 Hello friends. So we are going to continue with our revision courses. So this is for real numbers today. So we are going to do real numbers. So I'm just waiting for some more people to join in and then we will start. So today, not only the real numbers will be discussing will also be dealing with some of the problems. I hope you are able to hear me for, you know, so just for my confirmation, if you can put it in the comment section that you guys are able to hear me. Yeah, all of you are able to hear me guys. So I can see here another good present in the class. So let's begin guys. So today, we will be covering the real numbers in as much detail as possible. And towards the end, we will also be trying some of the problems which were asked, especially in the previous year board papers in 2018 in particular. And we'll see that if time permits, we'll also solve some of the papers which were asked some time ago, some years back. So let's start in real number chapters can be divided into three, four parts. The first part deals with Euclid's division lemma. Then we have fundamental theorem of arithmetic. Then we have application of Euclid's division lemma and fundamental theorem of arithmetic. So let me just write down what all we have studied. So we have studied Euclid's division lemma. So Euclid's division lemma is one thing which we have studied. Then we also saw its application. So application of Euclid's division lemma in finding. In finding let's say, HCF of two numbers, two positive numbers. HCF of two positive numbers. This is what we have studied. Then we have studied fundamental theorem of arithmetic. Then we have also studied then again application of this application of FTA, fundamental theorem of arithmetic in again finding HCF. Then we also saw problems related to irrationality of numbers. We have to prove whether numbers are rational or irrational. This is another. Then we also have something related to factors of cases where numbers are expressed in the powers of two and five. So that's one more thing which we have seen. Let me just... So basically it's related to and then yes, obviously the word problems around it. Now in my analysis I've seen in the last three, four years there has not been any case where they have given any word problems related to real numbers. Mostly questions have revolved around Euclid's division lemma application to find out HCF and the long division method to find out HCF and then fundamental theorem of arithmetic application of this. Or they have given you some certs and you have to prove that the number is irrational. So these are the basic types of question we'll see towards the end of this session. These particular topics, the questions have been asked from again Euclid's application of Euclid's division lemma and then application of fundamental theorem of arithmetic to find HCF and to prove some numbers are irrational numbers or not. So in last four years I have not seen a single word problem which has come from real number chapters. But nevertheless you will also focus on that. So let us start with our course. So first is Euclid's division lemma. Given positive integers A and B there exist whole numbers Q and R satisfying A equals BQ plus R and zero is less than R less than B. So in previous classes we have studied that while we divide we know that dividend is equal to divisor into quotient plus remainder. We have studied in previous classes. So now actually this is nothing but a manifestation of Euclid's division lemma only. So Euclid gave this lemma in somewhere around 300 BC. So now what does it mean? Let's say you have two numbers A and B. Then A and B for every pair of positive integers A and B you will get a pair of positive integer Q and R such that A is equal to B times Q plus R where zero is greater than less than equal to R and R is less than equal to B. So here Q is the quotient R is the remainder. For example if we have let's say A is equal to 30 and B equals 8. So A and B 30 and 8 can be expressed as 30 can be expressed as 8 into 3 plus 6. So if you see this is my A, A this is B, B this is Q and this is R. Clearly 6 is greater than 0 and less than 8. Okay another example could be let's say you have 50 A is equal to let's say 50 and B equals to let's say 22. So hence you can write A and B as 50 is equal to 50 is equal to 22 into 2 that is 44 plus 6. Again the remainder is 6. Okay basis this we now know that if you are dividing let's say if you are dividing A by B. If you are dividing A by B dividing both A and B are both A and B are positive integers. Both A and B are positive integers. Then we know that possibilities of remainders possibilities what all remainders can you get possibilities of remainders. Okay so if you see possibilities of remainders what all possibilities are there. So remainders could be anything from 0, 1, 2 till B minus 1. This is Euclid's division lemma. Now not only there exists Q and R Euclid division lemma also says that Q and R will be unique. We cannot be getting any other Q and R for same set of A and B. Okay so that is what is Euclid's division lemma. Now just a minute. Now next is let's go to the next slide. Next slide says now it is Euclid's division algorithm. So what is Euclid's division algorithm? It says in order to compute the HCF what is HCF? First of all we need to understand HCF is nothing but highest common factor. This is highest common factor. So hence we know that there are lots of factors of any particular composite number. So what is some definitions what is composite number? For example there was a question in I think last year where you have to find out the HCF of and will solve this problem. HCF of the lowest composite number and lowest prime number. So composite number is any positive integer which has more than two factors. So composite number is any positive integer which has more than two factors. Example being 4, 6, 8, 10 all even numbers except two are composite. 4, 6, 8, 10, 12 and all you know not all odd numbers but numbers like 15 in 21 then 25 these are composite numbers. And there is another you know the prime numbers what are prime numbers? Prime numbers are those positive integers which has only one or the number itself as the factor. Now this particular slide says Euclid's division algorithm in order to compute the HCF of two positive integers say A and B with A greater than B by using Euclid's algorithm we follow the following steps. So basically I will just take an example and then we will see what are you know how to find out how to compute basically an HCF. Let us take two numbers. Let us say we have 48 and 64. We have to find out HCF of this. So this particular notation is for HCF. This is nothing but HCF. Within brackets you write 48 and 64. So by Euclid's division lemma, the method is this simply express 64. So bigger number is 64. 