 So let's start the class with the discussion of that problem. So this was a homework problem that we had taken the other day, homework question. So what was the question? The question was actually that if there is a set having n elements, okay, so there is a set A which has got n elements. That is the card and cardinality of the set A is n and from this set A, if I make two subsets, okay, subset P and subset Q, okay, both are subsets made from the set A. The question was, find the probability, find the probability, find the probability that the intersection of these two sets will be a null set. That is to say that these two sets will not have any element in common between them, right? Okay. So we will start the discussion with a very brief recall of our probability of an event. So probability of an event, everybody would recall this is nothing but the number of favorable events upon the sample space, isn't it? So let us focus on sample space first, right? So let us find out the sample space. So tell me if I am making two subsets out of a given set, how many ways can I do it? Remember last class we had done the number of ways to make subsets from or one subset from a given set, right? In fact, we had basically taken one set and trying to make subset out of it, right? So that was two to the power n. But if I have to make two set of subsets, then what will I do? So for here you have to understand, let's say these are your elements of set A, okay? So these are your elements of set A, A1 to An. How many options will every element have? Let's say I talk about A1, how many options will A1 have? Now please understand this. If I have the element Ai, I could be anywhere between 1 to n, this will have or this has the following options. No, Arayan, you did not miss anything, we are just discussing the homework question. So the question was, as you can see on your screen, if there is a set A having n elements and I make two pair of subsets out of it, what is the probability that the intersection of these two subsets is a null set, okay? So I was just taking a scenario over here that if let's say the set A has got A1, A2, A3 till An elements, how many options does any element Ai have? So Ai has around four options. What are they? Ai could be present in P, right? And at the same time Ai could be present at Q also because you can take that element and it can include that in P and Q both subsets, right? Asadna, I'm talking about how many ways you can place these elements into these subsets, okay? So let's see how many possibilities are coming up. Second possibility could be that Ai goes to P and doesn't go to Q, right? Now you tell me what could be a third possibility? It could go to both, correct Zain, right? It could go to both P and Q, okay? Tell me what could be the fourth one? Neither, absolutely, neither. So it could go nowhere also, right? Okay, now. Isn't that the first option, first and third is the same thing? No, no. In the first one, first and third you have given same thing. In the first one, it is going to both of them. In the last one, it is going to none of them. First and third. First and third, so first and third, oh my, I'm so sorry, I missed out on scoring this off. Yeah, sorry about that. So every element now will have four choices each, four, four, four, four, till four. So in short, if I ask you what would be the sample space in which you can generate these pair of subsets, you say, sir, four into four into four, all the way n times and that would be nothing but four to the power n. So the denominator, at least I know it's going to be four to the power n. Is there anybody who is still doubtful about the denominator of this event or the number of sample space? Do let me know. I'm pausing here to take up your questions. If none, then we can proceed. Okay, no doubt coming up so far. Well and good. Sure, Raj, I will explain once again. I'll explain once again. No, no, no, I've not started the power set yet. This is pro array. Actually, you didn't read the question, maybe R and you came in late, right? You have to find the probability, my dear, right? So what is probability of any event? You have done this in class 9th and 10th, right? Yeah, so probability of an event is the number of favorable events upon the number of sample space, right? Yeah, so that is the speed, not the power set of the event. We haven't even started power set. You're going to do that immediately after this question. Okay, now what was my question? Question is, in how many, what is the probability that p intersection q, p and q are two subsets formed out of A, so that the intersection is an asset? So what did I do in order to know what are the total possibilities of such pair of subsets you can create and decided to see how many options each element will have, right? So think as if there are two buckets, p and q, right? And these are, let's say, items kept in the shop, right? So how many options does a particular item have? Number one, it can fall in both the buckets, okay? Now you must be thinking, sir, how can one item fall in two buckets simultaneously? See, assuming that you're creating independent subsets, right? So that item is again available for selection. So either you can put it in both the buckets or you put it in one and not in the other, right? Or you can put it in the first one, not put it in the first one and put it in the second one, or you can put in neither of the two, right? So this will decide how many options each item or each element of this set have. And then the total number of options that I will have for the entire, you know, subsets, the two subsets that has to be created will be the product of four into four into four into four. That will be four to the power n. So the denominator here will be four to the power n. This is what we have discussed so far. Got it, Raji? Okay, the next thing is I want the subsets P and Q to be null set. That means whatever element I'm trying to figure out, okay, if I want to find out how many ways can I create it? Can I say that element should either fall in one of the two or fall in neither of the two, isn't it? So can I say the first case can happen? No, you'll say if this cannot be a favorable event scenario, why it cannot be a favorable event scenario? Because if that element falls in both P and Q, then their intersection will not be the null set, right? Their intersection will have that AI element common to both of them. But that is not what we are looking for. Our favorable event is not that. So can second option be an option given to any element? Yes, because if it falls in P and not in Q, their intersection will never have that particular element common, and that's what we actually want. We don't want any common elements to come in both of them. Can it be third option as well? You'll say yes, sir. It can be third as well because it is falling only in Q, not in P, right? Can it be the fourth as well? Yes, because it is falling in neither of the two. So their intersection will never contain that particular element AI. So in order to calculate the favorable events, you are now giving three three options each to these elements. Isn't it? Yes or no? So yes, so the number of favorable event will be nothing but three into three into three N number of times, and that makes it three to the power N. Very good, Aditya. OK, so this numerator will become three to the power N. So here goes the answer. The answer to this question is three by four, whole to the power N. What is P and Q here? My dear, you're not reading the question here. Aditya Anand, P and Q asks, subsets to be formed from A. Oh, were you there in the last class when I gave this as a homework question? No, OK, OK, not to worry, not to worry. OK, anyways, as Chris says, I dropped off, could you please repeat what? What is three to the power N? Yes, three to the power N, Krish is nothing but the total number of favorable events. Now, how did it become three to the power N? If that is the question, then let me explain this again. Since you want, since you want, since you want a particular, no particular element to be present in both the sets. Yes or no? That means you are only allowing the element to be present in either P or Q or none of the two. Because if it present in both, then the intersection will have that particular element common to both of them, isn't it? But that's not what is my purpose. My purpose here is to make their intersection as null sets, right? So under no circumstance, any element can follow the first scenario. Because if it does, P intersection Q will have that element AI in common. And that defeats the purpose of the question, isn't it? So every element can exercise three options. Remember, every element will exercise four options when you are making a sample space. But in the favorable event, you are basically restricting every element to exercise only three options because it can't be in both of them. It can only be in one of the two or none of the two. Now make sense? Okay, thank you, no worries. So let's move on today. Our today's agenda is what? Our today's agenda is power set. So we have still not completed our types of sets. So I'll now be talking about power set of a set, okay? So in fact, I should write the topic name, not as power set. In fact, I should write it as power set of a set, okay? Power set of a set. So what is power set of a set? First of all, the representation that we use for power set of any set A is P. And within the argument, we put the set name, okay? Whose ever power set we are forming. So what is this as the name itself? It's a set which contains such elements which are subsets of A. So basically what is it? It's a set of subsets, exactly. So in short, PA is a set of subsets of A, right? So all you need to do is just figure out how many subsets are there for A, okay? And start writing those subsets within curly braces. And that would result into power set of A. Simple as that, right? Let me give you an example, a very simple example. Let's say I have a set which contains one, two, okay? And if I want to write down the power set of A, what I will do is, okay? First of all, I'll make the curly braces to represent that I'm going to write a set here. And within this, I'll start putting the elements which are actually the subsets of A. So what are the subsets of A? Let's start writing it within these curly braces. So first I will have a null set because as we all know from our previous class, null set is the subset of every set. Then I'll have a set containing just one. Then I'll have a set containing just two. And then I'll have the set containing the entire set A, okay? So this becomes the power set of A. This becomes the power set of A. Is it fine? Any questions, any concerns? Any questions, any concerns? Easy, correct. Now let's look into some properties of power set. Very important. Properties of power set, okay? The first property that is very trivial one, but you should be remembering it, power set of any set cannot be a null set. Please note, power sets are always non-empty sets. And why is that? Why is that? It is because even if your set A is a null set, it will have at least or it will have exactly one subset, which is the null set itself. So a power set of a null set will contain that null set as an element. So it can never be empty. It can never be empty. It will have a minimum one number of elements for sure. Okay? Second thing without any second thought about it is the total number of elements in a power set that is the cardinality of the power set will be what? Will be what? Will be what? 2 to the power n, right? So it'll be 2 to the power of number of elements in a, correct? So if let's say nA is the cardinality of A, so 2 to the power nA will be the cardinality of the power set of it. Any questions? Okay. Third thing which is actually important, if power set of A is equal to power set of B, okay? In fact, I should write it as if and only if statement or maybe the one implies the other. If power set of A is equal to power set of B, then it implies n is implied by the fact that A will be equal to B, okay? So now right to left is very easy to figure out because if two sets are equal, even their power sets will be equal. So that is a no-brainer actually. But if I say two power sets are equal, can we prove that that will lead to the fact that even set A and set B will also be equal to each other? Can anybody prove this? Or can anybody try proving this on your respective notebooks? Let me give you around one minute to do it. By the way, many people ask me, why do you keep proving this? Because proof will never be asked in the competitive exam. See, we are not learning for any particular competitive exam. My purpose is to make you learn for life, right? So when you prove something, when you derive something, you basically feel connected with the topic. Else it becomes very transactional. You remember it, you forget it after the UT. Again, you remember it for some test, again, forget it after a few weeks. So it doesn't solve the purpose, right? So let's prove it. And if at all you are a PU student by any chance, then of course this is going to be, you know, a possible question for you in your school exams as well, okay? Just try it, just try it if you can find a, you know, logical proof. See, by the way, when you are proving something, when you're saying something is true, you cannot prove things by citing example, right? Citing example for something which is true is like verifying it. It is not proving it. Understand this concept, okay? So proving something cannot be done by citing example, but disproving something can be done by citing an example. So if you feel that whatever I have written is not correct, then you can cite an example and say, see, this is the power set of A, power set of B, they're equal, but A and B are not equal. But that is when you want to disprove the statement, which actually you cannot because it's a true statement, right? So understand this because you're going to apply this in your, you know, future part of mathematics. If you have to prove, you have to give a generic proof. If you want to disprove, you can disprove by citing example, okay? So what I'll do is I'll prove this way around, okay? So let's prove one sided. So let me prove that. If PA is equal to PB, then A is equal to B. Let's try to prove this, okay? However, the question says it's a bi-direction, but I'm not going to prove from right to left because it's a very easy proof. In fact, let's see the method to prove it. So what I'm going to do is I'm going to take a set. So let there be a set taken from PA, okay? I'm calling that set as capital X. Now remember, dear students that set contains, power set contains elements which themselves are sets. Hence to represent a set, I have written it with capital alphabet. As you can see here, I have kept a capital X over here, okay? Because sets are normally written with capital English alphabets, right? So what I've done, I've taken a particular set capital X, okay? From the power set of A. So what does this mean? It means that capital X is a subset of A, right? And that is why it was present in the power set of A. Isn't it? Your capital X would have been a subset of capital A, then only it would have got the entry into the power set of A, right? Yes or no? Front of, makes sense? Front of bi-company? Does it make sense? Okay, okay, okay, yeah. Oh, you wanted me to wait for you to solve it? Oh, I'm so sorry, dear, so sorry. If you want, I can still wait. Because your solving is more important to me. Should I wait for some time? Okay, maybe let's say you can take this as a hint and move forward. See, I'll be more than happy to see the answer coming from your side. Okay, you think you got the answer. Okay, so let's see whether