 So, we will look at this in another way, we will take a bit of a digression and prove a more general result, but that result will stand us in good stead many lectures later when we deal with the second part of the course on Eigen values and Eigen vectors, but all in good time. So, if you permit me to take a seeming digression for a bit, then we shall look at something called Bezou's identity, also known as the Arya Bhatta Bezou identity, what does it say? It says that suppose for a, b belonging to the set of integers, if GCD the greatest common divisor or you might have come across this as the highest common factor HCF, the GCD of A, B is equal to G, then do you know what this identity says? It says that there exist x, y in the set of integers such that A x plus B y must be equal to G, you will always be able to find two integers x and y such that A x plus B y will lead you to that greatest common divisor of those two numbers. We will try to prove this and then we will see how we can use this to in fact prove the second part of this, that if it is a prime number then indeed Zn, if n is a prime number then Zn with those modulo operations is indeed an integral domain and therefore, a finite field. So, let us again try and see a sketch of this proof. So, since this result is by now known, let me take the liberty of erasing this. I am doing this Bezou's identity because later on as I said in the course we shall be using this for polynomials, now we are doing it for numbers, but maybe I will leave that to you as an exercise to prove it analogously exactly in the same fashion. Once you know how the mechanism fits in, you should be able to prove this for polynomials too, but we will come to that much later. So, here is again a sketch of the proof. Look at the set S which is a collection of all numbers that are positive and that can be represented as Ax plus By, in fact you can give it a name S, A, B because A and B are fixed Ax plus By for x, y being integers, positive negative both allowed, you should divide sorry, no if it is the greatest common divisor it divides both A and B and it is also the greatest common divisor. No, so what are you talking about this result, this equation, we will show that, we will show that, so that is the condition for any divisor, but then if it is a greatest common divisor if you have cooked it up in this manner, then this is indeed going to be the greatest that will be the last part of our proof, we will come to that. So, now we are looking at this set to start with, you can of course approach this proof from multiple aspects, you can actually carry out Euclid's division algorithm, I will just we will leave it to you to read that up and try as an exercise, but however this proof if you follow that is also fine. What we have to assure ourselves is that this set is after all non-empty, is it true that this set is non-empty, is it trivial to show, because this has to be a set of positive numbers. So, suppose A or B any of them is negative, can I say that again that dangerous word in mathematical proofs, obviously mod A mod B mod A plus mod B belong to S AB, is it obvious, it is not obvious, well you can choose one of them to be 0 and the other to be plus or minus 1 according to whatever the sign of A is, if A is negative choose X to be minus 1, Y to be 0, if A is positive choose X to be 1, Y to be 0, that way you have mod A, by the same token you will also have mod B if you flip it with Y and X and similarly you choose suitable plus and minus depending on the signs of A and B and you will have this, yeah. So, these are definitely elements, I have shown you more than one element in that set, yeah. So, S AB is non-empty and contains only positive integers, true, you agree, good. Next, if there is a set of positive integers by something called the well-ordering of these numbers, there must be a minimal entry, a minimum, the smallest entry in this set, yes, I am not allowing negative numbers, otherwise they could go to minus infinity, but I am only allowing positive numbers you see, so therefore, there exists S 0 such that it is S 0 is the smallest, so this smallest makes sense, right, because of the ordering among integers, smallest integer in, so that is why I needed the non-emptiness, if you have an empty set and you are cooking up a huge construction and then suddenly someone punctures the balloon and says, hey, you are talking about an empty set, all this construction is gibberish, so at least I should be sure that it is non-empty, that is why I checked if it is non-empty. Once I have assured myself it is non-empty, it makes sense to talk about a smallest entry in that set of positive integers, right. So, S 0 is a smallest entry, what else can I say? Because S 0 belongs to this set therefore, S 0 can also be written as some A X 0 plus B Y 0, ok, follow very carefully now, the next step. Also I or rather we may write and this is very important, A is equal to some quotient times S 0 plus some remainder with remainder strictly smaller than S 0 and greater than 0, just take time to think about this, it is a very important assertion, but if you follow this point, it is almost done. See, the only problem is that this A