 Hello students, very good morning to all of you. Today we are going to discuss this set a relation and functions chapter in that we are going to deal with the exercise based on relations that is exercise number two. So yes, let's begin our discussion. So this exercise to is mainly based on our relations topic. So let's begin with the first question. So here the question is saying there is one set a. Let me do it. Yes. So, there is one set a whose elements are two three and five, and there is one another set B whose elements are two five and six. Then a minus B cross a intersection B is so basically we have to take the Cartesian product of this a minus B set and a intersection B set right. So yes, let me write this set a. Our set a consists of this two, three and five. Okay. And our set B is having these three elements that is two, five and six. Okay, now what will be our set this a minus B, what will this set consists of. So basically it will consist the elements of set a. Which and we have from that set we have to subtract the elements which are in set B also. Okay, so the elements which are in set a are two three and five, and out of that this two and five are in set B also. So this set a minus B will consist of only one element that is three. So two and five are eliminated from set a because these elements two and five are in set B also right and what will be this set a intersection B. What will be the set a intersection B so it will consist of the elements which are present in both set a and set B because we are taking intersection of set a and B. So which which are those elements to is present in both of the sets so two will come three will not come and five will come right this five will come because these two elements two and five are present in both sets a and B. So yes, we are we are having these two sets. Now we have to take the Cartesian product of this, these two sets. So A minus B cross a intersection B we have to take the Cartesian product of these two sets. So what will be the Cartesian product. So in this, it will consist of the order pairs right order pairs in which the first element will come from this and this set right first element of these ordered pairs will come from these, this set a minus B, and our second element will come from this set right. So, first element of first element will come from a minus B set that is three. So, this three and second element will be two, one more element will be there in this, and that will be three comma five, right. So this will be the elements which will be present in this Cartesian product. Okay, so let's see which are the options available three comma two comma three comma five so this option C is correct, right. So option C is the correct option. So I hope it is clear to all of you, right. Now let's move to the next one. So this is our question number two. It is saying a number of elements in set A is four number of elements in set B is three and number of elements in A cross B cross C is 24. Then we have to find the number of elements in set C. So basically let me write the information whatever is given in the question. So it is saying number of elements in set A is how much for number of elements in set B is how much it's three. Okay, and number of elements in this A cross B cross C is given out to be 24. This is given to be 24, right. And we have to find this number of elements in set C. This is what we need to find, right. So how can we proceed with this question. So let me take this, this information. So number of elements in A cross B cross C, we can write it in this way also number of elements in A cross B, right. Number of elements or let me say in this way A cross B cross C, no. So number of elements in A cross B into number of elements in C, okay. In this way we can write. Now, we are having this information how many elements are there in this set it is equal to 24. Okay, and number of elements in A cross B that we can find anyhow from this information. Okay, so let me change the color of the pin. So from this given information we can write the number of elements in A cross B. That is how much number of elements in A into or let me write into in this way point number of elements in set B, is it okay. Now, number of elements in set A is four and number of elements in set B is how much three. So this is coming out to be 12 right I will I will be using this information here. So number of elements in A cross B will be 12 and this into number of elements in C. So from here we get the value of we get the number of elements in C as 24 upon 12 that is nothing but two. So yes we got this information that number of elements in set C is two. So this will be our answer. Is it available in the option. Yes it is there in option B. So yes this option B will be the correct answer. Right, so how we approach this question we we actually broke this thing number of elements in a cross B cross C is equal to number of elements in a cross B into. Let me write we can change it in this way also into number of elements in C. So from there, we got the value of number of elements in set C. So yes, let's move to the next one. So the question, this is question number