Let R be the relation on the set Z defined by xRy iff x − y is an integer. Prove that R is anequivalence relation on Z.

To prove that $R$ is an equivalence relation on $\mathbb{Z}$, we need to show that $R$ is reflexive, symmetric, and transitive.

1. Reflexive: A relation $R$ on a set $A$ is reflexive if every element is related to itself. That is, for all $x \in \mathbb{Z}$, $xRx$ must hold.

   • Given $x \in \mathbb{Z}$, we need to check if $x - x$ is an integer.
   • $x - x = 0$, and $0$ is an integer.
   • Therefore, $xRx$ holds for all $x \in \mathbb{Z}$, so $R$ is reflexive.

2. Symmetric: A relation $R$ on a set $A$ is symmetric if for all $x, y \in \mathbb{Z}$, whenever $xRy$, then $yRx$ must also hold.

   • Suppose $xRy$. By definition, $x - y$ is an integer.
   • We need to show that $yRx$. This means $y - x$ must be an integer.
   • Since $x - y$ is an integer, and the negative of an integer is also an integer, $y - x = -(x - y)$ is an integer.
   • Therefore, $yRx$ holds whenever $xRy$, so $R$ is symmetric.

3. Transitive: A relation $R$ on a set $A$ is transitive if for all $x, y, z \in \mathbb{Z}$, whenever $xRy$ and $yRz$, then $xRz$ must also hold.

   • Suppose $xRy$ and $yRz$. By definition, $x - y$ and $y - z$ are integers.
   • We need to show that $xRz$. This means $x - z$ must be an integer.
   • Since $x - y$ and $y - z$ are integers, their sum $(x - y) + (y - z) = x - z$ is also an integer.
   • Therefore, $xRz$ holds whenever $xRy$ and $yRz$, so $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, we conclude that $R$ is an equivalence relation on $\mathbb{Z}$.