Relation on \( \mathbb{N}\times\mathbb{N} \) Defined by \( a+d=b+c \)
📺 Video Explanation
📝 Question
Let relation \( R \) on \( \mathbb{N}\times\mathbb{N} \) be defined by:
\[ (a,b)\,R\,(c,d)\iff a+d=b+c \]
Then, \( R \) is:
- (a) reflexive but not symmetric
- (b) reflexive and transitive but not symmetric
- (c) an equivalence relation
- (d) none of these
✅ Solution
To check whether \(R\) is an equivalence relation, test reflexive, symmetric, and transitive.
🔹 Reflexive
For every \((a,b)\in \mathbb{N}\times\mathbb{N}\):
\[ (a,b)\,R\,(a,b) \]
Check:
\[ a+b=b+a \]
This is always true.
✔ Reflexive.
🔹 Symmetric
If:
\[ (a,b)\,R\,(c,d) \]
then:
\[ a+d=b+c \]
Rearranging:
\[ c+b=d+a \]
which means:
\[ (c,d)\,R\,(a,b) \]
✔ Symmetric.
🔹 Transitive
Suppose:
\[ (a,b)\,R\,(c,d) \quad \text{and} \quad (c,d)\,R\,(e,f) \]
Then:
\[ a+d=b+c \quad …(1) \]
\[ c+f=d+e \quad …(2) \]
Add (1) and (2):
\[ a+d+c+f=b+c+d+e \]
Cancel \(c\) and \(d\):
\[ a+f=b+e \]
Thus:
\[ (a,b)\,R\,(e,f) \]
✔ Transitive.
🎯 Final Answer
\[ \boxed{\text{R is an equivalence relation}} \]
✔ Correct option: (c)
🚀 Exam Shortcut
- Relations based on equality of expressions are often equivalence relations
- Check symmetry by swapping sides
- For transitivity, add equations and simplify