our processes are matching or not. Okay. So now let's say there is an arbitrary element X which belongs to this set capital X, okay? So what I've done, I've taken some arbitrary element, okay? Which comes from the set capital X, right? So now can I say if this arbitrary element has come from the set capital X, correct? It means this arbitrary element should have been in your set A, correct? Because it has come from the subset of A. So that element is definitely an element of A also. Then only it got its entry into one of the subsets, no? Isn't it? Now, please understand that since PA, let me write in mathematical language, since PA is equal to PB, can I say that capital X will also belong to PB, correct? Yes or no? Which means capital X is a subset of B as well, correct? Which means that arbitrary element which was belonging to X that will also be a member of B, yes or no? So this arbitrary element which was a member of X that will also be now a member of B. So overall what I have done from one end to what do I figure out? From one end to what do I figure out? I figure out that if an arbitrary element belongs to A, then that arbitrary element also belongs to B, thereby proving that A is a subset of B, yes or no? So that is half the job done because if you want A to be equal to B, you have to prove two things from it. A should be a subset of B which I have just now accomplished and I have to do another half that is B is also a subset of A which I'm going to do right now. But meanwhile, before I proceed, before I go any further on, I'm pausing here, I'm taking a small, small pause and a break for you to copy down and understand things before I move on. Done. Can a power set have subsets? Why not? Because at the end of the day it's a set tariff, right? So you can always create subsets, so any set and power set at the end of the day is a set. Ongar has a question. So could you explain how exactly A became, becomes a subset of BC? My very definition of a subset, I think you were there in the last class, no? Okay, I'll just give you a brief idea. If A is a subset of B, then it implies that any arbitrary element, any arbitrary element, X which comes from A will also belong to B, right? So if you're able to show that, if you pick any arbitrary element from A and conclude that that arbitrary element will also belong to B, then this leads to the fact that A is a subset of B. Got it now? And that's precisely what happened Ongar over here. So I started with X belonging to A, C here and I stumbled upon the fact, eventually that X will also belong to B. So when this happens, A becomes a subset of B. So just trying to say that if you pick any element of A, any arbitrary element of A, that will end up lying in B as well. So A is a subset of B, correct? Any questions? Any questions so far? Any questions so far? Can't you just directly just show that the largest element in P of A and P of B are equal? Largest element in? X belongs to P of A and X or cardinal number of X is equal to A and you do the same thing for B and if you show that X is equal to suppose Y, then doesn't it prove that A equals to B? Okay, so when you're trying to say that T, A and B, you see this proof covers that as well, right? This proof covers that as well because it is taking any arbitrary element of that power set. So X is any arbitrary set of that power setting. So your proof is also correct, but it is more like a specific one and mine is a more generic one. I'm not saying your proof is any less superior to mine, but what I'm trying to cover is, I'm trying to cover every possible element that is belonging to the power set of A. Okay. Does it sound good? Okay. Now, how do I prove the other way around? B is a subset of it. Now, see, again, you will have a similar proof. You'll say, let there be a element of power set of B. Okay, so let Y be, oh, sorry, Y be an element, my bad. I told element and I wrote a subset, slip off there. So let Y be a subset of B. That means Y belongs to the power set of B. Okay, so let Y be a subset of B, which means it comes from the power set of B. And let there be an arbitrary element which is belonging to Y. Okay, so this is an arbitrary element. Arbitrary element, okay? So what does this mean? It means that Y also is a member of B. Okay, let's call it as one. Now, since P A is equal to P B, can I repeat the same thing and say Y will also, or the capital Y will also be a subset of A. That means the same element Y will come from A also. Let's call it as two. So eventually, what are you trying to say? You're trying to say that if there was an arbitrary element of B, then that arbitrary element is also present in A, thereby making B the subset of A. Isn't it? So in light of these two, this and this, what do I conclude? That A and B are equal, right? So you did the same thing, but using Y and power set of B. Can you be a little louder, please? So you used the same thing, but then changed X to Y and then you took B. Yeah, it is just to keep things very generic, okay? So whenever you're writing such proofs, your proof should be most generic proof, not very specific one that, okay, mine is true only for that biggest element, or the names also should be very generic. So I'm trying to take a different name for an element. Okay, so what do we conclude is that A is a subset of B and B is also a subset of A, thereby resulting into the fact that A and B are equal, hence proved that's what we wanted, okay? Now, as a school level and as a competitive level exam, nobody's going to worry about these proofs, but when you do this proof, what will stay with you for lifelong is that this property P A equal to P B implies A equal to B and is further implied by, will always be there in your mind, okay? So that is the main purpose of doing the proofs in these class. It's not that they will be tested in any form. Yes, Krish, that is what exactly we did, we proved A is a subset of B and similarly B is also a subset of A and hence A and B are equal, okay? Let's see what Pranav has to answer, I didn't take the subset X and Y instead of saying, same thing, same thing. Yeah, that's absolutely fine, that's absolutely fine. Okay, any questions? Any questions, anybody? The school you said it's mostly proofs only for UTs, okay. See, let's learn the concept rather than worrying about, where it will be asked, where it will not be, fine. Okay, should we move on? Let's complete the property, now we are going to property number four. Property number four, maybe I will discuss once we, okay, let's take it because you already know union intersection from your school understanding also, right? Now I'm going to make a statement over here. P A union PB, do you think this is equal to P A union B? What do you think? This statement, is it true or is it false? Now, before you give out any answer, before you give out any answer, listen to me first. If you're saying it's true, then you have to give a generic proof. But if you're saying it's false, then it's sufficient for you to show me an example, where there is a set A, there is a set B. You found the power set of A, you found the power set of B, you took their union. And then you also found out the power set of A union B and then realize that they are not the same. Okay, so somebody is saying it's false. Then please give me an example, where I can see that this property is failing. Okay, take one minute, not in a hurry. Take one minute. Okay, very good Pranav. So people are giving me examples, very nice. Okay, so let's take a very simple example, a very, very simple example. And it cannot get simpler than this. Okay, let's say A has just one, okay? And B has just a two, fine? So what is the power set of A? So power set of A, as we all know, will contain all the subsets of this set one. So it'll only contain null set and only contain one, correct? What about power set of B? What will it contain? Null set and the set to itself? Yes or no? What about A union B? So A union B will contain one comma two. And here if you start writing P A union B, you would realize you'll end up getting null set. You'll end up getting set containing one, set containing just two and a set of one comma two, right? So what do you realize here is that if I do the union of P A and P B, that means the first two terms, if I take the union, I'll get a null set, I'll get a set one, I'll get a set two, but I won't be getting a set of one comma two, correct? So this clearly does not match with P A union B. So this statement is false, absolutely, right? I think all of you got it as false. Oh, is there anybody who's said it true for it? Okay, no issues. So in reality, please note this down. And this is a question which has been asked in the regional intense exams. Power set of A union power set of B is actually a subset of power set of A union B. So a direct question can be framed on this. So please note this down very, very important. So power set of A union power set of B is actually a subset of power set of A union B, okay? Now, can anybody tell me when would power set of A union power set of B become equal to P A union B or become an improper set subset of A union B? Can anybody tell me that? One of the sets is a null set. One of the set is a null set, okay? More generically? More generically? A is one of them is a subset of the other. Right, yes. Our, in my question was, when do you think, when do you think P A union P B will actually become equal to P A union B? Okay, so this holds, this holds when either A is a subset of B or B is a subset of A or A and B are equal, okay? So that's the only occasion or these are the only three occasions when you'll realize that the power set of A will be equal to, sorry, power set of A union power set of B will be equal to power set of A union B. Is it fine? But in general, in general, they are not equal. Is it fine? Any questions? Okay, great. So let's now move on to the fifth property. Let's now move on to the fifth property. Can I go to the fifth property? Have you all copied this down? Any questions? Sure, sure, take your time. Take your time. Done, okay, okay, let's move on. The fifth property says very important one. In fact, let me ask this as a question to you before I solve it. Power set of A intersection of power set of B, do you think this is equal to power set of A intersection B? What do you think? Is the statement true or is the statement false? Now mind you, if you're saying it true, you have to give me a generic proof and if you're saying a false, you can just give me a single example where this particular statement doesn't hold true. Take a minute. Any idea, anybody? At least you have an idea whether it's going to be true or false. Proof or definitely I will be able to do it. No, Aditya, if you're saying it's true, then an example will not work. You can't prove something by giving one example where it is working. Right? But even disprove something by giving an example. Okay, so Rn thinks it's going to be false. Anybody else? Okay, Sadhana also thinks it's false. Oka also thinks it's false. Okay, false. Okay, now, to everybody's surprise, this is actually true. Gugli Krishna, no worries, let's prove it. Okay, now when you're proving that it is true, you have to give a generic proof. That is what basically I told you. The previous approach is not going to help us here because in the previous approach, we were saying false. So one example was good enough. This time I'm claiming this is true. So I will give you now a generic proof. Now, how see? If you want to prove that PA intersection PB is equal to PA intersection B, please note that I have to accomplish two things. First, I need to prove that PA intersection PB is a subset of PA intersection B, okay? And also we have to prove that PA intersection B is also a subset of PA intersection PB. So when both of them are proved, then only I can conclude that these two sets are actually equal. So I'm coming from your understanding of power sets. Oh, sorry, I'm coming from your understanding of subsets. Zayan, can you hear me? Can you hear me? Hello, hello, hello, my test. Am I audible to everyone? Oh yeah, people can hear me. Zayan, am I audible to you at least? Okay, he's not able to hear me. Yeah, so there's no point asking him whether you can hear me. Okay, just to give me a second here, I think one of you is in trouble, so I'll just message him personally. I'll just message him to try to log in back again. You're right, yeah. So coming back, yes, so if I want to prove that one set is equal to the other, I need to ultimately prove what? I need to ultimately prove that one is the subset of the other, right? And let's see how do we do it? So I will be first proving the left side that PA intersection PB is a subset of PA intersection D. So what I'll do, I will take an subset of, or I can say I'll take a member of PA intersection PB. Please note, these are both power sets. So the intersection will also contain elements which are sets at the end of the day, right? So I should use a capital alphabet. It's a good practice. It's a good practice always to write sets in capital, okay? So what I'm doing, I'm saying, let there be an arbitrary element capital X, which is present in the intersection of these two. Now I'm sure you would have done this in school that if a particular element is present in intersection of two sets, it means that particular element is present in this also and this also. Am I right? Everybody knows this. Everybody knows this, right? Okay, so that means this capital X is a subset of A and this capital X is also a subset of the, correct, no, any question so far, all right? That means that capital X is also a subset of A intersection B. See, if you're saying capital X is a member of power set of A, that means it should be present as a subset of A, right? Correct? And if you're saying that particular capital X is present in the power set of E, that means capital X should also be a subset of B, correct? Now, there are two subsets of A and B, the same subset is of A also, the same subset is of B also, which means that subset is also a subset of A intersection B, because that subset would be comprised of such elements which are present in A and B simultaneously. Am I right? So that capital X will also be a subset of A intersection B. Everybody is agreeing so far. If anybody not, please ping me. Okay, Aditya agrees with me. Anybody else? Most of you should agree with me. Okay, now that means that, yes, I'll explain once again, see what I'm trying to say here is, first of all, what I did, I took one element from this left-hand side, okay? And I called it capital X, okay? So let there be a capital X, it's a set only, but I'm calling it as element for the time being. So let's say this is an element which comes from intersection of P A and P B. That means that element should be present in P A and should also present in P B, then only it is present in their intersection, which means that particular element is actually a subset of A. Hence it was present in the power set of A. Similarly, that particular element should have been a subset of B as well. That is why it was present in the power set of B. So that particular set now is actually a subset of intersection of A and B. Now this step is confusing to many people. This is a how. It is because if you take any arbitrary element, okay? Let's say I take any arbitrary element from X. So let's say there's a small X which belongs to capital X. Now since this arbitrary element is coming from capital X, which itself is a subset of A, that means that arbitrary element is also present in A. Similarly, the same arbitrary element would also be present in B. So what you're trying to say here that X is present in both A and B, which means X is present in A intersection B. Isn't it? So any element which basically