Let A and B be two finite set with same number of elements, and let f : A B be an everywhere defined functions : (i) If f is one-to-one, then f is onto. (ii) If f is onto, then f is one-to-one.

(i) If f is one-to-one, then f is onto.

To determine whether this statement is true or false, we need to understand the concepts of one-to-one (injective) and onto (surjective) functions.

A function f : A * B is called one-to-one (or injective) if each element in set A maps to a distinct element in set B. In other words, no two elements in set A can have the same image in set B.

A function f : A * B is called onto (or surjective) if every element in set B has a pre-image in set A. In other words, for every element in set B, there exists at least one element in set A that maps to it.

To determine the validity of statement (i), we need to show that if f is one-to-one, then f is onto.

Proof: Assume that f is one-to-one but not onto. This means that there exists an element in set B that does not have a pre-image in set A. However, since f is one-to-one, each element in set A must map to a distinct element in set B. Therefore, if an element in set B does not have a pre-image, it contradicts the assumption that f is one-to-one.

Hence, the assumption that f is one-to-one but not onto leads to a contradiction. Therefore, if f is one-to-one, then f is onto.

Therefore, statement (i) is true.

(ii) If f is onto, then f is one-to-one.

To determine whether this statement is true or false, we need to understand the concepts of one-to-one (injective) and onto (surjective) functions.

A function f : A * B is called one-to-one (or injective) if each element in set A maps to a distinct element in set B. In other words, no two elements in set A can have the same image in set B.

A function f : A * B is called onto (or surjective) if every element in set B has a pre-image in set A. In other words, for every element in set B, there exists at least one element in set A that maps to it.

To determine the validity of statement (ii), we need to show that if f is onto, then f is one-to-one.

Proof: Assume that f is onto but not one-to-one. This means that there exist two distinct elements in set A that map to the same element in set B. However, since f is onto, every element in set B must have a pre-image in set A. Therefore, if two distinct elements in set A map to the same element in set B, it contradicts the assumption that f is onto.

Hence, the assumption that f is onto but not one-to-one leads to a contradiction. Therefore, if f is onto, then f is one-to-one.

Therefore, statement (ii) is true.

In summary:

• Statement (i): If f is one-to-one, then f is onto.
  • Simplified: If a function is injective, it is also surjective.
  • Answer: True.
• Statement (ii): If f is onto, then f is one-to-one.
  • Simplified: If a function is surjective, it is also injective.
  • Answer: True.