64 can be expressed as 48 into 1 plus 16. Is it it? So express in Euclid's division lemma. So 64 is equal to 48 into 1 and 16. Now whatever B was, so in this case what is A? So this is A. This one is A. This one is B. This one is Q and this is R. Isn't it? Now this is step number one. In step number two what will you do? Step number two. In step number two you will take the previous B as A. So hence 48 comes here. And the previous R becomes the new B. So this is 16. Now into new Q is 3 plus remainder is 0. The moment you hit remainder is 0. The moment you hit remainder is 0, you stop the process. So hence we hit here R is equal to 0. Then what is the HCF? So the last B of this line item will be my HCF. So hence HCF is nothing but this 16. What is the method? The moment you hit remainder as 0, whatever was the value of B will be the HCF. Let us take another example. Let us take example like 24 and 80. You have to find out HCF of 24 and 80. So how to find out HCF using Euclid's division lemma? So here you have to do our algorithm. Division algorithm is nothing but 80 is equal to 24 into 3 which is 24 into 3 is 72 plus 8. 80 is 24 into 3 plus 8. So 8 is my, so this is my again, so what all values are there? This is A, this is B, this is Q and this is R. So then previous B becomes new A. So hence I can now write 24, 24 is equal to, and the previous R becomes my new B is equal to 3 plus 0. So hence this is my HCF. You hit R equals to 0. Just a confirmation, guys. Are you able to see the screen now? Just anyone can confirm on the comment section? Yeah, can you guys just confirm whether you are able to hear or not? Are you able to see the screen? Hello, can you hear the screen? Hello. Okay, give me a moment, guys. Yeah, I hope now you'll be able to see the screen. Tell me if you are able to see the screen. Just give me two minutes. Yeah, now I think you are able to see and hear. Are you able to see and hear? Yeah, are you able to hear me? Yeah, can you hear me? Yeah, cool. So, okay, so let us now go and continue our, okay. So, yeah, so the fourth point is we just discussed this and let us move to, okay, so for those who could not understand or let's say they were not able to see, so we will be discussing it once more so that you guys are at the same page, okay. Now, so what is it? Every composite number can be uniquely expressed as the product of powers of primes in ascending or descending order. So that's what I was explaining. This is in ascending or descending order. For example, 24, 24 can be expressed as 2 cube into 3. Similarly, 50 can be expressed as 2 into 5 square and there is no way that you can insert a 7 in between, right. 66 again is 2 into 3 into 1 and that is 2 into 3 into 11, sorry. And there is no way you can insert a 13 or anything else. Okay, so this is what this particular slide says. Every composite number can be uniquely expressed as the product of powers of primes in ascending or descending order. So you can reverse this order also. So, yeah, there is no problem in that. Okay, let's move on to the next slide, okay. Now, next is let A be a positive integer and P be a prime number such that P divides A square, then P divides A. This is a very common question. So let A be positive integer. So A is a positive integer and P is a prime such that P divides A square, then P divides A. Okay, so by fundamental law of arithmetic, I know that A can be expressed as P1, P2 in P1 to the power m1, P2 to the power m2, like that. And so on and so forth, Pn to the power mn, right, where P1, P2, dot, dot, dot, till Pn all are primes, all are prime numbers. Okay, P1, P2, Pn all are prime numbers, all are prime numbers and m1, m2, m1, m2 and mn. So let's say m1, m2 and till mn all are, all are positive integers, all are positive integers, right. So now, then what is, so let's say P divides A square, so what is A square? So A square will be, A square is equal to P1 to the power, this P1 to the power m1 into P2 to the power m2, dot, dot, dot, Pn to the power mn, whole squared. Okay, that means what, this is nothing but P1 into 2m1, right, and then P2 into 2m2 and then so on and so forth, Pn into 2mn, right. Now, what is, what does it say, P divides A square, P divides A square, that means P divides what P1 to the power 2m1 into P2 to the power 2m2 and till Pn to the power 2mn. Now clearly P is a prime number, right, so P is a prime number and P divides product of prime number, that means P has to be, P has to be, has to be, has to be P1, P2 or P3 or any of this till Pn. So Pn has to be either P1 or P2 or P3 or Pn, right, one of Pn has to be, we can write one off, one off, because, because, because P divides this entire thing and this expression is all primes, product of all primes, then what do we say, what do we infer, P has to be one of these P1, P2 till Pn, right. So if that is so, that means, let us say, you know, let us say P is equal to pr, P is equal to pr where, where r is anything from r is, r is, or you can say, one is less than equal to r, less than equal to n, right. So one of them is there, right, so that is pr, which is, let us say, if n is 20, so an r can be 16, 17, 18, whatever, 14, 12, like that, right, so P is one of them, right. So now if you see, if you see what was a, so a will be equal to nothing but P1 to the power m1 times P2 to the power m2 dot, dot, dot, there will be one pr also in it, which is, which has the power, let us say, mr and then Pn to the power mn, right. So if you see, pr exist in this, so hence we can say, pr divides a, that means, this implies P divides a, right, so whenever P divides a square, P will divide a, okay. So this is, yeah, next, let us move on to slide number 6, yeah, very important, there are infinitely many primes, there are infinitely many prime numbers. There are infinitely many prime numbers, guys, you know, just squeeze, excuse me for one minute, for just one minute, just one minute, guys. Yes, guys, I'm back, sorry for the delay, so there was some important, this thing, so never mind, so let us continue with our, this thing, I hope all of you are there and yeah, so we were trying to prove that there are infinitely many positive primes and let me just, yeah, so how to prove this, there are infinitely many primes, many positive primes, so we use, let's say, the concept of contradiction, right, so there are infinitely many positive primes, that means there is no largest, you can say, this can be also be expressed as, there is, there is no largest prime number, you can't find the largest prime number, so far we have not been able to and now we know that it is not possible to, it's also not possible to find the largest prime number, so how do I prove that there are infinitely many positive primes, so let us say, we have, you know, we will do it by contradiction, how will we do it, contradiction and the very method is this, that you assume that there is a largest prime number, okay, so let us say, let us, let us say, let us say that, let us say that pn is largest, largest prime number, let us say that pn is the largest prime number, that means, p1, p2, p3, all are prime numbers, still pn, pn-1, let's say, all are less than pn, right, pn is the largest prime number, now let us define our number, let us define a number p, such that p is equal to p1 into p2 into p3 into all of all the prime numbers till pn, we find a number which is a product of all the prime numbers, okay, now p is this much, that means, this is, this is the prime factorization of p, so this is the prime factorization of p, okay, so this is the prime factorization of p, now if that is so, then we have another number, let us say q, such that q is p plus 1, okay, so p is nothing but p1 into p2 till pn and we are defining a number q, such that q is equal to p plus 1, now clearly p1 does not divide q, right, why, because p1 divides p, obviously one particular prime number