came from A and B, both will be present in both A intersection B or not? Correct. That means what? That means your capital X is a subset of A intersection B. Clear? That is how this step actually came into picture. Is it fine? All right. Now, having said this, let's go back to this step, step number one and step number two. So what you're saying that there is an element which is present in intersection of these two, which is present in this and the same element is present in A intersection B also. Okay, let me write it in the next step maybe. Sorry, I should write this. Capital X is a member of B intersection B. Yeah. Yeah, sorry, I wanted to write this as my two. So what you're saying, one element that you have picked up, it's present in this set and that element is also present in this set. It's like saying your subset definition. If the small x belongs to A and the same element small x belongs to B, then A is a subset of B, right? In the same way you're trying to say is that this guy belongs to this and this guy also belong to this, which clearly implies that P A intersection P B is going to be a subset of A intersection B power set, which is what I wanted to prove here. So half the proof is covered over here. So this part of the proof is taken care of. Any question, any concerns? Right, Zeyan, you'll be surprised that if I want to prove the right side, this part, I will actually write the same steps in a reverse order, okay? So some of you would be saying, sir, this is camming, you are doing the same step. Yes, this is how it actually works, right? So in order to prove this, I will see, I'll exactly start from the last step. I will say, let there be an arbitrary element Y, I'm just keeping the name separate, that belongs to P A intersection B. So what I've written, I've written this last step. Then what I will say, then Y belongs to A intersection B, sorry, Y is a subset of A intersection B, right? Because Y is an element of power set of A intersection B. So the same Y should be a subset of A intersection B, which again means, Y is a subset of A and Y is a subset of B, which means Y is an element of power set of A and Y is also an element of power set of B, which is actually the second last step or the second step, which finally means Y is an element of intersection of power set of A and power set of B, right? So you started with this and you concluded with this. So what does this imply? So let me call this as three and let me call this as four. So from three and four, I can say that the power set of A intersection B is a subset of power set of A intersection power set of B. And in light of these two, in light of these two, what do we conclude that power set of A intersection power set of B will be equal to power set of A intersection B. Note this down. This is a very, very important property and direct questions have been framed on this. Please note this down. Okay. So yes, by reversing the steps of the first two, we actually got the second proof. I mean, second part of the same. Note this down and then we are going to talk about the next concept which is actually operation on sets. Take your time to copy. I'm not rushing here. If you have any questions, any concerns, please do highlight. What are the different super sets and power sets? A lot of difference. See, super set is basically a set from where a subset has been carved out. Whereas power set is a set of sub, all the possible subsets of a given set. Okay, I'll give an example to make that clear to you. See, if I say there is a set B, which contains AB, okay? If I say there's another set, A which is containing just only A, right? So here B is a super set of the set A or you can write it like this. B is a super set of A, okay? Got it? Now, if I say power set of A, then what it'll contain or power set of B, whatever you want to call it, AB. So power set of B or power set of A, whatever you want to call it, it's up to you, will contain all the possible subsets of A, right? So this is a super set of A and this is the power set of A, there's a difference. If A is a subset of B, then B will be a super set of A by default. It's like saying one is lesser than two, then two will automatically be greater than one. So at the same time, both of them will be applied. Isn't it? Okay, Aditya, let's move on now to operations on sets. Operation on sets. So we'll be coming across various type of operations under sets. So I'll be starting with union of sets. Okay, union of sets, okay? You have done this in school, so I'll be slightly faster here. So when you are taking union of two sets, how do we define it first of all? So what is the elementary definition of union of two sets? So I'll first write it as union of two sets and then I will extend it to more than two. So union of two sets will contain all such elements X such that X belongs to A or X belongs to B. Okay, so this particular operation can also be represented by a type of diagram which we call as the Venn diagram. Most of you would have heard of this word from your school teachers, Venn diagram. So what's a Venn diagram? Let me tell you a brief history about it. So long back ago in around 18th century, there was a famous Swiss mathematician called Euler, Leonard Euler. You must have heard of his name many a times. This fellow has got almost 90 references in the field of mathematics, highest by any mathematician so far. So the next highest is 60 actually. So the second highest person has got only 60 references in the field of mathematics and Euler has got 90 references. His name has been used 90 times in some or the other theorem or some kind of laws or something. What's the name of that with a dash in between? V and a dash in between. What is the name of the symbol that means for all? Huh, what is the name of it? It's a Greek symbol. I don't know the name of it right now. Maybe I can Google it out for you. It's not that important to know the name, but yes. You can Google it out. Yeah, so what did Euler do? He basically started representing some set of points by using closed figures. And later on in almost 19th century, there was an English mathematician called John Wynn. He started giving circles, rectangles, figures to represent the same set of points. So in order to honor these mathematicians, we call such diagrams as Euler Wynn diagram or sometimes just the Wynn diagram, okay? So if you start representing the elements of a set within these closed rectangular figures or these closed circles, we basically call those figures as Wynn diagrams, okay? Later on, we will learn that for Wynn diagram, universal sets Wynn diagram is represented by a rectangle. And for sets, we normally use circles, oh, sorry, circles of different shapes and sizes. So this is normally used for a set, okay? So if I talk about the Euler Wynn diagram for the union of two sets, how do I write it or how do I make it? So I make the universal set. I'll talk about universal set also in some time. So I call this rectangular shaped structure to be your universal set. And within this universal set, I'm going to represent my set A and set B, okay? So let's say A is this and B is the one which I'm making in orange, okay? So this is your set A and this is your set B. Now what is the region represented by A union B? So everything which falls under this zone, everything that falls under this zone, they will be considered to be a part of A union B. So it should be either in A or in B, right? Please note this or is inclusive or. Inclusive or means it could be present in both of them also. And hence that is the reason why I have shaded the overlap area as well. Is it fine? Any questions? Okay, so if I scale this up, this has scaled this up. If I talk about union of multiple sets, A1, A2, A3, da-da-da-da-da, till A n, what will it represent? It will represent a set of all X's such that that X comes from A1 or A2 or A3 and so on and so forth. Is that fine? Any questions? Very simple. Not to worry about. So the union must contain all the elements. Sorry? Present in A and B. Must it contain all the elements present in A and B or some specific elements would do? All the elements which are present in A and B. I'll give you an example. Okay. By the way, such kind of representation you can also write with elongated U. Let me write to the left. Yeah, elongated U, AI like this. Okay. This is just a short form rotation. You may have seen this notation while you're solving questions on sets. So this is just a notation which says I'm taking the union of all AIs. I going from one to it. Okay. Now just was in to give an example of union, let's say I have one, two, three as set A and set B contains let's say three, four, five. So when I have to write A union B I will only list down the elements which are present in A or B and if some element is getting repeated in both of them we can only make a mention of it once in their union. These do not write three, two times because three is present in A also and three is present in B also. Because it will go against our definition where we had learned that every element in a set is only written once. Is it fine? Any questions with respect to this? Any question with respect to this? Yes. So if somebody is asking me how do I write the union if there is more than two sets? Simple. All you need to do is shade everything which belongs to these sets. I'm just taking an example here. Let's say this is our universal set. This is set A, this is set B, this is set C. So I will end up shading everything over here. Is it fine? Any questions, any concerns? Do let me know. All fine? All right. Which intersection are you talking about? Intersecting is common, right? You're talking about intersection of all of them? We will come to it. Don't worry. Let's go step by step. Let's go step by step. Now, please note down the following. Please note down the following. Number one, if an element does not belong to A union B. Okay, if an element does not belong to A union B it implies that that element will not belong to A and that element will not belong to B as well. This logical statement is what we know as the D. Morgan's law, right? Which we are going to study in our laws of sets or LG law of sets. This says compliment of A union B is A compliment intersection B compliment. So this law is called the Morgan's law, okay? D. Morgan's law. Is it fine? So if there is something which is not lying in A union B means it's lying outside. So it is in the universal set, but it is lying outside. So if it is lying outside it cannot be in A and it cannot be in B either because if it was in A or in B exclusively then it would have appeared in A union B. Aditya, see this point. Okay, let's take a point over here. So if I take a point which is outside, let's say there's a point, there's an element which lies here. Okay, let's say call this element E1, right? Now, this element E1 is not present in A union B. And you can see from the Venn diagram that if it is not present in A union B, it cannot be in A and it cannot be in B. Okay, my dear Aryan, I have not yet reached the compliment of a set and my dear, your spelling of compliment is wrong. Thamba, wait, I will come to it. I'm not going to leave out anything. I'm going to cover more than what you expect in this chapter. Okay, let's wait. No, we have not covered it. We have not covered compliment of a set because we have not covered what is the universal set. Okay, I will come to it. Let's go step by step. Okay, next thing, very important. If A union B is equal to A union C, right? Let me drop this if because I'm going to use symbol. Right, last class itself I told I we should be using mathematical symbols rather than writing English, too much English. Then these note, this does not imply B is equal to C. And that's what I feared somebody will say. If A union B is equal to A union C, it does not imply B is equal to C. Now, let me give an example to substantiate this. Let's say, let's say A contains one, two. Okay, B contains, let's say, let's say, let's say three. Okay, and C contains, let's say, two comma three. Now, what do you realize from this example is that A union B, which is one, two, three, and A union C, which is also one, two, three, they are equal. Okay, so these two are equal. But what do you realize here that despite A union B and A union C is equal, our B and C sets were not equal. Right, so if somebody says A union B is equal to A union C, does it definitely mean that B is equal to C? The answer is no, B and C may be different sets also. Getting my point. So this is what we say as left-hand cancellation fails. Left-hand cancellation fails in case of union. So don't start cancelling things out, just the way you do with numbers, okay? It's a wrong, wrong, wrong practice to cancel or give the sets the same treatment as what you give to numbers, okay? So don't just cancel it off. Any question, any concerns? Similarly, even if I switch to the order, that means if I write B union A is equal to C union A, it does not imply B is equal to C. So this is what we say, right-hand cancellation fails. Right-hand cancellation fails, okay? So in a set, never start cancelling things on the right of something or left of something like the way you do it for numbers. Any questions, any concerns? Any questions, any concerns? All right, so should we move on now to, to, to, to, to the concept of universal set? So I was waiting for the union operation to be done and now we'll be talking about the universal set. Now, many people ask me, sir, why don't you do universal sets under types of sets? Why did you wait for a union to be done? That's, there's a reason for that, okay? So let me explain this to you, then you'll automatically understand the reason. By the way, anybody can tell me the definition of a universal set, anybody? So universal set is written by symbol xi or u, okay? Aditya, yes, more or less, you're right. So universal set is a set which contains all elements x which are in the context of the problem. X is in the context of the problem. Now, this is very important. This is very, very important. So notice now, a universal set is a set which contains all the elements which are present in the context of the problem unless until stated otherwise, unless stated by the definer, by the person who is defining the universal set for you. Let me give an example. What do I mean by context of the problem? See, let's say I talk about a set A which has got one, two, three. Let's say I have another set B which has got three, four, five, six. And I take another set C, let's say seven, eight, okay? Now, if I ask you, can you suggest a universal set for it? Right? So when you're suggesting a universal set for it, you will only incorporate those elements which are in the context of the problem. That means you can only have one, two, three, four, five, six, seven, eight, that's it. In short, what are you doing? You're taking the union of all the sets A, B, C which are present in the context of the question. Getting my point. So don't start taking the universal set as all natural numbers. No, that is not a right way to choose a universal set. Universal set is a set. I'm repeating again. Universal set is a set which contains the elements which are in the context of the problem. Getting my point. Unlisten to stated otherwise by the person. So there are many occasions when the person himself or herself will tell you that, see, this is the universal set. In that case, you have to follow what he or she is telling you. But if you want to make your universal set with respect to that given question, then all you need to do is take all possible elements which are present, right? Put them in a set that will be the universal set. In short, you are taking the union of all the sets which are present in the question, right? Now, from