Question

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

Answer 1

To determine the total possible relations from the set of speakers (S) to the set of judges (J), where each speaker can be evaluated by any judge, we follow these steps.

1. Definition of Relations:
A relation R from set S to set J is a subset of the Cartesian product \( S \times J \).

Thus:
$ R \subseteq S \times J $
The count of all possible relations equals the number of subsets of \( S \times J \).

2. Size of the Cartesian Product \( S \times J \):
- Cardinality of \( S = 4 \)
- Cardinality of \( J = 3 \)
Hence,
$ |S \times J| = 4 \times 3 = 12 $

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

Conclusion:
The total number of possible relations from set \( S \) to set \( J \) is \( \boxed{4096} \).

Answer 2

To address the problem, it is necessary to determine if the provided function \( f \) is bijective. A function qualifies as bijective if it exhibits both injective (one-to-one) and surjective (onto) properties.

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

2. Injectivity Assessment (One-to-One):
A function is considered injective if each element in the codomain corresponds to at most one element from the domain.
Observing the given function:- \( S_2 \) maps to \( J_2 \)- \( S_3 \) maps to \( J_2 \)Since both \( S_2 \) and \( S_3 \) are associated with the identical judge \( J_2 \), the function is demonstrably not injective.

3. Surjectivity Assessment (Onto):
A function is deemed surjective if every element within the codomain possesses at least one pre-image in the domain.
In this instance:- \( J_1 \) is the image of \( S_1 \)- \( J_2 \) is the image of \( S_2 \) and \( S_3 \)- \( J_3 \) is the image of \( S_4 \)As all elements in the codomain \( J \) are accounted for, the function is surjective.

Conclusion:
The function is not bijective due to its lack of injectivity (as evidenced by multiple domain elements mapping to a single codomain element), despite possessing surjective properties.

Answer 3

To address the problem, we must determine the count of injective functions from set \( S \) to set \( J \).

1. Definition of Injective Functions:
An injective function is characterized by the property that distinct elements in the domain map to distinct elements in the codomain. In other words, each element of the domain is associated with a unique element of the codomain.

2. Cardinality of Sets:
- Set \( S \) (domain) comprises 4 elements: \( \{ S_1, S_2, S_3, S_4 \} \)
- Set \( J \) (codomain) comprises 3 elements: \( \{ J_1, J_2, J_3 \} \)

3. Condition for Injective Functions:
The existence of an injective function from \( S \) to \( J \) is contingent upon the cardinality of the domain not exceeding that of the codomain:
$ |S| \leq |J| $
In this instance,
$ |S| = 4 > 3 = |J| $

Conclusion:
Given that the domain contains more elements than the codomain, it is impossible to construct an injective function from \( S \) to \( J \).

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

Answer 4

The objective is to determine the minimum number of ordered pairs to append to relation \( R_1 \) to render it reflexive while maintaining non-symmetry.

1. Definition of Reflexivity:
A relation \( R \) on a set \( S = \{S_1, S_2, S_3, S_4\} \) is reflexive if for every element \( x \in S \), the pair \( (x, x) \) is in \( R \). This requires the presence of:
\[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \]

2. Initial Relation:
The given relation is:
\[ R_1 = \{(S_1, S_2), (S_2, S_4)\} \]

3. Achieving Reflexivity:
To make the relation reflexive, the following pairs must be added to cover all elements in the set \( S \):
- \( (S_1, S_1) \)
- \( (S_2, S_2) \)
- \( (S_3, S_3) \)
- \( (S_4, S_4) \)
This constitutes a total of 4 ordered pairs.

4. Ensuring Non-Symmetry:
A relation is symmetric if for every pair \( (a, b) \) in the relation, the pair \( (b, a) \) is also present. Examining \( R_1 \):
- \( (S_1, S_2) \) is present, but \( (S_2, S_1) \) is not.
- \( (S_2, S_4) \) is present, but \( (S_4, S_2) \) is not.
Thus, \( R_1 \) is inherently not symmetric. To preserve non-symmetry, no pairs \( (b, a) \) should be added if \( (a, b) \) already exists and \( a eq b \). The addition of only reflexive pairs does not introduce symmetry.

5. Resulting Relation:
By adding only the necessary reflexive pairs, the relation becomes reflexive without becoming symmetric. The added pairs are:
\[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \]

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