 Welcome back. We are studying the properties of the convergence. We have proved that any real number theta has a continued fraction expansion constructed in a very natural way. Once you cut that continued fraction expansion at any finite stage, we get rational numbers, we call them convergence. They are denoted by the convergence P n upon Q n and towards the end of the last lecture we proved that the convergence to theta are the good rational approximations. So, in fact, these convergence satisfy the inequality that theta minus P by Q is less than 1 by Q square. The convergence P n upon Q n satisfy the inequality that theta minus P by Q is less than 1 upon Q square. We are going to prove that these are the best rational approximations, but before we prove that these are better rational approximations. We know what is a good rational approximation. Any rational P by Q to theta satisfying theta minus P by Q is less than 1 upon Q square. The difference is less than 1 upon, the difference is less than 1 upon the denominator square. These are called good rational approximations. And we are going to prove that our convergence are doing something better. So, these are not just good rational approximations, but they are better rational approximations. So, the result that we are going to prove is the result as follows. We have noted that our convergence are the good rational approximations, but we are going to prove that these are the better rational approximations in the following sense that if you take any P n upon Q n and P n plus 1 upon Q n plus 1. So, take these pairs of successive convergence, then one of them should satisfy that theta minus P by Q is less than 1 upon 2 Q square. So, we have already noted that the convergence are close enough to the real number theta in the sense that the difference is less than 1 upon Q square. But we prove that one of these two is going to be much closer to theta. The difference is going to be less than 1 upon 2 times Q square. This is what we are going to prove in this theorem. So, one of the two P n upon Q n and P n plus 1 upon Q n plus 1, one of these two should satisfy this better approximation. This is what we are going to do. Note here that P n upon Q n and P n plus 1 upon Q n plus 1, these are on two different sides theta. So, what does it say? We have these two numbers which are on two different sides of theta. That means, if I subtract each of them from theta, then for one of them we are going to get a positive quantity and the other will be a negative quantity. The difference will have two different signs. The differences of these two successive convergence from theta will have two different signs. This is the property that we are going to have. So, if one of them is positive, the other is negative. That is what we have. So, we now take their sum and put a mod. That is what we are going to do theta minus P n by Q n plus mod of theta minus P n plus 1 upon Q n plus 1. We consider the sum of their mods. Now, this quantity can be positive or negative. Therefore, the mod will be equal to that quantity or its negative. So, you have that this is plus or minus theta minus P n over Q n and this quantity is also plus or minus its modulus. But if this is positive, then this has to be negative. And if the first one for the nth one is negative, then the second has to be positive. Therefore, when we take the first one to be positive, the second is negative. First one to be negative, second is positive. So, this will give us, we take the plus minus sign common and we get theta minus P n upon Q n minus theta plus P n plus 1 upon Q n plus 1 which gives you mod P n by Q n minus P n plus 1 upon Q n plus 1. And we note that this is equal to 1 upon Q n Q n plus 1. This is something that we have already denoted because the numerator here will be P n Q n plus 1 minus Q n P n plus 1 that is plus or minus 1. So, under the mod it is going to be 1 and the denominator continues to be Q n into Q n plus 1. So, we have that this sum which is the sum of the two moduli is equal to this particular positive number. This is the thing that we have proved. Further we use one inequality for any alpha beta which are distinct we have 2 times alpha beta is less than alpha square plus beta square. This can be seen because the if you take 2 alpha beta to the other side of this inequality you are going to get alpha minus beta whole square but alpha is not equal to beta therefore alpha minus beta is a non-zero quantity. It can be positive or negative but its square will always be positive. So, we therefore get that 2 times alpha beta is always less than alpha square plus beta square. We use this to get our final inequality by putting the alpha to be 1 upon Q n and beta to be 1 upon Q n plus 1. So, we get that theta minus P n upon Q n in the modulus plus theta minus P n plus 1 upon Q n plus 1 in the modulus is strictly less than 1 upon 2 Q n square plus 1 upon 2 Q n plus 1 square. Now what we want to prove is that at least one of the pairs P n and Q n should satisfy this inequality. If you assume that this is not true that means P n by Q n and P n plus 1 by Q n plus 1 both do not satisfy this inequality. That would mean that for both of these we will have that this is bigger than or equal to 2 Q n square and this is bigger than or equal to 1 upon 2 Q n plus 1 square. So, the sum will be bigger than or equal to the sum of these 2 but that is a contradiction because it is strictly less than the sum. So, if you have 2 numbers if you let us say have 4 numbers and you have a plus b is less than c plus d then you should have that a is less than c or b is less than d. If you have a bigger than or equal to c and b bigger than or equal to d then a plus b is going to be bigger than or equal to c plus d. So, when you have the sum of 2 positive numbers to be strictly less than sum of 2 other positive numbers then at least one of the 2s on the left hand side should be less than at least one of the 2s on the right hand side. So, this is the inequality that we get and therefore at least one of the pair of successive convergence should satisfy this even better inequality. So, the