the union, from the concept of universal set comes out the concept of complement of a set. Okay, so let's now talk about complement of a set. Super set and universal set. Very good question asked by Sadhna. Super set, universal set acts like a super set of all the sets present in the context of the problem. So your psi is the super set of A, psi is a super set of B, psi is a super set of C. Got it? Sadhna, is that clear? So if I wanna write a universal set, you just write X is in the context of the problem. Yes, I mean, this is for the definition of you, definition for you to understand, but in reality, you have to take the union of all the elements present in that given set. So what will be the set builder form for this? Set builder form for, see, when you're writing the set builder form, you can write any definition by which you can produce that. So I can say all natural numbers, so one to eight. Okay, so, okay, yeah, okay, fine. Right, so you cannot write a set builder form for an operation. For the set, for the outcome, you can write a set builder form. So when I look at this, I can write a set builder form for it. See, set builder form is nothing but a statement by which you can produce that set without ambiguity. That is set builder form. Krish has a question, X belongs to union B. I didn't understand that, Krish, a large question. So then how does universal set change a set C is not in the context of the problem? Yeah, if C is not in the context of the problem, then the universal set will only contain one, two, three, four, five, six, seven and eight will not be there. Got it, Raj. Okay, okay, okay. Let's move on to complement of a set. So I'm covering the universal set and complement together. By the way, this is the spelling of complement. P-L-E, not P-L-I. P-L-I means praising somebody, giving somebody a compliment, right? So you give yourself, your parents a compliment if they, you know, let's say your mom cooked something which is very tasty to it. You give her a compliment, mom, you cook such a tasty food. Okay, that's a compliment, okay. Don't write P-L-I, okay. So what is complement of a set? So complement of a set is represented by symbol A bar, okay? You can write A with a small, you know, dash or A with a superscript of C. They all mean the same. So how is it defined? I normally prefer using the bar because I find it convenient to use. So it is basically a set which contains such elements which are present in the universal set and not present in A, right? So when I talk about difference, you'll, yeah, Krish, whatever you have used it's a very appropriate, you know. So Krish says it's xi minus A. So that is the complement of it. Okay, since I have not talked about difference operation, hence I was slightly skeptical of using it, but since you've already used it in school maybe you'll find, you know, convenience to understand this. So how do you represent it in a Venn diagram? So in a Venn diagram, let's say this is your xi. So complement of a set, let's say this is my set A, okay? So complement of a set will be nothing but everything which is outside A. So everything which is outside A, but within the universal set. So I'm just making like a sun ray coming out, okay? So this region represents A complement. Is it fine, any question, any concerns? All right, so if I go to this example and I ask you, hey, what is B complement? What will you say? What is B complement in this? Read the definition. Everything which is present in universal set xi, but not present in B. So it will be one, two, seven, and eight, absolutely right. Any question, any concerns? Okay, one important property here I would like you to remember, if A is a subset of B, then please note that B complement will be a subset of A complement. If A is a subset of B, then B complement is a subset of A complement, not the other way around. That's a mistake which people do. People think that if A is a subset of B, then A complement is a subset of B complement. No, no, that doesn't work. If A is a subset of B, then A complement will be a superset be compliment getting the proof getting this property okay very very important note this down note this down and for your homework you will prove this for me so give me a proof for this particular property okay easy one you know how to prove one is a subset of the other right so you have to take an element okay so please please give a proof to me I can either message me personally or put it on the group also it is up to you let's move on let's move on so we have already done these operation of union before and now we are going to do the operation of sorry we are going to do our operation on intersection will this proof come in the school ut's I can't say that depends on your teacher how you know your teacher is teaching in the class that depends there are some teacher who likes doing proof there are some teacher who do not so those who like doing proofs they will definitely ask you so prove this right yes you have to prove this this one which we have put an asterisk mark for your homework okay next we move on to the second operation the second operation is the operation of intersection of sets intersection of sets so intersection of two sets I'll start with so intersection of two sets means a set containing all such elements x such that x belongs to a and x belongs to b that means if an element is chosen from a intersection b then that element will be belonging to a and and this is a logical connector and it will also belong to b okay how do I represent this with a wind diagram so in a wind diagram if I have to represent it so let me make let me make the universal set and let me make the two sets a and b with a circle if we use a different color okay so this is our universal set xi this is set a this is set b so this portion this portion is where okay the elements of a intersection b will like is it fine and if you just scale this up if you want to do the intersection of all ais from one to n then it will contain all such x which actually belong to a1 and belong to a2 and belong to a3 is it fine any questions any concerns any questions any concerns okay so let's have the set of properties for it properties of intersection first note this down if you have any questions do let me know okay the first thing just like we had discussed in the previous union if an element does not belong to a intersection b what does this mean it means that the element does not belong to a or the element does not belong to b see very simple if I take any element if I take an any element which does not belong to a intersection b so it can either be here so in this case it doesn't belong to b so this part will be true correct right so this is false this is false this is true so overall since it is or this will be true right correct if your element is here then this will be true and this will be false correct and if your element is here then both will be true true so this will anyways be true in all these three situations right so when an element does not belong to a intersection b then either of the two or both of them may be happening that means it is not in a okay it may be in b but it is not in a or it is not in b it may be in a or it may be not in both of them got it and this is what forms the next de Morgan's law the second law of de Morgan which says is a intersection b complement is a complement union b complement okay this is what we call as the second of your de morgan's law okay we'll talk about the these laws as well in some time is it fine any questions any concerns anybody has okay in the previous scenario a intersection b if it is equal to a intersection c it does not imply b is equal to c okay similarly if you say b intersection a is equal to c intersection a this also doesn't imply b equal to c that means your left hand cancellation your left hand cancellation fails in case of intersection also and this means your right hand cancellation fails in intersection also now can i give an example where it is failing in fact you give me an example let's hear it out from you can you give me an example where there are three sets involved a b and c such that a union a intersection b and a intersection c are equal but still b and c are different give me an example anybody i'm waiting a equals one to say okay equals two three a k b is equal to two three okay and c is equal to two three four very good okay so what do you realize here is that a intersection b will be having two three only and that's the same as a intersection c also but b and c are not equal is it fine any questions any concerns okay if a intersection b is a null set it implies that a and b are disjoint a and b are disjoint that means as a when diagram if you make the when diagram then a and b would be separate separate from each other like this okay because they will have no element in common to each other okay so if a intersection b is a null set then a and b are disjoint like even numbers or even natural numbers and odd natural numbers disjoint or even numbers and odd numbers they are disjoint they cannot have anything in common okay so what about real number and imaginary number are they disjoint sets no they will have one element zero in common zero is both real and imaginary at the same time yes break will be at seven seven o'clock why are you just hungry now only is just one hour 20 minutes class but you are at home I mean okay okay all right so can we move on I have a question before we move on if a union b is equal to a union c and a intersection b is equal to a intersection c then b is equal to c do you think first of all is this true or false first of all what's your verdict okay so aditya says no it is not true okay there are two adityas who is who is the one who is saying aditya anand is there and aditya okay fine okay anybody else some people are saying it's true some people are saying it's false okay okay now those who are saying it's false give me an example give me an example of three sets a b and c where a union b and a union c are equal and so is a intersection b and a intersection c are also equal and b and c are not equal to each other so I basically I'm requesting you to give me an example of such three sets which satisfy these two but doesn't satisfy this b equal to c so I'll be giving you a proof for this basically that two will go through some some concepts which we are going to cover a little later on it's a futuristic concept for us today but we'll be covering in today's class only okay yes anybody who has this example people who are saying not true okay now people who are saying true you have to give me a generic derivation for it so is my is my you know offer to both these parties clear if you're saying it is false so false people this is for you give an example where this is not working out I mean despite this condition being fulfilled this condition will fulfill b and c are not equal to each other give me one example that is for the people who are saying false for the people who are saying true you have to prove it so I'll say sir I you we should have said false only but trust me you will not be able to find any example because it is true statement okay and it is not false so people who said false very sorry it is actually a true statement now let me prove it in fact whatever I'm going to use for the proof you're going to learn little later on but I'm sure you would have got an idea of it from your school teachers okay so let's start with a union b equal to a union c okay so we know that it's given it's a truth so what I'm going to do is I'm going to take intersection with b on both the sides okay so I'm going to take intersection with b on both the sides so now have you heard of a property called distributive law from your school teachers you have heard of distributive law no yes or no how many of you have heard of it okay some say no some say yes okay okay you learned it okay now people who have not learned it in fact I'll be taking that up in some time distributive law says like this so if you have something like this x intersection y union z then you can write it as x intersection y union x intersection z this is what we call as distribution of distribution of intersection over union and you can also do vice versa you can distribute union over intersection also like this okay so this is called distribution of union over intersection yes in multiplication also you have learned it exactly Narayan okay so I'm going to use this distributive law for this set now most of you would be thinking sir have you distributive law be you asked me no worries I will teach it eventually but since this concept was rhyming with our concept which we took up in union and intersection I thought I will make I will take this question before I move on because this has come in the past year papers so if I take the intersection with b on both the sides can I say the left side is b intersection a union b intersection b correct and this side will be b intersection a union b intersection c if an element belongs to a union b and a union c the elements should belong to b or c no not necessarily it could belong only to a only not not in b not in c is I am okay now see here this is as good as a b b intersection b is as good as a b okay and this is as good as a intersection b and this side again I'm writing it as a intersection b and this side I'm writing it as b intersection c okay so everything I've written same is just that b intersection b I've written it to be just a b is it fine everybody's fine with that okay now everybody please think a bit and answer is isn't a intersection b a subset of b isn't a intersection b a subset of b correct so can I say union of b with a intersection b will be as good as a b yes so so can I say left hand side becomes a b right hand side I will continue writing it as it is okay now since a intersection b is equal to a intersection c I can write the same statement as b is equal to a intersection c union b intersection c okay let's call it as the first result let's call it as the first result so I'm I'm taking a pause here any step that you have not understood which is right now on your screen feel free to ask me you're talking about the distributive law Omkar we will prove it when we take the law okay as of now take it with a pinch of salt when I'm taking the law I'll prove that as well third step yes with third step yes okay so b intersection b what is the intersection of karnataka with karnataka karnataka only so b intersection b is b only correct yes sir b intersection a is a intersection b so that is what I have written now since b is a superset of a intersection b see a intersection b is only this part right correct this part is a intersection b and this is a and this is b so this part is a subset of b can I say union of superset with a subset will be the superset itself yes or no and then in the next step what did I write I wrote a intersection b as a intersection c because it was given in the question look at the second condition a intersection b is equal to a intersection c so I use it yes clear everybody clear any question aja before I forget some people they're asking about their class pro marks not reflecting the test they have taken so please note that normally we take around a week to update it because this test is not only taken by you it is also taken by other centers that we have so please don't expect the marks to be reflected they immediately you're taking a test because the testing platform is lernist whereas the marks platform is class pro so it's a it's a transmission of marks from one platform to the other which is actually not done not done automatically it is manually done okay so once everybody has taken the test so give at least one week time then only your marks will be reflected okay