 Hello and welcome to the session. In this session we will learn about audit peer relation and mapping. First of all, let us discuss about audit peer. For this, first of all let us see an example. Now the table which is given here shows three numbers which appear with their cubes. Now this pairing will be shown by means of the following. The audit peer 2 and 8, the next audit peer will be 3, 27 and the next audit peer will be 4, 64. It is the cube of 2, 27 is the cube of 3 and 64 is the cube of 4. So these numbers are paired with their cubes and we have written it in the form of audit peers. Now in the first audit peer 2 is called the first component and 8 is called the second component of this audit peer. Similarly, in the second audit peer 3 is the first component and 27 is the second component. Similarly, in this audit peer also. So we can see these numbers which are 2 and 8, 3 and 27, 4 and 64 in a specific order in which the second component is the cube of the first component whose components see order separating them by a comma and enclosing the peer in parenthesis. We should remember one point that now let us see one example for this here. Let 8 is a search containing the elements 1 and 2. And now we have to form first by using the numbers 1 and 1, 2. Now let us see one more example here. Let A is a search containing the elements and B is a search containing 3. Now we have to form should be from the search A should be from the searching with each element of such B. And the elements of such B are 3 and 4. 1, 3. Now consider the following statements perpendicular to CD 75. 1 and 2 is a subset of sentences between two objects such as A B is perpendicular to CD 75 is greater than 25 is a subset. Relation is the association between two objects by a certain rule by which the first element that is this is a first element which is associated with this second element. Now to clarify this more let us consider some ordered pairs which are 1, 2 and 10s greater than. Now among these added pairs which are satisfied the first component is greater than the second component that means X is greater than Y. And also in which X is greater than Y that means first component is greater than the second component and also 4, 2 in which X is greater than Y the given sentence. The relation was given to us and by using this relation there is a relation components in the ordered pairs the domain of the ordered pairs. This will be equal to the set containing the elements 2, 3 and range will be equal to the set containing the second components. So this will be equal to the set containing the elements 1 and 2 between two sets. Let A and B be two non-empty sets then A to B let A is a set containing the elements and B is a set containing the elements 4, 9 and 36. Now let us see the relation from A to B, A to B. So in the ordered pairs the first component should be from the set A and the second component should be from the set B. And the relation between these two components should be is containing the ordered pairs. The first one will be 2 and the relation between these two is as 3 is the element of A, 9 is the element of B and the relation between these two elements is 3 is the square root of we will get all these ordered pairs. Now let us learn how to represent as we have discussed above which is a set of ordered pairs A to B should be paired from A to B. Let us consider the same example which we have mentioned above in which this is a set A, this is a set B and R be a relation from A to B is the square root of. Now here on one side we will list the elements of A and on the other side we will list the elements of B. Now the elements of A are 2, 3 and 6 and the elements of B are 4, 9, 16, 25 and 36. Now the relation from A to B is the square root of. So we will connect the relative elements that is 2 will be connected with 4 is the square root of 25 and 6 with 36. And since the relation is from A to B so the element of mapping of no element. Now consider an example is displayed by this. Now for this relation conditions should be satisfied but here the second condition is not satisfied of A is associated with the unique element of B except the element 25 which is the element associated. Therefore best relation in which the relation is the cube of is displayed by this arrow diagram. You can see that every element of a set A is associated with a unique element of set B that is 8, 27, 64 and 125. All of them are associated with the unique element of set B and no element of such A that is associated. And this completes our session.