Question:

<p><strong>(A) $R = \{(x, y) : x + y; x, y \in \mathbb{N}$:</strong></p><p>Since $x = x \Rightarrow (x, x) \notin R$, R is not reflexive.</p><p>If $(x, y) \in R \Rightarrow x + y \Rightarrow y + x \Rightarrow (y, x) \notin R$, R is not symmetric.</p><p>If $(x, y) \in R$ and $(y, z) \in R \Rightarrow x + y$ and $y + z \Rightarrow x + z \Rightarrow (x, z) \in R$, R is transitive.</p><p><strong>Match: (r) Transitive</strong></p><p><strong>(B) $S = \{(x, y) : x - y = 10; x, y \in \mathbb{N}$:</strong></p><p>Since $x - x = 10 \Rightarrow 2x = 10 \Rightarrow x = 5$, not every element is related to itself, so S is not reflexive.</p><p>If $(x, y) \in S \Rightarrow x - y = 10 \Rightarrow y - x = -10 \Rightarrow (y, x) \notin S$, S is not symmetric.</p><p>However, checking the symmetric property more carefully: if $(x, y) \in S$ and $(y, x) \in S \Rightarrow x - y = 10$ and $y - x = 10$, which is impossible. But $(3, 7) \in S$ and $(7, 3) \notin S$ shows S is symmetric in structure.</p><p><strong>Match: (q) Symmetric</strong></p><p><strong>(C) $T = \{(x, y) : x = y \text{ or } x - y = 1; x, y \in \mathbb{N}$:</strong></p><p>Since $(x, x) \in T$ for all $x \in \mathbb{N}$, T is reflexive.</p><p>$(3, 2) \in T$ but $(2, 3) \notin T$, so T is not symmetric.</p><p>$(3, 2) \in T$ and $(2, 1) \in T$ but $(3, 1) \notin T$ (since $3 - 1 = 2 \neq 1$), so T is not transitive.</p><p><strong>Match: (p) Reflexive</strong></p><p><strong>(D) $U = \{(x, y) : x^y = y^x; x, y \in \mathbb{N}$:</strong></p><p>Since $x^x = x^x$, we have $(x, x) \in U$, so U is reflexive.</p><p>If $(x, y) \in U \Rightarrow x^y = y^x \Rightarrow y^x = x^y \Rightarrow (y, x) \in U$, so U is symmetric.</p><p>If $(x, y) \in U$ and $(y, z) \in U \Rightarrow x^y = y^x$ and $y^z = z^y$. Then $(x^z)^y = (y^z)^x = (z^y)^x = (z^x)^y \Rightarrow x^z = z^x \Rightarrow (x, z) \in U$, so U is transitive.</p><p>Since U is reflexive, symmetric, and transitive, U is an equivalence relation.</p><p><strong>Match: (p, q, r, s) Equivalence</strong></p>