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Show that the relation R defined by $(a,b)R(c,d)\Rightarrow a+d=b+c$ on $A\times A$ where $A=\left \{ 1,2,3...10 \right \}$ is an equivalence relation. Hence write the equivalence [(3,4)]; $a,b,c,d\in A.$

Answers (1)

Here $(a,b)R(c,d)=a+d=b+c$ on $A\times A$ where $A=\left \{ 1,2,3...10 \right \}$

Reflexivity :

Let (a,b) be an arbitrary element of $A\times A$ . Then $(a,b)\in A\times A \; \; \; \forall \: \: a,b\in A$

So, $a+b=b+a\Rightarrow (a,b)R(a,b)$

Thus $(a,b)R(a,b)\: \: \forall (a,b)\in A\times A.$

Hence, R is reflexive.

Symmetry :

Let $(a,b),(c,d)\in A\times A$ be such that $(a,b)R(c,d)$ .

Then, $a+d=b+c\Rightarrow c+d=d+a\Rightarrow (c,d)R(a,b)$

Then, $(a,b)R(c,d)\Rightarrow (c,d)R(a,b)\; \; \; \; \forall \; (a,b),(c,d)\in A\times A$

Hence R is symmetric.

Transitivity :

Let $(a,b),(c,d),(e,f)\in A\times A$ be such that $(a,b)R(c,d)$ and $(c,d)R(e,f).$ Then $a+d=b+c$ and $c+f=d+e$

$\Rightarrow (a+d)+(c+f)=(b+c)(d+e)$

$\Rightarrow (a+d)+(c+f)=(b+c)(d+e)$

$\Rightarrow a+f=b+e$

$\Rightarrow (a,b)R(e,f)$

That is $(a,b)R(c,d)$ and $(c,d)R(e,f)\Rightarrow (a,b)R(e,f)$ $\forall (a,b),(c,d),(e,f)\in A\times A$

Hence R is transitive.

Since R is reflexive, symmetric and transitive, so R is an equivalence relation as well. For the equivalence class [3,4], we need to find (a,b) such that (a,b)R(3,4)

$\Rightarrow a+4=b+3\Rightarrow b-a=1$

So, $[(3,4)]=\left \{ (1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10) \right \}$

Posted by

Ravindra Pindel

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