it will not be reflected the moment you take a test now and and you go and check your class pro your marks will be sitting there no it takes a week's time you can check the previous week's assignment but for the current week you have to give it up you have to give seven days our time for it to reflect okay all right so I think some of you were you know slightly shaken up and you didn't see that marks appearing despite you have taken the test so give some time it will be reflected so it is manually done actually all right now I have to prove that b is equal to c so I have only reached this step right so let's do the similar set of steps now by taking intersection with c on both the sides so I'll be taking intersection with c on both the sides okay again here I will be using my distributive law so this will be c intersection a union c intersection b correct yes or no yes or no right and here also I will do the same operation c intersection by the way this is something which I can write in one shot why because a union c is a superset of c a union c is a superset of c so what is the intersection between a superset and a subset the subset isn't it so since a let me write that over here since a union c is a superset of c the intersection of a union c with c will be c it's like asking somebody what is the intersection of karnataka with India karnataka only isn't it and now since I know let me just rephrase it I can write it as a intersection c and this I can write it as b intersection c and now since since a intersection c is equal to a intersection b as per given by the question I can write this as a intersection b union b intersection c equal to c let's call it as 2 now dear students compare first and second compare first and second in fact let me write this as this yeah and let me call this as second rather than calling this as a yeah so compare this with this they're equal right so sure should be the left hand side of this and the right hand side of this correct so from one and two can I make this conclusion that since the right side of the first expression matches with the left side of the second expression so the left side of the first should match with the right side of the second as well so b becomes equal to c and that is what we wanted to prove is it clear any questions any concerns now this question has been asked multiple number of times in cognitive exams so the result of course it is important but even the proof can be asked to you in your school papers as well so please note this now and do let me know if you have any questions how they can ask this as an mcq format so they will say like this aditya so they'll say a union b is equal to a union c and a intersection b is equal to a intersection c then which of the following is true then they'll say b is equal to c or they'll say b is a subset of c or c is a subset of b none of these like that yes yes yes krish you can prove anything using when diagram okay so when diagram is another way to prove things but trust me at some places when diagram works miracles at some places the algebra of set works miracles okay so let's carry on intersection is done our next operation is difference of sets difference of two sets so difference of two sets is written as this what is the definition what is the elementary definition of difference of two sets so a minus b contains all such elements x says that x belongs to a and x does not belong to b right so as a matter of bend diagram as a matter of bend diagram let's say this is my a and this is my b and let me also make the universal set okay so let's say this is my a and this is my b so a minus b is represented by this zone so elements which are present in a and not in b got this point so this point is called a mine this region is called a minus b region in the bend diagram okay now please understand this please understand this by the very definition of a minus b you can also write it as a intersection with b complement see it should present in a and not to be right so a minus b is same as saying a intersection b complement very very important uh you know you can say identity which is used in several places several several places understood understood any question any question okay now many people ask me sir is it a minus b as good as saying remove from a all the members which are present in a intersection b also yes why not why not both mean the same thing okay so i'll give an example if you want to understand this through uh you know a set let's say a is one two three four okay and b is let's say three four five six when i say a minus b i have to only state those elements which are present in a and not in b so which elements will come you tell me write it on the chat box quickly it should be present in a and it should not be present in b right only one comma two will come absolutely right so for this particular example when i'm writing i will write one two here three four in the intersection zone and five six here as you can see from the definition so only one two will fall in the shaded area only one two will fall in the shaded area is it fine any questions similarly if i say b minus here what does it mean it means a set containing all those elements which are present in b and not present in a okay so which region will this represent it will be represented by this region okay so this is your b minus a area correct any question any concerns here so in the same example if i have to find b minus a what will be my answer what will be my answer for b minus a five comma six absolutely right simple question is this fine any question any doubt related to what is this operation of difference difference of two sets doesn't mean their intersection this is one one of the very i can say basic error that people make difference do not mean intersection see a minus b is one comma two but a intersection b is three comma four there's a difference between the two let's not get confused yes very good question what is a minus b if a and b are disjoint very good question you tell me the answer if a intersection b is a disjoint set what will be a minus b so it is as good as asking what will be a minus a intersection b which is as we're asking remove from a nothing remove from a nothing so it will be itself no absolutely okay all right so let's write down few properties of few important points let me write important points on note number one number one a minus b b minus a and a intersection b are pairwise disjoint are pairwise disjoint what is the meaning of pairwise disjoint pair by disjoint means if you take any two of them and take their intersection it will be a null set so if i take the first two let's say a minus b and b minus a this will be a null set if i take a minus b and a intersection b this will be a null set if i take a intersection b and b minus a this will be a null set is it fine arian has a question how would you highlight a intersection b complement a intersection b complement we want to come to it by a venn diagram okay so let me show it by a venn diagram see let's say this is a and let's say this is b okay now you are asking me arian to represent a intersection b complement correct so let's figure out b complement first so for b complement i will shade this zone which is outside b right okay now see the gray line that you have shaded outside b where does it intersect with a so what is the intersection part so this is your a part so where does the gray line intersect with a so the gray line intersects with a only in this zone correct which is actually your a minus b zone that's why a minus b and a intersection b complement are same clear okay next point to be written b is note this down second point a minus b union b minus a union a intersection b makes the complete a union b okay so basically it's trying to say that if you take this yellow region the blue region and this not shaded region together and take their union it will generate a union b that is a minus b union b minus a is nothing but a union b minus a intersection b both are the same things just written in different different way okay no need to remember it no need to remember it it is automatically evident from the graph it is automatically evident from the graph fine any questions any questions any concerns note down because you know uh some of you may not be as fast in copying it down so please make a note of it and please feel free to stop me in case you want in fact you all are asking you know questions a lot of questions which is very good in case anywhere anywhere you are stuck you want to know more please feel free to interrupt okay don't worry about that there are other people in the session they'll get disturbed because of me nothing like that we all are here for ourselves okay so you are the person most important person for yourself so please ask whatever doubts is coming in your mind is it fine any questions any concerns okay let's now move on to the symmetric difference uh one second so zhan has a question would it be i suppose all these tools to wind diagram see we can solve this using wind diagram also orient we can also use algebra offsets to get it okay we'll see it in some time once we learn the algebra offsets okay zhan that would be same as union zhan your doubt is clear sir if you're asked to put the second one how exactly do we disconnect we need to okay so if i want to prove the second one i will prove it later on by using algebra offsets okay so i will prove it by using algebra or laws of sets little later on wind diagram you can easily prove it wind diagram you can easily prove it if you want the other way around without a wind diagram i will prove it by using algebra offsets yeah i'll do it by that method okay before that i'll move on to the next operation which is called symmetric difference offsets symmetric difference of offsets so when i say symmetric difference of sets a and b it basically means it basically means a minus b union b minus a okay so when i make a wind diagram when i make a wind diagram let's say this is my set a and b okay and this is my this is my universal set so when i say a delta b what what do i have to do if i a delta b a delta b is the symmetric difference of two sets now why it is called symmetric difference because when i shade the region you will realize that the region is symmetrically placed almost not exactly about the intersection so this zone a minus b union b minus a this zone is what we call as a delta b okay read it as symmetric difference of a and b is it fine so you can also say it is a union b minus a intersection b clear what is this triangle this triangle is a delta triangle it's a symbol that we use for expressing symmetric difference a Greek symbol let's take an example if i say a has one two three four and b has three four five six then what is a delta b tell me tell me quickly a delta b one four five a minus b right the elements of this zone and b minus a which is five and six so one two five six one two five six absolutely is it fine any questions okay not that important in operation but yes if if they ask you this in the the cognitive exams you should be able to answer what is the symmetric difference of a and b is it fine okay with this i'm now going to take a in fact it's 647 what i'll do now is i'll give you a break because i don't want to start with algebra of sets because if i start with it it'll almost take me half an hour or something like that so let's take a break right now let's take a break as per my laptop the time is 647 we'll meet exactly at 7 or 2 p.m a 15 minute break and as you all know during the break the camera and my mic is muted so even i'll be taking a break so i may not be at my seat but see you exactly at 7 or 2 okay okay so let's proceed with our a second last leg of this chapter where we are going to discuss about algebra of sets algebra or laws governing sets okay now these laws they have all come from logic actually right so they are very fundamental and they have come from your understanding of logic and that's the reason why the same law you will find also used in logic circuits in electronics domain in computer science domain in logic circuits you will realize or logic gates what you call it these laws are basically used okay and the proof of these laws will be so so fundamental that many times people will say is that this is camp okay i mean they are they're just based out of logic nothing else all right so one of these laws let's let's start listing them down and let's start you know discussing them one by one so the first law that i'm going to talk about is idempotent law idempotent law what is an idempotent law the word idempotent means something which is applied to itself right so the two idempotent laws a union a gives you a and a intersection a gives you a okay now if i give you the proof for this you may you know laugh and fall off your chair also it is so trivial it is so trivial and obvious right so let's say i want to prove this okay so basically what i'll do is i'll take an arbitrary element which belongs to a union a okay so i'm i'm just proving the first part proof for it what does this mean it means x belongs to a or x belongs to a as per the definition of a union a which means x is just belonging to a so what does this mean it means a union a is a subset of a okay similarly write down the same set of steps in a reverse way so if you say there's an element let me name this time y let's say y is a member of a that means y is belonging to a or you can also write it y belonging there doesn't make any difference which means y belongs to a union a which means if an element is belonging to one set and it is also belonging to the other means this is a subset of this correct sorry a not b i don't know why did i write a b so from one end to you can say a is equal to a union a okay very trivial i mean there is nothing like you know which is uh beyond logic it is just logical base statement nothing else okay similarly you can prove the other one by the way since they are so trivial and so predictive i will not be proving them going forward okay i'll not be proving them going forward sir is that a or small a or both are capital a which one you're talking about okay you got it okay so idempotent law clear any question with respect to idempotent law next identity law maybe i'll write it on the same page because identity law so these laws can be used directly while solving any problem or proving any identity of sets without bothering about the proof for this because they are very fundamental it's as fundamental as why there is why does gravity exist okay it is as fundamental as why two like charges repel and two unlike charges attract right so they're very fundamental means very very basic and we basically use them as a building unit to prove anything so identity law has two laws under it it says union of any set with the null set is the same set whereas intersection of any set with the universal set is also the same set okay so these two laws are called identity laws note this down next law that we are going to talk about next law that we are going to talk about is the boundedness law boundedness law so boundedness law uses the same similar expression instead now you are saying a union null set you'll say a intersection null set so a intersection null set is always a null set whereas a union or universal set will always be a universal set what is this e looking thing e looking thing is the universal set over here by the way this is not called e looking thing it is actually called Xi Xi Xi okay I can I do see why I don't write you is because you may mix it with the union operation that's why okay so why has this been given the name boundedness law boundedness law is because no matter whatever is your set a its intersection with the null set it'll