three, it is saying that the relation are defined on the set of natural numbers. Okay, so R is defined on set of natural numbers. R is defined such that okay a relation is defined in this way a comma B such that a differs from be a differs from be by three. So basically this a differs from be you know from be I have to write from be so a differs from be by three. So this is the relation on the set of natural numbers, and the relation is defined in this way a comma be such that a differs from be by three. Is it okay. So, means a and B are related by this property a differs from be by three. Now we have to find this relation are. Let me find. So basically, let me say in this way are okay since R is defined on the natural numbers. So we can take the value of means the value of a and B will be from one. Okay, one, two, three, four all counting numbers, it may take a and B may take all counting numbers. Right. So the condition is a differs from be by three. So basically, if I take a as one, okay, we cannot take anything less than one because counting numbers to start or natural number it starts from one itself. So what will be the value of be be maybe for right be maybe for because the difference between them should be three. If I take a as two, our be will be five. If I take our a as three, our be will be how much six in this way, this series, these sets means these elements of the set will continue. So it will be four comma seven right, it will continue in this way. So this will be our relation, but there is one ambiguity right in this in this definition, if you see the definition of relation a differs from be by three. Okay. So basically, there is one ambiguity. So I can I can write this relation in this way also. Let me take this relation means, let me take this value of a to be four, we can we have, we can have one value, right. If let me take a as five, so we can have a BS two. So this is basically opposite of what we have written in the above set. So six comma three, seven comma four in this way also it may continue. So in options, if you see both these things are there one comma four, two comma five, it is in option a right it is in option a and if you see observe these elements, these are in option be okay. So, basically a differs from be by three, it generally means this thing. Right, a minus become more should be equal to three. Okay. So in that case, both this option a and b should be correct. But as far as I know, I have seen the answer key, they were saying this option be to be correct. So yes, option B is correct. But if I say option is correct, that also holds good. Okay. So in my opinion, both this option a and b should be taken. So yes, this is for all for this question. Right. I think these are you guys are clear on this ambiguous definition. So a differs from be by three. It generally means more of a minus be should be three. And from that, we can have both these options and be to be correct. So yes, moving to the next question that is question number four, let's observe this question. So let a be non void set of children in family. Let a be non void set of children in a family, then the relation, the relation x is a brother of why on a is a chat. So let me write it in more simpler way what is given in the question. So our set a is what non void set. Okay, very good. And it is the set of children in a family. So set a consist of children, right, set a consist of children in any family. So let me write in this way. So set a consist of a children in a family. Now there is one relation defined in this way. There is one relation defined in this way. The relation is x comma y such that such that x is a brother of why x is a brother of why. So this relation is defined in this way. Okay. And the relation is defined on set a. Right. The relation is defined on set a. Okay, set a is this thing. Set a is children in a family. Now we have to comment on this relation are whether it will be reflexive, anti symmetric, transitive or equivalence. So let's go with the reflexive first. So we will talk on the reflexivity, reflexivity of this relation. So what is reflexive relation mean, it will be basically x comma x, right, x comma x, and in that case our definition of relation as per the definition of relation x is a brother of x, x is a brother of x, right. This is what, what is coming from the relation definition itself. So, since we are talking about the reflexivity of this relation, we are taking the same element that is x comma x. So hope this is clear to all of you, like how do we normally take the elements for it to be reflexive symmetric and transited. So we have taken this x comma x element, and we are saying as per definition x is a brother of x, how can one be a brother of himself, right. So it's not possible. No, x cannot be the brother of himself. Okay, so this, this is coming out to be false, or let me write in this way. This is coming out to be false. Hence, this function is not reflective. Not reflexive. Not, I'm saying reflective, sorry for that. So hence this function is not reflexive, right. Let me talk about this symmetricity. So, let me talk about symmetry, symmetry, whether this function is