 Welcome back. I am now going to really give you the proof of the fundamental theorem of arithmetic. We have been talking about it in so many of our previous lectures and only now we are going to see a proof. The proof really required all the machinery that we have developed so far. You will see that the previous lemma that we have proved that whenever a prime divides product of 2 natural numbers then it should divide one of them. This lemma is very useful in the proof we will see that and further in the proof of the lemma itself we use the notion of the GCD which used the notion of the division algorithm and all that. This is how typically mathematics works when you have a nice result to be proved. There is often a very nice theory that one builds up to prove this and therefore while having the proof of the theorem as a bonus we also several we also have several interesting concepts a very nice theory. So let us begin with the statement of the fundamental theorem of arithmetic which is here in the slide. It says every natural number n bigger than 1 admits a unique factorization n equal to p1 power n1 p2 power n2 dot dot dot pk power nk where you have that p1 p2 pk are put in the increasing order. These are primes they are put in increasing orders and the powers are natural numbers. So there are two parts that we need to prove to prove this theorem. We need to prove existence uniqueness. We need to prove that such a factorization exists for every natural number n and once we have proven the existence we will need to prove that the factorization which we proved to exist is unique. So let us start we will use the method of induction quite liberally. So let n in n be fixed we fix a natural number n. Now there are two possibilities either so we also assume that this n is bigger than 1. So there are only two possibilities n may itself be a prime or it will have a prime factor. So we have already seen that every integer n bigger than 1 has to have a prime factor. If the if n itself is not a prime it will have a non-trivial prime factor of factor of prime p which is not equal to n. So if n is a prime then n equal to p this proves the existence part. If n is a prime then we have written n as p power 1 which has proven the existence part. Let us go to the second case if n is not a prime has a prime factor say p with 1 less than p less than n hence n by p while it is bigger than 1 because p is less than n and n by p is less than n. So we found a new natural number n by p which is not equal to 1 it is bigger than 1 but it is strictly less than n. Here we are talking about existence of factorization. The existence of factorization holds for n equal to 2. The existence of factorization holds for n equal to 2 where you will write n as 2 into 1 and the previous part n as n by p into p where this n by p is first of all bigger than 1 but is less than n. You can apply the induction hypothesis to n by p. So by induction hypothesis n by p admits a prime factorization okay and then you simply multiply by p then n is p 1 power m 1 p l power m l into p. Once we have such a factorization you can easily arrange these primes into increasing orders and then we are really done. Once we have a factorization as a product of primes order in factors the increasing order. So the factorization that we wanted for n as p 1 power n 1 p 2 power n 2 dot dot dot p k power n k can be easily arranged now because we have p 1 after this arrangement we have p 1 less than p 2 less than dot dot dot up to p k and the powers are of course natural numbers because the powers for the n by p where natural numbers if your prime p is a new prime if it was not 1 of the p 1 p 2 p l then the power of p will be 1 which is a natural number if p was one of them then you will simply increase one of those m i by 1. So we have a factorization as we had desired we now have to go towards the uniqueness. Now we go towards the uniqueness part. So what do and what does one mean to have uniqueness part we have proved that the factorization exists that means at least one way to write n as product of primes exists but it can of course happen that if I do a factorization in Mumbai I may have a factorization and if somebody else does a factorization somewhere else then the person may get different prime factors and may altogether obtain a different factorization. What we have to prove that such a thing cannot happen this is what is abstract about mathematics that a statement of the theorem should be true independent of the person independent of the place independent of the day independent of the time indeed independent of everything else it should simply depend on the assumptions the statement of the result if true should not depend on anything else. So we now have to prove that whenever there are two such factorizations then the primes occurring in both the factorizations are the same and then we have to prove that the powers are the same this is what we need to prove. If we have n as p1 power a1 pk power ak and q1 power b1 ql power bl then we must we also of course assume that p1 is less than up to pk and q1 is less than up to ql. Then we must prove that k is l, pi is qi and ai is vi quite a lot to prove. We should prove that the number of primes which occurs in both the factorizations is the same the number of primes occurring in both the factorization is the same. So we will need to prove that k is equal to l indeed this number can also be different. We need to first show that these two numbers are same then on both sides we have the same number of primes p1 p2 pk q1 q2 qk and we have of course ordered them in the increasing order. So we will then have that p1 is equal to q1 we will of course need to prove this p1 is q1 p2 is q2 so on up to pk is qk and after that we will prove that a1 is b1 a2 is b2 up to ak equal to bk. But we will not use all these things we will simply appeal to induction how do we do that so we will need to prove the very first step of induction which is that if n equal to 2 then there is a unique factorization. So we observe that n equal to 2 power 1 is the only factorization in terms of we observe that n equal to 2 so here of course we need to say for n equal to 2 for n equal to 2 we observe that n equal to 2 into 1 is the only factorization of 2 in terms of primes maybe there is a better way to write this for n equal to 2 we observe that n equal to 2 power 1 is the only factorization of 2 in terms of primes. Let us just think about it and let us just discuss how this is the only factorization. Can 2 have any other factor can 3, 5, 7, 11, 13 all the other primes can these be factors of 2? No these are all primes which are bigger than 2 remember here we are talking about a very explicit example n equal to 2 and so we can use all the things that we know up to now so we have known that whenever a divides b a has to be less than or equal to b. But the other primes are in fact odd primes and they are all bigger than 2 3, 5, 7, 11, 13 and so on so they cannot divide 2. So the only