can be negative, if it is positive then it is just obvious, you know, you can divide any positive number by another number and you get a remainder, it is a no-brainer, right, only if it is negative. So, even if it is negative, is it not possible for us to find a remainder like this? Let the quotient be negative, S 0 is positive, but let A, if A is negative then the quotient is negative and you want R to be positive. So, you have to be careful enough to choose that R to be positive, you can, for example, you can take say minus 10 and divide it by 7. So, you can write it as 7 into minus 1 minus 3, but that is not what I am asking for here. What I am asking for is 7 into minus 2 plus 4, but notice in either case, 4 is less than 7, right. So, I can always write this, please do convince yourself that there is nothing fishy going on here and that it is perfect. Once you grasp that, what am I going to say about R? What can we say about R? This means that R is going to be equal to A minus Q naught S naught is equal to A minus Q naught and what is S naught? A X naught plus B Y naught, sorry, not Q naught, right. Oh, I have called it Q naught. That is actually Q dot, right. Yeah, sorry, I confused with my own notation here, but what is this? Can you follow this? Is it visible from the back? Okay. So, what I can then write is this is 1 minus Q X naught times A 1 minus Q X naught times A minus Q Y naught times B minus Q Y naught times B. What is this form? Sum A X plus B Y. So, therefore, what can I say about R? R must belong to, belongs to S unless, but look, R I have claimed is smaller than S naught, the way I have constructed this and I know that S naught is the smallest number in the set S. Now, if R also has to belong to that set S, then it cannot be true, right. That is a contradiction. So, that is a contradiction, because if R exists in that set, then it must be the smallest, but I have already said that R is smaller than S naught and S naught is the smallest. So, therefore, this is a contradiction. So, what is the only possibility? R is positive or sorry, non-negative. So, the only possibility is here when I have considered R to be greater than 0, I must have missed out the possibility that R is also allowed to be 0. The only way that R cannot belong to S AB, this is S AB, right. Yeah, the only way that R does not belong to that set S AB is if R is equal to 0. So, this implies, no, it cannot. If it were equal to S naught, then it is 0, because then it would be taken in as a factor no remainder left, right. It has to be, because of the rule of division, it must be smaller. So, this of course, is an afterthought. Here, I could not have written this. This is ok, maybe that is creating a confusion. So, the only way this is possible is, only possibility is that R is equal to 0. But what is the consequence of R being equal to 0? What does that tell me? If R is equal to 0, then S naught divides A. Yeah, this implies S naught divides A. This is a notation for S naught divides A. Again, I am not going to repeat this, but starting from the same step. Similarly, we can show that S naught divides B, right. The same argument just start here with B. So, S naught divides B. So, S naught divides A and B. This combined result tells me, S naught is a common divisor of A comma B. Now, all that I have to show is that it is indeed the greatest common divisor. But that is again very straightforward. I am just going to use this narrow margin area for this. Suppose, C is another common divisor of A comma B, which implies A is equal to C alpha, B is equal to C beta for alpha beta being integers. But now, what do I know about this? This implies S naught is equal to C alpha A X naught plus C, sorry, C beta Y naught is equal to C times alpha X naught plus beta Y naught implies what? C divides S naught. So, anytime you find another common divisor of those two numbers A and B, that must also divide S naught and that is the definition of a greatest common divisor. One, it must be a common divisor, which is what we have shown here, that it is a common divisor. Now, once it is a common divisor, if you find any other common divisor, yeah, that common divisor must divide this greatest common divisor, which is what we have now shown. So, therefore, indeed the result that we have here, okay, it is that main result. But this is another way. We have not yet completed the main result. This is just a proof of Bezou's identity. But how is this Bezou's identity going to be used to prove the latter part of the original result, which we started with? Which is that if you have a prime number of numbers like 0 to n minus 1 where n is prime, then every element must have a multiplicative inverse. Once I believe in Bezou's identity, that becomes very straightforward. So, now back to the original result. Yes? This one? So, suppose you agree that this S naught divides both A and B. So, what is the definition of a common divisor? For example, you have A 4 and 24. 2 is a common divisor, but it is not the greatest common divisor. 