Question

Consider the relation \(R=\\{(x, y) \in \mathbb{R} \times \mathbb{R}: x-y \in \mathbb{Z}\\}\) on \(\mathbb{R} .\) Prove that this relation is reflexive, symmetric and transitive.

Short answer

The relation \(R\) is reflexive, symmetric, and transitive.

Step-by-step solution

Proving reflexivity

For all \(x \in \mathbb{R}\), the number \(x - x = 0\) is indeed an integer, since 0 belongs to the set of integers \( \mathbb{Z} \). Therefore, \(xRx\) for all \(x \in \mathbb{R}\), which means that the relation \(R\) is reflexive.

Proving symmetry

Assume \(xRy\), that is \(x - y\) is an integer. Then \(-(x - y) = y - x\) is also an integer, since the negation of an integer is also an integer. Therefore, \(yRx\). This shows that the relation \(R\) is symmetric.

Proving transitivity

Assume \(xRy\) and \(yRz\), that is \(x - y\) and \(y - z\) are both integers. Then, the sum \((x - y) + (y - z) = x - z\) is also an integer, because the sum of two integers is also an integer. Therefore, \(xRz\). This shows that the relation \(R\) is transitive.