$

$$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,
 Which one of the following is true ?

Accepted Answer

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

Solutions · 1