doesn't divide two consecutive numbers, so p1 doesn't divide q, similarly I can say p2 doesn't divide q, similarly p3 doesn't divide q, so on and so forth, pn doesn't divide q, none of the prime numbers will divide q, why, because the prime number, one of prime numbers are dividing p, then the consecutive number p plus 1, the exact, you know, the next number will never be divisible by any of the prime numbers, that is what we know, so that means, that means either, so what does it mean, it means that, it has this meaning that q is, you know, either q itself is a prime number, right, q itself is a prime number, this itself is a prime number, why, prime number, right, q itself is a prime, why, because there is no prime factors of it, because the total number of prime factors are only p1 to pn, so hence none of them are dividing q, that means q itself is a prime number, now clearly q is, if q is a prime number, then q is equal to p plus 1, which is, which is clearly greater than pn, which is clearly greater than pn, but our assumption was pn was the largest prime number, that means, that means our assumption that pn is the largest number is false, hence, hence it contradicts, it contradicts that, it contradicts the fact, contradicts the fact, the fact that, that pn is the largest prime number, prime number, that means you cannot really find any largest prime number, not just prime number, hence we conclude, hence, hence, hence there are, there are infinitely, infinitely many, many prime, prime numbers, there are infinitely many prime numbers, this is what the proof is all about, okay. So let's move to the next slide and slide number seven, right, okay, what does it say, so it says, let me just read all this, yeah, every positive integer different from one can be expressed as a product of non-negative power of two and an odd number, okay, every positive integer, right, different from one, so let's take an example first, so, yeah, can be expressed as a product of non-negative power of two and an odd number, very good, so how, so let's say we have any number, let's, let's 15, let's take 15, right, so it is different from one and can be expressed as a product of non-negative power of two and an odd number, okay, non-negative power of two means, power of two can, non-negative means zero, one, two and any positive integer, right, so these are all non-negative, these are all non-negative, non-negative numbers, so if you see 15 can be expressed as two to the power, can be expressed as two to the power of zero into three to the power of one into five to the power of one or two to the power of zero into 15, so non-negative power of, you know, two and 15 and odd number, similarly, 16 can be represented as non-negative power of two and an odd number, every positive integer different from one can be expressed as a product of non-negative power of two and an odd number, here how, what will you do, you will write two to the power of four into one, right, again you will, you will get this, the same, you know, expression, similarly, let's say we have 17 can be written as two to the power of zero into 17, then we can say 18 is equal to two to the power of two into three to the power of two or two into nine, so hence to the power of one into nine, similarly, 19 can be represented as two to the power of zero into 19, then 20 can be represented as two to the power of two into five, right, so all you will be able to, so if it is an odd number, if there is, if it is an odd number, then clearly there is no, the power of two will be zero, but if it is an even number, you know that, the two positive, how do you prove that, let's say the number is k, any number, any number is k, now if any number is k, it can be either even or odd, okay, if k is odd, if k is odd then there is no problem, then you can easily say it is two to the power of zero into that number k itself, okay, but if it is even, there are again two possibilities, either it will be purely a power of two, so what are the possibilities, so even using FTA, you can write k, definitely will be having one power of two, why, because it's an even, and then other prime numbers, let's say p one to the power m one into p two to the power m two, and let's say two to the power n is there, two to the power m is there, so hence k can be expressed as, because it's an even number, there will be some power of two in it anyways, and then all the other powers of, all the other powers of prime numbers will be there, so hence these are all odd numbers, so odd into odd anyways will be odd, so this whole is an odd number, and hence this is two to the power, some number, some non-negative power and odd number, that's what this particular theorem is saying, okay, fair enough, let's move on to the next one, the next one says, a positive integer n is prime if it is not divisible by any prime less than or equal to root n, this is very, another very important theorem, so again think about it, so it is saying that a positive integer n is prime if it is not divisible by any prime less than or equal to root n, what does it mean, let's take some examples first and understand, so what does it mean, it means that let's say you are testing whether 31 is a prime or not, okay, let's say you are testing whether 31 is a prime or not, now from our knowledge we know 31, but let's say if we were not aware of this number, let's say I will take another example to illustrate that, but to check whether 31 is a prime number or for that matter, to check whether any particular number is a prime or not, what do we do, we start dividing them by prime factor, isn't it, so we start first try divisibility by 2, then by 3, then by 5, then by 7, 11, 13 so on and so forth, till we reach 31, isn't it, so that is how we do, now this loss is this particular theorem is saying, you don't need to check the divisibility by all the prime numbers, all the prime factors less than 31, you can actually restrict it to prime factors less than root 31, so root 31 will be somewhere around 5 point something, right, 5 point something, it is, or you can say root 31 lies between 5 and 6, okay, that means they are saying if you check with all the prime numbers less than root 31, so what all prime numbers exist less than root 31, 2, 3 and 5, now if you divide, try to divide 31 by 2, 3 and 5, if you don't get any integer as the quotient, then you can clearly declare that 31 is a prime number, 31 is a prime number, let's say, if 31 is, if 2 doesn't divide 31, 3 doesn't divide 31 and 5 also doesn't divide 31, then you need not check and divide it by 7, 11, 13 and so on and so forth, this is what this theorem is saying, so let's say, if a number is not divisible by any prime, is any prime less than or equal to root n, then it will be a prime number, okay, let's take another example where it is the other case, let's say we have a number 64, right, so 64's root is 8, okay, then what they are saying is 64 clearly is not a prime number, 64 is not a prime number, that means root 64 is 8, that means there will exist a prime number less than 8 which will divide 64 and clearly we know that 2 divides 64, okay, so hence, hence let's say if you are not aware of the number 64, whether it is a prime or not, so what do we need to do, find first root 64, so let's say root 64 and find the nearest integer or greatest integer of root 64, so