symmetric or not. So basically, if you see, I'm going to take this x comma y, if x comma y is there, if x comma y is there in this relation, then y comma x would also be there in the relation. This is the definition of the symmetricity of any relation. So if x comma y is there, so what does it mean if x comma y means what x is a brother of y, x is a brother of y, right. And y comma x means what y is a brother of y is a brother of x, right. So if this thing, if x comma y is there, if x is a brother of y, does it imply that y is a brother of x? If x is a brother of y, does this imply that y is a brother of x? No, no, y may be the sister, right. Why may be sister? I am saying that why may be the sister of x? Why may be the sister of x, right. So this condition is coming out to be false, right. This condition, this thing is coming out to be false. So for symmetricity, if x comma y is there, then y comma x should mandatory present in the relation. So no, it's coming out to be false because why may be the sister of x also? Is it okay? So this function is not symmetric also. This is not symmetric also. And reflexivity we have already commented that it is not reflexive. Now let's talk about the transitivity. Let's talk about transitivity. Transitivity, okay. So now what does transitivity, transitivity relation means? So transitivity relation means if x comma y is there, okay. And if y comma z is there, then x comma z should be there in the relation. Now if x comma y is there, what does it mean as per definition of relation? x is a brother of y. x is a brother of y, right. And this y comma z, what does it imply? Means what does this defines? y is a brother of z. y is a brother of z. Then x comma z, then x comma, see if x is a brother of y, okay. x is a brother of y. y is a brother of z, okay. So obviously x will be the brother of z. Let me write in this way. Obviously this x will be the brother of z. If this condition is coming out to be true, because y, x is a, see x is a brother of y and we are further saying that y is a brother of z. So since x is a brother of y and y is a brother of z, so obviously this x and z are related, no? This x and z are related and by what relation? They are the brother of each other. Means x is a brother of z, okay. So this x will be the brother of z is coming out to be true. This is coming out to be true. So this relation will be transitive. This relation will be transitive. So it passed the transitive attest. So this will be a transitive relation, right. So I hope you guys are clear on this. So a reflexive, no, it is not reflexive. Transitive, anti-symmetric, no. It is transitive and equivalence, no. If a relation is all three, like for equivalence relation, it should be reflexive also, symmetric also, transitive also. So here the relation is not reflexive and not symmetric, then there is no question of equivalence relation. But this function, sorry, this relation is transitive obviously. So the correct option will be option C. So this was our question number four. Let's take this question number five. So let number of elements in set A is n, then the number of all relations on A is, okay. This is our question. Let number of elements in set A, let me record here. Number of elements in set A is how much n, then we have to find the number of all relations on A. So basically the theory part, if I go with the theory part, theory part of this question. So basically if you say this is our number of elements in set A is m, okay. And number of elements in set B is suppose n, okay. The number of relations, how do we use to define number of relations from A to B? Number of relations from A to B, how do we use to define this thing? This is coming out to be, means this used to come out to be 2 k power mn, right? Means number of elements in set A, 2 k power number of elements in set A, here in this case it's m, so 2 k power mn, this gives us the total number of relations, total number of relations. Now here, the relation is defined on the set A itself, right? Relation is defined on the set A itself, so total number of relations will be, total number of relations will be how much? 2 k power mn, so here that both the sets are A itself, so it will be n into n, right? 2 k power n into n, because the relation is defined on the set A itself, so it will be 2 k power n square. So this will be the total number of relations. This will be the total relations, total number of relations on A, okay. So yes, I hope that you guys are aware of this theory part, how this formula comes basically, 2 k power mn, how this thing usually comes, right? So I hope you guys are aware of this. So I'm not going in the derivation part, so if our set A and B are like different and one relation is defined from A to B, so this used to be our total number of relations. This used to be our total number of relations, since the relation is defined on set A itself, so it will be 2 k power n into n, that is nothing but 2 k power n square, okay. Now we will take this question number 6. So if set S is this from 1 to 20, set K is ABCD, set G comprises of BDEF, then the number of elements of this S, Cartesian product of S and K, union Cartesian