Let A= x :0 12 . Show that, R= (a, b):a, b , |a-b| is divisible b… - Vidyadip

Question

Let $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\}.$ Show that, $\text{R}=\{(\text{a, b}):\text{a, b}\in\text{A},\ |\text{a}-\text{b}|$ is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].

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Answer

R = {(a, b) : a, b ∈ A, |a – b| is divisible by 4}
Reflexivity: for any a ∈ A
|a – a| = 0, which is divisible by 4
(a, a) ∈ R.
So, R is reflexive.
Symmetry: Let (a, b) ∈ R
|a – b| is divisible by 4
⇒ |b – a| is also divisible by 4
So, R is symmetry
Transitive: Let (a, b) ∈ R & (b, c) ∈ R
|a – b| is divisible by 4
$|\text{a}-\text{b}|=4\lambda$
$\text{a}-\text{b}=\pm4\lambda\ \ \ .....(1)$
$|\text{b}-\text{c}|$ is divisible by 4
$|\text{b}-\text{c}|=4\mu$
$\text{b}-\text{c}=\pm4\mu\ \ \ .....(2)$
Add (1) & (2)
$\text{a}-\text{b}+\text{b}-\text{c}=\pm4(\lambda+\mu)$
$\text{a}-\text{c}=\pm4(\lambda+\mu)$
$\Rightarrow(\text{a, c})\in\text{R}$
So, Transitive
Hence, R is reflexive, Symmetry & Transitive so, it is an equivalence relation
Let x be an element of A such that (x, 1) ∈ R, then
|x – 1| is divisible by 4
x – 1 = 0, 4, 8, 12
⇒ x = 1, 5, 9
Hence, the set of all element of A which are related to 1 in {1, 5, 9}.

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