hence that will be, in this case it is clearly 8, now I will just check the divisibility of 64 by prime numbers less than 8, that is they are 2, 3, 5 and 7, these are the prime numbers less than 8, now if I, that is root 64, now if I divide 64 by 2 I get a quotient, right, that is 2 divides 64 clearly, hence we don't need to go further because now anyways we have got a factor, right, so all of you know that even numbers anyways are not prime, so it will be, you know, not correct to check for 64, let us check another number, let's say 91, okay, and we need to see, we need to see whether 91 is a prime or not, from our knowledge we know 91 is prime, but then how to check this, so let us find out root 91, so root 91 is something like this, 9 is less than root 91 is less than 10, why, because 9 square is 81 and 10 square is 100 and 91 lies between 81 and 100, so we know that 9 is less than root 91, now to check whether 91 is prime or not, we will only consider dividing 91 by prime factors less than 9 which are 2, 3, 5 and 7, we don't need to go even for 11 or anything else to check whether 91 is divisible by these numbers or not, so does 2 divide 91, clearly no, does 3 divide 91, clearly no, why because 9 plus 1 is 10 which is not divisible by 3, so 3 doesn't divide, does 5 divide 91, clearly no, why because it doesn't end either by 5 on 5 or 0, so this is also not true, but 7 does divide 91, why, because 13 times 7 if you see is 91, right, so hence we got a factor here itself, so hence we don't need to check any other factor, so that's what this particular theorem is saying, so if you want to check whether one particular number is prime or not, what is the step, find the root n value approximately and check the divisibility of that particular number by all the prime numbers less than equal to root of n, that's what this theorem is saying, let us try and prove this, prove is little interesting, so let me just delete all this, so we will start with this assumption, we will say that let n be a positive integer, so we are not talking about whether n is prime or not, we are saying let n be a positive integer, so we are not talking about whether it is prime or not, let n be a positive integer and what is the condition such that, such that, such that, such that, no prime, no prime less than, no prime less than root n, root n divides n, this is our starting point divides n, what are we saying, we are saying let n be a positive integer, forget about whether it is a prime or not, let n be a positive integer which is this character, what character, no prime less than root n divides n, all the prime which are less than root n are not going to divide n, so hence the question just boils down to proving that n is a prime, you started with saying that n is a positive integer such that no prime less than root n divides n, but if we somehow prove that n is a prime, then we are done, so let us say n is not a prime, so let us start with a contradiction and say that let us say n is a composite number, n is a composite number, what kind of a composite number such that no prime less than root n divides n, n is a composite number such that no prime, no prime less than root n divides n, this is my assumption n is a composite number, that means I can always find two factors a and b and can always be written as a and b okay now from our logic we know that we can say that let us say you know let us say a is less than b okay let us say a is less than b okay now we can also what we can say if a is less than b then clearly a is less than root n and b will be greater than root m isn't it right so a clearly will be one of them will be less than root n and another will be more than root n correct now let us say let us say let us say let us say p divides a p divides a where where p is a prime number p is a prime number that means a is n is equal to a into b and p is a prime number which divides a right if a happens to be prime number itself then p and a are same so p divide p is less than or p divides a where p is a prime number then or then then p is a prime factor so p divides a this implies p divides a b correct so if p is dividing a p will definitely divide a b that means p will definitely divide n p will definitely divide n right is it understood so p divides a that means p divides a b that means p divides n right so hence what is p a prime p was now prime p was less than root n right so p is p is what p is a prime so p is you know a prime number and p is dividing n and and p is clearly p can max b equal to a in this case if you see a is less than sorry p divides a then either p is a or p is less than a this is only two possibilities that means p is less than equal to a is less than a itself was a itself was less than root n so I can say root n that means sorry this is n so that means p is less than root n from here this it follows p is less than root n which means which means you know there is a prime there is a prime there is a prime less than root n root n that divides that divides that divides n okay now this contradicts our assumption that n is a positive integer such that no prime less than root n divides n okay then what what is it contradicting the fact that so if this particular conclusion says there is a prime there's a prime which is less than root n that divides n but then our assumption was that p n is a number such that no prime less than root n divides n so this contradicts the very first assumption hence hence it leads to hence it leads to you know contradiction and hence by contradiction we can say that and our assumption that n is a composite number was wrong and hence n has to be has to be a prime number right so hence n has to be a prime number so hence what do we conclude that a positive integer n a positive integer n is a prime if it is not divisible by any prime less than or equal to root n I'll again go through the proof please be careful and you know please you know understand it very thoroughly so we say that forget about being prime or not let n be a positive integer such that no prime less than or less than root n divides n this is our assumption and then we are saying that you know we just need to prove that if somehow we prove that n is a prime then we are done right to prove that we start with the contradiction saying that n is a non prime is a non prime that is n is a composite number the moment we say n is a composite number we'll get two factors a and b at least two factors now if that is there then n can be expressed as a and b where a let's say let's say a was less than b out of two they cannot be equal they will have to you know if they are equal then n is a perfect square but let us take a general case where n is not a perfect square so a is less than b if a is less than b and we also know that there are two factors then one has to be less than root n and another one has to be more than root n so let's say a is root less than root n then let us say there is a prime factor p which divides a correct now either a itself is a prime number or there will be a factor of a also let's us say that prime factor is p so p divides a where p is a prime number then if p divides a then p definitely divides a b then hence p definitely divides n that means p which is less than or equal to a is less than root n that means there is a prime number less than root n hence there is a