Question

How many relations are possible from Set \( A \to \) Set \( B \)?

Answer 1

Concept:
• A relation from set \( A \) to set \( B \) is any subset of the Cartesian product \( A \times B \).
• If \( n(A) = p \) and \( n(B) = q \), then the number of elements in \( A \times B \) is \( p \times q \).
• The total number of subsets of a set with \( m \) elements is \( 2^m \).

Step 1: Find the number of elements in each set
Set \( A \) contains 4 elements: \( n(A) = 4 \).
Set \( B \) contains 3 elements: \( n(B) = 3 \).

Step 2: Calculate the number of elements in the Cartesian product \( A \times B \)
\[ n(A \times B) = n(A) \times n(B) \] \[ n(A \times B) = 4 \times 3 = 12 \]

Step 3: Determine the total number of relations
The total number of possible relations is the total number of subsets of \( A \times B \). \[ \text{Total Relations} = 2^{n(A \times B)} \] \[ \text{Total Relations} = 2^{12} \] \[ \text{Total Relations} = 4096 \]

Answer 2

Concept:
• A relation is a function if every element in the domain has exactly one image in the codomain.
• A function is injective (one-to-one) if distinct elements in the domain have distinct images in the codomain.

Step 1: Check if the relation \( R \) qualifies as a function
In the given relation \( R \): The element \( G_1 \in A \) is associated with two different elements in \( B \), namely \( B_1 \) and \( B_2 \). \[ (G_1, B_1) \in R \quad \text{and} \quad (G_1, B_2) \in R \] By the definition of a function, an element in the domain cannot have more than one image. Therefore, \( R \) is not a function.

Step 2: Check the condition for injectivity
Even if we ignore the first step, let's examine the mapping of other elements: We see that \( (G_2, B_2) \in R \) and \( (G_3, B_2) \in R \). Two distinct elements \( G_2 \) and \( G_3 \) have the same image \( B_2 \). This violates the condition for a function to be injective.

Step 3: Conclusion and Justification
\( R \) is not an injective function because:
• It is not a function (since \( G_1 \) has multiple images).
• It is not one-to-one (since \( G_2 \) and \( G_3 \) share the same image \( B_2 \)).

Answer 3

Concept:
• Reflexive: \((x, x) \in R\) for all \(x \in A\).
• Symmetric: If \((x, y) \in R\), then \((y, x) \in R\).
• Transitive: If \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\).
• A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Step 1: Verify Reflexivity
For any student \( x \in A \), \( x \) and \( x \) are obviously from the same colony. So, \( (x, x) \in R \) for all \( x \in A \). Therefore, \( R \) is reflexive.

Step 2: Verify Symmetry
Let \( (x, y) \in R \). This means student \( x \) and student \( y \) are from the same colony. If \( x \) and \( y \) are in the same colony, then \( y \) and \( x \) are also in the same colony. So, \( (y, x) \in R \). Therefore, \( R \) is symmetric.

Step 3: Verify Transitivity
Let \( (x, y) \in R \) and \( (y, z) \in R \). This means \( x \) and \( y \) are in the same colony, and \( y \) and \( z \) are in the same colony. This logically implies that \( x \) and \( z \) must also be in the same colony. So, \( (x, z) \in R \). Therefore, \( R \) is transitive.

Step 4: Conclusion
Since the relation \( R \) is reflexive, symmetric, and transitive, it is verified that \( R \) is an equivalence relation.

Answer 4

Concept:

• A function is bijective if it is both injective (one-to-one) and surjective (onto).
• For a function \( f : X \to Y \) to be bijective, the number of elements in the domain and codomain must be equal: \( n(X) = n(Y) \).

Step 1: Compare the number of elements in both sets
Set \( B \) (Domain) contains 3 elements: \( n(B) = 3 \).
Set \( A \) (Codomain) contains 4 elements: \( n(A) = 4 \).

Step 2: Analyze the condition for Surjectivity (Onto)
For a function to be surjective, every element in the codomain \( A \) must have at least one pre-image in the domain \( B \).
Since there are only 3 elements in the domain \( B \), they can map to at most 3 distinct elements in the codomain \( A \).
Because the codomain \( A \) has 4 elements, at least one element in \( A \) will always be left without a pre-image.
Thus, no function from \( B \) to \( A \) can be surjective.

Step 3: Conclusion
A function must be both injective and surjective to be bijective. Since no function from \( B \) to \( A \) can be surjective (due to \( n(B) < n(A) \)), it follows that no such function can be bijective.