Givena relation R on N defined by aRb if 2a+3b is a multiple of 5
	Reflexivity: For aRa, check 2a+3a=5a, which is a multiple of 5. Thus, R is reflexive.
	Symmetry: If aRb (i.e, 2a+3b=5α), then to check bRa, consider
2b+3a=2b+((5α-2a)/3)⋅3=5((3α-b)/2)
	1/2=1/2
Simplifying shows that 2b+3a is also a multiple of 5, so R is symmetric
	Transitivity: If aRb and bRc hold, meaning 2a+3b=5α and 2b+3c=5β, then:
2a+5b+3c=5(α+β)
	Rearranging gives 2a+3c=5(α+β-b), showing aRc, so R is transitive. Since R is reflexive, symmetric, and transitive, it is an equivalence relation.


based on the above method solve below question

Q-1. Let R be a relation defined on the sef of integers Z as a R b if and only if  a2+b2 is even, where a b e Z Then R is 
A. Reflexive but not synmetric
B. Symmetric but not Transitive
C. Not Reflexive
D. an Equivalerce Relation


Let's analyze the given relation \( R \) on the set of integers \( \mathbb{Z} \), where \( aRb \) is defined as \( a^2 + b^2 \) being even. We need to determine which properties the relation \( R \) satisfies.

### 1. Reflexivity:
A relation \( R \) is reflexive if \( aRa \) holds for every \( a \in \mathbb{Z} \). For reflexivity, we check:
\[
aRa \implies a^2 + a^2 = 2a^2
\]
Since \( 2a^2 \) is clearly even for any integer \( a \), the relation \( R \) is **reflexive**.

### 2. Symmetry:
A relation \( R \) is symmetric if \( aRb \) implies \( bRa \). Let's check:
\[
aRb \implies a^2 + b^2 \text{ is even}.
\]
Symmetry requires:
\[
bRa \implies b^2 + a^2 \text{ is even}.
\]
Since \( a^2 + b^2 \) is the same as \( b^2 + a^2 \) (addition is commutative), \( R \) is **symmetric**.

### 3. Transitivity:
A relation \( R \) is transitive if \( aRb \) and \( bRc \) imply \( aRc \). Assume:
\[
aRb \implies a^2 + b^2 \text{ is even}, \quad \text{and} \quad bRc \implies b^2 + c^2 \text{ is even}.
\]
We need to check if this implies:
\[
aRc \implies a^2 + c^2 \text{ is even}.
\]
Consider counterexamples:
- Let \( a = 1 \), \( b = 1 \), and \( c = 2 \). 
  - \( aRb \) holds because \( 1^2 + 1^2 = 2 \), which is even.
  - \( bRc \) holds because \( 1^2 + 2^2 = 5 \), which is odd.
  - \( aRc \) does not hold because \( 1^2 + 2^2 = 5 \), which is odd.

This shows that the relation \( R \) is **not transitive**.

### Conclusion:
The relation \( R \) is reflexive and symmetric but not transitive. Therefore, the correct answer is:

**B. Symmetric but not Transitive**

---

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. What conditions make a relation an equivalence relation?
2. Can a relation be transitive but not symmetric?
3. How do counterexamples help in proving the failure of transitivity?
4. What is the significance of commutativity in checking symmetry?
5. How does the parity of integers (even/odd) affect relation properties?
6. Can a relation be both reflexive and transitive but not symmetric?
7. What are the common pitfalls in determining transitivity?
8. What are other common examples of symmetric but not transitive relations?

**Tip:** When working with relations, constructing counterexamples is a powerful method to test whether the properties like transitivity hold.

Explore the relation R defined on integers Z where aRb if and only if a^2 + b^2 is even. Learn about its properties, including reflexivity and symmetry. Discover why R is not transitive through examples and mathematical reasoning.

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