Relations And Functions

Function is onto but not one-to-one

The function is defined as:

Step 1: Prove that is onto

To show that is onto, we must prove that for every element , there exists some such that .

Consider any . We have two cases to consider:

• If , then and , so is mapped by or .
• If , then for , we have . Therefore, every natural number is the image of some , where .

Thus, the function is onto.

Step 2: Prove that is not one-to-one

To show that is not one-to-one, we need to find at least two different values and such that but .

Consider and . Here, , but clearly, .

Therefore, is not one-to-one because even though .

Conclusion: The function is onto but not one-to-one.

To prove that is an equivalence relation, we need to show that it satisfies three properties:

1. Reflexivity
2. Symmetry
3. Transitivity

1. Reflexivity:
A relation is reflexive if for every .
In this case, the number of pages in a book is always equal to itself. Thus, for every book , the relation holds: Hence, is reflexive.

2. Symmetry:
A relation is symmetric if whenever , we also have .
If the number of pages in book is equal to the number of pages in book , then the number of pages in book is also equal to the number of pages in book . Thus, if , we also have . Therefore, is symmetric.

3. Transitivity:
A relation is transitive if whenever and , we also have .
If the number of pages in book is equal to the number of pages in book , and the number of pages in book is equal to the number of pages in book , then the number of pages in book is equal to the number of pages in book . Hence, if and , we also have . Therefore, is transitive.

Conclusion:
Since the relation is reflexive, symmetric, and transitive, we conclude that is an equivalence relation.

Step 1: An equivalence relation on a set must be reflexive, symmetric, and transitive.

Step 2: In the set , we start by considering the given relations and .

Step 3: Reflexive property requires that each element is related to itself, i.e., must be included.

Step 4: Symmetric property requires that if is in the relation, then must also be in the relation. So, if exists, must also exist.

Step 5: Transitive property ensures that if and are in the relation, then must also be in the relation.

Step 6: The equivalence relations on that satisfy these conditions are: Thus, the number of equivalence relations is 2.

The given relation is .
For each pair:
For : and .
Since is not satisfied, .
For : and . Since , .
For : and .
Both conditions and are satisfied, so .
For : and .
Since , .
Thus, the correct pair is .

Step 1: A relation is reflexive if for every element , . In this case, since , the relation is reflexive.

Step 2: A relation is symmetric if implies . In this case, if , then , so the relation is symmetric.

Step 3: A relation is transitive if and implies . In this case, it holds that if and , then , so the relation is transitive.

Thus, the relation is reflexive, symmetric, and transitive, so the correct answer is (A). \bigskip

Step 1: A function is one-one (injective) if for every distinct pair , . In this case, , , and , so each element in maps to a distinct element in . Thus, the function is one-one.

Step 1: Determine the number of elements in sets and .
The set has 3 elements, and set has 2 elements. The number of functions from set to set is given by the formula , where and represent the number of elements in sets and , respectively.

Step 2: Apply the formula.
Substituting the values and , we get:

Step 3: Conclusion.
Thus, the total number of functions is , which is the correct answer.

Step 1: Reflexivity: For reflexivity, we need , i.e., . Clearly, , so is reflexive.

Step 2: Symmetry: For symmetry, we need that if , i.e., , then . Since , we have , thus symmetry holds.

Step 3: Transitivity: For transitivity, if and , then we need . From and , we get , confirming that transitivity holds.