(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.