This lecture provides a masterclass in foundational rigor, stripping away the intimidation of abstract algebra through systematic and accessible logic. It is an essential bridge for any student looking to transition from basic calculation to formal mathematical reasoning.
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2nd puc relations and functions from zero levelAdded:
Hello students, myself Jabi Kasha, mathematics lecturer. So through this channel I'm coming to teach you uh the PC first PC and second PC board C and JW level questions from zero level which I mean to say even the students don't know nothing about the concept. So from that level to the higher level board, C and JW concept by concept complete syllabus will be discussed through this channel.
So in today's con concept I'm starting with relations and functions. In today's concept is relations and functions. See to deal uh the chapter the relations and functions from a zero level. So we need to understand few basic concepts. The basic concepts which we have studied in first PC which we have studied in first PC such as ordered pair equal ordered pair and then cartition product of two sets followed by that relations then the types of relations functions types of functions. This is all about relations and functions which we have. So let us start with basics of relations and functions. If we have some basic knowledge then we can easily understand the concept which is there in relations and functions of second QC syllabus. So here the basic concepts are like as I said the first one is ordered pair.
Second one is equal ordered pair.
And the third concept we wanted to know is cartesian product of two sets.
Cartesian product of sets and then relations followed by the types of relations.
In types of relations we have a few types such as the first one is empty relation.
Second one is universal relation.
The third one is most important that is reflexive relation.
Fourth one is symmetric relation.
Fifth one is transitive relation and the sixth one is equivalence relations.
So in this class we are about to discuss these are the concepts. Okay. So one by one one by one we will try to understand how exactly these concepts will work.
Let us start with ordered pair. What do you mean by an ordered pair? An ordered pair is an ordered pair can be represented as open interval of a comma b. An ordered pair can be represented as open interval of a comma b where a is called as first concept I'm discussing ordered pair an ordered pair is represented as an open interval of a comma b where a is called first element and b is called second element in a simple words we can say like that or else we can say that a is absisa and b is ordinate I repeat again ordered pair is nothing but an ordered pair is represented as open interval of a comma b where a is called as absisa and b is called as ordinate. If you don't remember the word absisa and ordinate you can use simple words like a is first element a is first element and b is second element. I hope everyone can remember these two because we don't remember some mathematical names. So instead of that we can use first element and the second element easily easily we can remember this. Okay. So this is what we call it as an ordered pair. I repeat again an ordered pair is represented as an open interval of a comma b where a is called as first element or absis of and b is called as second element or ordinate. So this is about ordered pair which we have completed. Now let us move on to the next concept that is equal ordered pair. When do we say that the two ordered pairs are equal.
So to say that any two ordered pairs are equal, it should satisfy one of the condition. For example, for example, assume we consider an ordered pair a comma b is equal to c comma d.
Assume we have taken the two ordered pair a comma b and ca d. Now if I want to say that a comma b is same as c comma d, it should satisfy one of the condition. The condition says that the first element of the first ordered pair must be equal to first element of the second order pair and second element of the first order pair must be equal to second element of the second order pair.
Which I mean to say when a comma b must be equal to ca d means a must be equal to c and b must be equal to d then only we can say that the given two ordered pairs are equal ordered pair. You may have observed in first in the relations and functions the first problem will be based on equal ordered pair only. Like the question will be given like find the value of x and y if x + 3 y - 1 is equal to 4a 5 which means so here they had given two equal ordered pair two ordered pairs is given in between the two ordered pair they had given equal sign which means we have to equate the first element of the first ordered pair with first element of the second ordered pair and we need to equate second element of the first order pair with second element of the second order pair. then we will obtain the required values of x and y.
So let us see how we will obtain the values of x and y. So I'm equating first element of the first order pair with first element of the second order pair that is x + 3 = 4. Then when I simplify the same x= 4 I'm bringing + 3 to the right hand side it becomes minus 3. Then we get the value of x= 1. Similarly when I equate the second element of the first order pair with second element of the second order pair we get the value of y which is of the following form that is y - 1 is equal to 5. When I bring minus1 to the right hand side it becomes + one.
So 5 + 1 is equal to 6. So this is how we find the value of x and y when the two equal ordered pairs are given. So this is all about equal ordered pair. I hope you have understood. So this is also we have done with this. So based on this we have a number of questions will be given to you. So uh similar to this you can solve any number of problems.
Now moving on to the next concept that is cartesian product of two sets.
Cartesian product of sets. So here the cartition product of sets are denoted by a cross b. A cross B is equal to A cross B is equal to it is all set of all ordered pair A B such that every first element is belongs to set A and every second element is belongs to set B. I repeat again. Now we are about to discuss cartition product of two sets.
Cartition product of two sets are denoted by A cross B is equal to it is all set of all ordered pairs A comma B such that every first element is belongs to set A and every second element is belongs to set B. We can say let us take an example. Assume we have set A is equal to 1 2 3. Assume we have set A is equal to 1 2 3 and set B is equal to only one element that is X. Now to find the cartition product of two sets we write the set of ordered pair which is of the following form that is A cross B is equal to as I said that cartition product of two set is nothing but it is a collection of ordered pair in such a way that the first element should be always belongs to set A and the second element should be always belongs to set B. So here we are about to write the elements like we can write 1 is related to x and then 1 2 3. So like this we'll get 1 comma x. We can write an ordered pair like 1 x 2 x and 3 x. When we observe the ordered pairs so here you can observe every first element. What are first elements here? The first elements are 1 2 3. See these 1 2 3 I mean first elements are belongs to set A. As we already written in the definition that A cross B is nothing but the cartition product of two sets A and B is set of all ordered pair A comma B such that every first element is belongs to set A. See here A 1 2 3 R belongs to set A and second element second element is nothing but here X that X is belongs to set B like this we write. Similarly to find B cross A B cross A to find B cross A here A first you can write the B element set B elements followed by that 1 2 3. So here if you want an arrow diagram you can write like this X followed by that 1 2 3 X will be relating to one relating to two relating to three. So when we represent the same in terms of ordered pair we get b cross a = b cross a = x 1 x 2 and x 3. So this is how we get the collection of ordered pair according to the given two sets. B cross a is equal to x 1 x2 x 3. So this is how uh the cartition product of two sets will work out. And now followed by this we need to know the number of elements the total number of elements in cartitionian product of two set that also we'll understand here only. So here we have the two sets I have already taken the elements of set A and set B as 1 2 3 and X respectively. So here you can observe the number of elements in set A are three. So I can write number of elements n of A is equal to 3 and N of B is equal to we have only one element. Then the total number of elements in cartition product of two sets. We already written the cartition product of two set that is a cross b that is 1 x 2 x 3 x. How many ordered pairs we have? Only three ordered pairs.
Which are those? 1 x 2 x and 3x. We have only three elements. I mean three ordered pair. Which means the number of elements in cartition product of two sets are denoted by n of a cross b. The formula says n of a into n of b. The number of elements in set A into number of elements in set B. So that is nothing but number of elements in set A is three and the number of elements in set B is one. Accordingly 3 into 1 will become three only. So the total number of elements in the cartition product of two sets A and B are three. Even if you find the number of elements this one the number of elements in cartition product of B and A. So n of b cross a can also be calculated by using same formula. So that is nothing but n of b into n of a.
So here the number of elements in n of b is one and the number of elements in n of a is three. So the total number of elements will become three only. So n of b cross a. So from this what we can conclude is the number of elements in a cross b is always same as number of elements in b cross a that is number of elements in a cross b is equal to number of elements in b cross a that is nothing but n of a into n of b or n of b into n of a. So this is how we calculate the total number of elements in cartition product of two sets. So this is all about the concept of cartition product of two set. I hope you can solve some more and more number of problem based on cartition product of two sets. So we have done with ordered pair, equal ordered pair and cartition product of two sets. Now we are entering to the actual concept which is there in second PC syllabus that is relations. So let us start with the definition of a relation.
So what is relation? So in our daily life we come across with many relation.
For example, when we are staying at home, mother, father, mother, son, mother, daughter and then brother and brother, sister and sister like this, we have many more relations in our daily life. Like that only when it comes to relations in mathematical functions.
Then what is relation? Let us understand with the definition. Defin see dear students, if you understand the definition properly, we can easily understand the concept very easily. Now we are looking for relations. So relation is nothing but relation is nothing but it is a subset of cartitionian product of two sets. I can say that a relation is relation is a subset of relation is a subset of cartesian product of two sets.
Relation is subset of cartesian product of two sets. I repeat again relation is subset of cartisionian product of two sets and the total number of relations from set A to set B can be calculated by using formula. Total number of elements.
The total number of elements total number of elements from set A to set B can be calculated.
can be calculated by using formula 2 to the power n of a into n of b. This is a formula. Okay, I repeat again. If set a has a elements and set b has b element. If set a and set b are two sets and when we are looking for the total number of elements from set a to set b then the total number of relation from set a to set b can be calculated by using formula 2 to the power n of a into n of b. Why we are considering 2 to the^ n of A into N of B? Because the definition of relation says that it is a subset of cartition product of two sets.
In first QC we have already studied that the total number of subsets from a given set can be calculated by using formula 2 to the power n where n stands for number of elements. So like that only here um we are looking for the total number of elements in the relation and the definition of relation says that it is a subset of cartition product of two set and cartition product of two sets can be calculated by using formula n of a into n of b that is n of a cross b is equal to n of a into n of b. So 2 to the power n of a into n of b where n of a into n of b is nothing but the number of elements in set a uh number of elements in cartition product of two sets that is n of a into n of b. accordingly we use this formula to find the total number of relations. Okay, this is all about the definition of relations. Now let us move on to the next concept. The next concept is very important. So in second P you see the actual concept will start from will begin from types of relations only.
So till now whatever we have discussed it is all about uh introduction or the previous first PC basic concepts to understand the concept of types of relation. I have given uh basics of this concepts such as ordered pair equal ordered pair cartition product of two sets and the definition of relation and the total number of relation from set A to set B formula I have given. Now let us discuss about types of relations. As we already mentioned there are six types of relations we have. One is empty relation. Second one is universal relation. Third one is reflexive. Fourth one is symmetric. Fifth one is transitive and the sixth one is equivalence relation. What do you mean by empty relation? Let us begin from empty relation. When do we say that the given relation is empty? The word itself will give you the clue that empty relation which means there is no ordered pair in the given relation. So here we can say that the meaning of empty relation is nothing but no element of set A is related to any element of set B is called as empty relation. That means no element of set A will be relating to any element of set B. So here I say the definition of empty relation. No element of set A, no element of set A is related to is related to any element of set B. Any element of set B is called set B is called empty relation is called empty relation. And here the relation is denoted by capital letter R.
We can consider where R is equal to empty relation. No elements will be given or this one can also be considered as a five which is null empty. So this is the definition of empty relation we can check and then let us move on to universal relation. What do you mean by universal relation? The meaning of universal relation let us understand now.
Universal relation. The meaning of universal relation is nothing but if the given relation R is same as cartition product of two sets. If the given relation is if the given relation R is same as cartesian product of two sets then we can say that the given relation is universal relation. For example for example assume we have set A is equal to 1 2 and set B is equal to 3 and 4. If I write a relation R, if I write a relation R like 1A 3, 1 4, 2a 3 and 2a 4, then this relation is to be considered as universal relation. How can we say that the given relation is a universal relation? I told you very clearly that the meaning of universal relation is nothing but if a relation R is same as cartesian product of two sets then we can say that the given relation is universal relation. So here we have already written the elements of the relation are that is 1a 3 1 4 2 3 and 2a 4. If these elements are same as cartition product of two sets then we can say that the given relation is universal relation. Let us cross check whether these elements are same as cartition product of two sets or not.
