1. **Problem Statement:** We need to show that the relation $R = \{(a,b) \mid a \equiv b \pmod{m}\}$ is an equivalence relation on $\mathbb{Z}$ and verify that if $x_1 \equiv y_1$ and $x_2 \equiv y_2$ (mod $m$), then $(x_1 + x_2) \equiv (y_1 + y_2)$ (mod $m$). 2. **Definition of Equivalence Relation:** A relation $R$ on a set is an equivalence relation if it is reflexive, symmetric, and transitive. 3. **Check Reflexivity:** For any integer $a$, $a \equiv a \pmod{m}$ because $m$ divides $a - a = 0$. So, $R$ is reflexive. 4. **Check Symmetry:** If $a \equiv b \pmod{m}$, then $m$ divides $a - b$. This implies $m$ divides $b - a$ (since $b - a = -(a - b)$), so $b \equiv a \pmod{m}$. Hence, $R$ is symmetric. 5. **Check Transitivity:** If $a \equiv b \pmod{m}$ and $b \equiv c \pmod{m}$, then $m$ divides $a - b$ and $b - c$. Adding these, $m$ divides $(a - b) + (b - c) = a - c$, so $a \equiv c \pmod{m}$. Thus, $R$ is transitive. 6. **Conclusion:** Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation on $\mathbb{Z}$. 7. **Addition Property:** Given $x_1 \equiv y_1 \pmod{m}$ and $x_2 \equiv y_2 \pmod{m}$, then $m$ divides $x_1 - y_1$ and $x_2 - y_2$. Adding these, $m$ divides $(x_1 - y_1) + (x_2 - y_2) = (x_1 + x_2) - (y_1 + y_2)$, so $(x_1 + x_2) \equiv (y_1 + y_2) \pmod{m}$. This shows the relation respects addition modulo $m$. **Final answer:** $R$ is an equivalence relation on $\mathbb{Z}$ and addition is compatible with this relation.