product of S and G is number of elements of this set we have to calculate. So yes, given information is this is our set S in which there are 20 elements from 1 to 20, so 1, 2, 3 up to 8 is going up to 20, okay. There is one another set K that is comprises of 4 elements, this A, B, C and D and there is one another set this set G, okay. Set G is there and that is what that is also comprises of 4 elements and those elements are B, D, E and F, okay. Now we have to find the number of elements in this Cartesian product of S and K. So basically let's start with this Cartesian product of S into K, okay. So if I ask you what will be the number of elements in this set, number of elements in the Cartesian product of S and K will be how much? It will be number of elements in S into number of elements in K, means number of elements in set S into number of elements in set K. Now we are aware of this number of elements in set S, how many? 20, okay. And number of elements in set K, it's 4, so it's coming out to be 18, is it okay? In the similar way I'm going to find this number of elements into Cartesian product of S and G, right. S and G, so that will be number of elements into S into number of elements in this G set. So number of elements in S is 20 and number of elements in G is how much? 4. So here also it's coming out to be 18. Now we have to take the union, the question is asking to find the number of elements in this S into K, union S into G, right, we have to find this thing. So if you observe here, if you observe here, see S into K, S into K, this means Cartesian product of S and K. So here we are having these set sets, means these will consist of ordered pairs. I'm talking about this, I'm talking about this Cartesian product, okay, S and K. So this Cartesian product will have the first element from set S, okay, and the second element from set K, right, and this Cartesian product of S and G will have first element from set S and the second element from set G. Now if you observe here, there are two elements common in K and G. What are those elements, this B, okay, B here, and what else, this D here, okay. So basically there will be some common elements, there will be some common elements or we can say common ordered pairs, Cartesian product have elements of ordered pairs, right. So common elements or we can say these elements will be ordered pairs between common elements between these two Cartesian product, S and K and Cartesian product of this S and G. Now, how many common elements will be there? How many common elements will be there? Forty no, because anything let's take, let's say I'm taking this one of the element from S is one and I'm taking the other element is B. So one element of S cross K will be 1, B, and in the same way if I take S cross G, we can have 1, B, right. So in this way, the common elements will be basically 20 into 2 because 20 elements are there in set S and two elements are common that is 40, right. And so since these are the common elements, so these will repeat, these 40 elements will be there in this 80 also and these 40 elements will be in these 80 elements also, okay. So basically number of elements, number of elements in this Cartesian product S cross K union S cross G will be how much? Will be this 80 means 40, 40 that is separated that is not common, right. That is not common between S cross K and S cross G and this 40 elements which is common, this 40 elements which is common. So I am arrow marking also for better understanding and this 40 elements which are only in S cross G, right. So there will be total number of elements will be 120. Total number of elements in this set will be total number of elements in this set will be 120, right. So this will be our answer. So I hope you guys are clear on this, okay. So 120 will be the correct answer to this question, okay. It will be neither, it will be not 80 plus 80, 160, okay. So yes, there is one option C, okay. So that will be our correct answer, 120, okay. Now let's move to the next question. This is our question number 7. So it is saying that the relation R is defined on the set of natural numbers, okay. So R is defined on the set of natural numbers, very good. And relation is defined in this way, relation is defined in this way A, B such that such that A is equal to 2B, okay. So relation is defined on natural numbers and relation is defined in this way that is it will consist of ordered pair A, B such that A is equal to 2B. Then we have to find the R inverse, right. We have to find R inverse. So it's simple only. Let's first find the R, okay. So what will be our R? So this A and B should be our natural numbers only. So let me take, since A is equal to 2B, so if I take A as 1, okay, if I take A as 1, what will be our B? It will be 1 by 2, right. Then only this A will be 2 times of B, okay. But this here, this B is coming out to be fraction, no, it's not a natural number. So it will be not our part of the relation. Let me take A as 2. Let me take A as 2. So yes, 2 will be there. Then in that case our B will be 1, okay. And if we take 3, if we take 3, the value of A as 3, then it will be, what will be the value of B? It will be 3 by 2. So this will also be not in our relation. So basically, after 2 comma 1, we can have 4, 4 comma 2, right, 4 comma 2, then 5 will again not work. Then we can have this 6 comma 2, okay. So basically the first element will be even number, right. First element will be even number. 