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# Show that each of the relation $R$ in the set $A=\{x\in Z: 0 \leq x \leq 12\}$, given by $(i) R = \{(a,b) |a-b|$ is a multiple of 4 $\}$ is an equivalence relation. Find the set of all elements related to 1.

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com.This question has appeared in model paper 2012.

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Toolbox:
• A relation R in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
• A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
• A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
• A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
• $|a-b|=|b-a|$
Given a set $A=\{x \in Z:0 \leq x\leq 12\}$ and relation $R=\{(a,b):|a-b| \;is\;a\;multiple\; of \;4\}$:
For every $a \in A (a,a) \in R$, $|a-a|=0$, which is a multiple of 4. Therefore R is reflexive.
If $(a,b) \in R$ then $|a-b|$ is a multiple of 4. Similarly, if $(b,a) \in R$, then $|b-a|$ is also a multiple of 4.
Since $|a-b|=|b-a|$, if $(a,b) \in R$, then $(b,a) is also \in R$. Hence R is symmetric.
Let $(a,b) \in R$ and $(b,c) \in R$.
$\Rightarrow$ $|a-b|$ and $|b-c|$ are both multiples of 4.