Let `A={a_(1),a_(2),a_(3),a_(4),a_(5)}andB={b_(1),b_(2),b_(3),b_(4)}`, when `a_(l)'s and b_(l)'s` are school going students. Define a relation from a set A to set B by x R y iff y is a true friend of x.
If `R={(a_(1),b_(1)),(a_(2),b_(1)),(a_(3),b_(3)),(a_(4),b_(2)),(a_(5),b_(2))}`
Is R a bijective function?

Answer

To determine whether the relation \( R \) defined from set \( A \) to set \( B \) is a bijective function, we need to check if it satisfies the properties of being one-to-one (injective) and onto (surjective).

### Step-by-Step Solution:

1. **Identify the Sets and the Relation**:
- Let \( A = \{a_1, a_2, a_3, a_4, a_5\} \)
- Let \( B = \{b_1, b_2, b_3, b_4\} \)
- The relation \( R \) is given as:
\[ R = \{(a_1, b_1), (a_2, b_1), (a_3, b_3), (a_4, b_2), (a_5, b_2)\} \]

2. **Check for One-to-One (Injective)**:
- A function is one-to-one if different elements in the domain map to different elements in the codomain.
- In \( R \):
- \( a_1 \) maps to \( b_1 \)
- \( a_2 \) also maps to \( b_1 \)
- \( a_3 \) maps to \( b_3 \)
- \( a_4 \) maps to \( b_2 \)
- \( a_5 \) also maps to \( b_2 \)
- Here, both \( a_1 \) and \( a_2 \) map to \( b_1 \), and \( a_4 \) and \( a_5 \) map to \( b_2 \). Therefore, the same element in \( B \) has multiple pre-images in \( A \).
- **Conclusion**: \( R \) is **not one-to-one**.

3. **Check for Onto (Surjective)**:
- A function is onto if every element in the codomain has at least one pre-image in the domain.
- In \( R \), we have:
- \( b_1 \) has pre-images \( a_1 \) and \( a_2 \)
- \( b_2 \) has pre-images \( a_4 \) and \( a_5 \)
- \( b_3 \) has pre-image \( a_3 \)
- **However**, \( b_4 \) has no pre-image in \( A \).
- **Conclusion**: \( R \) is **not onto**.

4. **Final Conclusion**:
- Since \( R \) is neither one-to-one nor onto, it cannot be a bijective function.
- Therefore, \( R \) is **not a bijective function**.