Now I'm finding a cross b. How do we find a cross b? First let us write the elements of set a. Elements of sets a are 1a 2 cross. Elements of set b are 3a 4. When we find the cartition product of these two sets we get the elements which is of the following form that is every first element is belongs to set a and second element is belongs to set b. 1a 3 and 1 4. we will get if I pair from one and then the next elements are like 2a 3 and 2a 4 only we will get 2a 3 and 2a 4 by observing by observing the elements of relation are and the elements of relation and the elements of cartition product of two set it is very clear that these two are one the same these two are one other the same you can observe here see a cross b is nothing but 1a 2 will be relating to or else I will Say I will write in more simple way. Assume we have one and two and three and four. So one will be relating to three, one will be relating to four. Similarly two will relate to both 3 and four. 2a 3 and 2a 4 like this. If I start from 1 1 3 1 4 will get. Yes, we have written.
Similarly if I start from 2 2a 3 2a 4 will get that we have written. So from this by looking at the equation number one and two where the equation number one is stands for relation R equation number two is stands for A cross B. So the definition of definition of universal relation says that if a relation R is same as cartition product of two set then we can say that the given two sets are I mean the given relation is universal relation we can say since R is equal to since R is equal to A cross B. Therefore the relation R is universal relation. We can say universal relation. Finish. This is about the definition of universal relation. I hope you have understood. So let us move on to the next type which is reflexive relation. Reflexive, symmetric and transitive relation. Okay. Let us see the definitions of reflexive symmetric and transitive relation quickly and then we will move to the problems which is given in NCERTT and you will be very surprised that the problems I will not take as in an order which is given in the textbook NCERTT.
I will be teaching in my own way so that every student every student can understand the concept as easy as possible. And after discussing three or four problems, every student will solve the next upcoming problems. So that is the way of uh uh solving in my class. Okay. So now let us see the definition of reflexive relation. What is reflexive relation? Reflexive.
Reflection. You may have heard the word reflection. Okay. Reflexive relation.
Since you all are science students, you may have already heard the word reflection. What do you mean by reflection? I will give one simple example. Assume we have one light. The light is coming and falling on a one surface and obviously the same light will get reflect coming and going. So whatever the income is coming and the same outgoing is going. So that is what we call it as a reflexive relation. In a simple word to say that the given relation is reflexive it should satisfy the condition if a comma a belongs to r for all elements of a is belongs to set a for every element of set a means if we have all the identical ordered pair for the given elements of set a then we can say that the given relation is reflexive relation. I repeat again for all elements of set A if you have identical ordered pair the meaning of identical order pair is nothing but the first element and the second elements both should be same a comma a which means both are same first element and the second elements both are same 1a 1 2a 2 3a 3 and so on so let us understand with an example to say that what exactly the meaning of reflexive relation I'm taking a set assume set a is equal to 1 2 3 4 5.
Now if I write a relation R, if I write a relation R which is of the following form. Assume we have set A is equal to 1A 1 R1 and R2 is equal to R2 is equal to 1A 1 2 and 3A 3.
R3 is equal to 1a 1 2a 2 3a 3 4a 4 5a 5 R4 is equal to 1a 1 2a 2 2a 3 4a 4 5a 5 R5 is equal to 1a 1 2 2 3 3 4a 4 and 5a 5 and even R six is equal to we write empty R7 is equal to R7 is equal to 1a 1 2a 2 3A 3 4a 4 5a 5 2a 5 and 3a 5. Okay, these are the ordered pairs from the relation R1 to R R seven. I have given seven example to understand to understand what exactly the reflexive relation. Now let us understand the definition of reflexive relation. So first as I said that reflexive relation is nothing but we should have all identical ordered pair for the given elements of set A. For all the elements of set A we should have identical ordered pair. The meaning of identical ordered pair as I've already said it is it should be in the form of 1 comma 1 2 that mean both first element as well as second element should be same. Now here I had written the elements of set A. Set a is equal to 1 2 3 4 5. How many elements are there?
There are five elements are there. Now to say that the given relation is reflexive we should have five identical ordered pair which are those 1 comma 1 2a 2 3a 3 4a 4 and 5a 5. Among these ordered pair if any one ordered pair is missing immediately we can say that the given relation is not reflexive relation. I repeat again for the given elements of city for all elements of seta we should have all identical order pair then only we can say that the given relation is reflexive relation. So now you can observe the relation R1. What is R1? R1 has only one ordered pair that is 1 comma 1. Can we say that the given relation is reflexive relation? No, it is not a reflexive relation. Of course we have identical ordered pair that is 1 comma 1. But the definition says that we should have all the identical ordered pair A, A for all elements of set A. For all elements of set A, we have only one ordered pair 1 comma 1. We don't have 2a 2 3a 3 4a 4 5a 5. Therefore we can say that and among these five ordered pair if any one ordered pair is missing then immediately we can say that the given relation is not reflexive relation. But here there are four ordered pairs which is missing that is 2a 2 3a 3 4a 4 5a 5.
Therefore immediately we can say that R1 is not reflexive relation. Moving to the second relation R2. Can we say that R2 is reflexive relation?
So we have ordered pair 1 comma 1 2a 2 3a 3. Here also we can say that the given relation is not reflexive relation because we don't have 4a 4 and 5a 5.
Since we don't have 4a 4 and 5a 5.
Therefore we can say that R2 is also not reflexive relation. Now moving on to R3.
Observe the elements of R3. The R3 ordered pairs are 1a 1 2a 2 3a 3 4a 4 5a 5. We have all identical order pair for all elements of set A. So in set A there are five elements are there that is 1 2 3 4 5. As I said the definition of reflexive relation means we should have all identical ordered pair for the elements of given set. Accordingly we have 1a 1 2a 2 3a 3 4a 4 even 5a 5.
Therefore we can say that R3 is a reflexive relation. Now moving on to the next uh example R4. R4 is equal to 1a 1 2a 2 2a 3 4a 4 and 5a 5 we have can we say that the given relation is reflexive relation until unless we don't have all identical ordered pair for the given elements of set A we can never say that the given relation is reflexive relation now by looking at this ordered pair these ordered pairs we can say that here 3a 3 is missing we have 3 in a given elements of set A but we don't have that identical order pair 3 since 3A 3 does not belongs to R4. Therefore R4 is not reflexive relation. Clear? So we can conclude that R4 is not reflexive relation. Now moving on to fifth example that is R5 relation. R5 is equal to 1a 1 2a 3 3a 3 4a 4 5a 5. Now just focus on identical ordered pair. According to the definition of reflexive relation, the definition of reflexive relation says that we should have all identical ordered pair for the given all elements of set A. Set A has five elements. We should have five identical ordered pair.
You may observe that in this relation we have one nonidentical order pair as well. What is that nonidentical ordered pair? Which is not in the form of A, A format. See here the first element and the second element is different from this. Nothing to worry. We will never bother about non identical ordered pair from the given relation or we only bother about whether we have all identical ordered pairs or not for the given elements of set here only we observe identical ordered pair. We have five elements we have 1 comma 1 for the first element 2a 2 second element third element 3a 3 fourth element 4a 4 fifth element 5a 5. We have all identical ordered pair according to the definition of reflexive relation we have all identical ordered pair. Even though if you have non identical ordered pair still we can say that the given relation is reflexive relation. We will never bother about non identical ordered pair.
Our intention is whether we have all identical order pair or not for the given elements of set A that we will cross check. Accordingly the relation R5 is also reflexive relation we can say.
Now moving to R six. Obviously R six can never be the reflexive relation because there is no ordered pair. So this is to be considered as an empty relation.
There is no element empty relation. But we say that the given relation is reflexive only if you have all identical order pair for the given elements of set A. Now moving to the last example that is R seven. R seven has a certain set of elements that is 1 comma 1 2 3 4 5. See from the first five elements whatever we wanted to wanted that we have to say that the given relation is reflexive relation. We have all five identical ordered pair 1A 1 2 3 4a 4 and 5a 5.
These two ordered pairs are non identical ordered pair. These two ordered pairs are non identical ordered pair. As I said in the example five, we will never bother about the nonidentical ordered pair. In the reflexive relation, we should bother only about identical ordered pair for the given elements of set A. Accordingly, it is satisfying. We have all identical ordered pair. 1A 1 2 3 4 5. Therefore we say that the given relation is reflexive relation without bothering about nonidentical ordered pair. So among the given examples among R1 to R seven which are reflexive relation means we can say the reflexive relations are R3, R5 and R7 respectively. We can say only these three relations are reflexive relation among seven examples. So this is all about reflexive relation. I hope you have understood and there is no confusion at all.
So this is all about reflexive relation and even we can calculate the total number of reflexive relation. I will say the total number of reflexive relation can be calculated by using formula 2 to the power n into n minus one. We can say the total number of reflexive relation can be calculated by using formula 2 the power n into nus1. So the question may be asked in the MCQs. So to be frank this is not there in the board syllabus.
The number of reflexive relation no nowhere it is given. So here this uh the the question the total number of reflexive relation how many total number of reflexive relation can be obtained from the elements of set A. They will give set A elements accordingly we can calculate. Assume set A has certain number of elements. Assume we have set A is equal to two elements. Then how many total number of reflexive relation can be obtained from the elements of given set A. So since set A has two elements, we can calculate the total number of reflexive relation as 2 to the power 2 into 2 - 1 which is nothing but 2 to the power 2 into 2 - 1 is nothing but 1. 2 into 1 is nothing but 2 only we will get. So that is nothing but four. So we can find out there are four reflexive relation which we can obtain. Similarly when we have set as three elements assume we have seti as a three elements when we have seti as three elements then the total number of reflexive relation can be calculated by using formula 2 ^ 3 into 3 - 1 3 - 1 is nothing but 2 so 2 ^ 3 into 2 is 6. Calculate the total number of elements as a 2 the^ of 6.
What is the value of 2 the^ of 6? 2 are 4 2's are 8 8 2 are 16 16 2 are 32 32 2 are 64. the total number of elements will be 64 clear accordingly we calculate the total number of reflexive relation still I'm saying the number of reflexive relations we don't have in board syllabus but it can be asked in MCQs it may be in C or uh in JW competitive or even in MCQs also it can be asked so this how we calculate the total number of reflexive relation now moving on to the next concept that is symmetric relation What is symmetric relation and how exactly it works? Let us understand now.
All right. The next concept is symmetric relation.
So what is symmetric relation? Again to say that the given relation is symmetric it should satisfy one of the condition.
Let us observe what is that condition and how exactly it works. Symmetric relation to say that the given relation is symmetric it should satisfy one of the condition. The condition says that if a comma b belongs to r we are saying non identical ordered pair. If a comma b belongs to r then b comma a should belongs to r. I repeat again if a comma b belongs to R then B comma A should belongs to R. See try to understand the meaning here a B belongs to R. Implication this sign is called it as implication that implies is nothing but then clear instead of writing the word then we use mathematical sign implies. If a comma b belongs to R then B comma A belongs to R. This is how we read the given statement. So definition of symmetric relation says that if a comma b belongs to r then b comm a should belongs to r then only we can say that the given relation is symmetric relation that means when we have non identical ordered pair reverse of that non identical order pair should be there in the given relation then only we can say that the given relation is symmetric relation we can say now let us have an example to understand the concept of symmetric relation. Assume we have set AI is equal to 1 2 3. Assume we have set a equal 1 2 3. And moreover whenever we are discussing about the types of relation especially reflexive, symmetric, transitive and equivalence without the set element without any set, we cannot give any example directly. We cannot write a relation R without any set. So we have to consider a set based on the given set only. We will write the relation R. Okay. Now let us see the examples for symmetric relation. Now I'm writing relation are r1 is equal to 1a 1 2a 2 1 2 r2 is equal to 1 2a 1.
R3 is equal to 1 2 1 and then 3A 1.
R4 is equal to 1A 2 2A 1 3A 1 1 3 R4 is equal to I mean R5 is equal to 1 2 2a 3 3A 1 2a 1 1 3 Similarly R six is equal to an Ordered pair 1 comma 2 2a 3 3a 1 2a 1 3 3 2 and 1 3.
