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Question: Let A= { 1,2,3......9 } and R be relations in \(A\times A\) defined by (a, b) R (c, d) if \(a+d=b+c\...

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

Explanation

Solution

To solve this question, we will first follow the definition of equivalence relation. A relation R is an equivalence relation if it is a reflexive, symmetric, and transitive relation. Then, we will proceed with the definition of a relation being reflexive, symmetric, and transitive to proceed further.
A relation R’ is a reflexive on set B if (x,x)∈R′\left( x,x \right)\in R' for all x∈Bx\in B
A relation R’ is said to be symmetric on set B if (x,y)∈R′⇒(y,x)∈R′\left( x,y \right)\in R'\Rightarrow \left( y,x \right)\in R'
A relation R’ is said to be transitive on set B if (x,y),(y,z)∈R′⇒(x,z∈R′)\left( x,y \right),\left( y,z \right)\in R'\Rightarrow \left( x,z\in R' \right)
At the end of the solution, we will use the fact that all the elements of a set that are equivalent are in the same equivalence class.

Complete step-by-step solution:
Given that, A=\left\\{ 1,2,3......9 \right\\} and R is a relation defined on A×AA\times A as (a, b) R (c, d) if a+d=b+ca+d=b+c
A relation R is an equivalence relation if it is a reflexive, symmetric, and transitive relation.
A relation R’ is reflexive on set B if (x,x)∈R′\left( x,x \right)\in R' for all x∈Bx\in B
Here, in this case, we have relation R defined as
(a, b) R (c, d)=a+d=b+c . . . . . . . . . . . . (i)\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)=a+d=b+c\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}
Take (a, b) R (a, b) holds true\left( \text{a},\text{ b} \right)\text{ R }\left( \text{a},\text{ b} \right)\text{ holds true} where a,b∈Aa,b\in A
Then using equation (i) we get
(a, b) R (a, b)=a+b=b+a\left( \text{a},\text{ b} \right)\text{ R }\left( \text{a},\text{ b} \right)=a+b=b+a
Clearly as a+b=b+aa+b=b+a
So, (a, b) R (a, b) holds\left( \text{a},\text{ b} \right)\text{ R }\left( \text{a},\text{ b} \right)\text{ holds} So, R is reflexive relation on A×AA\times A
A relation R is said to be symmetric on set B if (x,y)∈R′⇒(y,x)∈R′\left( x,y \right)\in R'\Rightarrow \left( y,x \right)\in R'
Here, let us assume that, (a, b) R (c, d) holds\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\text{ holds} where a,b,c,d∈Aa,b,c,d\in A then if (c,d)R(a,b)\left( c,d \right)R\left( a,b \right) also holds then R will become symmetric.
(a, b) R (c, d) holds\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\text{ holds} then a+d=b+ca+d=b+c
⇒a+d=b+c\Rightarrow a+d=b+c
Reverse both sides of equation we get
b+c=a+d⇒c+b=d+ab+c=a+d\Rightarrow c+b=d+a
(c,d)R(a,b) holds true\left( c,d \right)R\left( a,b \right)\text{ holds true}
Hence the relation R’ is symmetric on A×AA\times A
A relation R is said to be transitive on set B if (x,y),(y,z)∈R′⇒(x,z∈R′)\left( x,y \right),\left( y,z \right)\in R'\Rightarrow \left( x,z\in R' \right)
To show R is transitive let (a, b) R (c, d) and (c,d) R (e,f) holds true\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\text{ and }\left( c,d \right)\text{ R }\left( e,f \right)\text{ holds true}
Where, a,b,c,d,e,f∈Aa,b,c,d,e,f\in A
We have to show that, (a,b)R(e,f)\left( a,b \right)R\left( e,f \right) holds true then as (a, b) R (c, d)\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right) holds true
⇒a+d=b+c . . . . . . . . . . . . . . (ii)\Rightarrow a+d=b+c\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}
And as (c,d) R (e,f) holds true\left( c,d \right)\text{ R }\left( e,f \right)\text{ holds true} holds
⇒c+f=d+e . . . . . . . . . . . . . . (iii)\Rightarrow c+f=d+e\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}
Consider equation (ii) we have
a+d=b+ca+d=b+c
Rearranging terms we have
a−b=c−d . . . . . . . . . . . . . (iv)a-b=c-d\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)}
Consider equation (iii) we have
c+f=d+ec+f=d+e
Rearranging the terms, we have
c−d=e−f . . . . . . . . . . . . . (v)c-d=e-f\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (v)}
Now, the LHS of equation (iv) and equation (v) are same. Then, we have from (iv) and (v) that,