7 will also not work. In that case, the B will be 7 by 2. That is not, that will not fit in this relation, right, because R is defined on natural numbers. So our first element will be basically even number, because A is equal to 2B. So 8 comma 4, okay, then 10 comma, 10 comma, how much? 5, yes, this will fit. In this way, it will continue, right. So basically our R is coming out to be this, 2 comma 1, then it will have, can be 4 comma 1, then it can have 6 comma 3, then 8 comma 4, okay. And then 10 comma 5, in this way it will continue, okay. But yes, this is our R, but we need to find R inverse. So what is R inverse basically? In R inverse, in R inverse the domain of R becomes a range of R and the range of R becomes the domain of R. So basically we will, what you say, we will swap the ordered pairs. It will give us the R inverse. So R inverse will be 1 comma 2, 2 comma 4, then 3 comma 6, then 4 comma 8, okay. Then 5 comma 10, in this way it will continue. So this will be our R inverse and this is what we were asked to find in the question. So R inverse will be 1 comma 2, 2 comma 4, 3 comma 6, yes, this option B is correct. This option B is going to be our correct option, okay. So I'm writing here also this option B will be correct. So this is all in this question. Let's move to the question number 8. So one relation is given here comprising of these three, these elements and this relation is defined on set A and we have to comment about this relation, okay. We have done similar type of questions earlier also. Our set A is what? Set A is 1, 2, our set A consists of three elements, 1, 2 and 3. And this relation is already given here in row straight form. So no need to worry. It's already given. So 1 comma 1, 2 comma 2, 3 comma 3, then 1 comma 2 is there, then 2 comma 3 is there and 1 comma 3 is there, okay. So 1, 2, 3, 4, 5, 6 elements are present in this relation. Now we have to talk about reflexive. So let me see the reflexivity of this function. So reflexive, since set A is the comprising of three elements 1, 2 and 3 and all the elements are related to itself in this relation. So it will be reflexive. Why? Because this 1 comma 1, okay, 2 comma 2 and 3 comma 3, all are in relation. All are there in the relation, right? So it will be a reflexive relation. This is the definition of reflexive. All the elements of this set A should relate to itself and it is happening here in this relation. So it will be a reflexive relation. Now let's talk about symmetric, symmetricity. So whether this relation is symmetric or not? So if you observe here, 1 comma 2 is there. 1 comma 2 is there in this relation. But as per symmetricity, 2 comma 1 should be there in the relation. For this relation to be a symmetric, 2 comma 1 must be one of the elements of this relation. But 2 comma 1 is nowhere present in this relation, right? So it will be not symmetric. This relation will be not symmetric, right? Now let's talk about the transitivity of this relation, okay? Let's talk about whether this relation is transitive or not. So if you see 1 comma 2 is there, right? 1 comma 2 is there. Then if you see, is there any terms starting from 2? So 2 comma 3 is there. Yes, there is 1 term 2 comma 3, okay? 2 comma 3 is there. Then 1 comma 3 should be there in the relation to be transitive. So yes, obviously 1 comma 3 is also there. So yes, it is transitive. Let's take one another example. I'm taking this 2 comma 3. So 2 comma 3 is there, okay? Is there any element that starts with 3? Yes, there is 1 element 3 comma 3, okay? So 3 comma 3 is there. So this 2 comma 3 should be there in the relation. And yeah, it's there, no? It's there. 2 comma 3 is there, okay? So let's take this element. 1 comma 3 is there. 1 comma 3 is there. And is there any element that starts with 3? Yes, 3 comma 3. So 3 comma 3 is there. Now we have to check whether 1 comma 3 is there or not. So yes, 1 comma 3 is there. So yes, this function is going to be transitive. This function is going to be transitive. So this function is reflexive also and transitive also, right? So now let's check the option. Reflexive but not symmetric. Yeah, it is correct. Reflexive but not transitive. No, this option is wrong because it is transitive. Symmetric and transitive, no. It is not symmetric. Neither symmetric nor transitive. Neither symmetric nor transitive but it is transitive, no. So it seems to be like this option A is correct only but there is no mention of transitivity. So function is reflexive but not symmetric. Yes, it is there. This is, this will be the correct option. Okay, so this option, question number 8, option A is the correct answer to this question. Okay. Now let's move to the next question. Question number 9. Okay. The question is asking the number of equivalence relations defined in the set is which are having these three elements A comma B comma C. Okay, so let's find out. So there is one set is