contradiction we have started with the assumption that there is no prime less than root n which divides n but here we are getting one prime which is less than root n which divides n hence we say that our assumption that n is a composite number was wrong hence n has to be a prime number this is our you know this proof is done here okay great so let's go to the next question you know this discussion yeah if p is a positive prime then root p is an irrational number for example root 2 root 3 root 5 root 7 root 11 etc irrational number you if at all they however do not give such questions to prove that root 2 is a prime number they will always start with saying that if root 2 sorry not root 2 is a prime number root 2 is an irrational number so they will give you problems which will be discussing a little later that if root 2 is irrational prove that some particular third is also irrational problems of that kind will be given problems like proving that root 3 itself is a prime number is not asked in board exams because you have already done this in ninth grade so this is you know so hence any prime number if put under root is always irrational number so what does it mean so root of 31 will always be irrational root of 97 will always be irrational root of 43 is always irrational and things like that so you put a prime number under root it will always give you an irrational number okay let's let's let us not prove that because you already know if you need I can you know give you this proof the proof is let's say you can start with this and I'll quickly try and prove this for you so how do you prove that root 3 or root 2 is a prime number so let us we again we start with contradiction and we say that let us say let us say let us say that root 2 root 2 is a rational number let it let it not be a irrational number let us say that it is a rational number then if it is a rational number we know rational numbers can all will always be expressed as p by q form so hence I can always say p by q is root 2 root 2 where p and q belong to the set of integer and q is not equal to zero q is not equal to zero right so this is what rational numbers definition means that means squaring both sides I'll get p square by q square is equal to 2 is equal to 2 is equal to 2 so hence p square is equal to 2 q square that means that means p square is an even number right if p square is an even I know p is an even number okay and one more criteria in this is that p and q are co-prime p and q are co-prime right there is no common factor between p and q right then only be expressed so three three criteria of a rational number p and q must be it should be expressed in the form of c by q where p and q are integers q cannot be zero and the numbers p and q are co-prime that means there is no common factor between p and q okay so p is an even number from this analysis similarly if p is an even then when I can say p is equal to let's say 2k okay p can be expressed as 2k then hence I can now write p square by q square is equal to 2 then that means 4k square is equal to 2 q square that means now I can say q square is 2k square hence q square is even hence q is even now here is what you know we are seeing a contradiction so p is an even and q is an even that means p and q are the cf of p and q is not one why because both are even that means at least two is a common factor so it contradicts our this assumption that p and q were co-prime so hence why this contradiction because we assumed what hence our assumption hence our assumption assumption hence our assumption that root 2 is a rational number root 2 is a rational number is wrong is wrong so hence we say root 2 is an irrational number perfect let's move on and then see what is there up for us next so slide number 10 let x be rational number whose decimal expansion terminates then x can be expressed in the form of p by q where p and q are co-prime and the prime factorization of q is of the form 2 to the power m into 5 to the power n where m and n are non negative integers very interesting again so let us go back to yeah yeah so let x be a rational number so x is a rational number whose decimal expansion terminates that means it has a terminating decimal expansion then x can be expressed in the form of p by q where p and q are co-prime okay so let's let us take an example first let's say I have 0.42 right so there's a terminating decimal expansion is it so 0.42 terminates and there's not no repetition of any digits here right so clearly you can see 4.42 can be written as 42 upon 100 okay and you can reduce it to as much as possible so that p and the numerator and denominator do not have any common factor so let us reduce it it is divisor by 2 those is 21 by 50 so I reduce it to the most you know simplest form of fraction now if you see p is equal to 21 here and q is 50 and p and q are co-prime so p q is equal to 1 at cf of p and q is 1 so hence p and q are co-prime right so if you see q what is q q is 50 which can be expressed as powers of 2 and m so let's say it is 2 into 5 square right so powers of 2 and 5 right 2 and 5 so here 2 to the power 1 and 5 square right so m is equal to 1 and n is equal to 2 this will always happen let us take another one let us take 5 20s 52.6 52.6 is a terminating decimal representation so 52.6 can be written as 5 26 upon 10 okay and then you can eliminate a 2 2 is there in denominator and numerator so you will see this is 2 6 3 upon 5 right so if you see 5 the denominator this is again 263 at cf of 263 and 5 is 1 so they are co-primes co-primes right so once you have done it you will know that 5 the denominator q is equal to 5 can be expressed as 2 to the power 0 into 5 to the power 1 so m is equal to 0 and n equals to 1 right this will always be there I have not seen a question where they have asked for proving and all so let us skip the proof okay now next okay this is a converse of the previous thing so so what is it this is a converse of the previous thing and this is now what they are saying is if x equals to p by q be a weight so it is just a converse of the previous case so x equals to p by q be a rational number such that the prime factorization of q q of the denominator right prime factorization of q is of the form of 2 to the power 5 into 5 to the power sorry 2 to the power m into 5 to the power n where m and n are please mind it non negative integer so it can be 0 and any positive integer then x has a terminating decimal expansion that means you can expect the decimal expansion to be terminating right which terminates after k places of decimal where k is the larger of m and n let us see an example okay so let us say we have 3 30 or not 30 let's say 21 upon 21 upon I will write 2 to the power 1 and 5 square that is 50 yeah 2 with 2 1 to the power 1 into 5 square which is 50 right so hence if you see the decimal expansion of 21 by the decimal expansion of 21 by 50 is nothing but 0.42 okay 0.42 there and it is terminating if you see it is terminating and and the number of decimals after the number of decimal places how many decimal places 1 and then this is 2 which is nothing but this 2 here okay it's very easy to understand what do I do is we complete the multiples of 10 here so hence 21 upon 2 to the power 1 into 5 square can be written as 21 upon 2 let's let's complete the multiples of 10 so if you see it is 5 square if I