Question

Let A = \left\\{ {x \in z:0 \leqslant x \leqslant 12} \right\\}. Show that R = \left\\{ {\left( {a,b} \right):a,b \in A,\left| {a - b} \right|is{\text{ }}divisible{\text{ }}by{\text{ }}4} \right\\} is an equivalence relation. Find the set of all elements related to $1$. Also write the equivalence class $\left[ 2 \right]$.

Answer 1

First we have to prove that the given relation is reflexive, symmetric and transitive. A function is said to be Equivalence only if it is reflexive, symmetric and transitive.

Complete step-by-step answer:
Given, A = \left\\{ {x \in z:0 \leqslant x \leqslant 12} \right\\}
\Rightarrow A = \left\\{ {0,1,2,3,4,5,6,7,8,9,10,11,12} \right\\}
R = \left\\{ {\left( {a,b} \right):a,b \in A,\left| {a - b} \right|is{\text{ }}divisible{\text{ }}by{\text{ }}4} \right\\}
To show that the relation is an equivalence relation, we show it is reflexive, symmetric and transitive.
Check reflexive:
For every $a \in A$
$\left| {a - a} \right| = 0$, which is divisible by $4$.
then $\left( {a,b} \right) \in R$ $\left( {a,a} \right) \in R$
$\therefore R$ is reflexive.
Check symmetric:
We know that $\left| {a - b} \right| = \left| {b - a} \right|$
Hence, if $\left| {a - b} \right|$ is a multiple of $4$, then $\left| {b - a} \right|$ is also a multiple of $4$.
Hence, if $\left( {a,b} \right) \in R$, then $\left( {b,a} \right) \in R$
$\therefore R$ is symmetric.
Check transitive:
If $\left| {a - b} \right|$ is a multiple of $4$ $\Rightarrow \left| {a - b} \right| = 4m$,$m \in Z$
Similarly, if $\left| {b - c} \right|$ is a multiple of $4$ $\Rightarrow \left| {b - c} \right| = 4n$,$n \in Z$
Since $a,b,c$ are integers , so we have:
$\left| {a - c} \right| = \left| {a - b + b - c} \right| = \left| {a - b} \right| + \left| {b - c} \right| = 4m + 4n = 4\left( {m + n} \right)$, $m\& n \in Z$
Hence, $\left| {a - c} \right|$ is a multiple of $4$.
Therefore, if $\left| {a - b} \right|$ and $\left| {b - c} \right|$ are the multiples of $4$, then $\left| {a - c} \right|$ is also a multiple of $4$.
i.e., if $\left( {a,b} \right) \in R\& \left( {b,c} \right) \in R \Rightarrow \left( {a,c} \right) \in R$
$\therefore R$ is transitive.
So, we have shown that the relation $R$ is reflexive, symmetric and transitive. Therefore, the relation is an equivalence relation.
The set of elements related to $1$ is \left\\{ {\left( {1,1} \right),\left( {1,5} \right),\left( {1,9} \right),\left( {5,1} \right),\left( {9,1} \right)} \right\\}, since
$\left| {1 - 1} \right| = 0$ is a multiple of $4$,
$\left| {1 - 5} \right| = 4$ is a multiple of $4$,
$\left| {1 - 9} \right| = 8$ is a multiple of $4$,
$\left| {5 - 1} \right| = 4$ is a multiple of $4$and
$\left| {9 - 1} \right| = 8$ is a multiple of $4$.
Let $\left( {x,2} \right) \in R$; $x \in A$
$\left| {x - 2} \right| = 4k$, where $k$ is a whole number and $k \leqslant 3$
$\therefore x = 2,6,10$
So, the equivalence class $\left[ 2 \right]$ is \left\\{ {2,6,10} \right\\}.

Note: Multiples of $4$ are $0,4,8,12$; so $\left| {a - b} \right|$ can be $0,4,8,12$ only. Also noted a point that the sum of multiples of $4$ is also a multiple of $4$. For ex- if $\left| {a - b} \right| + \left| {b - c} \right|$ is a multiple of $4$, then $\left| {a - c} \right|$ is also a multiple of $4$.