Question

A school wants the students of class XII to do a project on ‘Sustainability’ keeping the world environment in mind. They select the student participants on the basis of an essay writing competition.

7 students out of 80 are selected for the project and are categorized into two sets such that:

Girl students belong to Set A = {G₁, G₂, G₃, G₄}

Boy students belong to Set B = {B₁, B₂, B₃}

Based on the above information, answer the following questions:

1. How many relations are possible from Set A → Set B? (1)
2. Let R be a relation from A → B such that
   R = {(G₁, B₁), (G₂, B₂), (G₃, B₂), (G₄, B₃), (G₁, B₂)}
   Is R an injective function? Justify your answer. (1)
3. 1. Let the relation R from A → A be such that
      R = {(x, y), x, y ∈ A, x and y are students from the same colony in the city}
      Verify if R is an equivalence relation. (2)
      OR
   2. Verify if any function f: B → A is bijective. Give reason to support your answer. (2)

Answer 1

i.

A relation from Set A to B is any subset of the Cartesian product A × B.

Number of elements in Set A, n(A) = 4

Number of elements in Set B, n(B) = 3

Number of elements in A × B = 4 × 3 = 12

The total number of possible relations is given by `2^(n(A) xx n(B))`

Total relations = 2¹² = 4096

ii.

Let R = {(G₁, B₁), (G₂, B₂), (G₃, B₂), (G₄, B₃), (G₁, B₂)}

To be considered a function, each element in the domain (Set A) must map to exactly one element in the codomain (Set B).

• Not a function: Notice that G₁ is related to both B₁ and B₂. R is not a function because it has two different outputs for a single input.
• Not injective: Even if we ignore the first point, G₂ and G₃ map to B₂. An injective (one-to-one) function requires unique inputs and outputs.

Conclusion: No R is not an injective function (it is not a function at all).

iii. (a)

R = {(x, y), x, y ∈ A, x and y are from same colony} to be an equivalence relation, it must satisfy three conditions:

• Reflexive: Every student x lives in the same colony as themselves. So, (x, x) ∈ R for all ∈ A.
• Symmetric: If student x lives in the same colony as student y, then student y necessarily lives in the same colony as student x. So, if (x, y) ∈ R, then (y, x) ∈ R.
• Transitive: If students x and y are in the same colony, and y and z are in the same colony, then x and z must be as well. so, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

Conclusion: Since it is reflexive, symmetric, and transitive, R is an equivalence relation.

iii. (b)

A function is bijective if it is both injective (one-to-one) and surjective (onto).

n(B) = 3 (Domain)

n(A) = 4 (Codomain)

For a function to be surjective, the number of elements in the domain must be greater than or equal to the number of elements in the codomain (n(B) ≥ n(A)). Here, 3 < 4. This means that at least one element in Set A will always be missing a corresponding element from Set B. Therefore, the function cannot be onto.

Conclusion: No, no function from B → A can be bijective because the sets have different cardinalities.