multiply the denominator by 2 I will get a multiple of 10 here so let's do that so if I do this I will get multiple of 10 in the denominator but I have to also adjust it by multiplying to the numerator so hence what do we see we see 21 into 2 upon 10 square okay so 2 was the maximum of the powers of 2 and 5 so 10 square and I know anything divided by 100 will have 2 decimal places after the decimal point right so this is what you know it's very easy to understand hence the value is 0.42 let's take another example let's say we have 4 23 divided by 2 to the power 2 and 5 to the power 1 5 to the power 1 or let's say we take 2 to the power 3 here so that it becomes a little different and this is 5 square okay so hence we know to come to make the denominator as a multiple of 10 I have to multiply this by 5 why because once 5 is getting multiplied in the denominator I'll get 2 to the power 3 and 5 to power 3 it will become 10 to the power 3 correct but I have to also multiply the numerator by 5 so that the fraction doesn't change okay so hence if you see if you multiply 4 23 into 5 you will get 2 1 1 5 I hope I am right 5 4 yeah yeah 2 1 1 5 divided by how much it is 10 to the power 3 which is 1000 now whenever you divide anything by 1000 you know after decimal there will be 3 points is it it this is what this particular point says okay fair enough let's now quickly move to problem solving so yeah yeah and the last this point is if we P is a rational number P by Q is a rational number such that the prime factorization of Q is not of the form of this where m and n are non-negative integers then x has non-terminating repeating decimal expansion that means if you simplify any of the fraction and you get any other factor in the denominator apart from 2 and 5 then the rest assured it will have a non-terminating non-terminating decimal expansion now Pritika says if it is given that root 2 is irrational how do we prove that 5 plus 3 root 2 is irrational I'll come to that question Pritika we have some questions like that now let X be so hence what is it let X be P by Q rational number so hence what I'm saying is let's say if you have a number like 341 divided by 2 into 3 okay now in the prime factorization of denominator you have a 3 which is extra from 2 and 5 right so there is no the denominators are not denominator is not restricted only to 2 and 5 factors or multiples of 2 and 5 then we say that clearly this will end up in non-terminating non-terminating and repeating non-terminating and repeating decimal representation and repeating right example let's take very simple examples let's say 1 upon 3 you all know that it is nothing but 0.3 bar yeah so if it were 2 and 5 or a combination of 2 and 5 you will still get a this thing but here you didn't get it right similarly 1 upon 7 you will never get a what is what is 1 upon 7 if you see 1 upon 7 is 1 upon 7 is 0.1 and then 4 and then 2 and then you know I think 8 yeah so 1428 and then 5 then 7 142 like that so this is how it will be it will start repeating after some time yeah so 1.14285714 like that it will go on right again so if you see 1 by 7 is nothing but 0.142857 bar I hope I'm doing the calculations correctly you guys can pinpoint if it's not done properly right 1.14285714 right so like that it will be so if you see had it been some 2 and 5 and you know multiples of 2 and 5 only here then you would have got a terminating decimal representation fair enough so let's move on to our most important part of for this this thing and this is question solve problem solving so let's solve this is our cbse 2000 so if you see this was asked in this was asked in cbse 2018 I'll write it here cbse 2018 this question was asked there okay if I remember correctly so what is the hcf of smallest prime number so what is smallest prime number smallest prime number this was one market question I believe so smallest prime number is two and smallest composite number smallest composite number composite number is four why because one two three anyways are not one is neither prime nor composite two and three are prime so the smallest composite number is four you have to find out this you know so you can use Euclid's division algorithm to find out the hcf so hence you can write four is equal to two into two plus plus zero so hence two is the hcf so hcf is equal to two correct this is cbse 2018 okay let's move on to next question question is yeah so this is what Krithrika is also asking so let's prove this let's solve this question this is again cbse I think cbse 2018 2018 problem given that root two is irrational prove that five plus three root two is an irrational number okay so root two so how will you solve this is how you should approach the paper okay yeah so this is how you should approach the paper so you write a given given root two is an irrational number root two is an irrational number right you have to prove to prove what to prove what five plus three root two is an irrational number as well irrational number not very simple it's you know typical real number problem so what do you say you say solution proof you will start with let us assume let us assume that five plus three root two is equal to p and and p is let's say rational number or instead of p you write something which is unique let's say a okay is a right and you write a is rational so let me just write it a little bit more clearly yeah so what you need what you say is a and let a be rational let let a be a rational rational number okay if that is so then a can be expressed as a is equal to p upon q where where p and q are integers integers and q is not equal to zero right that means I can write p upon q is equal to five plus three root two five plus three root two then I can just rearrange them to write p minus five sorry p by q minus five is equal to three root two then I can say p minus five q upon q is equal to three root two this implies is p minus five q by three q is equal to root two correct root two now LHS here so what you will write LHS right hand side is equal to p minus five q upon three q which is an it is a rational number why why is a rational number is equal to a rational number because because p minus five q is an integer and three q is an integer integer integer if p and q are integers right but RHS is equal to root two which is an irrational number right so hence LHS cannot be equal to RHS so hence it leads to a contradiction so hence it leads to contradiction hence we can say okay hence we can infer hence our assumption was wrong hence five plus three root two is not a rational number and hence it is a something which is not rational is always an irrational number okay that's how you will have to go over here I think this was our two marker question this was of two marks in the board okay fair enough let's move to the next question question is find hcf and lcm of 404 and 96 and verify that hcf into lcm is product of two given numbers okay so let us do it find the hcf and lcm so how will you find out hcf and lcm let us find out so we will use the fundamental theorem of arithmetic in this way you will also learn and this will be revised for you as well okay so hence how do we find out hcf and lcm using fta so 404 let us express 404 what is the prime factorization of 404 so it is clearly two square into 101 isn't it right two square into 101 now let us say usually what