& a-b=c-d=e-f \\\ & \Rightarrow a-b=e-f \\\ \end{aligned}$$ Rearranging the terms, we get: $$\begin{aligned} & a+f=e+b \\\ & \Rightarrow \left( a,b \right)R\left( e,f \right)\text{ holds true} \\\ \end{aligned}$$ So, R is a transitive relation. Hence, we have R is reflexive, symmetric, and transitive. Hence, R is an equivalence relation on $A\times A$ Finally, we have to obtain equivalence class of [(2, 5)] we have $$A=\left\\{ 1,2,3,4,5,6,7,8,9 \right\\}$$ An equivalence class is a name that we give to the subset of a set S which includes all elements that are equivalent to each other. Here [(2, 5)] is equivalent to all those elements (p, q) of set $A\times A$ such that $$\left( 2,5 \right)R\left( p,q \right)\text{ holds true}$$ $$\left( 2,5 \right)R\left( p,q \right)\text{ holds true}$$ when 2-5 = p-q So, $$\begin{aligned} & \left( 2-5 \right)=p-q \\\ & \Rightarrow p-q=-3 \\\ & \Rightarrow q=p+3 \\\ \end{aligned}$$ So, all those elements (p, q) of $A\times A$ satisfying the relation q = p + 3 are in equivalence class of [(2, 5)] So, we will check all possibilities of (p, q) from $A\times A$ such that q = p + 3 Consider (1, 4) then 4 = 1+3 so $$\left( 1,4 \right)\in \text{ equivalence class of }\left[ \left( \text{2},\text{ 5} \right) \right]$$ Similarly, (3, 6), (4, 7), (6, 9), (5, 8) also satisfy the property q = p + 3 Now, as R is an equivalence relation hence, symmetric. So, if $$\left( 2,5 \right)R\left( 1,4 \right)\text{ holds true}$$ $$\Rightarrow \left( 1,4 \right)R\left( 2,5 \right)\text{ also holds true}$$ All the symmetric relation of (3, 6), (4, 7), (6, 9), (5, 8) are also present in equivalence class of [(2, 5)] So, we have equivalence class of [(2, 5)] $$\Rightarrow \left\\{ \begin{aligned} & \left( 2,5 \right)R\left( 1,4 \right),\left( 2,5 \right)R\left( 3,6 \right),\left( 2,5 \right)R\left( 4,7 \right),\left( 2,5 \right)R\left( 6,9 \right),\left( 2,5 \right)R\left( 5,8 \right), \\\ & \left( 1,4 \right)R\left( 2,5 \right),\left( 3,6 \right)R\left( 2,5 \right),\left( 4,7 \right)R\left( 2,5 \right),\left( 6,9 \right)R\left( 2,5 \right),\left( 5,8 \right)R\left( 2,5 \right), \\\ \end{aligned} \right\\}$$ **Note:** The biggest possibility of confusion here in this question can be the point where we are considering $\left( a,b \right)R\left( b, a \right)$ to show R is reflexive. Remember that, in the definition, we have used that R' is reflexive on a set B when $\left( x,x \right)\in R'$. Here, in this question, we had R' relation on $A\times A$ and not on a single set A. Therefore, we will consider $\left( a,b \right)\in A\times A$ and not only $a\in A$ Similar is the case when R is shown to be symmetric and transitive $\left( a,b \right)R\left( c,d \right)$ is taken in place of (a, b). This is because we are considering R on $A\times A$ and not on A.