Question

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

Answer 1

To solve the problem, we need to find the total number of possible relations from set \( S \) (speakers) to set \( J \) (judges), where each speaker can be judged by any judge.

1. Understanding Relations from One Set to Another:
A relation \( R \) from set \( S \) to set \( J \) is defined as a subset of the Cartesian product \( S \times J \).

That is:
$ R \subseteq S \times J $
The total number of possible relations is equal to the number of subsets of \( S \times J \).

2. Calculating the Size of \( S \times J \):
- Number of elements in \( S = 4 \)
- Number of elements in \( J = 3 \)
Therefore,
$ |S \times J| = 4 \times 3 = 12 $

3. Calculating the Number of Subsets:
The total number of relations = Total number of subsets of \( S \times J \) = \( 2^{|S \times J|} \)
$ \Rightarrow \text{Number of relations} = 2^{12} = 4096 $

Final Answer:
The number of relations from \( S \) to \( J \) is $ \boxed{4096} $.

Answer 2

To solve the problem, we need to determine whether the given function \( f \) is bijective. A function is bijective if it is both injective (one-to-one) and surjective (onto).

1. Understanding the Function:
The function is defined as:
\[ f = \{ (S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3) \} \]
The domain is the set of speakers \( S = \{ S_1, S_2, S_3, S_4 \} \)
The codomain is the set of judges \( J = \{ J_1, J_2, J_3 \} \)

2. Checking Injectivity (One-to-One):
A function is injective if every element of the codomain is mapped by at most one element of the domain.
In the given function:
- \( S_2 \rightarrow J_2 \)
- \( S_3 \rightarrow J_2 \)
Both \( S_2 \) and \( S_3 \) map to the same judge \( J_2 \), so the function is not injective.

3. Checking Surjectivity (Onto):
A function is surjective if every element in the codomain has at least one pre-image in the domain.
Here:
- \( J_1 \) is mapped by \( S_1 \)
- \( J_2 \) is mapped by \( S_2 \) and \( S_3 \)
- \( J_3 \) is mapped by \( S_4 \)
Since all elements in the codomain \( J \) are covered, the function is surjective.

Final Answer:
The function is not bijective because it is not injective (two inputs map to the same output), although it is surjective.

Answer 3

To solve the problem, we need to find the number of one-one (injective) functions from set \( S \) to set \( J \).

1. Understanding One-One Functions:
A function is one-one (injective) if no two elements of the domain map to the same element of the codomain. That is, every element in the domain maps to a unique element in the codomain.

2. Analyzing Set Sizes:
- Set \( S \) (domain) has 4 elements: \( \{ S_1, S_2, S_3, S_4 \} \)
- Set \( J \) (codomain) has 3 elements: \( \{ J_1, J_2, J_3 \} \)

3. Applying Injective Function Rule:
A one-one function from \( S \) to \( J \) is possible only if the number of elements in the domain is less than or equal to the number of elements in the codomain:
$ |S| \leq |J| $
But here,
$ |S| = 4 > 3 = |J| $

Conclusion:
Since the domain has more elements than the codomain, it is not possible to form a one-one function from \( S \) to \( J \).

Final Answer:
The number of one-one functions from \( S \) to \( J \) is $ \boxed{0} $.

Answer 4

To solve the problem, we need to determine the minimum number of ordered pairs that must be added to the relation \( R_1 \) so that it becomes reflexive but not symmetric.

1. Understanding Reflexive Relation:
A relation \( R \) on set \( S = \{S_1, S_2, S_3, S_4\} \) is reflexive if every element is related to itself. That means the relation must include:
\[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \]

2. Current Relation:
Given:
\[ R_1 = \{(S_1, S_2), (S_2, S_4)\} \]

3. Add Ordered Pairs to Make it Reflexive:
We need to ensure that all 4 elements have reflexive pairs. So we must add:
- \( (S_1, S_1) \)
- \( (S_2, S_2) \)
- \( (S_3, S_3) \)
- \( (S_4, S_4) \)
That is 4 ordered pairs in total.

4. Ensure Relation is Not Symmetric:
A relation is symmetric if for every \( (a, b) \in R \), the pair \( (b, a) \) is also in \( R \).
Currently, we have:
- \( (S_1, S_2) \) but not \( (S_2, S_1) \)
- \( (S_2, S_4) \) but not \( (S_4, S_2) \)
So, the relation is already not symmetric, and we must avoid adding symmetric pairs.

5. Final Relation:
To make it reflexive but not symmetric, we add only the missing reflexive pairs. That means adding:
\[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \]

Final Answer:
The minimum number of ordered pairs to be added to make \( R_1 \) reflexive but not symmetric is $ \boxed{4} $.