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

Hint: 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 $\left( x,x \right)\in R'$ for all $x\in B$
A relation R’ is said to be symmetric on set B if \[\left( x,y \right)\in R'\Rightarrow \left( y,x \right)\in R'\]
A relation R’ is said to be transitive on set B if \[\left( x,y \right),\left( y,z \right)\in R'\Rightarrow \left( x,z\in R' \right)\]
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 $A\times A$ as (a, b) R (c, d) if $a+d=b+c$
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 $\left( x,x \right)\in R'$ for all $x\in B$
Here, in this case, we have relation R defined as
\[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)=a+d=b+c\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Take \[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{a},\text{ b} \right)\text{ holds true}\] where $a,b\in A$
Then using equation (i) we get
\[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{a},\text{ b} \right)=a+b=b+a\]
Clearly as \[a+b=b+a\]
So, \[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{a},\text{ b} \right)\text{ holds}\] So, R is reflexive relation on $A\times A$
A relation R is said to be symmetric on set B if \[\left( x,y \right)\in R'\Rightarrow \left( y,x \right)\in R'\]
Here, let us assume that, \[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\text{ holds}\] where $a,b,c,d\in A$ then if \[\left( c,d \right)R\left( a,b \right)\] also holds then R will become symmetric.
\[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\text{ holds}\] then \[a+d=b+c\]
\[\Rightarrow a+d=b+c\]
Reverse both sides of equation we get
\[b+c=a+d\Rightarrow c+b=d+a\]
\[\left( c,d \right)R\left( a,b \right)\text{ holds true}\]
Hence the relation R’ is symmetric on $A\times A$
A relation R is said to be transitive on set B if \[\left( x,y \right),\left( y,z \right)\in R'\Rightarrow \left( x,z\in R' \right)\]
To show R is transitive let \[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\text{ and }\left( c,d \right)\text{ R }\left( e,f \right)\text{ holds true}\]
Where, $a,b,c,d,e,f\in A$
We have to show that, \[\left( a,b \right)R\left( e,f \right)\] holds true then as \[\left( \text{a},\text{ b} \right)\text{ R }\left( \text{c},\text{ d} \right)\] holds true
\[\Rightarrow a+d=b+c\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
And as \[\left( c,d \right)\text{ R }\left( e,f \right)\text{ holds true}\] holds
\[\Rightarrow c+f=d+e\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
Consider equation (ii) we have
\[a+d=b+c\]
Rearranging terms we have
\[a-b=c-d\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)}\]
Consider equation (iii) we have
\[c+f=d+e\]
Rearranging the terms, we have
\[c-d=e-f\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (v)}\]
Now, the LHS of equation (iv) and equation (v) are same. Then, we have from (iv) and (v) that,
\[\begin{align}
  & a-b=c-d=e-f \\
 & \Rightarrow a-b=e-f \\
\end{align}\]
Rearranging the terms, we get:
\[\begin{align}
  & a+f=e+b \\
 & \Rightarrow \left( a,b \right)R\left( e,f \right)\text{ holds true} \\
\end{align}\]
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, \[\begin{align}
  & \left( 2-5 \right)=p-q \\
 & \Rightarrow p-q=-3 \\
 & \Rightarrow q=p+3 \\
\end{align}\]
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)]
\[\Rightarrow \left\{ \begin{align}
  & \left( 2,5 \right)R\left( 1,4 \right),\left( 2,5 \right)R\left( 3,6 \right),\left( 2,5 \right)R\left( 4,7 \right),\left( 2,5 \right)R\left( 6,9 \right),\left( 2,5 \right)R\left( 5,8 \right), \\
 & \left( 1,4 \right)R\left( 2,5 \right),\left( 3,6 \right)R\left( 2,5 \right),\left( 4,7 \right)R\left( 2,5 \right),\left( 6,9 \right)R\left( 2,5 \right),\left( 5,8 \right)R\left( 2,5 \right), \\
\end{align} \right\}\]

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.