question_answer

Let A = {1, 2, 3,..., 9} and R be the relation in \[A\times A\] defined by (a, b)R(c, d), if  \[a+d=b+c\]for (a, b), (c, d) in \[A\times A.\] Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

OR

Prove that the relation R in set A = {1, 2, 3, 4, 5} given by \[R=\{(a,\,b):|a-b|\]is even}is even} is an equivalence relation.

Answer:

Given, A = { 1, 2, 3,..., 9} and a relation R in \[A\times A\] defined by (a, b)R(c, d), if  \[a+d=b+c.\]

Reflexivity Let \[(a,\,\,b)\in A\,\times A\] be any arbitrary element. Now,\[a+b=b+a\]

[\[\because \] addition is commutative on the set of real numbers]

Therefore, (a, b) R (a, b).

\[\because \]       \[(a,\,\,b)\in A\times A\] was arbitrary.

\[\therefore \]      R is reflexive.

Symmetricity Let \[(a,\,\,b),(c,\,\,d)\in A\times A\] such that

(a, b)R(c, d).

\[\because \]       (a, b)R(c, d)

\[\therefore \]      \[a+d=b+c\]

\[\Rightarrow \]   \[c+b=d+a\]

[\[\because \] addition is commutative on the set of real numbers]

\[\Rightarrow \]   (c, d) R (a, b)

So, R is symmetric,

Transitivity Let \[\left( a,\,\,b \right),\,\,\left( c,\,\,d \right),~\,\,(e,\,\,f)\in A\times A\] such that (a, b)R(c, d) and (c, d)R(e, f).

Since, (a, b)R(c, d) and (c, d)R(e ,f).

So. We have      \[a+d=b+c\]                      ...(i)

and                   \[c+f=d+e\]                    ...(ii)

On adding Eqs. (i) and (ii), we get

\[(a+d)+(c+f)=(b+c)+(d+e)\]

\[\Rightarrow \]   \[a+f=b+e\]

\[\Rightarrow \] (a, b)R(e, f)

So, R is transitive.

Hence, R is an equivalence relation.

Now, equivalence class of (2, 5) is given by

\[[(2,\,\,5)]=\{(a,\,\,b)\in A\times A:\,(a,\,\,b)R(2,\,\,5)\}\]

\[=\{(a,\,\,b)\in A\times A:a+5=b+2\}\]

\[=\{(a,\,\,b)\in A\times A:a-b=-\,3\}\]

= {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8),

(6, 9)}

Hence proved.

OR

The given relation is \[R=\left\{ (a,\,\,b):|a-b| \right.\] is even} defined on se A = {1, 2, 3, 4, 5}.

To show that R is an equivalence relation, we show that it is reflexive, symmetric and transitive.

(i) Reflexive \[As|x-x|=0\] is even, \[\forall \,\,x\in A.\]

\[\Rightarrow \]  \[(x,\,\,x)\in R,\] \[\forall x\in A.\]

Hence, R is reflexive.

(ii) Symmetric  As \[(x,\,\,y)\in R\,\,\,\Rightarrow \,\,\,|x-y|\] is even

[by the definition of given relation]

\[\Rightarrow \] \[y-x|\] is also even   \[[\because \,\,\,|a|\,\,=\,\,|-a|,\,\,\forall \,\,a\in R]\]

\[\Rightarrow \] \[(y,\,\,x)\in R,\] \[\forall \,\,x,\] \[y\in A\]

We have shown that \[(x,\,\,y)\in R\]

\[\Rightarrow \] \[(y,\,\,x)\in R,\] \[\forall \,\,x,\] \[y\in A\]

Hence, R is symmetric.                    (1)

(iii) Transitive As \[(x,\,\,y)\in R\] and \[(y,\,\,z)\in R\]

\[\Rightarrow \] \[|x-y|\] id rven and \[|y-z|\] id even

[by using definition of given relation]

Now, \[|x\,-y|\] is even

\[\Rightarrow \]  x and y both are even or odd and  \[|y-x|\] is even

\[\Rightarrow \]  y and x both are even or odd

There are two cases arise

Case I When y is even.

Now, \[(x,\,\,y)\in R\] and \[(y,\,\,z)\in R\]

\[\Rightarrow \] \[|x\,-y|\] is even and \[|y\,-z|\] is even

\[\Rightarrow \] x is even and z is even

\[\Rightarrow \] \[|x\,-z|\] is even

[\[\because \] difference of two even numbers is also even]

\[\Rightarrow \] \[(x,\,\,z)\in R\]

Case II When y is odd.

Now, \[(x,\,\,y)\in R\] and \[(y,\,\,z)\in R\]

\[\Rightarrow \] \[|x\,-y|\] is even and \[|y\,-z|\] is even

\[\Rightarrow \] x is even and z is odd

\[\Rightarrow \] \[|x\,-z|\] is even

[\[\because \] difference of two even numbers is even]

\[\Rightarrow \]   \[(x,\,\,z)\in R\]

So, we have shown that

\[(x,\,\,y)\in R\] and \[(y,\,\,z)\in R\]

\[\Rightarrow \]   \[(x,\,\,z)\in R,\] \[\forall \,\,x,\] \[y,\,\,z\in A\]

\[\therefore \] R is transitive.

Since, R is reflexive, symmetric and transitive. So, it is an equivalence relation.