Q.Let $A = \{1, 2, 3, \ldots, 9\}$ 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 $[(2, 5)]$.

Concept: Equivalence Relation Proof — Show reflexivity, symmetry, and transitivity.

Step 1 – Reflexivity: For any $(a,b) \in A \times A$, $a+b = b+a$, so $(a,b)R(a,b)$.

Step 2 – Symmetry: If $(a,b)R(c,d)$, then $a+d = b+c$. This implies $c+b = d+a$, so $(c,d)R(a,b)$.

Step 3 – Transitivity: If $(a,b)R(c,d)$ and $(c,d)R(e,f)$, then $a+d = b+c$ and $c+f = d+e$. Adding: $a+d+c+f = b+c+d+e \Rightarrow a+f = b+e$, so $(a,b)R(e,f)$.

Thus $R$ is an equivalence relation. …

The core idea here is that the condition $a+d = b+c$ can be rearranged to $a - b = c - d$. So two ordered pairs are related precisely when they have the same difference between the first and second components. This immediately suggests that the relation groups pairs by a common number — the difference $a-b$. Proving it's an equivalence relation then becomes a matter of checking that "having the same difference" is reflexive, symmetric, and transitive.

Let’s go step by step.

1. Reflexive: For any $(a,b) \in A \times A$, we need $(a,b) \, R \, (a,b)$.

   Check the condition: $a + b = b + a$ is always true (addition is commutative). So $R$ is reflexive.

2. Symmetric: If $(a,b) \, R \, (c,d)$, then $a + d = b + c$. We need $(c,d) \, R \, (a,b)$, i.e., $c + b = d + a$.

   But $c + b = b + c$ and $d + a = a + d$, so the equality $a+d = b+c$ directly gives $c+b = d+a$. Hence $R$ is symmetric.

3. Transitive: Suppose $(a,b) \, R \, (c,d)$ and $(c,d) \, R \, (e,f)$. Then:

Add the two equations:

Cancel $c$ and $d$ from both sides (they appear on both left and right):

That is exactly $(a,b) \, R \, (e,f)$. So $R$ is transitive.

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

Method: Proving an equivalence relation and extracting a class

Use this for 'prove $R$ is an equivalence relation and find $[a]$' questions, especially relations on $A\times A$ defined by an equation.

Steps

Step 1: Rewrite the defining condition into a cleaner invariant.

A condition like $a+d=b+c$ rearranges to $a-b=c-d$: two pairs relate exactly when they share the same value of some quantity (here the difference of components). Spotting this invariant makes all three proofs almost automatic.

Step 2: Verify reflexive, symmetric, transitive on the condition.

• Reflexive: the pair shares its own invariant — always true.
• Symmetric: the defining equation is unchanged when the two pairs are swapped.
• Transitive: add/combine the two defining equations and cancel the shared terms to recover the required relation.

Step 3: Build the equivalence class $[a]$ by fixing the invariant. …

Common Mistakes

Mistake 1: Listing class members without keeping both components inside $A=\{1,\dots,9\}$.

Why it's wrong: $(7,10)$ satisfies $a-b=-3$ but $10\notin A$, so it is not in the class. Correct approach: generate candidates from $b=a+3$, then discard any pair with a component outside $A$.

Mistake 2: Sign slip when converting $a+d=b+c$ to a difference.

Why it's wrong: for $(a,b)R(2,5)$ the condition is $a+5=b+2$, giving $a-b=-3$ (so $b=a+3$), not $a-b=3$. Correct approach: substitute the fixed pair carefully and solve for the exact difference. …