R7 is equal to 1A 1 2a 2 3A 3 and R8 is equal to 1a 1 2a 2 3a 3 2a 1 excuse 1 comma 2 okay so these are the eight examples I have written here now let us understand among these examples from R1 R1 to R8 which relation can be considered as a symmetric relation. So first you understand dear students what exactly the definition of symmetric relation when do we say that the given relation is symmetric relation to say that the given relation is symmetric it should satisfy one of the given condition. The condition I have already written here the condition should be if a comma b belongs to R. If we have nonidentical ordered pair the reverse of that nonidentical order pair should be there in the relation R. If a comma b is there in the relation r then the reverse of that b comma a should also be there in the relation r then only we can say that the given relation is symmetric relation and now we have the set of examples let us identify among these examples which are symmetric relation.
Now when we look at the first example r1 is equal to 1a 1 2a 2 1 2 we have can we say that the given relation is symmetric relation.
Can we say that the given relation is symmetric relation 1 comma 1 2a 2 1 comma 2 we have see here as I said that when we have nonidentical ordered pair reverse of that non identical order pair should be there so we are bothering about nonidentical ordered pair if we have nonidentical ordered pair reverse of that nonidentical ordered pair should be there if you have identical ordered pair we have never bothered about identical ordered pair in the definition of symmetric relation. We didn't discuss anything about identical ordered pair. Even though if you consider identical ordered pair, reverse of that identical ordered pair will remain same. For example, 1 comma 1.
What is the reverse of 1 comma 1?
Obviously, it is same 2a 2. Reverse of 2a 2 is 2 comma 2 only you will get.
That means I'm saying the position of first element and the second element. If I interchange the position of first and second element, it remains the same. For example, if I say 1 comma 1 and if I interchange the position again, it becomes the same 1 comma 1 only we will get. So if you have identical ordered pair, nothing to worry directly we can say that the given relation is symmetric as well and the definition we need to cross check only if we have non identical order pair then the reverse of that nonidentical order pair is present or not that we need to cross check that's it. So in the relation R1 we have one non identical order pair one only one non identical order pair which is that that is 1 comma 2. What is reverse of this 1 comma 2? Obviously 2 comma 1.
But do we have the 2 comma 1 in the given relation r1? Since we have a non identical ordered pair 1 comma 2 but we don't have reverse of this 1 comma 2 that is 2a 1 is not belongs to relation r1. Therefore this is not symmetric relation.
I hope it is very clear. And let us move to the second example that is R2. See here we have both the ordered pairs are non identical ordered pair 1 comma 2.
What is the reverse of 1 comma 2?
Obviously 2a 1 only. So we have non identical ordered pair and reverse of that. So therefore we can say that the given relation R2 is symmetric relation.
Now moving to the next relation R3. What is R3 given? R3 is given as 1a 2 2a 1 3a 1. See for the first non identical ordered pair we have reverse that is 1 comma 2 reverse of 1 comma 2 is 2a 1 that is there but we have 3a 1 that is also non identical ordered pair but the reverse of 3a 1 is 1a 3 which is missing so therefore since we don't have 1a 3 does not belongs to r r3 since 1a 3 does not belongs to r3 therefore the given relation is not symmetric relation When can we say that the given relation R3 be a symmetric relation? When we have 3 comma 1 reverse of that 31 should be there. So that is what I had written in the next example. You can observe here R4 is equal to 1a 2 2a 1. Reverse of 1a2 is 2a 1. Similarly reverse of 2a 1 is 1 2 vice versa and 3 comma 1 we have n identical order pair 3 1. Reverse of 3a 1 is 1a 3. Similarly 3 1 comma 3 can be reversed as 3a 1 vice versa we have if a comma b is there then b comma a should be there accordingly it is there so we can say that the given relation is symmetric relation and then fifth example what is fifth example we can observe 1a 2 2a 3 3a 1 we have three nonical ordered pairs 1a 2 2a 3 3a 1 2 reverse of 1a 2 is 2a 1 which we have similarly 2a 3. We don't have reverse of 2a 3. What is reverse of 2a 3? 3a 2. But we don't have 1a 3 is there and 3a 1 is also there. 3a 1 1 3 we have but 2a 3 is there. Reverse of this 2a 3 is not there in the given relation r.
Therefore r5 will not be considered as a symmetric relation. This r5 will become symmetric relation only if we have along with these ordered pair 3a 2. Then we can say that the given relation is symmetric relation. You can observe R six we have 1a 2 2a 3 3a 1 and 2a 1 3 2 1 3 we have you can observe 1 comma 2 2a 1 we have the reverse of these two.
Similarly 2a 3 and 3a 2 we have reverse of each other and 3a 1 and 1 3 both we have accordingly we can say that the given relation is symmetric relation. So R six is symmetric relation. Then R7 1A 1 2 3. Can we say that this R seven is symmetric relation?
Yes dear students. Can we say that R seven is symmetric relation? Think on.
We have discussed about six examples with the proper definition of symmetric relation. To say that the given relation is symmetric, it should satisfy that. If you have nonidentical order pair, the reverse of the nonidental ordered pair should be present. Then only we can say that the given relation is symmetric. Or in a simple words, if a comma b belongs to R, then B comma, A should belongs to R. Then only we can say that the given relation is symmetric relation. In such definition, you can observe here we have identical ordered pair. In the beginning itself I told you very clearly that if we have identical ordered pair the given relation automatically will become symmetric relation. For example we have 1 comma 1 2 3 reverse of these elements are itself. Therefore we can say that the given relation is symmetric relation. So R7 is also symmetric relation. And the last example R8 1A 1 2 3 1 2 2 3. See here the reverse of the first three ordered pairs are itself.
The reverse of first three ordered pairs are itself. So and the last two are nonidentical ordered pair which is 2a 1 and the reverse of 2a 1 will be 1a 2.
Since we have all ordered pair in the form of a comma b and b comma a format.
Accordingly we can say that the given relation is symmetric relation. So r8 is symmetric relation. Finish. So from these examples I hope it is very clear that the definition of symmetric relation is understood to you all. Now let us move on to transitive relation.
This is most important relation because many of the students will get confusion in transitive relation. How exactly this transitive relation will work out? Let us understand.
Okay. The definition of transitive relation says transitive relation.
Transitive relation.
When do we say that the given relation is transitive? Again here to understand the concept of transitive relation we should have a proper definition.
Obviously we have a proper definition.
If the uh to say that the given relation is transitive it should satisfy one of the condition. The condition says if a comma b belongs to r and b comma c belongs to r then a comma c should belongs to r.
Repeat again.
Excuse if a comma b belongs to R and B comm C belongs to R then A C should belongs to R then only we can say that the given relation is transitive relation. So try to understand only if we have the first two ordered pairs only if we have these two ordered pairs in the given relation or then only we will check whether the third order pair is present or not.
Until unless we don't have first to two ordered pairs, we will never think about the third ordered pair. And if you don't have the first to two ordered pairs, automatically we can say that the given relation is transitive relation. Try to understand the definition of transitive relation. If you don't understand the definition of transitive relation properly, you will never understand how to solve the problems related to transitive. I'm saying again we check the third ordered pair. That is if we have a comma b and b comma c in the given relation r then only we will check whether a comma c is there or not. So assume if a comma b and b comma c is not there then obviously it becomes transitive relation only. How? Let us understand with one of the example. Now assume we have set a= 1 2 3.
Set a= 1 2 3. Now let me write the relation R.
R is equal to I will write 1 comma 1 2a 2 and then 1a 3 R2 is equal to 1a 2 2a 3 R3 is equal to 1 3 3A 1 R4 is equal to 1A 1 2 3 R5 is equal to R5 is equal to 1A 2 and 2A 3 and 1 3 Similarly 2A If you have like an order pair R six is equal to 1 only 1A 1.
If you have an ordered pair like 1, 2 and 3, 3.
Okay. So these are the seven examples I have considered to cross check the definition of transitive relation. I told you very clearly in the beginning.
We say that the given relation is transitive only if a comma b and b comma c belongs to r. Then a comma c should belongs to r. According to that according to that look at the first example the first r1 is 1a 1 2a 2 1 3.
Do you have any two ordered pair which is in the form of a comma b and b comma c format? You just observe it.
>> If so a comma c should be there. See if I observe 1 comma 1 and 1a 3. These two ordered pairs are in the form of a comma b and b comma c format.
You can observe a comma b is 1 comma 1 and b comma c is c. It is ending at one and starting at one. So we can consider b as a one here.
So we have a comma b and b comma c. When you have a comma b and b comma c obviously a comma c should be there.
Then only we can say that the given relation is transitive relation. So what is a and c from these two ordered pair?
Obviously the first element is a and the last element is c. So a comma c will be 1a 3 only we will get. So from this relation we have 1 comma 1 and 1 3 and we should have 1a 3. So that 1a 3 is already here. Therefore we can say that the given relation is transitive relation. Is it here? So we don't bother about 22.
Yeah. So accordingly we can say that the first relation R1 is transitive relation. Now moving on to the next example R2 1A 2 2A 3. We have a two ordered pair which is in the form of a comma b and b comma c format. A comma b and b comma c format.
Now we need to cross check whether a comma c is also belongs to the relation r or not. See here you can observe a comma b 1a 2 a b respectively. A is 1 b is 2. Similarly 2a 3 again b is 2 c is 3. We have a comma b and b comma c format. Then what ordered pair should be there as a a c a c should be 1a 3. But that 1a 3 does not belongs to r2.
Therefore the given relation is not transitive relation. So this is not transitive relation. If we had 1a 3 in the relation R2 then we would have say that along with these two ordered pair along with these two ordered pairs if we had 1a 3 then we would have say R2 is transitive relation but we don't have 1a 3 in the relation R2. Therefore we say say that the given relation is not transitive relation.
Moving to R3 what is R3 order are given 1A 3A 1 is given. Again you can observe the elements are in the form of a comma b and b comma c format where a is 1, b is 3, b is 3, c is 1. So when can we say that the relation r3 is transitive? R3 will become transitive only if a comma b and b comma c is there in the given relation r then a c should also be there. Here a c will become 1 comma 1.
Since 1 comma 1 does not belongs to r3.
Therefore the relation R3 is not transitive relation because here we have a comma b and b comma c but we don't have a comma c. Accordingly we can say that r3 is also not transitive relation.
Now moving to R4 fourth example 1a 1 2a 2 3a 3.
See here we don't have any ordered pair which is in the form of a comma b and b comma c format. In the beginning itself I told you very clearly if we don't have an ordered pair a comma b and b comma c in the relation r then obviously the given relation will become transitive relation accordingly here we don't have a comma b and b comma c format so obviously this relation will be considered as transitive relation on which basis we are saying that it is transitive relation to say that the given relation is transitive relation a comma b if a comma b and b comma c belongs to r then a comma c should belongs to r that is the definition we are But without referring that definition we are saying that the given relation is transitive. The reason is if we had a comma b and b comma c it may satisfy a comma c as well. based on that assumption we are saying that the given relation will be transitive relation.
Yeah. So here we don't have a comma b and b comma c format ordered pair. So therefore we are saying that the given relation is transitive relation. Moving to the next example that is R5 1 2a 3.
See here we have an ordered pair in the form of a comma b and b comma c format.
To say that the given relation is transitive we should have 1a 3. So that 1a 3 is nothing but here a comma c we have already accordingly we can say that the given relation r5 is transitive relation. Yeah. If I consider if I consider 1 comma 2 as a a comma b and b comma c will be b comma c will be I will consider the other example to cross check whether the given relation is transitive or not as of the following form. I will consider a comma b as a 1 comma 2 and b comma c as 2a 2. See here ending at two starting at two.
Accordingly we are considering a comma b and b comma c. Even this ordered pair also we can consider to cross check whether the given relation is transitive or not. So here a comma b is 1 2 b comma c is 2a 2 then a comma c will become 1 comma 2 only that 1 2 is already we have written in the first place itself.
Therefore whether you consider in anyhow the given relation will satisfy accordingly we can say that the given relation is transitive relation. And the next one is R sixth example. Can we say that the given relation is transitive?
Yes, of course we can say because we don't have an ordered pair a comma b and b comma c format. Therefore, we can say that the given relation is transitive relation. 1 comma 1 is also transitive.
R7 1 2 3 uh 3a 3. So here we don't have any two ordered pair in the form of a comma b and b comma c format. Therefore this r7 is also a transitive relation.
