 We were discussing the structure of the real number system and in the last class we listed various properties of the real numbers and let me recall that I mentioned one of the most important property of the real numbers is what we had called the order completeness property or also known as LUB action. So, let me state that once again what it simply says is that every non-empty subset of R that is bounded above has least upper bound. And we have already seen that once we say that it has a least upper bound then that least upper bound has to be unique. And by the similar argument we have we can also see that if this is accepted then we can in a similar way using this we can also prove that if a set is non-empty and bounded below then it has the greatest lower bound or what we call infimum. Also the way in which this LUB or supremum comes into picture in various proofs is as follows that is suppose you say that A is non-empty and suppose alpha is the LUB of A. Then what we have seen in the other class is that for every epsilon bigger than 0 the alpha minus epsilon is not an upper bound. So, you can always find the element in A which is bigger than alpha minus epsilon. So, for every epsilon there exists x in A this x may depend on epsilon this x for different epsilon you may need to choose different x. So, if you want you can write x suffix epsilon which is what some books do, but not very important right term. So, there exists x in epsilon such that alpha minus epsilon is strictly less than x this means that alpha minus epsilon is not a lower bound and alpha itself is a sorry alpha minus epsilon is not an upper bound and alpha itself is an upper bound. So, x less than or equal to alpha this will be always true and this one particular axiom is the one of the is the most important axiom of the real number system which distinguishes real numbers from all other number systems. And we have seen also in the last class that any with this axiom real number system becomes what is called a complete ordered field and there is only one complete ordered field which has all these properties. Also the way in which the this axiom comes into picture in the various well done theorems in real analysis. Let me again remind you in the last class perhaps I had mentioned this article by Professor S. Kumarisan role of L U B axiom in real analysis. We shall see some instances of this. In fact what we will notice that is what we want to emphasize is that most of the important properties of the real line either follow directly from this axiom or in some chain of logic that is you prove some theorem using this axiom then something else using that theorem etcetera and at the last stage you prove whatever that particular property. So, every major property of the real numbers depends either directly or indirectly on this particular property. Let us see few instances of that to begin with for example we can show this that the set of all natural numbers n is not bounded above there is no real number which is bigger than or equal to all all natural numbers. So, how does one prove that when suppose n is bounded above we shall see what happens we will get a contradiction suppose n is bounded above by the way instead of bounded above if we had talked about bounded below is that true it is bounded below that part is trivial. So this is only important n is bound suppose n is bounded above then what should happen in view of this n is already non-empty we know several natural numbers. So, it should have a least upper bound it should have a least upper bound. So, suppose n is bounded above then n has least upper bound say alpha suppose I call that least upper bound alpha then use this we know that for every epsilon this should exist some x in a such that alpha minus I will take this epsilon as 1 I will take this epsilon as 1. So, suppose I take this epsilon as 1 I can say there exist x in n such that alpha minus 1 is strictly less than x and this is less than or equal to alpha. Now, take this part if alpha minus 1 less than x from this can we say that alpha less than x plus 1 alpha less than x plus 1. In fact, since this is a natural number let us use a different notation. So, that you will understand what is happening instead of x I will say small n small n belonging to n such that alpha minus 1 less than n. So, this proves that alpha less than n plus 1, but what is n plus 1 n plus 1 is a natural number because n is a natural number and alpha is less than n plus 1. So, that contradicts that alpha is an upper bound. So, that is not possible. So, n is not bounded above then similarly there is another very well known property of real numbers which depends on or which relates national numbers and real numbers it is called Archimedean property. Archimedean this name may have come from this Greek mathematician Archimedes, but apparently he was not the one who discovered this. Anyway, we will not go into those historical points if you those of you who have interest can look at the book. What this property says is the following suppose you take two positive real numbers. So, suppose suppose x y in R or such that 0 less than x and 0 less than y you take two positive real numbers then you can always find a natural number such that n times x is bigger than y that is given any two real positive real numbers you can take any of those positive numbers and multiply that by a suitable natural number. So, that that multiple becomes bigger than the other real numbers that is called Archimedean property then of course that number n may depend on both x and y that number n. So, then there exist a natural number n such that n times x is bigger than y can you see that this follows immediately from this see suppose this is false suppose this is false then what does that mean n s that means n x less than or equal to y x is bigger than 0. So, does it mean that n less than or equal to y by x right see if this is false that is if not n less than or equal to y by x for every n in n that is clear. So, if there is no such n it means that for every n n must be less than or equal to y by x, but can that happen because that will mean that again that will mean that n is bounded above. So, that is not possible now many of you would have seen this proof of this famous thing that root 2 is irrational and how does the proof go suppose it is rational then you say that root 2 is m by n and then you remove the common factors from m and n then you will write that 2 is equal to m square by n square. So, that will give 2 n square is equal to m square etcetera and then since we already. So, this will mean that 2 divides m square that will give that 2 divides m. So, removing 2 again you will get that m is even and similarly it will prove that n is also even by after you remove 2. So, fine that proof is quite well known to you right, but does this say that there exists root 2 as a real number how do you know that there exists a real number see all that this proves is that there is no rational number all that this proves there is no rational number whose square is 2 that is what this proves does it prove that there exists a real number whose square is 2 it does not it does not that. So, that needs a proof separately that there exists a real number which in fact without that even this notation root 2 is meaningless without proving that there exists a real number whose square