Consider the non-empty set consisting of children in a family. State giving reason whether each of the following relations is (i) Symmetric (ii) Transitive
x likes y

Answer

To determine whether the relation "x likes y" is symmetric and transitive, we can analyze the definitions of these properties in the context of the set of children in a family.

### Step-by-Step Solution:

1. **Understanding the Relation**:
We define the relation \( R \) as \( (x, y) \in R \) if "x likes y". Here, \( x \) and \( y \) are children in a family.

2. **Checking for Symmetry**:
- A relation is symmetric if for all \( x, y \) in the set, if \( (x, y) \in R \) then \( (y, x) \in R \).
- In our case, if child \( x \) likes child \( y \), it does not necessarily mean that child \( y \) likes child \( x \). For example, one child may like another without the feeling being mutual.
- Therefore, we can conclude that the relation "x likes y" is **not symmetric**.

3. **Checking for Transitivity**:
- A relation is transitive if for all \( x, y, z \) in the set, if \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \).
- In our case, if child \( x \) likes child \( y \) and child \( y \) likes child \( z \), it does not imply that child \( x \) likes child \( z \). For instance, child \( x \) may like child \( y \), and child \( y \) may like child \( z \), but child \( x \) may not have any feelings towards child \( z \).
- Therefore, we can conclude that the relation "x likes y" is **not transitive**.

### Final Conclusion:
The relation "x likes y" is neither symmetric nor transitive.

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