Among these examples, among these example, from example number 1 to 7, which are transitive relations means we can consider R1, R5, R six and R seven.
These are all transitive relation. We can say I hope you have understood the method of um transitive relation. So the last relation is equivalence relation.
When do we say that the given relation is equivalence relation? So let us explain here only the meaning of equivalence relation is nothing but it is a combination of all the relation which we have discussed in the previous three types.
A relation R is said to be equivalence relation only if only if R is reflexive, symmetric and transitive.
symmetric and transitive then only we can say that the given relation is equivalence relation. I repeat again relation R is said to be equivalent only if only if the given relation R is reflexive, symmetric, transitive. That means it should be a combination of the last three relations. If a given relation R is reflexive, symmetric and transitive, we can say that the given relation is transitive relation. For example, assume we have set A is equal to 1 2 3. And if I write the few relation such as R1 is equal to 1A 1 and R2 is equal to 1A 2 and 2A 1.
R3 is equal to 1A 1 1 2 1 3.
and R4 is equal to 1a 1 2a 2 and 3a 3.
So let us cross check whether these relations from R1 to R4 which one is equivalence relation.
So let us look at the first relation R1 we have only one ordered pair that is 1 comma 1. Now cross check whether this R1 is reflexive, symmetric, transitive all the three then you can say that the given relation is equivalence relation.
So here R1 is not reflexive because set has three elements. Set has three elements. To say that the given relation is reflexive, it should have all identical ordered pair for all elements of set A. Set a has three elements. That means we should have three identical order pair that is 1a 1, 2a 2 and 3a 3.
Then only we can say that the given relation is reflexive relation. Since the relation r is not reflexive, the given relation r is not reflexive.
Therefore, this is not equivalence relation. Among the three types of relation, if any one relation is not satisfying, then and there only we can say that the given relation is not equivalence relation. Now moving on to the second example. R2 is equal to 1A 2 2A 1 we have can we say that the given relation is equivalent relation first of all the given relation is not reflexive because we don't have identical order pair 1 comma 1 2 3 so immediately we can say that since the relation R is not reflexive therefore this is not equivalence relation and we don't think about symmetric and transitive no need to think because among the three relation if any one relation is not satisfying immediately we can say that the given relation is not equivalence relation. Now moving moving to the third example R3 1A 1 1 2 1 3. So see here also we don't have um reflexive relations. We don't have an element to say that the given relation is reflexive until unless we don't have all the identical order pair 1 comma 1 2 3. We can never say that the given relation is reflexive. Here we have only one element that is 1 comma 1 and we don't have 2a 2 and 3a 3. Therefore, even this relation is also not transitive relation.
Moving on to example number four. R4 is equal to 1A 1 2 3 V we have. Can we say that the given relation is equivalent relation? Think once. So here R4 is is it reflexive? Yes, of course it is reflexive because we have all identical ordered pair for the given elements of set A. Set A has three elements. Three identical order we have 1 comma 1 2a 2 3a 3. Now cross check about symmetric.
Does the relation is symmetric?
Of course we can say that the given relation is again symmetric as well because we know the definition of symmetric relation that is if a comma b belongs to r then b comm a should belongs to r. Reverse of these identical ordered pairs are itself. So we can say that the given relation is symmetric as well. Now the last case is to check is transitive relation. Can we say that the given relation is transitive? Of course, we can say the given relation is transitive. Here only I have not erased about the transitive relation. When we have an elements which is of the form 1a 1 2 and 3a 3. Such example under the relation under the set A is equal to 1 2 3. We can say that the given relation is transitive relation. Here also we have the same set and the same order pair. So this is transitive as well. Since the relation R is reflexive, symmetric and transitive. Therefore we can say that the given relation R4 is transitive relation. I will give few more example to understand the transitive relation.
Assume if you have R5 is equal to 1a 1 2a 2 and 3a 3. Along with that if you have 1a 2 and 2a 3 can we say that the given relation is transitive relation?
Let us cross check. See here 1 comma 1 2 3 based on these three ordered pairs we can never say that the given relation is uh equivalence relation okay because even we have the two more ordered pair that is 1a 2 and 2a 3 if you don't have these two ordered pair we would have say that the given relation is equivalence relation as we already concluded in the previous example but along with these three ordered pair we have two more ordered pair that is 1a 2 and 2a 3 and here if I check the reflexivity symmetric and transitivity Does all the three definitions are satisfying or not that we need to cross check. Obviously the given relation is reflexive because for the given three elements we have all the three identical ordered pair. And the next one is symmetric relation. To say that the given relation is a symmetric we should have the non identical ordered pair and the reverse of the same. We have non identical ordered pair 1 comma 2. But we don't have reverse of this that is 2a 1.
Do we have 2 comma 1 here? Since 1 2 belongs to r but 2a 1 does not belongs to r. Therefore the given relation is not symmetric. Since the given relation is not symmetric it is not necessary to think about the transitive next type of relation. Since this is not symmetric immediately we say that the given relation is not equivalence relation. So this how we discuss if you want I will give one more example to understand the same when do we say that this relation is transit I mean equivalence relation.
Let us conclude that we had 1a 1 2a 2 3a 3 that is symmetric 1a 2 and 2a 3 we had. So if I write 1 comma 2 reverse of that one will be 2a 1 2a 3 reverse of that one will be 3a 2 and cross check whether the given relation is equivalence or not.
We have reflexive uh the given relation is satisfying reflexive relation because we have all identical order pair for the given elements of CTA 1 comma 1 2 3 and the symmetric condition also it is satisfying as we have nonidentical order pair and the reverse of the same 1 2 reverse of the 1 2 is 2a 1 2a 3 reverse of this 2a 3 is 3a 2 and now let us cross check about transitivity let us cross check about transitive relation if I consider the first ordered pair and the second ordered pair 1 2 and 2a 3. 1 2 and 2a 3. Observe these two ordered pairs. These two ordered pairs are in the form of a comma b and b comma c format. Correct? a comma b and b comma c format. But do we have a comma c that is nothing but 1 3. We don't have 1a 3.
If I include that 1a 3 will it become transitive relation? Of course for these two elements the given relation will become transitive. But if I consider some other ordered pair, some other ordered pair. Consider 2a 3 and 3 2 3 and 3a 2. These two ordered pair. A comma b is 2a 3 and b comma c is 3a 2 respectively. Then a comma c will be 2a 2 only. We will be yes it will satisfy.
Now check once any other ordered pair is not satisfying.
So from this we can say that the given relation is transitive as well. Correct?
You consider any ordered pair any two ordered pair in the form of a comma b and b comma c obviously it will satisfy a comma c as well. For example here 1 comma 2 and 2a 1 2 and 2a 1 a comma b and b comma c format then a comma c will become 1 comma 1. Obviously that 1 comma 1 is already we have similarly if I consider 2a 3 3a 2 that we have already observed 2a 3 3a 2 and 2a2a 2 should be a comma c that 22 is also there. If I consider 1 comma 1 and 1 3 again a and c will be 1a 3 only that we have and then yes of course even if you consider any two ordered pair 2a 2 2a 1 again 2 comma 1 we have yes it will satisfy so r3 is a transitive relation since the given relation r 6 is reflexive symmetric and transitive therefore it is transitive relation among the example number r1 to r six the transitive two relation I mean equivalence relations are R4 and R six only these two relations are equivalence relation since it satisfy reflexive symmetric and transitive all so this is all about the types of relations excuse me we have discussed ordered pairs equal ordered pairs cartition product of two sets relations types of relations in types of relations empty relations universal relation reflexive relation symmetric relation transit relation and equivalence relation all the types of relations we have discussed. So now it's a time to solve the problems based on these concepts. Let us see one by one.
Yeah dear students uh till now we were discussing about all the types of uh relations. Now we are solving the problems under those types of relations which is there in our NCERT book and I'm not solving the problems which is in such order which is given in the NCERT books. As I said in the beginning uh I am solving in my own way that if the problem is I'm solving under problems on empty and universal relation in the entire textbook if you search the problem on problems on empty and universal relation is only one problem.
So that is the given one and if we had some more problems which is based on the same I would have discussed these problems one below the other so that every student can understand the concept of problems under empty and universal relation clearly. So the my way of teaching is if I consider any one type of the problem in the entire textbook I will search if there is any such type of problems are there then immediately I will discuss the remaining all other problem which is similar to the previous so that every student every student I'm saying even the student is topper or the average students or below average students or the student doesn't know anything about the concept then also the student can able to understand in such a way I'm going to teach so that is what my uh specializer in uh this uh video through this channel and that is what I'm teaching. If I solve any one problem of one type then I will gather all the problem which is there in the NCERTT and I will discuss one below the other solving one or two problem the student will understand and I'm damn sure that after that the student will eagerly waiting to solve the problem by his own or her own upcoming next problem which is similar type of the same. Now the question is based on the problems on empty and universal relation. We have only one problem. When we are about to solve this problem, we need to recall the definition of empty relation and universal relation. Now let us understand what exactly the given question and how to solve the given problem. So the given question is let a be the set of all students of a boy school. So here we are considering set elements are like boys school. Show that the relation R in A given by R is equal to an ordered pair A comma B such that A is a sister of B is empty relation. We need to prove that the relation R is empty relation under the given condition. What is the given condition?
We consider any two students from the boys to school and we need to prove that the given relation R is empty relation.
The condition is A is a sister of B. A is sister of B. First you understand the question then we will move to the solution. And the second part of the given problem is Rdash is equal to an R pair A B such that the difference between the height of A and B is less than 3 m is universal relation which means we need to prove that Rdash is universal relation. So this is what we have the two part in the given question.
So the first part is to prove that the given relation R is empty relation and the second part is to prove that the given relation Rdash is universal relation. Okay. Now let us understand what exactly the given question they are saying that set is a boy school. In a boy school assume in a boy school if you refer any two students in the entire school we may have a number of students in the boys school. Boy school remember the word boys school. In a boy school assume I am picking any two students randomly. Can we have the relation of those two students as a sister? Will it be possible?
the relation of any two students in a boy's school. Will it be possible that the relation between those two students will be a sister which is not possible because from a boy's school no first student will be a sister of the other student in a boy's school. So therefore in such a way we will not even get one ordered pair that is a comma b which means we will not get a single element according to the given condition. So what you can write in your own words you can write the solution in your own words we can write a solution from a boy school from a boy school here we can say that r is equal to an empty relation directly we can say it is five because what is the reason since from a boy's school no two student will be a sister of each other we can say clear no two students can have a relation as a sister in a boy's school.
So that's what we can conclude mathematically if you want to write write simple since r is equal to since r is equal to an empty relation that is equal to five. Therefore the relation R is empty relation in a simple words you can conclude or else empty relation empty relation we can say or if you want to write an explanation for the same from a boy school from a boy's school no student is a sister of any other student any other student.
Therefore, therefore R is an empty relation. Finish.
We are done with first part. Any doubt?
It is very clear that from a boy school if I construct any two students randomly the relationship between those two students will never be a sister.
Therefore we can say that the given relation is empty relation. Done. We are done with first part. Now let us move on to the second part. What is second part?
We need to say that the relation Rdash is universal relation. What is the condition? The difference between height of A and B is always less than 3 m. They are saying. Now let us understand the meter in terms of feet. So 1 m is equal to 3 something m we can say 1 m is equal to 3 something ft we can say approximately 3.125 we will get okay accordingly 1 m is equal to 3 something we can say then 3 m will be equal to almost approximately 10 ft we will get here 3 m is nothing but 10 ft almost approximately I'm saying not exactly approximately in that case if I consider the difference between the height of two student is always less than 3 m means we can say that the given relation is universal relation. Let us see. So here rdash is equal to 1 by a comma b such that such that the difference between the difference between height of height of a and b is less than 3 m. they are seen.
Let us cross check whether the given relation is universal relation or not.
See assume uh in our yeah in our daily life we may have come across with the many people having very short very tall and average average height. Yeah. Assume assume the height of A and B is always less than 3 m that they are saying according to the given condition. If I consider any two students height is same. Assume uh the first person height is uh some xt.