be bounded to means it will just drop down to the null set so it's bounded by a null set right intersection of any set with a null set will go as less as a null set not you know not any further than that similarly union of a set with a universal set will go to the universal set so it's bounded by null and the universal set so as to say thank you all right so now the next property that we are going to talk about is the complementation law complementation law so another complementation law we are going to talk about five sub-laws number a complement of a null set is the universal set complement of a universal set is the null set they are very obvious okay complement of a complement gives you the same set back okay and this is very important a intersection a complement will always be null and a union a complement will always be the universal set so these two are very important okay we'll be using this or this will be seen many a times when we are solving problems on algebra of sets so c is complement of a complement right right c is complement of a complement so double complementation gives you the same set back double slash means this yeah this is read as complement of a complement i mean that's not a relevant question to be asked we do bar more see it depends on question to question okay all right all right so we will now move on to our next law which is called the commutative law commutative law you would have already you know seen it in your addition subtraction when it says that the order in which you are keeping your operands doesn't matter so whether you do two plus three or three plus two doesn't matter whether you do two into three or three into two doesn't matter so those kind of operations are called commutative operations or commutative law so here even in case of union and intersection it doesn't matter whether you take a union b or b union a it means the same and it doesn't matter whether you take a intersection b or b intersection a it is the same result so there's no difference created by changing the position of the sets with respect to that union or intersection operation okay next law that we are going to talk about is associative law i hope i'm not going fast in case you need some time to copy things down do let me know okay so what is associative law associative law says that if you are doing a union b and then taking a union with c it actually doesn't matter the order of association that means a union b taken first and then union with c or b union c taken first and then union with a or for that matter a union c taken first and then union with b all of them give you the same result so the order in which you are taking the union is immaterial same goes with intersection also if you do a intersection b then intersection with c or you take b intersection c first and then intersection with a or you do a intersection c and then intersection with b they do not you know make any difference to your final result okay so this is called associative law because they are associating with each other very very freely very very freely okay please make a note of it then we can move on to the distributive law can we move on distributive law let's write it in yellow the name distributive law distributive law you already have got a glimpse of it but i'll be proving this law since somebody was asking the proof for it so you can distribute intersection over union and union over intersection so i'll be just proving this one first see again it comes from logic and you will again feel that you have been this has been scanned for proving okay so if i want to prove this law what i have to show i have to show two things so if i have to prove this i have to show two things somewhere one i have to show this guy is a subset of this guy correct and at the same time i have to show that this guy is also a subset of this guy is it fine okay now what i'll do is i will only prove one of them because for the next one you already know that you have to write the same set of steps reverse manner okay so i'll just do this part the next part you can easily do it by writing the same set of steps in a reverse fashion so here i will say let there be an arbitrary element x which belongs to a intersection b union c what does it mean it means x belongs to a and x belongs to b union c it's all logic it's all logic there is nothing new about these proof correct oh sorry c okay so i'm just writing this in terms of logic in terms of and or connectors so now this is as we were saying x belongs to a and x belongs to b or x belongs to a and x belongs to c see what are you trying to say here is that let's say i'm giving a very simple analogy let's say there is a party okay so doubt yes tell me so for x belongs to a so you've written and x belongs to b union c so shouldn't be x belongs to a or x belongs to b union c because it's intersection so the second step first step basically x belongs to a and so shouldn't be or because it's intersection intersection is and not or it should be oh yeah some are bad no worries okay now see it is like saying that that element x must definitely be in a and it could be either in b or c or in both of them right so think as if there is a you know there is a party okay and there are three friends a b and c right a says i will definitely go to the party right let's say that's a party that is given by a himself so he has to be in the party but b and c they say that either b can come or c can come not both or both of them see only b can come or only c can come or both of them can come so how many possibilities are there so the possibility that can arise is a and b are there in the party correct a and c can be there in the party correct or all of them can be there in the party so this or that you're seeing is an inclusive or every time that or we use is an inclusive or so that party could be attended by a b a c or all the a b c okay that is what it is trying to say from our logic which further boils down to saying that x belongs to a intersection b or x belongs to a intersection c which means x belongs to a intersection b union a intersection c so what did you start with you started with this and you concluded with this right so what does this imply it implies a intersection b union c is a subset of a intersection b union a intersection c right and for the second part i would not do anything i will say in the second part please write the steps written on the left side in reverse order that's it so i will not waste your and my time writing it again so can you show the left side once again like yeah for a minute yeah it's completely logic that's why they are called fundamental laws so you kind of use distributed law in the proof itself sorry so you kind of use distributive law while proving this right that's what many people will say sir you use the law while proving the law but that is so fundamental we basically support the support this by saying that it's a logically driven statement okay any questions okay so let's now move on let's now move on all right so this is the second this is the first part of the distributive law second part of the distributive law i'll write it over here same thing it's a distribution of union over intersection so it goes like a union b intersection a union c okay not only that you can also distribute union over union as well right see union over union always follows associative law which we already did but you can also apply distributive law as well and you can also apply distributive law of intersection over intersection like this also so even i include c and d also in my list okay so this is called distribution of distribution of intersection over union that is your a b is called distribution of union over intersection okay and c is called distribution of union over union and d is called distribution of intersection over intersection is it fine any question one of the most most widely used laws that you will be requiring to solve problems okay and all these laws you can always prove by wind diagram also so they can all be proved all can be proved or proven by using wind diagram okay i'll give you as a homework now try proving try proving prove b for homework okay so just make a wind diagram for the left scenario make a wind diagram for the right scenario and see are you getting the same shaded area for both of them is it fine yes yes aran you can please copy this down okay last but not the least d morgan's law morgan's law very very important d morgan's law okay so i've only given you d morgan's law while i was talking about union and intersection so in d morgan's law there are two primary laws and two derived laws so what are the primary laws primary law says that the intersection sorry the union of a and b complement is as good as intersection of their complements and this can be generalized this can be generalized so if you have if you have union of ai's okay i going from one to n and you're taking the intersection of this whole lot it is as good as intersections of their complements i going from one to n right so if i want to write this in a lengthy way i mean this is just a notational way of writing it so in a lengthy way if you have to write you'll say a1 union a2 union a3 da da da da da a1 union a2 union a3 da da da da da let's say till a n if you're taking full complement okay it is as good as a1 complement intersection a2 complement intersection a3 complement and so on is it fine so is it necessary to note a generalized pattern as well like could i present a question in the general it can have a question where you can use this generalized pattern we'll take one question in some time okay now many people ask me sir proof for d morgan's law very simple if you say x belongs to a union i'll just give you one sided proof one sided proof means i'll give you that a union b complement is a subset of the right side vice versa you already know how to do it by reversing the steps okay so it's pointless to write the same things i know in a reverse fashion so if you say x belongs to a union b what does it mean x does not belong to a union b and remember when i was giving you the fact that x does not belong to a union b that means x doesn't belong to a and x doesn't belong to b which means x belongs to a complement and x belongs to b complement which clearly means x belongs to a complement intersection b complement so what did you do you started with this and you stumbled upon this right so that clearly implies that a union b whole complement is a subset of a complement intersection b complement similarly you may prove which i'm not writing as i told you you may prove this is a subset of this by writing same steps in reverse fashion in reverse order okay and hence and hence from these two you can conclude a union b whole complement is equal to a complement intersection b complement which part you did not understand are you this to this or this to this see if x doesn't belong to a union b sorry if x belongs to a union b complement it means x doesn't belong to a union b correct no if something belongs to the complement it will not belong to the actual set right now if it doesn't belong to this and as i showed this in your union if if doesn't belong to union of a and b it cannot belong to a and it cannot belong to b because it has to be outside that you know circular part so that part you understood okay okay so this is your first of demorgan's law okay i hope all of you have noted down the steps let's move on to b part b part demorgan says opposite if you do a intersection b whole complement it is a complement union b complement and you can generalize this you can generalize this by saying that the you the intersection of ai's the intersection of ai's whole complement is same as union of their complements okay again proof is same i will not be doing it now there are two derived demorgan's law what is the derived demorgan's law it says that a minus b union c is a minus b intersection a minus c umka which one you wanted to see here b or a this is b oh you want to see a okay this is your a you want to see the last part or the top part clear okay great let's let's continue yes so this is called a derived demorgan's law okay this is called a derived demorgan's law now how do i derive this result of course it'll be using the basic primary demorgan's law for the proof let's prove it let's prove it let's start with the left hand side a minus b union c now remember when we had done the difference of two sets a minus b or let's say let's say x minus y x minus y can be written as x intersection y complement isn't it yes or no do you remember that fact x minus y could be written as because x minus y could be written as x intersection y complement so in the light of this i'm treating x to be my a y to be my b oh sorry b b b union c to be my y okay so now i'm going to apply demorgan's law here so i can write it as b complement intersection c complement okay now use the distribution of intersection over intersection correct so this is by demorgan's law okay and this is by distribution of intersection over intersection so this is nothing but a minus b and this is nothing but a minus c and hence the result is it fine any questions any concerns so mind you very very important there are some students who basically try to make their own set of rules so the other day some people said sir can't be used distributive law of you know this difference over union no there is no such distributive law don't make your own set of laws beyond the ones which i stated you if you are using any result make sure you have proven it so don't start generating your own mathematical laws okay so a minus b union c is a minus b intersection please note there is a union here and this becomes intersection over here very very important in a similar way and again this is something that you will be proving for homework a minus b intersection c is a minus b union a minus c okay so please prove it for homework same way as what how i did the c part same way as how i did the c part is it fine any questions should we move on okay so with this we are going to take a few problems okay one by one we'll take a few problems and then we'll move on to the last part of this chapter which is the cardinal number properties offsets also called as the practical problems on sets but before that at least three two three questions we'll take on this algebra offsets okay shall we okay so here comes our first question i'll be starting with a very simpler one if a and b are two sets then a intersection a union b complement is which of the following yes yes we should be able to finish the chapter today itself because next class i'm planning for relations and functions okay sake very good i hope you have given your response on the poll as well very simple question i'm getting two responses we'll try enough we'll see last one minute okay almost everybody has answered okay now i'm ending the poll out of 19 of you uh 13 say option c but six of you say something other than c also okay so six of you uh is non-c option and uh 13 of you say c okay great see it's a super easy question it should not deserve more than 10 seconds also a union b complement right if i if i talk about a union b complement and roughly speaking it's everything which is beyond a and b so let's say this is your a and b right so a and b a union b complement will be everything outside this correct so i didn't realize that was complement the apostrophe oh yeah this is a complement apostrophe now what is the intersection of a with something outside a nothing there's no intersection so clearly option number c is right okay so when diagram sometime works miracles when you're solving the question sometimes it may not so use whichever is more convenient at that particular point of time now let's see