4 is the greatest common divisor. So, the property of a greatest common divisor is it is a common divisor. But if there is any other common divisor, it is greater than the common divisor, which is to say that the common divisor, other common divisor must divide the greatest common divisor. So, let us assume that there is another common divisor, which is C. We have to show that the C is a factor of S naught or C divides S naught. So, I have written that if C is another common divisor, then A is written as C alpha, B is written as C beta where alpha and beta are integers correct. But S naught because of the way I have defined it here it is an element of that set SAB. So, S naught can be represented as C times alpha X naught plus C times beta Y naught. Then you take the C common out which means it is C times something else and this is also an integer. That means, C also divides or C is a factor of S naught. So, any other common divisor that I find out in the form of C must be a factor of S naught. And therefore, S naught the way I have figured it out from there must be not just a common divisor not just A common divisor, but B greatest common divisor. So, that completes the proof of Bezos identity here. But how is this Bezos identity useful to us in proving that main result? So, here how? Suppose you take consider any element in Z n where n is prime. Then so, any element say alpha suppose we have alpha in Z n where n is prime. Then what can you say about the GCD of alpha and n? n is a prime number alpha is a number smaller than n. It is one that is why I am using the primeness of that number that property earlier we were showing it by division modulo. Now, we are using it here. So, remember in a normal case it would be assume that you know Bezos identity. So, then I would have started from here. So, do not think that the second method of proof is a long winded method only thing I have only proved a very useful result Bezos identity that is why it took longer otherwise the starting point of the proof would have been here. So, what am I saying here? This means based on Bezos identity there exists X comma Y in a set of integers such that when I say set of integers I do not necessarily have to confine myself to this, but I can take the modulo operation later does not matter right such that alpha X plus n Y is equal to 1. Can you guess what I am going to do next? Let us take taking modulo n on both sides. What does this turn out to be modulo n? 0 because it is a multiple of n right. So, then I have alpha X is equal to 1 both sides modulo n what does that mean? Therefore, any alpha that I have picked out I will find some X such that alpha times X is 1. Therefore, X may be modulo n is equal to alpha inverse. So, take any maybe I will just add non-zero that is it. So, if you have a prime number n and you want to show that it must have a multiplicative inverse if n is if you pick out a number smaller than n in that set and this is the proof right. So, you have seen at least two methods of showing that proof the less second one maybe a little more circuitous because we have done this Bezos identity, but please keep this at the back of your mind for at least the next couple of months which is when the course will be running because we will be revisiting this Bezos identity again. In the next lecture we shall talk about something a little more interesting because it is going to be related to directly to the problem that we have discussed so far which is solving X is equal to b just to give you a preview of what is to come. And so far we have looked at these row reduced echelon form and we have tried to see when solutions exist in terms of the fact that this B coming in in the augmented matrix tampering with the rank of the overall matrix that is original matrix A right. Now, we are going to think of the solution as at least in the three-dimensional Euclidean space you might think of it like this if you have a matrix A think of its columns as A 1, A 2 till A n and think of these unknowns as X 1, X 2 upon X n and let us say this is some vector B. Immediately you will notice that A i comma B are both coming from the same vector space R m. What it means pictorially is that if m is 3 at least I can draw the following and I can say that take these vectors this is a vector this is another vector maybe take another vector like so likewise and let us say there is this vector which is say B. What I am essentially asking for in this question is can you find me scalars which when scaling these different different columns of the matrix A lead to the resultant vector which is on the right hand side. This is a second way of viewing that same question and it is from this perspective that we shall see that construction such as those of a vector space allow us to handle this question in a more sophisticated way in the sense that allows us to generalize this problem to not just looking for scalars over n tuples of numbers but to more abstract vector spaces as such. Side by side we will also see the utility of defining this field well enough to another allied problem with the second part of the course that is diagonalizability of a matrix through some similarity transformation or some basis change but those are still fancy terms for us we have not defined any of them yet but that is why we will take off from in the next lecture. Thank you.