Okay. The second person is also xt. Then what will be the difference of these two? Obviously the difference between these two will be zero. Which is lesser than 3 m. As I told you that 3 m is nothing but approximately 10 ft we will get. When the two students having the same height then the difference of those two students will be equal to 0 ft we will get which is lesser than 3 m. So it will come under the same. Now assume the first person may have 5 ft. The second person can have 2 ft. The difference between these two will be considered as that 3 ft. We can consider the 3 ft is again lesser than 3 m. 3 m is nothing but 10 ft. Again if I consider 7 ft and 3 ft. 7 ft and 3 ft. The difference between 7 ft and 3 ft will be equal to 4 ft. This 4 ft is again lesser than 10 ft which is again true. Then you consider any any height any height randomly you can consider any height. The difference of u those two height will be always lesser than 10 ft or lesser than 3 m only we will get accordingly we can say that the given relation is universal relation and moreover when do we say that the given relation is universal relation if the relation r is same as if the relation r is same as cartitionian product of two sets then we can say that the given relation is universal relation so I can say r is equal to a cross b since rdash is equal to a cross b therefore Rdash is universal relation we can conclude because the difference between height of any two students will be always lesser than 10 ft only we will get that means every student will come under the given condition. Therefore we can say that the given relation is empty relation. So this is all about the problem based on empty and universal relation. Dear student I'm saying we have only one question based on empty and universal relation that is this problem. Finish. We have done this. Now let us move on to some other problems.
Let us see. Yes, dear students, as we have completed the previously the problems on empty and universal relation. Now we are about to discuss the problems on equivalence relations.
This is very important uh concept in relations and functions because these type of problems will be asked for three marks in our board syllabus. These type of questions will be asked for three marks. Very very important. We have almost 15 to 20 plus problems which is based on the problems on equivalent solution. Among these problems any one question will be 100% one question will be given to you for your annual examination for three marks. One compulsory question and I have been observing from the past 13 to uh 13 years. So every every year one or the other question will be given among the problem based on problems on equivalent solution. Let us see what is the question. So here the given question is and moreover the given question is example number two if you refer NCH textbook. The question says that let T is set of all triangles in a plane with R a relation in T given by R is equal to an ordered pair T1 TS2 such that T1 is congrent to TS2. T1 is congrent to TS2.
Show that R is equivalence relation. We need to prove that the given relation R is equivalence relation. Now to prove that the given relation is equivalence relation, we need to prove that the given relation is reflexive, symmetric and transitive. If the relation R is reflexive, symmetric and transitive, then only we can say that the given relation is equivalence relation. Now to cross check this or to prove this, the given relation is equivalence relation.
Now let us prove one by one. See dear student, I am teaching only one way. As I'm saying that there are 20 plus problems are there under equivalence relation and I teach one way and the entire problems each and every problem under this equivalence relation the same methodology will be taken for each and every problem so that no student will get confused that one one problem is in one one way no I will stick on with only one method. So kindly stick on with this method so that every student can solve each and every problem in the same and easy way. Let us see how to solve the given problem. So here we need to prove first of all the given relation is equivalence relation. To prove that the given relation is equivalence relation it should be a reflexive symmetric and transitive. First I'm considering reflexive relation. First I'm considering reflexive. When do we say that the given relation is reflexive? To say that the given relation is reflexive. We should have all identical ordered pair for the given elements of set A. That is A belongs to R for all elements of A is belongs to set A. Then only we can say that the given relation is reflexive. But here the given question is not in terms of a and b. It is given in terms of t_1 and t2. So we need to prove according to the same way.
So I will assume. So just observe the methodology that I prove that the given relation is so. First for reflexive relation let us assume let t1 comma t1 belongs to r and write the meaning of this t1 comma t2 belongs to r from the given question only. No need to buy hard, no need to memorize anything. You can write the solution from the given question only.
What is the given question? R is equal to T1, T2 such that T1 is congrent to T2. See here they had given T1 and T2 and they are saying that the relation T1 is congrent to TS2. Since they had given T1 and T2, they are writing in terms of T1 and T2. If you're writing in terms of t_1 and t_sub_1 then you can write the meaning of t1 comma t1 belongs to r is t_1 is congrent to congruent to t1 only the meaning of congrent is nothing but similar or equal we can say in a simple words I'm saying almost equal we can say congrent is nothing but almost equal so you can refer the meaning of congrent is nothing but equal the meaning of congrent is nothing but equal almost equal we can consider. So here t1 is equal to t1 which is true correct now that is what we'll conclude. Let t1 comma t1 belongs to r which implies that t1 is congrent to t1 which is true which is true.
Therefore therefore the relation r is reflexive relation finish.
There is no much difficulty to prove that the given relation is equivalence relation. Just observe the methodology.
to say that the given relation is so and so. So I hope you have understood to say that the given relation is reflexive relation first we assume that t1 comma t1 belongs to r and we say the meaning of t1 comma t1 in words t1 comma t1 belongs to r is nothing but t1 is congrent to t1 which means t1 is similar to t1. We all know very well that t1 is similar to itself or t1 is similar to t1. Therefore we can say therefore which is true therefore r is reflexive relation. Now moving to the next relation that is symmetric relation.
When do we say that the given relation is symmetric.
We know the definition. If a comma b belongs to r then b comma a should belongs to r. Then only we can say that the given relation is symmetric relation. I repeat again. If a comma b belongs to r then b comma a should belongs to r. But here we need to prove in terms of t1 and ts2. First I will assume let t1 comma t2 belongs to r. I will say what does it mean? It means that t_1 is question in the question only they had given the meaning of t1 comma t2. What is the meaning of t1 comma t2? T1 is congrent to t2 that is what I'm writing in words. So t_1 is congruent to t_sub_2 which means we are saying that t_s_1 is similar to t2 we are saying that means if t_sub_1 is similar to ts2 can't we say that ts2 is also similar to t1 of course we can check so that's what I'm writing in the next line so t_sub2 is also congrent to t1 I had written this one in words. If I compute the same mathematically, how we can write? We can say that t_sub_1 comma I mean t2 comma t1 belongs to r. We can say yeah and we write a conclusion since t_sub_1 comma ts2 belongs to r implies t_sub_2 t1 belongs to r.
Therefore the relation r is symmetric relation. Finish. Very simple.
And the way I'm teaching this particular problem the same way for all the problems which is there under equivalence relation I will follow and even you all can um follow the same methodology so that everyone will get three marks 100% you will get three marks. So I repeat again first I have assumed t1 comma t2 belongs to r. We are assuming that the given relation is I mean t1 comma t2 belongs to r. The meaning of t1 comma t2 I have written in words that is t_s_1 is congrent to ts2 which means t1 is similar to ts2. If t_sub1 is similar to t2 we can say that ts2 is also similar to t_1 which is true. So I had written this in words and converting the same mathematically that is t2 t1 belongs to r and we writing the conclusion since t1 comma ts2 belongs to r. Ulta. Therefore, E stands for since.
Okay. Since T1, TS2 belongs to R implies T2, T1 belongs to R. Therefore, the relation R is symmetric. We are done with this. Now the last one is transitive relation. Transitive relation. When do we say that the given relation is transitive? If A B belongs to R and B, C belongs to R, then A, C should belongs to R, then only we can say that the given relation is transitive relation. Now to check that let us prove this in terms of t1 and t2.
So we assume let t_1 comma ts2 belongs to r and ts2a t3 belongs to r.
What does it mean? You should write the meaning of t1 and t2. t2 and t3 inverse.
Meaning of t1 and t2 which is already given in the question. What is the meaning of t1 comma ts2? t_1 is congrent to t2. Write on the same. This implies that T1 is congruent to TS2 and TS2 T3. In place of T1 and T2 just write T2 and T3. Then you can write the same sentence like T2 is congruent to T3.
What is the meaning of this one? If T1 is similar to T2 and TS2 is similar to T3, can't we say that T1 is also similar to T3? That's what we will write in the next sentence. This implies that T_1 is also congruent to T1 is also congruent to T3. Okay, I have written this in terms of sentence in words. So convert the same in terms of mathematical expansion. We can write T1 T3 belongs to R. So we assume that T1 T2 belongs to R and T2 T3 belongs to R. And we have concluded that T1 and T3 is also belongs to R. Now we will write the conclusion. What is the conclusion?
Since t_sub_1 ts2 belongs to r and ts2a t3 belongs to r implies t1a t3 t1 comma t3 belongs to R.
Therefore the relation R is transitive relation. Finished. We have proved that the given relation is transitive relation. So what you can do final conclusion we should write what was our intention? Our intention was to prove that the given relation is equivalence relation. We should have proved that the given relation is equalence relation.
Till now we have proved the given relation is reflexive. We have proved that the given relation is reflexive symmetric and transitive. But if you don't write the proper conclusion that since the given relation is reflexive, symmetric and transitive, the given relation is transit equivalence relation. So let us conclude it properly. Since R is reflexive, symmetric and transitive and transitive.
Therefore the relation R is equivalence relation. Therefore the relation R is equivalence relation.
Hence proved.
So this is how we prove that the given relation is transitive relation. I hope you have understood each and every problem which is there just below I mean under the concept of equivalence relation. Each and every problem I will solve in the same method. Even I want you all to stick on with the same method so that every student can easily prove that the given relation is so and so relation. Clear? Now let us move to the next problem. Let us consider the next.
Okay dear students, now we have the next question. Um that is again the problem based on equivalence relation only. The given question is let T be the set of all the triangles in a plane with R a relation in T given by a relation R is equal to T1 T2 such that T1 is similar to TS2.
Then further they asking you to prove that show that R is an equivalence relation and consider the three right angle triangles T1 with the sides 3 4 5 and T2 with the sides 5 12 13 and T3 with the sides 6 8 10 which triangles among T1 T2 T3 are related. So this is what we need to check it. Okay. So here there are two parts are there to prove that the given relation is given relation R is equalence relation A is to be considered as the first part and the second part is among the three sides of the right angle triangle which three triangles are related that is what we have to cross check. Okay. Now let us prove the first part of the given problem. The first part is to prove that the given relation is equivalence relation. to say that the given relation is equivalent relation already we have observed one problem just before to this and in the same methodology I am following so let us try to solve the given problem as quick as possible so first one is I am considering reflexive to say that the given relation is transitive relation I mean equivalence relation we should prove that the given relation is reflexive symmetric and transitive so under that the first one I'm considering reflexive methodology same method only I'm following As I have done the previous question, let T1 comma T1 belongs to R. What does it mean? It means that T1 is similar to TS2 I mean T1 is similar to itself. That is T1 is similar to T_1 which is true which is true.
Therefore the relation R is reflexive relation.
Therefore the relation R is reflexive relation. Finish.
First one is done. Now moving to the next one that is symmetric relation.
When do we say that the given relation is symmetric? If a comma b belongs to R then B comma A should belongs to R.
Accordingly I'm considering let T1 comma TS2 belongs to R. And I'm writing the meaning of this one in words. The meaning of T1 comma T2 belongs to R is T1 is in the question only they have given T1 TS2 means T1 is similar to T2 that's what I'm writing T1 is similar to TS2 t1 is similar to T2 if T1 is similar to TS2 if T1 is similar to TS2 can't we say that TS2 is also similar to T1 that's what I'm writing ts is also similar to also similar to T1 and And we are writing the conclusion and before that the meaning of this one mathematically can be written as ts2a t1 belongs to r. Now we write the conclusion since t_sub_1 ts2 belongs to r implies ts2a t1 belongs to r.
Therefore the relation r is r is symmetric relation.
Now moving to the next one that is third one transitive relation. When do we say that the given relation is transitive?
If a comma b belongs to r and b comma c belongs to r then a comma c should belongs to r and we need to prove that in terms of t1 and t2. So we assume let t1 comma t2 belongs to r and ts2a t3 belongs to r.
I'm writing the meaning of t1 comma t2 and t2a t3 inverse. that is t_s_1 is similar to tsub_2 and tsub2 is similar to t3.