if i want to solve this by using my laws of sets so i can use i can use my i can use my d Morgan's lawyer so it's a complement intersection b complement so if i apply my distributive property of intersection over intersection what do i realize this is a null set which law law of law of complementation right so intersection of intersection of a null set with any set gives you a null set which law which law boundedness right so from here to here it's boundedness law right so from here to here it's the distributive law of intersection over it intersection and from here to here it is law of complementation so whenever you're writing these answers in your exam also make sure you are mentioning which law is being used oh no problem aria i will take more questions not to worry is it fine should we go on to the next one all right next question this question i will write it down a union b whole complement union a complement intersection b is which of the following a complement b complement just a or just b i'm relaunching the poll okay once you're done please press on the poll button if you are unable to see the poll button don't panic you can always write your answer on the chat box i can i can acknowledge it okay guys and girls almost three minutes gone i'm now putting the poll to an end okay so anybody who'd okay so poll is already ended most of you have said option number a but i can see a very mixed response seven three three five okay let's discuss this now if you want to solve this by using when diagram then maybe it will become a little lengthy i have not tried it using when diagram but what i feel is without when diagram it can be solved in a very very super easy way by using your algebra offsets how let's see the first expression a union b whole complement i can write it as a complement intersection b complement okay and let's copy the other terms as it is right now if you see this very closely it is actually the the right side of the distributive law okay so write down the distributive law from right to left so when you do that it is as good as you doing this operation correct me if i'm wrong okay so you are as if applying distributive law like this so i'm just using the reverse of the distributive law this is why it becomes very important to know our law not only from left to right but also from right to left right so this is as we were saying a complement intersection now what is this b complement union b universal set isn't it law of complementation and intersection of any set with universal set will be the set itself which is clearly option number a is it clear any question any concerns okay should we move on one should we move on okay let's take one last one before we move on to the law to the cardinal number properties of sets a union b union c intersection with a intersection b complement intersection c complement whole complement intersection c complement is which of the following option a intersection c option b b union c complement option c b intersection c complement and option d none of these none of these okay so question is clear a union b union c intersection with complement of a intersection b complement intersection c complement intersection c complement can be simplified to which of the following whole is odd aditya for the previous question the answer was a complement yeah option a correct take your time not a hurry one more minute okay five four three two and one so please vote now because i'm not going to stop the poll okay so most of you have gone with option numbers c other options have got significantly lesser votes as compared to c okay so let's see so first of all i'll use a generalized version of d morgan's law and i would write this term the term which is in the middle as a complement union b complement complement union c complement complement okay can i write that so many complements i've never got in a single day oh you found it easier by writing i go for it all the best okay so now so i can write this as a complement union b union c intersection c complement now pay attention i see this term sitting here also and i see this term sitting here also so can i now apply can i now apply reverse of distributive law only on this so i can write it as b union c union a intersection a complement can i say that and of course we have intersection c complement waiting outside can i do this can i do this do let me know anybody who's confused about this let me know i can write it in a simple way see let's take this guy separately i'm writing it in maybe a green color yeah so when you're saying a union b union c intersection a complement union b union c take b union c as x okay so it is like writing a union x intersection intersection a complement union x right or not so it is as good as saying x x union a intersection a complement or not right that is what exactly i've done so this guy is my x this guy is my x where did the double complement go after the first step my god double complement means you're back to the same set no a complement complement is a only know law of complementation forgot clear umka you it sift out of your mind no issues okay now this is as good as saying a null set because a intersection a is a null set correct so you are taking you are taking union of a set with a null set so doesn't it become this only so from this operation i can only get b union c is it fine makes sense yes proud of so far so good whatever you want to copy please do it now let us use distributive property so distributive property you can write it as you can also write it this as this way so if you use distributive property it becomes c complement intersection b union c complement intersection c so this is as good as b intersection c complement commutative law and this is anyways a null set again so union of any set with a null set is the same set which basically gives you this as your answer which is proper which is option number c is correct is it fine any question any concern take your time i'm pausing right now here okay whatever is on your screen just have a good look at it tell me if you need assistance anywhere all right so there you go law in the third last step right sorry yes yes so i use distributive lawyer from here to it is distributive law and this is your this is your complementation law that's how i got a null setting and union with a null set is going to be the same set that's your identity law yes makes sense yeah guys and girls here we move on to the last part of this chapter which is called the cardinal number properties of six cardinal number properties of six please note the name of this topic is the properties applied to cardinal number okay not the sets themselves but applied to their number of elements in the set many people also call it as practical problems on sets so basically there are some set of laws which the cardinal number follow and please take care of this please make a note of these laws because you're going to use this law even in your probability chapter so whatever law i'm going to write here exactly the same set of laws will now be used and will later be used in your probability concept under addition theorems of probability okay so let me start with the first property which says number of elements in union of two sets is nothing but number of elements in a plus number of elements in b minus number of elements in their intersection of a and b now how does this work very simple and you can easily you know verify this by using our vent diagram okay so let's say this is a vent diagram situation of a and b now when you want to find out the number of elements in union of a and b you're looking for how many number of elements lie in this zone so i'm just you know showing it with a zigzag okay okay i mean and removing it also so this zone let me call k number of elements are lying in this zone okay then how many number of elements lie in this zone i hope you can see the movement of my marker okay in this zone let's say l number of elements lie and how many number of elements lie in their intersection part which is this zone and let's say o number of elements lie here okay so when i'm looking for n a union b i want to actually get the sum of k plus o plus l right now see oh cool it becomes cool now see if i do n a it gives me k plus o because it'll cover the whole part the whole circle so k elements lie in this part that is only in a o elements lie in intersection of a and b so for entire a it will be k plus o similarly if n b if you do it will be l plus o isn't it now if you add them you would realize you'll end up getting k plus o plus l plus an extra o also isn't it now k plus o plus l is n a union b correct that's what i wrote over here from one correct and this o is nothing but n a intersection b because o is the number of elements lying in their intersection correct so that means i can easily say n a union b is n a plus n b minus n a intersection b very easily proved by using my diagram is that fine any questions any concerns now this basic law that you see this can be extended to any number of sets okay even if you have three sets or four sets or let's say n number of sets you can extend this particular property how let's see that a second last step second last step this step aren't so in this step what did i do aren't i replace this guy with n a union b c here k plus cool is n a union b and this o i replace with n a intersection b because o is the number of elements in a o is the number of elements in a intersection b right so i took this to the left side and the proof automatically came up clear everyone okay now how does this property work when you are generalizing it so if you take three sets if you take three sets the result becomes n a plus n b plus n c minus intersection two at a time plus n of intersection taken three at a time now how does this come again the proof is similar to the proof of the previous one so i'll now make three sets for you just to show you please observe the pattern in these results that is more important okay let's say these are my three sets a b c okay and i want to find out let's say i call it p q r okay and let's take orders k l m and this is o i hope you're you're aware of which region i'm calling as containing p elements so p is only in a okay correct l is only in a and b o is in all the three k is only in a and c like that okay so when you're trying to find out n a union b union c it is basically k or you can say p plus q plus r plus k plus l plus m plus o this is what we want okay now let us start with n a n a is p plus l plus k plus o what is b what is b q plus l plus m plus o what is c c is r plus k plus m plus o let's add the three let's add the three so when you add the three you will end up getting something like this p plus q plus r please pay attention very very important p plus q plus r plus k plus l plus m they occur twice isn't it yes or no and o will occur twice because there's o o o any question anybody so far any question anybody so far okay now what i'm going to do is i'm going to write separately the number of elements in a intersection b a intersection b will be l plus o how many elements are there in b intersection c b intersection c m plus o correct and how many elements are there in c intersection a k plus o yes or no yes or no yes or no and how many elements are there in a intersection B intersection C that is O from the diagram. Okay, now see what we are going to do here. Maybe I have to copy this result in the next space because I don't have more space here. Oh, I can manage here also. Done everybody. Copied things down, copied things down. Now see what do I need? I need P plus Q plus R plus K plus L plus M plus O. So can I say it is NA plus NB plus NC? Note that in that you have P plus Q plus R plus one set of K plus L plus M. So you need to remove one set of K plus L plus M. And at the same time you need to remove two O also because this has got three O's and I only want one O. Yes, I know. So I need to remove two O's from it. Yes, I know. Now can I not write this as NA plus NB plus NC? Now please note here, if you had added these three, you would have got something like this. You would have actually got L plus M plus K plus three O. So what I'm going to do is I'm writing this term as NA intersection B and B intersection C and C intersection A minus three O. So K plus L plus M, I'm writing it in terms of the intersection two at a time and I already have a minus two O sitting outside. So in light of this, if I open this particular bracket, everything I open up, what do I realize? I will end up getting something of this sort. You'll have plus three O minus two O which is actually a plus O, isn't it? And what is this plus O? What is this plus O? I have already written O is an intersection, N of A intersection B intersection C. So this will become NA, NB, NC minus this, minus this, minus this and plus O means NA intersection B intersection C. Let me shift it to this side because it is blocking my way. Is it fine? But what is more important, my dear, is to see what is happening while writing these formulas. What is the pattern? Why minus three O? Minus three O because I only needed K plus L plus M. So this three O went to the left side. So instead of this term here, REN, I took this to the other side and wrote the whole thing. That is why minus three O. Okay. And when I opened the bracket, that minus three O became plus three O Pratik and that got adjusted with minus two O to become an O. Is it fine? Krish, what's your question? I didn't get your question, Krish. Okay. So we will generalize it even further. So if you have, let's say four terms, A union, B union, C union, D, four sets, then what will happen? So in this case, you take some of one at a time. So you will end up getting something like this. See the pattern NA, NB, NC, ND. So one at a time, you added them. Sorry. I wanted to write a D. I wrote a B, my mistake. Okay. Then subtract two at a time. So two at a time means take all the intersections of two possible. So A intersection B, B intersection D, D intersection A, then then A intersection C. Then B intersection D. Oh, I think I missed out. This was C by the way. So D intersection C. Yeah, I missed out one more. C intersection D. Okay. So one, two, three, four, five, six terms will come as a subtraction. Then add intersection of three at a time. So A intersection B intersection C. B intersection C intersection D. C intersection D intersection A. And D intersection A intersection B finally. Okay. And then again subtract four at a time. The only option that you will get is something like this. Okay. So this is the generalized version. So look at the pattern. This is how you are going to generalize it even for even for for the steps. So here what you're doing, you are adding single single subtracting double double again adding triple triple subtracting quadruple. So one at a time you add two at a time intersection you subtract three at a time again you're adding four at a time you're subtracting. So this continues on and on getting the point. So this can go on and on. So this particular principle is given a name in mathematics. This is called this law is called the principle of inclusion and exclusion. And this is very much used in the field of combinatorics. You realize that the use of this is used in permutation combination. It is used in probability. Is it fine? Any question? Any concerns? So two minutes. I'll just take it down. Absolutely. Yes, please. Just for homework figure out if there were five sets, how would their union look like? So let's say n of a union b union c union d union e. Okay, so for homework, try writing the expression or let me write it like this and write the expression for expression for n a union b union c union d union e. Okay. And just send it across. Copy to