On the set of integers Z,a relation S is defi | vedclass

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(A) The condition $|x - y| < 1$ for integers $x, y \in Z$ implies that $|x - y| = 0$,because the absolute difference between two integers must be a no

Vedclass · Accepted Answer

It is an equivalence relation.

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(A) The condition $|x - y| Thus,$|x - y| = 0 \implies x = y$. Therefore,$S = \{(x, x) : x \in Z\}$,which is the identity relation on $Z$. $1$. Reflexive: For any $x \in Z$,$|x - x| = 0 $2$. Symmetric: If $(x, y) \in S$,then $x = y$,which implies $y = x$,so $(y, x) \in S$. Thus,$S$ is symmetric. $3$. Transitive: If $(x, y) \in S$ and $(y, z) \in S$,then $x = y$ and $y = z$,which implies $x = z$,so $(x, z) \in S$. Thus,$S$ is transitive. Since $S$ is reflexive,symmetric,and transitive,it is an equivalence relation. Note: The provided option $B$ is incorrect based on the definition of the identity relation. The correct classification is an equivalence relation.

On the set of integers $Z$,a relation $S$ is defined as: $S = \{(x, y) \in Z \times Z : |x - y| < 1\}$. Which of the following is true about $S$?

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