If t1 is similar to ts2 and ts2 is similar to t3 then we can say that t_sub_1 is also similar to also similar to t3. This means t_1 comma t3 belongs to r.
Since t1 comma ts2 belongs to r and ts2a t3 belongs to r implies t_sub_1 t3 belongs to r.
Therefore the relation r is transitive relation. finish and I'm writing a conclusion for equalence relation since R is reflexive symmetric and transitive and transitive. Therefore the relation R is equivalence relation equivalence relation. Finish. We have done the first part of the given problem to prove that the given relation t1 comma t2 is equivalence relation. So we have done it. Now let us enter to the second part of the given problem. What is second part of the given problem? We need to consider the right angle triangles.
Right angle triangles.
Consider the three right angle triangles. T1 with the side 3 4 5. T2 with the side 5 12 13 and t3 with the sides 68 10. Which triangles among t1 t2 t3 are related? We need to identify which triangles among T1, TS2, T3 are related. Now to cross check this. Now to cross check whether T1 is related to TS2 or TS2 is related to T3 or T3 is related to T1, we need to take the ratio of the sides of the triangle.
Whichever the ratio of the sides of the triangle are equal then those triangles are related to each other. We can conclude which I mean to say let us consider T_1 and TS2. I am taking the ratio of T_sub1 and TS2. What are the sides of the triangle? T1 3 4 5 is given. So 3 4 5. Okay. And the ratio and the T2 sides are T2 sides are 5 12 13 5 12 13. Does they are equal? Cross check.
They are not equal. 5 3x 5 is not equal to 4x2 and the 4x2 is not equal to 5x3.
Therefore t1 and t2 are not related to each other. Now moving to the next two triangles TS2 and T3 ratio I will take ratio of TS2 and T3. What are the sides of triangle T2 5 12 and 13. Let us take the uh T3 sides. T3 sides are 68 10. 6 8 10 they are equal cross. Since 5x 6 is not equal to 12x 8 that is not equal to 13x 10. Again TS2 and T3 are also not related. Now we are left with last one that is T_sub_1 and T3. T1 and T3 or T3 and T1. Anyhow you can consider. Now I'm taking the sides of the triangle T1.
What are the sides of the triangle T1? 3 4 5 3 4 5 and sides of the triangle T3.
sides of the triangle T3 are 68 10 6 8 10 If I simplify this what we will get you can cross check 3x 6 is nothing but 1 by 2 right 1x2 and 4x 8 is also equal to 1 by 2 5x 10 is also equal to 1x2 that means all these three sides of the triangles are equal so from this what we can say is since ratio of t1 and t3 are equal to each other therefore for t_1 and t3 are related.
So among t_1 and t2 t3 we should have identified which triangles are related to each other that we need to cross check. See here among the uh which triangle among t1 t2 t3 are related.
That was the second sub question. So now we have after checking the ratio of t1 and t2 t2 t3 and t3 t1 and t1 t2 we can conclude that t1 and t3 are related to each other we can conclude. So this is how we can solve the given problem. I hope you have understood. Now let us move on to the next question.
Okay, the next student is I mean the next problem is let L be the set of all the lines in XY plane and R be the relation in L defined as R equ= to L1 L2 such that L1 is par to L2. Show that R is an equivalence relation. We need to prove that the given relation is equivalence relation and find the set of all the lines related to the line y = 2x + 4. So here also we have two part of the given problem. The first part is to prove that the given relation r is an equivalence relation and the second part is to find we need to find out the set of all the lines which is related to y = 2x + 4. Okay. Now first we will prove that the given relation is equivalence relation. Now to prove this the given relation is equivalence relation. We should prove that the given relation is reflexive symmetric and transitive. So let us prove one by one. First I'm considering reflexive relation. So we all know very well to prove that the given relation is reflexive we assume let l1 comma L1 belongs to R. Meaning is L1 is par to L2 I mean L1 is par to L1 L1 is par to itself L1 which is true a line is par to itself which is true.
Therefore, therefore the relation R is reflexive.
Reflexive is done. Now we are moving to symmetric relation. When do we say that the given relation is symmetric? If A B belongs to R, then B, A should belongs to R. So we assume let L1 L2 belongs to R. What does it mean? It means that L1 is paral to L2. Clear? We can write this in words.
If L1 is par to L2, we can say that L2 is also parallel to also parallel to L1. This implies that TS2A T1 belongs to R. Since L1, L2 belongs to R implies that L2A Okay, this is not T1 and T2.
This should be L1 and L2.
L2 and L1 L1 L2 belongs to R. Then L2 L1 also belongs to R.
Therefore the relation R is symmetric relation.
Next one is transitive relation. Next one is transitive relation. When do we say that the given relation is transitive? If a comma b belongs to R and B c belongs to R then A c should belongs to R then only we can say that the given relation is transitive relation. We need to prove in terms of L1 L2 L3. So let us assume let L1 L2 belongs to R and L2 L3 belongs to R.
meaning is L1 is parallel to L2 and L1 is par to L2 and L2 is par to L3. If L1 is par to L2 and L2 is par to L3 then we can also say that L1 is also par to L1 is also parallel to L3. This implies that L1A L3 belongs to R. And we write the conclusion since L1A L2 belongs to R and L2A L3 belongs to R which implies L1A L3 belongs to R.
Therefore the relation R is transitive relation transitive relation. Now I will write the conclusion for equivalence relation.
Since relation R is reflexive, symmetric and transitive and transitive.
Therefore, the relation R is therefore the relation R is equivalence relation.
Equivalence relation. Finish. We have proved that the given relation is equivalence relation. So, first part of the given problem is done. Now, we need to prove the second part. What is second part? Second part is to find the set of all the lines which is related to the line y= 2x + 4. Which means we need to find out uh the set of all the lines which is related to the line y= 2x + 4. Now to find the set of all the line which is related to the given line. So first we need to find the slope of the line. The given equation of the line is in the form of y = mx + c format in place of m whatever the value will be there accordingly we can find out. and we can find it parallel lines or the set of the lines which is related to the given line. So here if I consider the given equation of the line is y = 2x + 4 is given y = 2x + 4 is given and this equation is in the form of is in the form of y = mx + c y = mx + c. So see here in place of m if I write just below to this you you can easily compare y = mx + c format. Here m is nothing but slope of the line. M is nothing but slope of the line which we have learned in first QC. M is nothing but slope of the line. When I compare the given equation first equation with the standard equation of the line y= mx + c.
We can easily identify that the value of m will be equal to 2 which is coefficient of x. Meaning of coefficient of x is nothing but the value of just beside x what we have that is to be considered as a coefficient. So just beside r the coefficient of x is 2. So I can say that the value of m = 2. Then I can write the required set of all the line which is related to y = 2x + 4 is y = 2x + c where c can be any constant.
Then we write therefore the required the required set of all the lines related to related to y = 2x + 4 is y = y = y = 2x + c. This is our required answer to the second sub question. Clear? So this is how we conclude the required set of all the lines which is related to y= 2x + 4 is y= 2x + c. Which I mean to say we need to find the parallel line for the given equation of the line y= 2x + 4. To find the parallel line for the given equation of the line. First we need to find the slope and c can be anything. Here constant can be anything.
So we can say that this equation of a line is parallel to y = 2x + c where c can be any real number. It can be any real number. So this is how we solve the given problem. I hope you have understood the question and the solution as well. Now let us move on to the next problem.
Okay dear students, next problem is the fourth question that is let L be the set of all the lines in a plane and R be a relation in L defined as R= L1 L2 such that L1 is par perpendicular to L2. Show that R is symmetric but neither reflexive nor transitive. So I hope you have understood the question. In the previous question we had L1, L2 such that L1 is parallel to L2 was given. But in this question the relation R is given as L1 L2 such that L1 is perpendicular to L2. From the given condition we need to prove that the given relation is symmetric but it is not reflexive and not transitive. This is what we have to prove. Now to prove this again we follow the same procedure. I will consider the first one is symmetric relation because we need to prove that the given relation is symmetric and we need to prove that it is not reflexive and not transitive.
Okay, in an order only I will try to prove I will consider the reflexive condition.
Let L1 comma L1 belongs to R. I will write what does it mean? It means that L1 is perpendicular to L1 is perpendicular to L1 only which is not true.
Which is not true or false also we can say which is not true or false. Therefore the relation R is not reflexive. I hope you have understood because line is parallel to itself but line is not perpendicular to itself. We can never say that the line is perpendicular to itself. Therefore we say that L1 and L1 does not belongs to R. We can say L1 comma L1 does not belongs to R or L1 L1 belongs to R which implies that L1 is perpendicular to L1 which is not true or false. Therefore R is not reflexive. We have proved that the given relation is not reflexive. Now let us move to symmetric condition. What is the definition of symmetric? If a comma b belongs to r then b comma a should belongs to r. We need to prove in terms of l1 and l2. So we assume let l1 l2 belongs to r. What does it mean? It means that l1 is perpendicular to l2. If l1 is perpendicular to l2 and we say that l2 is also perpendicular to l1.
That is what I'm writing in the next line. So L2 is also perpendicular to also perpendicular to L1. This implies that L2A L1 belongs to R.
Conclusion since L1A L2 belongs to R which implies that L2A L1 belongs to R.
Therefore the relation R is Therefore the relation R is symmetric relation finished. We have proved that the given relation is symmetric and the last one is transitive. You should prove that the given relation is not transitive. Let us see how it is not transitive. Now to prove the given relation is transitive or else for this one you can consider the geometrical representation as well.
Assume this is L1 and this is L2. We can easily say that L1 is perpendicular to L2 and L2 is perpendicular to L1. Based on this, the given condition will satisfy that the relation R is symmetric relation. But now we are about to prove that the given relation is not transitive. Why it is not transitive?
Let us see. First I will assume the same methodology as we used to prove that the given relation is transitive. So let L1 comma L2 belongs to R and L2 L3 belongs to R. What is this? What is meaning? The meaning of this order pairs are L1 is perpendicular to L2 and L2 is perpendicular to L3.
So here directly I cannot say that L1 is also perpendicular to L3. So to understand the concept let us have geometrical representation. Assume we have a line L1. This line L1 is perpendicular to L2. Okay. And this line L2 is perpendicular to some other line L3. So try to observe carefully the horizontal first horizontal line is to be considered as a L1. Vertical line is to be considered as a L2. So L1 is perpendicular to L2. That is correct.
Then L2 is perpendicular to L3. This vertical line is perpendicular to another horizontal line L3. Then we need to think about L1 and L3. Now you can observe from the figure geometric representation only we can understand this L1 and L3 are not perpendicular.
They are parallel. So in the next line what we can say this implies that L1 is not perpendicular to L3 R. This implies that L1 L3 does not belongs to R. So I'll write the reason since L1A L2 belongs to R and L2A L3 belongs to R. But this implies that L1A L3 does not belongs to R. Therefore the relation R is not transitive relation. Therefore the relation R is not transitive relation.
Hence we prove that the given question is so and so. So in this case we have proved that the given relation is not reflexive. The given relation is not reflexive because a line can never be perpendicular to itself. Therefore the given relation is not reflexive. If L1 is perpendicular to L2 and L2 is perpendicular to L3 then L1 and L3 are not perpendicular. Instead they are parallel. Therefore R is not transitive.
And we know that uh symmetric it is symmetric because when L1 is perpendicular to L2 then obviously L2 is also perpendicular to L1. From this conclusion we can say that the given relation R is only symmetric but it is not reflexive and transitive. This is about the given question. Let us move on to the next question.
Okay. Uh the next question is we need to prove that we need to prove that the given relation is again equivalence relation. Show that the relation r in the set z of integer given by r is equal to an ordered pair a comma b such that 2 divides a minus b such that 2 divides a minus b is an equivalence relation. Let them give any problem based on equivalence relation.
We just follow the same procedure as we have already proceeded. Clear?
Yeah. We need to prove that the given relation is an equivalence relation. Let us follow the same procedure. So I will start with reflexive relation.
So here we assume we need to prove that the given relation is reflexive in terms of a and b. We need to prove. So we know the definition of reflexive relation. If a comma a belongs to r for all elements of a.