convergence give us a better approximation to the real number theta that we began with. So, we proved first of all that convergence to theta which we have obtained through a natural continued fraction expansion. The convergence give good approximation but they satisfy even this better result therefore they are better approximations. Of course, you have that one of the 2 will have this property. So, whenever you take p0 by q0 p1 by q1 one of these 2 should satisfy this better one p1 by q1 and p2 by q2 again one of these 2 should satisfy and so on. So, you get a sequence of rational numbers converging to your theta giving you a better approximation than many good approximations. We can improve this result even further. We can in fact prove that the 2 that we had in the denominator with q square can be replaced by root 5. So, let us see what we have done so far. We have proved that there is a sequence given by the convergence satisfying that theta minus p by q is less than 1 upon q square. This is true for all convergence. Then we proved that one of the pair of successive convergence should give even better approximations. And now we are proving that if you take the triples pn by qn pn plus 1 by qn plus 1 pn plus 2 by qn plus 2 then at least one of these should give even better approximation that these 2 can be further increased to be to root 5. So, at least one of these three successive convergence should satisfy this very strong inequality. That means these convergence are really close to the real number theta they are not less than just 1 upon q square not just less than 1 upon 2 q square but one of them has to be less than the distance from theta has to be less than 1 upon root 5 q square. So, this proof is slightly more complicated than the earlier proof which we proved for 2. So, we are going to skip this proof for the moment. But this proves one result which was proved by Hurwitz way back in 1891. So, let me just quote this result for you. It says that if your theta is an irrational number. So, this theta is an irrational number then there is a sequence of rational satisfying this very strong inequality. This was proved earlier by Hurwitz and what we have discussed so far if you prove this result it will give us another proof of Hurwitz's result. Now, you may ask can you make this root 5 even better? Is it possible to increase this constant root 5 and maybe get something better than this? And you may take one of the 4 consecutive or one of the 5 consecutive or whatever you want. Can you improve upon this constant root 5 that we get? And mathematics is interesting in both the ways. If you could improve this that would be even better or if you could prove a result saying that no, you cannot improve further this root 5 is the best constant that you get then that would also be an interesting result. Here we have that this constant root 5 is the best possible constant that we have. So, let me just explain what we mean by this. This means that consider the following statement. If theta is irrational then there is a sequence p by q of rational numbers satisfying mod theta minus p by q is less than 1 upon c q square. Suppose we prove such a result for some c. So, let us call this statement c then the statement holds for every c up to root 5. Of course, we have proved it for root 5 and therefore, it holds for every smaller c up to root 5 but it does not hold for any c bigger than root 5. And what does it mean to say that it does not fold for anything bigger than root 5? It means that there is one particular irrational for which the statement c bigger than root 5 does not hold. Because our statement is for all irrationals if you wanted to fix a c bigger than root 5 and you wanted to see whether that statement holds my claim is that the statement does not hold. So, I should give you an irrational for which that statement for a c bigger than root 5 should not hold and we are going to give that irrational. I will give you the irrational right away but to prove that the statement for c bigger than root 5 does not hold for that irrational will have to wait. So, we have not computed any nice genuine continued fraction expansion yet. Let us do that and compute the convergence or the continued fraction expansion to the golden ratio. Let me explain the term golden ratio here for you first. So, this was the question which was posed by some of the ancients to the other of the ancients that if you take a rectangle suppose of the sides the side is a and b and if a is let us assume that a is less than b and you cut out this square. So, this is length a and this is b minus a b is taken to be bigger than a. So, the remaining is a positive quantity. Here we get another rectangle then we assume that this rectangle the new rectangle is congruent to the earlier rectangle. That means the sides of these new rectangle their ratio which is a upon b minus a is congruent. So, the ratio is same as the new ratio which is b upon a. So, remember here b is this full length and so the question is what can this ratio be if I wanted to construct the rectangles with this property that when I cut out a square obtained by taking the smaller side. So, we put this thing here therefore this is equal to the length a and take it out take that square out the remaining rectangle should be congruent to the earlier rectangle. That means you expand the rectangle by a fixed quantity that means you expand both the length and breadth of the new rectangle by a fixed quantity multiply that by a fixed quantity and you get your original rectangle. Then what should the ratio be and the actually the question is does it hold again for the smaller rectangle. Now, here b minus a is smaller than a. So, you cut this square from here and what you have obtained here is a smaller length. So, we have this new rectangle it will turn out that this new rectangle will also be congruent to the first rectangle which we had obtained and also congruent further to the original rectangle. Then the question is what this ratio can be and this ratio turns out to be 1 plus root 5 by 2 and therefore it is called golden ratio it is very useful in many