Question

Show that the relation \( R \) in the set of natural numbers \( N \times N \) defined by \( (a, b)\, R\, (c, d) \) if \( a + d = b + c \) is an equivalence relation.

Hint:

To prove equivalence relation, always verify: Reflexive + Symmetric + Transitive.

Answer 1

Concept: A relation is an equivalence relation if it satisfies three properties: \begin{itemize} \item Reflexive \item Symmetric \item Transitive \end{itemize} Given relation: \[ (a, b)\, R\, (c, d) \iff a + d = b + c \] Step 1: Reflexive property.
A relation is reflexive if every element is related to itself. Check: \[ (a, b)\, R\, (a, b) \] Condition: \[ a + b = b + a \] This is true for all natural numbers. Hence, \( R \) is reflexive.
Step 2: Symmetric property.
A relation is symmetric if: \[ (a, b)\, R\, (c, d) \Rightarrow (c, d)\, R\, (a, b) \] Given: \[ a + d = b + c \] Rewriting: \[ c + b = d + a \Rightarrow c + b = d + a \] Which is equivalent to: \[ c + b = d + a \Rightarrow (c, d)\, R\, (a, b) \] Hence, \( R \) is symmetric.
Step 3: Transitive property.
A relation is transitive if: \[ (a, b)\, R\, (c, d) \text{ and } (c, d)\, R\, (e, f) \Rightarrow (a, b)\, R\, (e, f) \] Given: \[ a + d = b + c \quad \cdots (1) \] \[ c + f = d + e \quad \cdots (2) \] Add (1) and (2): \[ a + d + c + f = b + c + d + e \] Cancel common terms \( c + d \): \[ a + f = b + e \] Thus, \[ (a, b)\, R\, (e, f) \] Hence, \( R \) is transitive. Conclusion:
Since the relation is reflexive, symmetric, and transitive, \( R \) is an equivalence relation.