So let a comma a belongs to r which implies that what does it mean?
The meaning of this a comma a belongs to r is 2 divides 2 divides a minus a which is true just to follow the same methodology as we have done from the previous few problems which is true. Therefore the relation r is reflexive. That's it. No more calculation you will do and you will never take the elements by your own own.
You should never take the examples. If you take examples, you are proving that the given relation is so and so for uh the elements which you are taking for as an example. But we need to prove that the given relation is equivalence for all elements for all integers. So you should not never take the examples. You should prove in general only. So this is about reflexive. Let us move on to symmetric.
To say that the given relation is symmetric, we assume let a comma b belongs to r.
It implies that the meaning of a b belongs to r is 2 divides 2 divides a minus b.
If 2 divides a minus b, then we can also say that 2 also divides b minus a. We can say so we can say 2 also divides 2 also divides 2 also divides b minus a.
which implies that b comma a belongs to r.
Since a comma b belongs to r implies b comma a belongs to r.
Therefore the relation r is symmetric relation. Finish. We have proved that the given relation is symmetric as well. Now we need to prove that the given relation is transitive.
If so we can say that the given relation is equivalence relation. Now to prove that the given relation is transitive, let us assume a comma b belongs to r and b comma c belongs to r. Meaning of a comma b and b comma c we should write in words. So the meaning of a comma b belongs to r is 2 divides 2 divides a minus b and 2 divides 2 divides b minus c. And in the next line we should never write two also divides a minus c. We need to write one mathematical expansion that is when we add a minus b and b minus c we will get a minus c. Then we can conclude that is a minus b. This is very important. A minus b plus b minus c will give you a minus c. When you open the bracket we will get the same minus b plus b. Both will get cancelled. We will get a minus c. From this we can say this implies that 2 also divides two also divides a minus c. This implies that a comma c also belongs to r. I will write the conclusion. Since a comma b belongs to r and b comm c belongs to r implies that a comma c belongs to r.
Therefore the relation R is therefore the relation R is transitive relation transitive relation now we write the conclusion for the entire problem what is the conclusion our intention is to prove that the given relation is equalence relation so I'm writing since the relation R is reflexive symmetric and transitive and transitive therefore the relation R is equivalence relation.
Equivalence relation. Hence true. Clear.
Very simple. Dear students, the chapter is very simple. Only the thing is you should understand only one problem based on equalence relation and you will stick on with that method and you will solve all the other problems of the same method. I hope you have understood this method and the problems the solution of the same. Let us move on to the next question.
Okay dear students, you can see the fifth question which is again based on equivalence relation only. The given question says that show that the relation R in the set A of all the books in a library of a college given by R is equal to an ordered pair X Y such that X and Y have same number of pages is an equivalence relation. We need to prove that the given relation is again here equivalence relation. So again we will follow the same procedure. First we'll prove that the given relation is reflexive, symmetric and transitive and we conclude that the given relation is equivalence relation. So I'm considering the first one that is reflexive relation. So let us assume let x belongs to r. What does it mean? Write the meaning of x belongs to r in words.
The meaning of x belongs to r is x and x have same number of pages. Same number of pages. Assume we have one x book.
Clear? If x has certain number of pages then obviously the same book can have the same number of pages. So that is what we are writing. X and x have same number of pages. Which is true?
Which is true.
Therefore, therefore we can say that the given relation R is reflexive. Finish.
We are done with the first one. Now we are moving to symmetric relation. When do we say that the given relation is symmetric? If a comma b belongs to R then B comma A should belongs to R. So let us assume let X Y belongs to R.
Right? The meaning of this ordered pair inverse. The meaning of X Y belongs to R is X and Y have Same number of pages.
Same number of pages. If X and Y which means two books, two books X and Y have same number of pages. Can't we say that Y and X also have same number of pages?
Yeah, that is what we have to write in the next line. So this implies that Y and X also have same number of pages. Same number of pages.
This implies that y x belongs to r.
Then we write a conclusion since x y belongs to r which implies that y x also belongs to r. Therefore the relation r is symmetric relation.
symmetric relation. Now we are moving to the last one that is transitive relation. Let us see the transitivity.
So we assume let x y belongs to r and y z belongs to r and we write the meaning of x y and y comma z in words.
So that we can say X and Y have same number of pages.
Same number of pages and Y and zed have same number of pages.
When X and Y have same number of pages and Y and zed have same number of pages.
Can't we say that X and Zed also have same number of pages? X and Z also have same number of pages.
So from this we can conclude that the value of X belongs to R. And now we write the conclusion. Since x y belongs to r and y z belongs to r which implies that x z also belongs to r. Therefore the relation r is therefore the relation r is transitive relation transitive relation. Now we write the conclusion for the entire problem because our intention is to prove that the given relation is an equivalence relation. So I'm writing the conclusion since relation R is reflexive symmetric and transitive.
Therefore the relation R is Therefore the relation R is R is an equivalence relation equivalence relation hence prove hence prove I hope you have understood the method and the solution of the given problem.
Now let us move on to the next question.
Okay. And dear students, next question you can observe. We have a question like show that the relation R in the set 1 2 3 4 5 given by relation R is equal to an order A B such that modus of A minus B is even is an equivalence relation. So first part of the given problem is to prove that the given relation is an equivalence relation. Then we need to show that all the elements of 135 are related to each other and all the elements of 2 and 4 are related to each other. But no elements of 1 35 is related to any elements of 2 and four.
So there are three parts of the given problem. The first part is to prove that the given relation is equivalence relation. And the second part is to prove that all elements of 135 are related to each other. And the third part is all elements of 2 and 4 is also related to each other. But no elements of 135 are related to any element of 2 and four. This is what we have to prove. Now to prove this one by one let us consider for equivalence relation. So I will prove that the given relation is reflexive symmetric and transitive.
Accordingly we can say that the given relation is equivalence relation. So first I will assume uh for reflexivity.
So let us assume uh for reflexive I will write a heading here for reflexive.
For reflexive we can assume let a comma a belongs to r. What does it mean? It means that modulus of a minus a is even number which is true isn't it because a minus a is 0. 0 is to be considered as the even number. So the given condition is true which is true. Just follow the procedure dear students. Never think about it. Whenever they asking you to prove that the given relation is an equivalence relation blindly you follow the procedure as we did it for the previous few questions.
Okay. 100% you will get the right uh full marks. If the question is wrong then nothing to worry. you will get a grace marks just pro according to the given question never think about it is really true or not 100% it will be according to the given question only you can never think about the reverse okay so we can say that here let a comm a belongs to r which implies that a minus a is I mean modulus of a minus a is even which is true therefore the relation r is reflexive finish we have proved that the given relation is reflexive relation now we are moving to the next method I mean next type of relation that is symmetric relation When do we say that the given relation is symmetric? If a comma b belongs to r then b comma a should belongs to r. So first let us assume that a comma b belongs to r. What does it mean? Write the meaning of this order pair inverse. The meaning is modus in the question only it is given the order pair. The meaning of order pair a comma b is modulus of a minus b is even.
Modulus of a minus b is even. Now assume the difference between any two number is even. Then obviously the difference between the reverse of those two number will be always even only we will get. So we we are need not to give any example.
As I told you in the previous example if we give an example we are concluding or we are trying to prove that uh the given relation is so and so only for those elements which you are taking as an example. But we are not supposed to do because we need to prove that the given relation is true for all elements and that two for under the given condition.
It may be for under real number or it may be under integers or they may have given set of elements accordingly we have to prove that nothing else. So never take an example by your own. So here let a comma b belongs to r which means modulus of a minus b is even. If modulus of a minus b is even then we can say that modulus of b minus a is also an even. So from this you can conclude that b comm a an ordered pair b a belongs to r. We'll write the conclusion since a comma b belongs to r which implies that b comma a also belongs to r. Therefore the relation r is therefore the relation r is r is symmetric relation. Therefore the relation r is symmetric relation.
Now let us move on to the transitive relation. The meaning of transitive relation is nothing but if a comma b belongs to r and b comma c belongs to r then a comma c should belongs to r. Let us try to prove accordingly. So we are assuming the first two ordered pair that is let a comma b belongs to r and b comma c belongs to r. What does it mean?
Write the meaning of ordered pair a comma b and b comma c uh in words. So we can say that modulus of a minus b is even and modulus of b minus c is even.
Now before b write directly modulus of a minus b is also even. We need to write one mathematical expansion as we have written in the previous question. So addition of a minus b and b minus c we should write. So we need to prove that modus of a minus b plus modulus of b minus c. modulus of b minus c is equal to modus of a minus c because plus b minus b both will get cancel accordingly we get a minus c. From this you can conclude this implies that modulus of a minus c is also is also an even number.
From this we can conclude that a comma c belongs to r and we write the conclusion since since um an ordered pair a comma b belongs to r and b comma c belongs to r which implies that a comma c belongs to r.
Therefore the relation R is transitive relation transitive relation. And then we can write a conclusion for equivalence relation. Since R is Since R is reflexive, symmetric and transitive and transitive. Therefore the relation R is an equivalence relation.
So we have done with the first part of the given problem. Now let us enter to the remaining parts. What are the remaining parts? The second part is to prove that we need to show that all elements of 135 are related to each other. Which means when we consider the elements of the set 135 that is the difference between any two elements of the set 135 are always even number. So I will conclude. So here uh all the elements of the set 135 are related to each other. You can say I'm writing the conclusion just in one sentence. I'm concluding that how it is related to each other. All the elements of All the elements of the set 1 35 are related to each other are related to each other to each other because because the difference because the modus of a minus b modus of a minus of any elements of any elements of the set 135 is always is always even for example here dear students you can observe 1a 3 modulus of 1 - 3 is nothing but minus2 and modulus of two will become 2 which is even number similarly 1a 5 if I consider 1 - 5 1 - 5 I'm writing here 1 - 3 modulus of 1 - 3 is equal to -2 modulus of -2 is equal to 2 which is even 1 - 5 modul of 1 - 5 is modulus of -4 which is equal to 4 similarly if you write for 3 and 5 modulus of 3 - 5 is equal to -2 modulus of -2 is equal to 2 again it is an even number so you can consider any two elements and the difference between any two elements with the modulus it value will give always even number accordingly we can say that All elements of the sets 1 35 are related to each other.
Similarly, similarly you can say that all elements of the set all the elements all the elements of this set of the set two and four are related to each other are related to each other because to each other since the difference between a minus b of the set since the difference between modulus of a minus b of the set of the set 2a 4 is always even is always even I hope you have understood this one so here if I take the difference between 2 - 2 is zero difference between 4 - 4 is also zero so 0 is to be considered as the even number similarly the difference between 2 - 4 and 4 - 2 if you consider 100% you will get uh the even number only we will get here accordingly you can say that the given uh all the elements of the set 2 and four are related to each other. And the last part is last part is we need to prove that no elements of this set 1 3 5 is related to any elements of the set 2 and four. Because when we take the elements first element from the first set and the second element from the second set for example 1 comma 2 the difference between 1 2 what we'll get 1 - 2 is equal to minus1. Modulus of - 1 is nothing but 1 which is not an even number. Similarly 1 - 4 modulus of 1 - 4 is nothing but three we will get. So which is again not an even number. So if I consider the first element from the first set and the second element from the second set the difference between odd elements and the even elements is always odd only we will get. Therefore we can conclude that no elements of the set 35 is related to any element of the set 2 and four. We can write and you can write the same inverse. I repeat again.
No elements of the set 135. Better I will write it. So no elements no elements of this set 135 is related to is related to is related to any elements of the set any elements of this set 2 and four.
Since the difference between difference between even numbers and odd numbers odd numbers is always an odd number.
Hence we say that the given relation is so and so. Hence prove.
and prove. I hope you have understood the problem and even this problem also I have done as I have explained the previous problems. So dear students, see all other problems are of the same type.
Only the thing is you should have the proper basics. As I told you through this channel, I am taking from zero to hero level because without knowing not even a single definition you just go through with my videos 100% you will understand the definition and the problems of the given problem. You can easily understand the solutions. I hope you have understood the problem with the solution. And now I am moving to the next question. Let us see the next question.