comprising of three elements A, B and C. Okay. And the relation is defined on S, relation is defined on S. And we have to find the number of equivalence relations. So basically, number of equivalence relations, there is no formula for this. Okay. In case you have gone through the theory part, we have seen that there are, there is a fixed formula for finding the number of reflexive relations. Okay. If it is defined on set S, if there is defined, we are having a fixed formula for finding the number of this, what do you say, symmetric relations. Okay, the number of symmetric relations can also be found out by using the formula, but there is no fixed formula for finding the number of equivalence relations. So we have to go through what you say, we have to define the relations and see whether how many relations are possible. We have to make relations in such a way that the relation should be equivalence relations, right? This is the question, equivalence relations. So let me take, let me define relation one in this way. I am going to define relation one in this way in which I am going to relate all the elements to itself. So A comma A, then we can have this B comma B and we can have this C comma C. This is basically identity relation. This is what this relation R1 is the identity relation. Okay. So yes, identity relation is our, what do you say, it will be reflexive, right, symmetric and positive. So obviously this R1 will be our equivalence relation. Okay. Now let me define one more relation R2 in this way. That will consist of these three elements, no doubt. Okay, B comma B, C comma C. Now I am going to take first two elements of this. So A comma B, okay, and what? Now for making it equivalent, so suppose I'm trying to make it a symmetric relation. So since A comma B is there, I should include B comma A also, right? So obviously this will be a reflexive because these three elements are there. A comma B if I have taken, I have to take B comma A also in this way it is coming out to be symmetric also and guys if you observe it, it will be a transitive relation also because if A comma B is there, B comma A is there. So A comma A should relate, A comma A should be there for it to be transitive and obviously it is there, no? A comma A is there. So it will be a equivalence relation. So in this way, I can define two more relations in this way. A comma A will be there, B comma B. These three elements will be no doubt there. Now I'm going to take the second and the third element. So B comma C and then C comma B. Okay. This will also be a reflexive relation. Sorry, equivalence relation. And I'm going to define one more relation taking the first and the third element. That is, that is A comma C and C comma A, right? A comma C and C comma C here, one comma C is there. So C comma A is there. It is a symmetric. Then if A comma C is there, C comma A should be there and it is there. So yes, this will be, these are the four relations which are equivalence relation. Now guys, there will be one more relation that is universal relation. There will be one more relation that will be universal relation. That universal relation will also be in one equivalence relation. So it will consist of all the elements of this Cartesian product, S cross S. Okay. So it will have A comma A. Okay. Then A comma A, then what? A comma B, then A comma C. Okay. Then we will have B comma A, B comma B, then B comma C. Okay. Then C comma A, C comma B and C comma C. So this relation, this relation is basically a universal relation. Okay. This will be a universal relation consisting of all the elements of S cross S. Okay. And this will also be the equivalence relation. So there are possible, there are these five relations possible on set S so that all these relations can be equivalence relations. So total number of equivalence relations defined in set S comprising of these three elements will be five. So this option A is correct answer for this question. Okay. So I hope you guys are clear on this. If yes, can we move ahead? Okay. Okay. Okay. So let's take this next question. I think this is the last question of this exercise. Okay. Let's see what is, what this question is asking. So if R be a relation from A to B. Okay. And A set A is listed here. And A comma B belongs to R such that A is less than B, then we have to find this R composite R inverse. Right. We have to find R composite R inverse. So let me first list down our set A. So set A is what? Comprising of four elements and these elements are one, two, three and four. Okay. And our set B is this which is having three elements one, three and five. Now once a relation is defined from A to B, one relation is defined from A to B and the how this relation is defined relation is defined in this way. A comma B such that A should be less than B. A should be less than B. Is it okay? This is the question what it is saying. So let's first define our, let's first find out our relation R. Okay. So what will be our relation R? So this A, first element of this relation will come from A, set A. Right. And the second element of this relation will come from B. Okay. And A should be less than B. So let's find the elements of