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Let $R$ be the real line. Consider the following subsets of the plane $\boldsymbol{R} \times \boldsymbol{R}$

$S=\{(x, y): y=x+1$ and $0<x<2\}$
$T=\{(x, y): x-y$ is an int eger $\}$

Which one of the following is true?

1) Neither $S$ nor $T$ is an equivalence relation on $R$
2) Both $S$ and $T$ are equivalence relations on $R$
3) $S$ is an equivalence relation on $R$ but $T$ is not
4) $T$ is an equivalence relation on $R$ but $S$ is not
in Mathematics by Diamond (75,914 points) | 9 views

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Answer

(4)

Explanation

$T=\{(x, y): x-y \in Z\}$
as $0 \in Z T$ is a reflexive relation
If $x-y \in Z \Rightarrow y-x \in Z$
$\therefore T$ is symmetrical also If $x-y=Z_{1}$ and $y-z=Z_{2}$
Then $x-z=(x-y)+(y-z)=Z_{1}+Z_{2} \in Z$
$\therefore T$ is also transitive.
Hence $T$ is an equivalence relation Clearly $x \neq x+1 \Rightarrow(x, x) \notin S$
$\therefore S$ is not reflexive

by Diamond (75,914 points)

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