Okay. Dear students, you can observe the next question is almost similar to the previous one. The given question is let R be the relation defined in the set A is equal to 1 2 3 4 5 6 7 by R is equal to an ordered pair A, B such that both A and B are either even or odd. Show that R is an equivalent relation. Show that R is an equivalence relation. Further show that all the elements of the subset 1357 are related to each other and all the elements of the subset 2 46 are related to each other but no elements of the subset 1357 is related to an element of the subset 2 46 almost the same I mean similar type of question as we have observed the previous one in the previous one we had the modulus of a minus b is even we had but here they are saying that the uh both a and b will be either even or odd. This is what we have to prove. Again, we follow the same procedure. Let us solve the given solution as quick as possible because this is also similar to the previous question. So to first part, I will prove that the given relation is equivalence relation. To prove that the given relation is equalence relation, it is necessary to prove that the given relation is reflexive, symmetric and transitive. So I'm considering the first one reflexive. Let a comma a belongs to R. The meaning of this ordered pair a comma a I'm writing in words that is already given in the question. So I can say both a and a are either even or odd which is true which is true.
Therefore the relation r is reflexive.
You can say I will take one example only to make you understand. Okay, only to make you understand. Assume I'm taking a= 5. A= 5.
If I take a equal to 5. Can't I say that both a and a 5a 5? Both a and a is either even. It will be any one of uh this either or even. It will be either even or it will be odd. So if I take the value of a is equal to 5, it will satisfy odd condition. If I take the value of a as a four then it will satisfy even condition. So 100% whatever you take the value of a that a will be both either even or odd only will satisfy accordingly we can say that the given relation is reflexive and here never take an example because we need to prove that the given relation is so and so in general only. Next we go for symmetric relation. To say that the given relation is symmetric we need to prove that the given relation is I mean if a comma b belongs to r then b comma a should belongs to r. Accordingly I will take an ordered pair. Let a comma b belongs to r. Right? In general only the meaning of a comma b belongs to r is nothing but both a and b are either even or odd.
If a and b both are either even or odd then obviously we can say that the reverse of that one is also both uh are either even or odd only we can say this implies that both b and a are also are also either even or odd which implies that b comma a belongs to r. So conclusion since a comma b belongs to r implies that b comma a belongs to r therefore the relation r is symmetric relation. We have proved that the given relation is symmetric. Now let us consider the transitivity transitive condition. We are assuming the first two ordered pair let a comma b belongs to r and b comma c belongs to r.
The meaning of these two ordered pairs I'm writing in words. Meaning is both a and b are either even or odd. And now the meaning of B comma C we are writing inverse that is both B comma C B and C are either even or odd.
From this we can say that From this we can say that both A and C are either even or odd. From this we can conclude that an ordered pair a comma c is also belongs to r. Now finally we writing the conclusion since a comma b belongs to r and b comma c belongs to r which implies that a comma c is also belongs to r.
Therefore the relation r is transitive relation. Yeah transitive relation.
Again we write the conclusion for the equivalence relation. Since R is reflexive, symmetric and transitive and transitive. Therefore, R is an equivalence relation. Therefore, R is an equivalence relation. Finished. The first part of the given problem is done.
Now, let us move on to the second part.
What is second part? Further show that all the elements of the subset 1357.
1357. That means you take any two elements from the subset of 1357 357 both a and b will be always odd number only we will get because 1357 these are all odd numbers only. So from the given four odd numbers if I choose any two elements as a and b that both a and b will be always odd number only we will get. So from that we can conclude that all the subsets of the elements of 1 3 5 7 are related to each other because both a and b will be always odd number only we will get. So we can say that uh here the subsets of the set the subset the subset of this set of the set 1357 1357 are related to I mean are related to each other are related to each other each other because our Since both A and B are always are always odd.
Okay. Now if I take the next order uh next set what is the next condition? All the elements of the subset 2 47 are also related to each other. Again we write um all the subsets all the subsets of of 2 4 6 2 4 6 are related to each other are related to each other related to each other. What is the reason? Since both A and B both A and B are always are always an even number are always an even number. Finish. And the last one, what is the last one? We need to prove that no elements of the subset 1 3 5 7 is related to any subset of the element 2 4 6. Because when we consider the first element from the first set, first set is odd number, second set is even number. A and B both will not be either even or it will be odd. It will be I mean it will be both because A we are taking A from the first set 35 7 which is an odd number and B is from the second set.
Second set is an even number. So A is odd, B is even. So here it will become both A and B are odd and even. But it has to be either odd or even. So from this we can conclude that but no elements no element of this subset of 1357 is related to is related to subsets of the elements subsets of the ele element elements subsets of the elements 2 and 4 six.
Since both will both A and B both A and B is not either even R hence proved.
Hence proved. Yeah. So this is all about the given question. I hope you have understood because even this question is also almost similar to the previous question only. There is no huge difference. I hope you have understood the question. Let us move on to the next problem.
Okay dear students. Next question you can observe. So here we have a question.
Show that each of the relation R in the set A is equal to X belongs to integer such that the value of X is in between 0 to 12 where 0 and 12 is included given by the first sub question is R is equal to an ordered pair A B such that modulus of A minus B is a multiple of four and the second sub question is R is equal to an ordered pair a comma B such that a= B is an equivalence relation as we already discussed almost seven question based on equivalence relation. This is the eighth question. I hope you all can solve the given problem by your own. So I am leaving this question as a homework to you. But I will give you the solution for the second part of the given problem because the student can able to solve the first two sub questions by saying that the given relation is equivalence relation. Everyone can prove it after seeing my previous questions. Now the second part of the given problem is we need to find we need to find all the set of the elements which is related to one in each case for the both first sub question as well as second sub question we need to identify one is related to which other element we need to identify. Now to identify the same let us observe the first sub question. What is the first subition given? First sub question is given as relation r equal to unordered pair a comma b such that the modulus of a minus b is multiple of four is given multiple of four is given and you should remember that remember that for the given elements of set a set a elements are in between 0 to 12 including 0 and 12 here I can take the elements of set a as 0 1 2 3 4 5 6 7 8 9 10 11 and 12. Only these many elements we can take according to the given condition. Okay. X belongs to inteious where the value of X should be from 0 to 12 including 0 and 12. I have taken the elements of set here. Now I am saying that we are looking only the second part of the given question because the first part I said very clearly that as a homework you try by your own to say that the given relation is an equivalence relation. Now we are looking for uh the one such element is related to what other element. So first of question is given very clearly that the modus of a minus b is a multiple of four is given modulus of a minus b must be multiple of four. Suppose if I take one along with this another one what is 1 - 1 obviously we get zero only modulus of 1 - 1 is 0 is multiple of four which is true. Similarly from among these elements one is related to what other elements so that the resultant element is a multiple of four that we need to identify even I can say modus of modulus of 1 minus 5 also 1 - 5 which is 1 is related to five also because it is the difference between 1 and five will be equal to four modulus is given therefore accordingly we'll get four similarly 1 and five also we can take which is equal to 8 which is again multiple of four so I can say that here for the first sub question One is related to 1 is related to 1 5 and 9. This is for the first sub question answer. Okay. Now let us move on to the second sub question. What is second sub question? Second sub question is given very clearly that if r is equal to unordered pair a comma b such that a is equal to b. I need to consider one such element with one in such a way that it should be equal to some other element. Suppose if I take an element 1 this 1 is equal to itself right I cannot say 1 is equal to 0 1 is equal to 2 1 is equal to 3 no I cannot say I can never say so I can say only in one case one is equal to only one so that in the second sub question an element one is an element one is an element one is related to itself related to itself or one related to one or itself clear. So this is how we find the second part of the given question. I hope you have understood. Let us move on to the next question.
Okay dear students, I'm giving two more uh problem as a homework because even these two problems are also based on equivalence relation. I will go through the given question. The first question is show that the relation R in the set A of a point in a plane given by R is equal to an ordered pair P comma Q such that such that the distance of the point P from the origin is same as the distance of the point Q from the origin is an equivalence relation which means you have to prove that the given relation is equivalence relation.
Further show that the set of all the points related to the point P is not equal to 0 comma 0 is a circle passing through the point P with the origin as a center. Which means the second part you have to prove that second part we have to prove that the set of all the points related to P which is not equal to 0 is a circle which is passing through the point P with the origin as a center.
Which means here the condition is given very clearly that an ordered pair P comma Q such that the distance of the point P the distance of the point P from the origin is same as the distance of the point Q from the origin.
So when you move to the second part when you move to the second part by considering one fixed point as a 0 comma 0 the distance from origin to any point and if you move circularly 100% we consider any two point on the circle P and Q so that we can say that the distance from the origin is same as uh distance from origin to P is same as the distance from the origin to Q. We can say this is possible only when you consider a circle and the center as a 0 comma 0. That is the second part of the given problem. I hope the first part you can easily prove it. And the second question says that show that the relation r defined in the set a of a polygon as r is equal to p1 comma p2 such that p1 and p2 have same number of sides is an equivalence relation. Here also we need to prove that the given relation is an equivalence relation. And what is the set of all the elements in A related to right angle triangle T with the sides 3 4 5. Try these two problems as a homework. And these are all the problems which is based on equivalent solution. Now let us see some more problems which is based on reflexive, symmetric, transitive and like the question will be check whether the given relation is reflexive, symmetric or transitive. This type of problem will be given to you. Let us see how to solve those type of problem.
Okay dear students here you can observe the next question we have like to show that the relation r in the set 1 2 3 given by r is equal to unordered pair 1 comma 1 2a 2 3a 3 1 2 2a 3 is reflexive but neither symmetric nor transitive see usually these type of problems are very simple once we understood the definition of reflexive symmetric and transitive once we understood the definition of reflexive symmetric and transitive so let us solve how to solve the given problem. So here the given question is very simple to prove. So let us write the given relation R. What is the relation R given? Relation R is given as 1 comma 1 2 2A 2 3 3A 3 1A 2 2a 3. Okay, these are the ordered pairs are given under the relation under the set A under the set A is equal to 1 2 3. Okay, as I told you very clearly when we discussed about the definition of reflexive, symmetric, transitive and equivalence relation, I have told you very clearly that whenever we are writing a relation R, that relation R will be under one of the set that set is given. The set elements are 1 2 3. Here we should prove that the given relation is reflexive but neither symmetric nor transitive we should prove. So here we can say very clearly that since since we have all identical ordered pair for the given elements of the set 1 comma 1 2 3 see here we can observe here the given relation R is so we can say R is reflexive R is reflexive since it has it has 1a 1 2a 2 and 3a 3 For all set a we have all identical order pair for the given elements of set A.
Therefore we can say that the given relation is reflexive. Can we say that the given relation is not symmetric? Of course we can say you can write in the next line R is not symmetric.
Why it is not symmetric? Definition of a symmetric says that if we have nonidentical order pair it's a reverse must be there. R is not symmetric. What is the reason? Since 1 2 belongs to R but 2a 1 does not belongs to R. Finish.
And we say that the next one R is again not transitive. Again we say that R is not transitive.
Why it is not transitive? You can observe here reason I'm writing since a relation R has 1 2 belongs to R and 2a 3 belongs to R but 1a 3 does not belongs to R. Definition of a transitive relation says that if a comma b belongs to R and B comm belongs to R then A C should belongs to R but we have A B and B comma C ordered pair but A C is not there. Therefore, we can say that the given relation is not transitive. It is very simple. Moving to the next problem.
Okay. Dear students, you can observe the next question. Show that the relation R in the set 1 2 3 given by R is equal to an ordered pair 1 2 and 2 1 is symmetric but neither reflexive nor transitive is given. We need to prove that the given relation is symmetric but it is not reflexive and it is not transitive. This is what we have to prove from the elements of the set 1 2 3 and the relation R. So I'm writing the relation R.
Relation R is let us see the relation R.
The given relation R is R is equal to an ordered pair 1 comma 2 and 2A 1 and set elements are 1 2
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