this relation. So if I take A as one. Okay. And B. No, we cannot take A as one because B will be from the set B and okay, it will be there. So A comma B means we can, we cannot take this one, but we can take this three. Okay. So this element fits in this relation because A is less than B. Obviously one is less than three. Okay. So let's try to find the another element will be one comma five. Okay. Now I'm going to take the second element of set A that is two. So two, then two comma one. No, it will not fit in this relation. Two comma three will fit. So I'm writing two comma three. Two comma five will also fit. So I'm writing that also. Now I'm taking three. So three comma one will not going to help us. Three comma three will also not going to help us. Yes. Three comma five will be there in this relation. And let me take this fourth element of set B. So four comma one. No, four comma three. No, four comma five will be there. So yes, this is our set. This is our relation comprising of six elements. So there will be six elements in this relation. Now the question is asking R. This is basically composite relation. R O R inverse. Right. R O R inverse. So let me first write R inverse. So what will be our R inverse? It will, we can just, we will just swap the elements of these ordered pairs. So R inverse will be three comma one, five comma one. Okay, then three comma two. Okay, then five comma two, then five comma three, and then five comma four. This will be our R inverse. Okay, now coming to coming to the question, coming to the question, question is asking to find this R O R composite R inverse. Right. So it will be basically, let me draw the arrow diagram for this. Okay. So this is our R inverse. Okay. No, not in this way. What is the domain of R inverse in this way we can see what is the domain of R inverse domain of R inverse is three and five. Right. Three and five domain of R inverse is three comma five. What is the range of R inverse range of R inverse is one, two, three, four. One, two, three, four. Okay. And what is our R? R is basically this R. If you see the domain of R, it's one, two, three, four, and its range is three comma five. Okay. So this is basically what you say R inverse. Okay. And this is basically R. This is our R. Okay. Now we have to find this R composite R inverse. So this will be actually R composite R inverse. This will be R composite R inverse. Now we have to find the elements of these, this relation composite relation. Okay. So let me write this R inverse on the top of this arrow. So this will be R inverse. Okay. And this is R and this from here from here to here, it will be R O R inverse. Okay. So if you see three comma three R composite R inverse will consist of what? Three comma three, three comma five, then five comma three and five comma five. This will be the R inverse. So basically this will be three comma three from here and going to here. So three comma three, then three comma five. Okay. Then five comma three and five comma five. These will be the four elements. These will be the four elements of this set, of this set. This you can also visualize in this way. This you can also visualize in this way. So suppose R inverse I am taking. So the set, the first element is three comma one. So three is mapped with this one. Okay. Now I have to see this mapping of one in relation R. So if you see one is mapped with three in R. Okay. And one is mapped with five also one is mapped with five also. Right. One is mapped with five also. So basically this composite relation R composite R inverse is directly linking this three is directly linking this three to this three and five. Right. So this is direct linkage. This is directly directly linking three to three and five. So hence these two elements are there in this composite relation. Okay. Now let me take one more example. This five comma one. Let me take this five. So five is getting linked with one. Okay. Five is getting linked with one. Now we have to see the element in R, which is getting mapped from one to which element. So here also from one you see three and five are getting mapped. So this five is getting linked with this five. So there are three elements. If you further take this element in R inverse three comma two. Okay. So three is getting linked with two. Three is getting linked with two. Now I have to see the relation R the linkage of two to which element like two is getting linked to which. Now here also if you see two is getting linked with three and five. So three and five. So basically this is nothing but what I have drawn here. So this R O R inverse is linking these two elements with this. So obviously only four elements will come in this composite relation. So I hope this is clear to all of you. So if you observe the options, we are having this option C. I'm seeing this option C to be correct. Okay. So this option C is correct. So I think this is all because there were only 10 questions in this exercise. So yes. Thank you all of you. Okay. So let me draw one big smiley here. Okay. Thank you. Thank you. Thank you all of you. And yes, we will be back with the next exercise of this chapter. So till then, good bye. Take care.