prime that can divide 2 is 2 can it come with any other power can it come with a power 2? No because 2 square is 4 and then you have gone ahead if you were able to write n which is 2 as power of 2 with any higher number then you get a contradiction because all higher powers of 2 are bigger than 2 they cannot divide 2. Therefore n equal to 2 has only one factorization which is n equal to 2 into 1. So after this we are going to use induction remember once again that we have assumed that n equal to p 1 power a 1 p 2 power a 2 p k power a k is one factorization for n and q 1 power b 1 q 2 power b 2 q l power b l is another factorization. We want to apply induction we have just now observed that the beginning step of the induction is done. So once we reduce the case of n to anything smaller than n but bigger than 1 then we apply induction hypothesis to that and get our result. So this is what we are going to do. So since p 1 divides n because you were able to write n as p 1 into some natural number p 1 divides n we have that p 1 divides q 1 power b 1 q 2 power b 2 dot dot dot q l power b l. So p 1 divides the product of these finitely many integers. What did we prove in the lemma in the last lecture we proved that whenever p divides whenever p is a prime and p divides product of two natural numbers it should divide one of them. So here I have the following thing that p the p 1 that we have started with which is a prime should divide one of the prime power that we have here. So using the previous lemma we get p 1 divides q 1 power b 1 or p 1 divides q 2 power b 2 up to q l power b l. One of these two should hold because I will write m as q 1 power b 1 and n as the remaining prime powers product. So p 1 must divide q 1 power b 1 or p 1 should divide the product of the remaining prime powers. In the first case if p 1 divides q 1 power b 1 which I will write as q 1 into q 1 power b 1 minus 1 then p 1 divides q 1. In fact here we apply the lemma again to get that p 1 divides q 1 or it will divide q 1 power b 1 minus 1. If it divides the latter you apply the lemma again to say that p 1 divides q 1 or p 1 divides q 1 power b 1 minus 2 and so on. Ultimately you will reach a state where p 1 must divide q 1. So we have that p 1 which is a prime. So p 1 is bigger than 1 divides q 1 which is another prime. Now q 1 being a prime cannot have any non-trivial factors, p 1 is not 1 so p 1 must equal q 1. This is the conclusion that p 1 is equal to q 1. We started with the least prime dividing n in the first factorization that was p 1 and we started with the least prime in the second factorization and in one case we have proved that p 1 is equal to q 1. What is the case? The case is here. The case is this case. In this case we have proved that p 1 must be equal to q 1. Now it is quite possible that this case occurs that p 1 divides q 2 power b 2 dot dot dot q l power b l. This can also occur. Then what do we do? If p 1 divides q 2 power b 2 dot dot dot q l power b l then as above p 1 divides q i power b i for some 1 less than i less than or equal to l. How do we see this? We will again write this product q 2 power b 2 up to dot dot dot up to q l power b l as q 2 power b 2 into the remaining things. Now p 1 should divide q 2 power b 2 or it will divide the remaining products. If it divides the remaining products you again write it as q 3 power b 3 into the remaining products. By doing this a finitely many times ultimately we reach that p 1 must divide q 2 power b 2 or q 3 power b 3 or q l power b l all the way. It should divide one of the q power b. But then as we have seen in the previous case again p 1 will be equal to that particular q i because p 1 must divide q i and p 1 being a prime is bigger than 1, q i being a prime cannot have any non-trivial factor. So, p 1 must be equal to q i. But this is a contradiction because what we have done by taking with one prime factor in one decomposition going to the other factorization can also be done in the other way. So, this says that q 1 is strictly less than p 1 which is further strictly less than p 2 and so on up to p k. Remember p 1, p 2, p k these were the primes occurring in the first factorization. If your p 1 which is the smallest prime occurring in the first factorization happens to be equal to a q i which is 2 onwards then q 1 which is smaller than q 2 has to be smaller than p 1 smaller than p 2 smaller than p and so on up to p k. But reversing the argument we will see. So, since q 1 divides p 1 power a 1 up to p k power a k q 1 is p j for some j. Repeating the same argument that we have done for p 1 can be done for q 1 and this gives a contradiction. This gives a contradiction if p 1 is not dividing q 1. So, what we have there observed. So, thus p 1 is q 1 n by p 1 equal to n by q 1. Now n by p 1 n by q 1 this has a smaller factorization. We will in fact have that here we have n applying induction hypothesis call this n 1 gives us the result. So, we have indeed proved that whenever we are given any natural number it can be written as product of primes in a unique way provided you write the factors in the increasing order. Once again let me quickly in very brief go over the proof. There were two parts of the proof existence and uniqueness. The existence theorem part was proved by appealing to induction. So, was the uniqueness part done. So, what is the beginning step of the induction? We observed that 2 has a prime factor decomposition which is 2 equal to 2 into 1 and later we also observed that this decomposition is the unique decomposition for 2 as prime product of primes. Now we go to a general n if the n was a prime you would already have a factorization n equal to p if n is not a prime then you would look at n by p which we call n 1 induction hypothesis gives you a factorization for n 1 multiplying that factorization by p gives you a factorization for n. So, the existence of factorization is proved using the induction in this way. Now we want to prove the uniqueness. So, for the uniqueness we assume that there are two factorizations p 1 up to p k power a 1 up to a k q 1 up to q l power b 1 up to b l then we start with the smallest prime dividing n on the left hand side factorization we prove that it has to be equal to one of the primes on the other side and since this can be done on both sides we must have that the smallest one p 1 has to be equal to the smallest one q 1 since they are same cancel them out you get n 1 which will have a reduce which will have a smaller factorization appeal to the induction hypothesis to get that k equal to l a i equal to b i and p i equal to q i and therefore we have the proof of uniqueness using the induction method. Induction method is a very powerful method and we have seen one very good application of the method. So, the fundamental theorem of arithmetic is proved in the next lecture we will see some comments about primes and then we will go to the theory of congruences which is going to be the next part of our course. Thank you and I hope to see you again.