geometric configurations. In fact, there are these geometric polyhedral the regular solids and in their construction also this golden ratio is very useful you should perhaps see and search on internet for more material on this we are not going to spend more time on this information about golden ratio. But we will try to now compute the continued fraction expansion for our theta which is 1 plus root 5 by 2 theta is 1 plus root 5 by 2 we want to find the continued fraction expansion. So, we should compute a 0 first which is the integral part of theta. So, that should be the largest integer which is less than or equal to the theta how do we compute it. So, note that our root 5 sets between 2 and 3 root 5 is strictly bigger than root 4 and is strictly less than root 9. So, this is between 2 and 3 we add 1. So, we get that 3 is less than 1 plus root 5 less than 4. So, at least the numerator of this quantity is between 3 and 4 divide the equation by 2 and you get 3 by 2 is less than 1 plus root 5 by 2 is less than 2 which is 4 by 2. And of course, we have that 3 by 2 is strictly bigger than 1. Therefore, a 0 is equal to 1. So, we have that our theta which is 1 plus root 5 upon 2 is 1 plus 1 upon theta 1 where theta 1 is certainly bigger than 1. But we should be able to compute the quantity theta 1 from whatever we have obtained so far. So, theta 1 its reciprocal is 1 plus root 5 upon 2 minus 1. So, this is 1 plus root 5 by 2 minus 1 reciprocal. So, we compute theta 1 we are going to simplify this this becomes 1 plus root 5 minus 2 whole thing divided by 2. So, you have 2 in the numerator and root 5 minus 1 in the denominator. So, this root 5 has gone in the denominator. We want to simplify it further by bringing the root 5 to the numerator and the standard trick to do that is to multiply both sides by what is called the conjugate of the denominator. So, numerator is 2 times 1 plus root 5 and denominator is of the form alpha minus beta into alpha plus beta. So, that is alpha square minus beta square alpha square is root 5 square. So, that is 5 minus the square of 1 is 1. So, this is 4 we have 2 in the numerator we get it to be 1 plus root 5 upon 2 which is the theta we started with. So, theta 1 is equal to theta. Therefore, theta which was 1 plus 1 upon theta 1 is in fact equal to 1 plus 1 upon theta. And we just keep putting these values and we get the continued fraction expansion for the number theta. So, this is going to be 1, 1, 1, 1 dot dot dot. The continued fraction expansion for the golden ratio is indeed one of the most beautiful continued fraction expansion that we should have which is 1, 1, 1, 1, 1 and since these ones are going to just continue as they are it is customary to denote it by putting a bar on the top of the 1. So, whenever you have an expression in the continued fraction expansion where the some certain subset of the integers that you get keeps repeating after a stage then you put a bar on the top. Of course, here we have taken the first one out because that was the integral part but you may also write it in this way. So, the golden ratio has this very beautiful continued fraction expansion. It is simply given by 1 plus 1 upon 1 plus 1 upon 1 plus 1 upon 1 dot dot. You can also check it by observing the quadratic equation that this golden ratio satisfies that will also give you this particular continued fraction expansion but you have to be slightly careful that quadratic equation will have two roots being a quadratic equation it will have two roots one of them is the golden ratio which is a positive quantity the other is going to be a negative number. In fact, it is one minus root 5 upon 2 that is a negative number and that does not quite have this particular continued fraction expansion. So, do not get confused although we will it may seem that you are using the quadratic equation to get this expansion it does not mean that any root of the quadratic expansion any root of the quadratic equation gives you the same continued fraction expansion. Here we have that our quadratic continued fraction expansion consists of only positive numbers and therefore it is going to converge to a positive quantity which will therefore be the positive root of that equation which is our golden ratio. I will just quickly tell you one more thing about what are called as Haimachandra numbers or which are more commonly known as the Fibonacci numbers. So, if you wanted to compute the convergence to this 1 plus root 5 by 2 we have noted that this is simply 1 1 1 1 repeated the convergence are the first one will simply be 1 next one will be 1 plus 1 by 1 which are 2 comma 1 and the one after that will be 1 plus 1 upon 1 plus 1 upon 1 which is 1 plus 1 by 2 which is 3 by 2 the next one is let us be patient and solve this out dot dot dot the reason for this is that P n by Q n is going to be 1 plus 1 upon P n minus 1 and here you have Q n minus 1. So, you are going to get P n minus 1 plus Q n minus 1 and P n minus 1. So, the next ones the 2 that you have obtained here is 1 plus 1 the 3 that you obtained here is 2 plus 1 5 is 3 plus 2 the next one would therefore be 8 upon 5 after that you are going to get 13 by 8 21 by 13 and so on and these this sequence of numbers is very commonly known as Fibonacci numbers but you should note that the Indian mathematician Haimachandra had discovered this at least 10 years before Fibonacci and it is said that Fibonacci thought about these numbers by looking at the rabbits and the increasing population of rabbits whereas Haimachandra is supposed to have come to these numbers by observing the matras in the vrutas that we have in classical music and so on. So, Manjul Bhargava the Fields Medalist from Indian Origins has talked many times on this and I urge you to go to YouTube and find search for Manjul Bhargava Fibonacci and you will see video where he explains very nicely how the attributes should really be given to Haimachandra and not to Fibonacci. We stop at this level but we will continue with getting the better and better approximations by the convergence and we will be doing that in the next lecture. See you then. Thank you very much.