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(B) For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive.$1$. Analysis of $S = \{(x, y) : y = x + 1, 0 < x <

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$T$ is an equivalence relation on $R$ but $S$ is not

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(B) For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive.$1$. Analysis of $S = \{(x, y) : y = x + 1, 0 - Reflexivity: For $S$ to be reflexive, $(x, x) \in S$ for all $x \in R$. This requires $x = x + 1$, which implies $0 = 1$, a contradiction. Thus, $S$ is not reflexive.- Symmetry: For $S$ to be symmetric, if $(x, y) \in S$, then $(y, x) \in S$. If $(x, y) \in S$, then $y = x + 1$. For $(y, x) \in S$, we need $x = y + 1$. Substituting $y$, we get $x = (x + 1) + 1 = x + 2$, which is impossible. Thus, $S$ is not symmetric.- Since $S$ is neither reflexive nor symmetric, it is not an equivalence relation.$2$. Analysis of $T = \{(x, y) : (x - y) \in \mathbb{Z}\}$:- Reflexivity: $(x - x) = 0$, which is an integer. So, $(x, x) \in T$.- Symmetry: If $(x, y) \in T$, then $(x - y) = k$ for some $k \in \mathbb{Z}$. Then $(y - x) = -k$, which is also an integer. So, $(y, x) \in T$.- Transitivity: If $(x, y) \in T$ and $(y, z) \in T$, then $(x - y) = k_1$ and $(y - z) = k_2$ for $k_1, k_2 \in \mathbb{Z}$. Adding these, $(x - z) = k_1 + k_2$, which is an integer. So, $(x, z) \in T$.- Since $T$ is reflexive, symmetric, and transitive, it is an equivalence relation.Therefore, $T$ is an equivalence relation on $R$ but $S$ is not.

Let $R$ be the real line. Let the relations $S$ and $T$ on $R$ be defined by $S = \{(x, y) : y = x + 1, 0 < x < 2\}$ and $T = \{(x, y) : (x - y) \text{ is an integer}\}$. Then:

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