happens is people get stuck that whether 101 is a prime or not but then you can always do a rough work and check we just learned what did we learn and you just need to divide the if you let's say you have to check whether 101 is a prime or not because if you don't remember the tables usually people forget that so hence what we do is let's check whether 101 is a prime or not so hence what should I do I should find out root 101 approximate value and so hence root 101 will be less than or let's say greater than 10 isn't it because root 100 is 10 right so you just need to check with what all two three five and seven yeah I just need to check the divisibility of 101 by two three five seven if it is divisible by any one of them then it's anyways not a prime if it is not then you don't need to check any other number and you can declare that 101 is a prime number is a prime number so clearly two doesn't divide 101 it's an odd three doesn't divide 101 because digits don't add up to multiple of three five doesn't divide 101 because the unit space is not neither zero nor five seven also doesn't divide 101 why because if you if you divide you can do the division little division because you know if you divide 101 by 7 3 will be the remainder 7 divides 98 but 7 doesn't divide 101 right so hence clearly 101 is a prime factor so hence 404 has this prime factorization then 96 is a smaller number you can always 16 into 6 is 96 right in fact 32 into 3 so hence it is 2 to the power 5 32 into 3 is my prime factorization now if you see we have how do I find out hcf and lcm so when you have to find out hcf you first so how do I find out hcf hcf hcf is nothing but you write down all the prime factors which you see in both the expressions so I can see two one one is two another is three and third is 101 okay these are the three distinct prime factors which you can see in both the expressions all put together not necessarily that they should appear in both of them so two three and 101 now you see between the powers of two which is minimum so when you are dealing with h highest you have to check the minimum power so minimum power between five and two is two so you will write two here now minimum power here it is three to the power one here it is three to the power zero so hence between one and zero what is minimum zero so you write zero yeah here it is 101 to the power one here it is 101 to the power zero so between zero and one which is minimum zero so zero so hence hence hcf will be plain and simple four this is how hcf is found out now how to find out lcm exactly reverse process write down all of them all of them once again and now pick up the maximum between two and five maximum so when you're doing with dealing with lcm you have to find out maximum two and five what is maximum five what is maximum between zero and one here it is zero here it is one so maximum is one and what is maximum between one not zero again for 101 it is one so hence lcm will be 32 into 3 into 101 which is 9 6 9 6 right this is lcm now let us check whether it is happening or not so hcm into lcm hcf into lcm is nothing but four into nine six nine six which is nothing but if you do four six start 24 carry two four nine yet 36 38 carry three four is it ready four again and three added to 27 carry two four nine yet 36 and 238 so 38784 is my hcf into lcm and now let us say what is this uh you know uh 404 into 96 is so either you can use your distributive law what is it 400 plus four into 100 minus four either you can do that you have to simplify or you can simply you know multiply uh on the sidelines right and the rough page and you can see 404 into 96 is also 38784 so if you want to check with here itself so 400 into 100 is 40 000 and 400 minus 4 is minus 1600 and then this is plus 400 and this is minus 16 so if you see 1600 minus 1600 plus uh 48 is minus 1200 so minus 1200 16 you have to subtract what from 40 000 you have to subtract um this is minus 1200 and minus 216 you have to so one two one six you have to subtract so if you subtract it this you'll get eight and then again um seven four nine one eight nine two seven nine one is eight and this is 38 38784 so both both of them matches so hence product of two given numbers is verified this is product of two numbers product of numbers product of two right so this is verified this was another previous year paper fair enough next let's go to question number 16 oh cool so this is another uh another previous year paper so right whether two root 45 plus three root 20 divided by two root five on simplification gives an irrational or a rational number let us do this so two root 45 can be written as two in two again you used fta and uh you know uh find the prime factor so it's clearly three square nine into five and plus three into root of 20 can be written as two square into five and then it is two root five okay so hence you will get two into three into root five plus three into two into root five so three square three root five root five and plus three this two comes out so three into two yeah three this two will come out and then root five divided by two root five so clearly it is nothing but six root five plus six root five it is 12 root five upon two root five so this root five and this root five will get cancelled why because they are not zero so hence this is six answer is six so it is a rational number all integers are rational number right very good next next question given that root three is a then again similar type of problem so let's solve it given that root three is an irrational number prove that two plus root three is an irrational number so what do we do we start with assumption again so let us say let us say let us say that a is equal to two plus root three two plus root three and a is a rational number now if a is rational number then a can we express this p upon q where p and q are integers integers q not equal to zero and you can also write that p and q are co-prime that is in here it will not be used by the way but anyway so hence it is two plus root three is equal to p upon q hence i can say root three is equal to p upon q minus two which is p minus two q upon q now if p and q are integers then then p minus two q is an integer right and and q so anyways is integer q is an integer right that means p minus two q upon q is a rational number is a rational number but but p minus two q by q is equal to root three which is an irrational number now this cannot be true simultaneously so hence hence our assumption so hence a rational number rational number cannot be equal to irrational hence our assumption hence our assumption assumption that that two plus root three is rational is wrong and hence two plus root three is an irrational number okay this is similar to whatever we just did some time back okay next so using Euclid's division algorithm find the hcf of the number 867 and 255 let us solve this as well for a practice so what Euclid's division says you you have to express this as a into bq plus r and you have to keep on doing till you get r equals zero that was the algorithm so let's start so 867 is equal to how much i think you know 250 into three so 255 into 3 it will be and it is 750 and the remainder will be plus 117 just inform me if i do any calculation mistake but i think 867 and 255 right so 255 into 3 plus 117 i'm sorry oh it's 255 sorry it's wrong