everyone. Okay, is that done? Zeyan? Yes. Yeah, okay, okay. All right. So here we are with the next, the third property. In fact, first, yeah, first only we had generalized the second one. See now many times, you know, people ask like questions like if there are three sets. Okay, I'm just making a diagram here. Let's say these are three sets a, b and c. Okay. a, b and c. Right. How many elements lie in exactly two of the sets? a, b, c. Okay. Now express your answer or let me write the question in this way. Let's say this expression is x times n of a intersection b, y times n of b intersection c, z times z times n of c intersection a and let's say w times n of a intersection b intersection c. Okay. Where x, y, z and w, they are integers. Okay. Find x, y, z, w. So instead of giving you the direct law, I'm just giving you as a form of a question. So please note, first of all, which part in the Venn diagram will represent the number of elements lying in exactly two of the sets. So this will be the part which will contain elements lying in exactly two of the sets. So I'm just helping you out shading that area. Okay. So tell me how do I get the number of elements lying in this wide shaded area in terms of n, a, u intersection b, n, b intersection c, n, c intersection a and n, n, a intersection b intersection c. Recording of this, of the, this thing is already uploaded on the learners platform because we don't share it on the group because it may get circulated. Okay. So any recording that you will be, you want to access will be on the learners platform. So I hope you have all created a profile for yourself on learners. You can access this from there. Okay, Aditya. Yes. Sorry, somebody was saying something. So x, y, z is one and w is minus three. Okay. So please write down your answer on the chat box because everybody else will get influenced by your answer. Okay. Fine. You can ping me. I'll send you. But I cannot put it on the group. There are videos. Yes, there are videos. Okay. Let's see this. Okay. Now, if I talk about this area and let me name it, let me name it as p, q, r and o. Right. Now, what I exactly need is p plus q plus r. Right. Now, when I do, this is what I need. This is what is required. Okay. This is the number of elements in exactly, exactly two of the sets a, b, c. No, no, no, you don't have to pay anything for the videos. Those, those price tags are put because I don't, don't want anybody else to, you know, look at them other than our own students. So if you have enrolled with us for that particular bundle, so the bundle is the JEE mathematics bundle. So in that, you will find out all the videos. So further inquiries, you can always touch base with the research. He will guide you. Okay. Okay. Now, see the moment you do n, a intersection b, you not only get p, but in addition to that, you'll get o as well. Right. So this will give you p plus o. Similarly, n, b intersection c, what will it give you? B intersection c will give you r plus o. Okay. And if you do c intersection a, it'll give you, it'll give you q plus o. Correct. Now, the moment you add them all, the moment you add them all, what you realize is you end up getting p plus q plus r, but in addition to that, you get 3o as well. Okay. But I don't need the 3o. So I set the 3o through the other side. So p plus q plus r will be nothing but this minus 3o. And what is 3o? And what is, in fact, what is o? o is the number of elements in the intersection of all these sets, isn't it? So o is nothing but n of a intersection b intersection c. Clear. Right. So clearly means that, let's say if I put a one here, one here, one here, it clearly means that x is one, y is one, z is one, and w is minus three. Is it fine? Simple. So in the same way, in the same way, I'll give you one more homework question so that, you know, we can move ahead with the other laws. Anything that you would like to copy, please do so. Sure. Sure. Take your time. Take your time. Yes. Yes. LMS has been activated. So I have a homework question. If the number of elements in exactly one of the sets, a, b, c, is equal to, is equal to n a plus n b plus n c plus, let's say x times n a intersection b and b intersection c and c intersection a plus y times n a intersection b intersection c. Question is, find x and y. Question is clear to everybody. So question is the number of elements which lie in exactly one of the three sets a, b, c is given by this expression. Okay. Where x is multiplying, x is multiplied to the entire set here, and y is only multiplied to n a intersection b intersection c. You have to find out the value of x and y. Good. Zayan is trying to find the answer now. It's a homework question. It's a homework question. All right. So we'll move on to the other properties. I think we have already done one, two, and now third one. Now, see, I'm not going to give you all the set of properties. It all comes from your Venn diagram. So if somebody asks you, what are the number of elements in a minus b? Now, just imagine the Venn diagram of a minus b and tell me what expression could you possibly write for n a minus b? I'll just make a diagram for everybody to recall that a minus b part. So let's say this is a and this is b. Okay. So a minus b is this one. So how will you write this? How will you write this? Yeah, Arthur will discuss it in the next class. Section b problem. Right. See, no, number of elements I'm asking you that you're telling us expression in terms of properties of sets. I'm asking you how many number of elements will be sitting in this shaded area. Right. So you'll see number of elements in a and from there you subtract the number of elements present in a intersection b. Isn't it? Okay. Similarly, if I ask you how many elements will be there in b minus a, what will you say? Number of elements in b minus the number of elements lying in the intersection of a and b? Yes or no? Yes or no? Similarly, if I ask you how many number of elements are there in a delta b, what will you say? So what did delta mean again for this? Delta means union of this part and this part. I mean the entire shaded area which is in white and yellow combined. So n a plus n b minus n a intersection b minus 2, n a plus n b minus twice of n a intersection b. Isn't it? Because in both n a intersection b comes in n b also n a intersection b comes, but I don't want to n a intersection b at all. So I have to subtract twice of it. Yes or no? Yes or no? Can I also say it is as good as n a union b minus n a intersection b? That also can be written. In fact, both are same formula. It's written in different way. Is it 5? So this list. So in this one, in the second one also want to be twice n a intersection b. In the second one, okay? Like a union b, n into a union b minus twice n a intersection b. Why? Because a union b only has this intersection part once in it. Okay. Yeah. Okay. Now we'll see how it is applied to solving questions. Let's take few questions based on it. I'll take a few simpler examples. Let's take this one. Okay. I'm putting the poll also for this. Read the question very carefully. The question says a survey shows that 63% of the people watch our news channel whereas 76% watch another news channel. If x% of the people watch both the channels, then which option is right? Okay. Anybody else who wants to answer? Okay. Let's look into this. I'm putting the poll to an end because it's almost three minutes gone. Okay. And I want to do a few more questions before we end today's session. It's a very confused response. B and D have got most number of votes and C has got the second highest vote. Okay. Fine. See, many of you have interpreted this question in a wrong way. And that's the result why you have gone, got different answers. See, there are two channels. A, B and there could be more channels also. Right? The question is not trying to say that A and B are the only two channels. He's saying one channel is watched by 63% of the people. Another channel is watched by 76% of the people and x watch both the channels. Right? Now, see, this is x. This is 76 minus x. Oh, sorry, 63 minus x. Okay. And this is 76 minus x. Correct. But please realize, many of you, what you have done, you have added this three and equated it to 100% while it is less than equal to 100% because you don't have any information about the other channel existing. What's the point? So when you solve this, you end up getting this is less than equal to 100. That means x is greater than equal to 39. Okay. So x is greater than equal to 39. Wait. It is not equal to 39 necessarily. So people who went for a B, that is not going to be right. Right? So this C still has a chance of being right. Now see, many of you would actually realize this inequality that the number of elements in union of two sets will be, will be lesser than, sorry, will be greater than equal to max of the number of elements in either A or B. Correct? And at the same time, the number of elements in A intersection B will be lesser than min of the elements present in A and B. Getting this point. See, the union of two sets will always contain more elements or you can say greater than equal to number of elements, which is more of the two. So out of N, A and NB, whichever is the greater of the two, NA union B will have more elements or greater than equal to number of elements than that. And at the same time, NA intersection B will have an element lesser than equal to the lease of the two. This important inequality should be kept in mind. I didn't tell you purposely because I wanted you to think. So now this x, this x that you see, it is NA intersection B and it has to be minimum of NA and NB. So NA and NB minimum is 63, 76, which is minimum of the two, 63. So x should be lesser than 63. So if you combine this first and the second, the result that you will get is x should be between 39 to 63. So option number C, which unfortunately was the least chosen option out of BCND is actually the right answer. So one more like this. Yes, yes, we will take one more question like this. Done any questions here? Okay, let's take this question. A is 15, NB, sorry NA is 10, NB is 15, NC is 20, NA intersection B is 8, NB intersection C is 9. Okay, then the possible value or values of NA union B, union C is which of the following? Option A, 26 only. Option B, 27 only. Option C, 28 only. And option D, it could be 26, 27, 28, either of this. So all are, this all are possible. Or let me write, all in front of it are possible. So all 26, 27, 28 are possible. Let's solve this. Okay, let's discuss this. Okay, one answer I have got from Sake. Let's discuss this. See, if you see the information here, whatever information I have at hand, I will try to use that information. So this is nothing but NA, which is 10, NB, which is 15, NC, which is 20, minus NA intersection B, NB intersection C, NA intersection C, which is not given to me. Plus NA intersection B intersection C. So this term, these numbers, they will add up to how much? 328. Now pay attention. Pay attention, this is very important. What I'm going to do is I'm going to take this minus sign common and I'm going to write it as NA intersection C, minus NA intersection B intersection C. Okay, now you tell me from your common sense, is this more or this more? Is NA intersection C more or NA intersection B intersection C more? This is more or this is more? This is more than this. Please note this down that the number of elements in this will always be more than the number of elements in this because more intersection means lesser number of elements it will have. So which means NA intersection B minus, sorry, why did I write B? It's C, no? Yeah. NA intersection C minus NA intersection B intersection C will be a positive term. So if I say this term is X, then this X is actually positive, isn't it? What does it mean? You are subtracting a positive term from 28, which means NA union B union C will be lesser than equal to 28. Let's call this as 1. So from this information, at least I can say that my NA union B union C will be lesser than equal to 28. Now I'm still not done. Hold on. Can I say NA intersection B intersection C will be minimum of the three, this or this or this. Correct? Sorry, this is going to be lesser than, lesser than equal to minimum of these three, because intersection cannot, intersection of all of these cannot exceed the least value of these three, whichever has the least value, it cannot exceed that value. Correct? Now out of these three, this information is not provided to me. I only know this two information. So it is less than minimum of NA intersection B. NA intersection B is 8. Correct? Oh, sorry, I have to write union. Oh, so sorry, one second, one second. This is union, no? So in case of union, this will be, this will be greater than and this will be max. Sorry, my mistake wrote an intersection there. It is a union. Okay. So now, now I can see that I have been provided indirect information to get NA union B. This will also become union. So I have information indirectly to find the first two, but not the last one. So I will only work with the first two only. So A union B, how do I find it out? A union B is NA plus NB minus NA intersection B, which as per our given information, NA is 10, NB is 15, and NA intersection B is 8. So 10 plus 15 minus 8, which is nothing but 17, I believe. And what about NB union C? NB union C will be NB plus NC minus NB intersection C. How much that will be? NB is 15, 15 plus 20 minus 9. 15 plus 20 minus 9. How much is this? This is 26. So NA union B union C will be greater than equal to max of the two. So 17 and 26, which is max, 26. So as per the data which I have, I can only conclude these two things. It is less than 28, but more than 26. In short, NA union B union C is a number which can be a natural number between 26 and 28. So what all possible values can it take? 26 also it can take, 27 also it can take, 28 also it can take. So option number D becomes the right answer. Is it clear? This question is not easy because many people will not be able to think of all these inequalities. So what is important is not only the theorems, not only the equality theorems, but also the inequality theorems. Is it fine? Any questions? Scroll up means still what position is enough? Is this good enough? So when I initially looked at the question, I had a feeling that the information that's been provided is not enough to get the exact value. I wasn't able to think about all this and find the exact solution, but I sort of knew that maybe you might not be able to get an exact value. So I picked the last option. So is that a valuable shortcut or do you think that many times in the competitive exams you could go wrong if you move in that direction? See if you have not utilized all the information in the question, that means something may be missing in your answer. So many times the information that is in provided, they are just enough for you to crack the problem and even in that if you're missing out using something, then it will definitely lead to some kind of an error coming in your result. That is a thumb rule which I normally use. So in this particular problem, if you have used just the upper part of the data, that means only this part less than equal to 28, then the chances of you making a mistake may happen. Here you may be lucky to escape with a D, but not always. Yes. Can you show the question just once? I want to try solving by myself. Okay, fine enough. Just note this down. Take a snapshot. N A, N B, N C, N A intersection B, N B intersection C, they're given. We have to find the possible values of N A union, B union C. So that's all from my side. Thank you so much for joining in. Bye bye, take care, good night. So have your dinner and do take good care of yourself. Sir, I had one last question. Yes.