Problem Statement
Show that the relation $R$ in the set $A = \{x \in \mathbb{Z} : 0 \le x \le 12\}$, given by:
$R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$
is an equivalence relation. Also, find the set of all elements related to 1 (i.e. the equivalence class $[1]$).
Verified Solution & Marking Scheme
Check Reflexivity
For any $a \in A$, $|a - a| = 0 = 0 \times 4$, which is a multiple of 4.
Hence, $(a, a) \in R$ for all $a \in A$. Thus, $R$ is reflexive.
Check Symmetry
Let $(a, b) \in R$.
Then $|a - b|$ is a multiple of 4 $\implies |a - b| = 4k$ for some $k \in \mathbb{Z}_{\ge 0}$.
Since $|b - a| = |-(a - b)| = |a - b| = 4k$, $|b - a|$ is also a multiple of 4.
Hence, $(b, a) \in R$. Thus, $R$ is symmetric.
Check Transitivity
Let $(a, b) \in R$ and $(b, c) \in R$.
Then $|a - b| = 4k_1$ and $|b - c| = 4k_2$, which means:
$a - b = \pm 4k_1 \quad \text{and} \quad b - c = \pm 4k_2$
Adding these two equations:
$(a - b) + (b - c) = a - c = \pm 4k_1 \pm 4k_2 = 4(\pm k_1 \pm k_2) = 4m$
Since $a - c$ is a multiple of 4, $|a - c|$ is a multiple of 4 $\implies (a, c) \in R$.
Thus, $R$ is transitive.
Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation.
Find the Equivalence Class [1]
The set of elements related to 1 is:
$[1] = \{x \in A : (x, 1) \in R\} = \{x \in A : |x - 1| \text{ is a multiple of } 4\}$
Testing elements of $A = \{0, 1, 2, \dots, 12\}$:
- $|1 - 1| = 0$ (multiple of 4) $\implies 1 \in [1]$
- $|5 - 1| = 4$ (multiple of 4) $\implies 5 \in [1]$
- $|9 - 1| = 8$ (multiple of 4) $\implies 9 \in [1]$
Hence, $[1] = \{1, 5, 9\}$.