it's not correct it's not correct yeah so 255 into 3 will be 765 sorry so 102 so it will be 102 i believe yeah so it is 102 yeah just check once again so calculation becomes important because if you miss out on calculation then everything in the later steps will be wrong so hence this was my a this is my b this is q and r now next step the b becomes a so 255 will come here is equal to r becomes new b so 102 into 2 clearly why because 200 approximately 204 and then it is 255 so hence remainder will be i think 51 all right 102 into 2 is 204 plus 51 is 255 right 51 so again this is again r is not equal to 0 so you have to repeat the process this 102 will come here and this 51 will come here as my new b so b is 51 into 2 plus 0 51 into 2 is 102 and 0 so hence you achieve r is 0 so hence your your hcf is 51 right hcf is 51 isn't it yeah so you see 51 into 17 is 867 and 51 into 51 into 5 is 255 okay this is what this is how you have to do with euclid's division algorithm next question question number okay show that n square minus 1 is divisible by 8 if n is an odd positive integer again use of euclid's division lemma so let's start so how do you approach this so that n square minus 1 is divisible by 8 if n is an odd positive integer so n is odd they have themselves have given so if n is odd if n is odd then n can be written as 2 k plus 1 where where k is a positive positive positive integer okay then let's do this so n square minus 1 is equal to 2 k plus 1 whole square minus 1 which is equal to 4 k square plus 4 k k plus 1 minus 1 which is nothing but 4 which is nothing but 4 times k square plus 4 k which is nothing but 4 k times k plus 1 right now k can be either k can be even so if k can be even or odd let's say when k is even k is even then k can be written as 2 m for for some positive positive integer m then then 4 k square plus 4 k is equal to 4 into 2 m into 2 m plus 1 correct which is 8 into within brackets you write m into 2 m plus 1 then clearly 8 divides 4 k square plus 4 k or 8 divides n square minus 1 correct now if k is odd k is odd then k can be written as 2 m plus 1 then 4 k into k plus 1 becomes what 4 into 2 m plus 1 into 2 m plus 2 so again you can write 4 into 2 into 2 m plus 1 into m plus 1 clearly all these m is an integer so hence hence you can write this as 8 into within brackets within brackets 2 m plus 1 into n plus 1 right so 8 clearly divides so again in this case also see i'm writing here because of the lack of space so 8 clearly again divides so in this case also 8 divides n square minus 1 so in both the cases 8 divides n square minus 1 okay so this is question number 19 and let us now go to the next question okay so this question says that the show that the square of any positive integer is of the form 3m or 3m plus 1 for some integer m right what is what does it say square of any positive integer right is of the form 3m or 3m plus 1 for some integer m right so let us say let us say let us say n let n let n be let n be an integer n be an integer integer let n be an integer okay now n there are three possibilities of n n can be of the form of n can be of the form of the form a 3k or b 3k plus 1 or c 3k plus 2 right by Euclid's division lemma isn't it so a is equal to b q plus r right and 0 less than equal to r less than b so here b is equal to 3 so values of r possible is 0 r could be 0 r could be 1 or r could be 2 okay so this is how right so when let's say case a so take case a so square right squaring it so you'll get 3k whole square is nothing but 9k square which is nothing but 3 into 3k square hence you can write this is equal to 3m where m is equal to 3k square right so it's all the form of 3m now b case b is 3k plus 1 whole square is equal to 9k square plus 6k plus 1 a plus b whole square is a square plus twice of a b right so hence you can write this as 3 into 3k square plus twice of k plus 1 and hence you can write 3 times m plus 1 where m is equal to 3k square plus 2k right and case c is 3k plus 1 whole sorry 3k plus 2 whole square so if you square it let me write it a little better yeah so let us say it is 3k plus 2 whole square squaring it you'll get 9k square plus 6 to the 12k 12k why 3 into 2 into 2 6 into to 12k plus 4 which can be written as 9k square plus 12k plus 3 plus 1 which can then be written as 3 into 3k square plus 4k plus 1 plus 1 which again can be written as 3 into m plus 1 right so hence in all the cases we saw either it is of 3m form or it is of 3n plus 1 form or again it is of the form 3m plus 1 okay so this is how you should be approaching all these problems and with this we come to the last slide of our this thing so I don't have any further questions on them on real numbers so I hope we could cover all the important topics of real numbers my suggestion to you would be all these questions what we have solved around eight questions we solved today you just solve them once again in boards I have a reasonable amount of confidence that questions will be of same nature there will not be any question which will be beyond this but nevertheless don't neglect any other topics especially the world problems so try and solve them as well some some of them in the last few years I have not seen a problem a world problem coming from this real numbers chapter word problems are mostly from typometry quadratic equations or pair of linear equations or you know your mensuration chapter but mostly real numbers word problems I have not seen as I told you earlier either it will be of the form of proving something to be real or non real or you know is rational sorry real and irrational not real and non real so whether you have to use for example the cases we just dealt here with problems like these given that root three is an irrational number prove that two plus root three is an irrational number and problems like these as well where you have to simplify and see whether this is a rational or irrational or using the application of division lemma so like that what is this it's cf of smallest prime number and the smallest composite number you can expect up one marker to marker on at max three market question from real number chapter okay so we will stop here if you have any query I will be more than happy to take it up and if you don't want to ask it here you always have my number you can always reach out to me offline and if you have any other queries here please post it so that we can discuss otherwise we will call it a day I hope the session was useful for you guys and yes so if you have any other question either you can just post it here or you can always reach out to me a little later is there anything you want to discuss now you can always post it here in the comment section so I don't think there is any doubt or any other question which you guys are going to put it here so thanks guys for watching this session and this will be edited and put it uploaded on the youtube so in case you want to have a slow paced revision again you can always log on to our channel youtube.com slash center academy and you can always go through these sessions over and above that we are also posting a lot of previous year questions video solutions so please make sure to go through them so you can subscribe to the channel so that whenever there is a new solution posted you will get a notification so thanks a lot guys and all the best see you in the next class