Question

How many relations can be there from \( S \) to \( J \)?

Answer 1

Relation and Cartesian Product

A relation from \( S \) to \( J \) is any subset of the Cartesian product \( S \times J \).

Step 1: Compute the Number of Elements in \( S \times J \)

The Cartesian product \( S \times J \) contains all ordered pairs \( (x, y) \) where \( x \in S \) and \( y \in J \).

\[ |S \times J| = |S| \times |J| = 4 \times 3 = 12 \]

Step 2: Compute the Number of Possible Relations

Each subset of \( S \times J \) is a possible relation, so the number of relations is:

\[ \text{Total relations} = 2^{|S \times J|} = 2^{12} = 4096 \]

Final Answer:

• Total relations = 4096

Answer 2

Function Analysis

Step 1: Check if \( f \) is a function

Each element in \( S \) is mapped to exactly one element in \( J \), so \( f \) is a function.

Step 2: Check if \( f \) is one-to-one (Injective)

A function is injective if different elements in \( S \) map to different elements in \( J \). Here, \( S_2 \) and \( S_3 \) both map to \( J_2 \), meaning \( f \) is not injective.

Step 3: Check if \( f \) is onto (Surjective)

A function is surjective if every element of \( J \) is mapped by at least one element of \( S \). Here, \( J_1 \), \( J_2 \), and \( J_3 \) are all mapped, so \( f \) is onto.

Final Conclusion:

Since \( f \) is not injective but it is surjective, \( f \) is not bijective.

Answer 3

Injective Function Analysis

A function is one-one (injective) if every element of \( S \) maps to a unique element in \( J \).

Step 1: Count Valid Assignments

Since \( |S| = 4 \) and \( |J| = 3 \), we must assign 4 elements uniquely among 3 elements, which is not possible.

Thus, no injective functions exist.

Final Answer:

No one-one functions exist from \( S \) to \( J \).

Answer 4

Step 1: Make \( R_1 \) Reflexive.
A relation is reflexive if every element \( x \in S \) satisfies \( (x, x) \in R_1 \). The elements in \( S \) are \( S_1, S_2, S_3, S_4 \), so we must include: \[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4). \] Step 2: Ensure \( R_1 \) is Not Symmetric.
A relation is symmetric if \( (a, b) \in R_1 \Rightarrow (b, a) \in R_1 \). Since \( (S_1, S_2) \in R_1 \) but \( (S_2, S_1) \notin R_1 \), it is not symmetric. Final Answer:
The minimum ordered pairs to include are: \[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4). \] \[ \text{to ensure reflexivity while keeping the relation asymmetric}. \]