is 2, even this notation root 2 is very less. And to prove that, you will need this. Let us see how that can be done. So, let us just take that as a theorem. And then we shall subsequently see that there is nothing very particular about 2. But let us prove it for 2, since this is a familiar proof. Let us say, I will say there exists x, in fact one can say something more. There exists x, unique x bigger than 0. This is unique x bigger than 0, such that x square is equal to 2. Of course, I can say it is unique because of this. Suppose I remove that condition, then that is not unique. Because minus x square is same as x square. So, as far as the uniqueness is concerned, there is nothing much to prove. See, suppose there are 2 x and y, we satisfy both. You will get x square is equal to y square. And that will give that x minus y into x plus y is 0. See, suppose there exists x y in R such that x square is equal to 2 and y square is also 2. Then this will give that x minus y into x plus y is equal to 0. Of course, this is important. Suppose there exists x y in R such that x square equal to 2 and y square equal to 2 and both are positive. Now, if that is the case, x plus y is also positive. So, you can multiply both sides by 1 by x plus y and you will get x equal to y. So, if we do not assume that x is equal to 0, then x plus y also can be 0 and that will give y equal to minus x. But if x and y both are positive, x and y must be equal. Now, let us go to the existence. Let us say I will take the set A as the set of all x which are bigger than 0 and x square less than or equal to 2. Is this a non-empty set? Obviously, 1 belongs to A. That is clear. 1 square is less than or equal to 2. So, 1 belongs to A. So, that is non-empty. Is it bounded above? How do you prove that it is bounded above? Yes? No. How do you show that? 2 is an upper bound. Fine. How does 1 prove that? That will only show that 2 does not belong to A. How does it show that 2 is an upper bound? What you say is right? 2 square is 4 and that is not less than or equal to 2. But that only means that 2 does not belong to A. How does it show that 2 is an upper bound? To show that something is an upper bound, what is needed to show? You have to show that if you take any element in x, then that must be less than or equal to 2. And how do you show that? Let us do. Let x belong to A. Then what we know is that these two things we know. x is bigger than 0 and x square is less than or equal to 2. We want to prove that x is less than or equal to 2. If we want to say 2 is an upper bound, we want to prove that. Now, tell me how does one proceed after this? We are not sure any such thing that taking square roots preaches the order. Just use whatever we have said so far about real numbers. Let us let me just give a hint. Once you realize that, that will be clear. See, suppose x is less than or equal to 1. If x is less than or equal to 1, then 1 is less than 2. So, there is nothing to be proved. So, then obviously x is less than or equal to 2. If not 1 is strictly less than x, if not 1 is strictly less than x. If 1 is strictly less than x, can you say that that will give x is less than x square? Remember x is bigger than 0. Remember x is bigger than 0. If 1 is less than x or even less than or equal to 1, this will give that x is less than or equal to x square. We already assume that x square is less than or equal to 2. So, that shows that a is bounded above. Now, use this action, since a is bounded above, it has a least upper bound. Let us call that least upper bound as alpha. Let alpha be an u b of t. Obviously, alpha is a real number. We have said that every non-empty set that is bound above has a least upper bound. This least upper bound is a real number. Alpha is a real number. What we want to say about this alpha? That alpha is or alpha square is equal to 2. This is what we want to say. So, alpha square is equal to 2. Now, we use this property of this least upper bound. We have said that if alpha is a least upper bound of everything, then if you take any epsilon, then alpha minus, there should exist some x in a such that alpha minus epsilon is strictly less than x. We shall choose that appropriate epsilon little later, but let us start with any epsilon. So, let epsilon be bigger than 0. Then, there exists x in a such that alpha minus epsilon is strictly less than x. Of course, this is less than or equal to alpha. We shall choose this epsilon in such a way that this alpha minus epsilon is also positive. That we can always do. It is because suddenly alpha is an upper bound. So, alpha is bigger than or equal to any element here. Alpha is bigger than or equal to any element here. So, in particular alpha is bigger than 1 also. Alpha is bigger than 1 also. For example, if I can choose epsilon as half, alpha minus epsilon will be strictly positive. So, let me say here, let epsilon be bigger than 0 such that alpha minus epsilon is also bigger than 0. This we can do always. So, this is 0 less than alpha minus epsilon less than x. Since, alpha minus epsilon is less than x, now we can say that we will multiply this by alpha minus epsilon, this by x, then this inequality will remain unchanged because both are positive numbers. So, what we will get is alpha minus epsilon whole square is less than x square and less than or equal to this alpha square. So, suppose we expand this, what we will get is alpha square minus 2 alpha epsilon plus epsilon square is less than or equal to alpha square. So, what follows from this? Of course, I want this also. Let me forget about this part. This is strictly less than x square and since x square is in A, we can say that x square is less than or equal to 2. x square is in A, we will say x square and that is what we want to use. x square is less than or equal to 2. Now, does it follow from here that alpha square is less than or equal to 2? Remember, this is true for every epsilon. So, I can make further choice of epsilon such that this number becomes strictly positive. See, this is something I have mentioned right in the beginning. That is, suppose you want to show that something is, suppose you want to show x is less than or equal to y, you can show that x is less than y plus epsilon for every epsilon or which is same as saying that x minus epsilon less than or equal to y or x minus epsilon less than y for every epsilon. This is what we have shown here. Whatever epsilon you take, this is always going to be less than or equal to 2. For every epsilon, that is important. Since this happens for every epsilon, we must have alpha square less than or equal to 2. We can say that since this holds for every epsilon bigger than 0, we have alpha square is less than or equal to 2. Is that clear? Alpha square is less than or equal to 2. One more thing we should not notice right here. Since alpha is an upper bound of A, alpha is bigger than or equal to every element in A and every element in A is bigger than 0. In particular, alpha is bigger than 0. In fact, we have also noticed that alpha is bigger than or equal to 1 actually. So, alpha is strictly bigger than 0. Now, we have proved that alpha square is less than or equal to 2. But what we wanted to prove? We want to prove that alpha square is equal to 2. So, that means what we need to prove next? That is alpha square is also bigger than or equal to 2. Suppose we prove that alpha square is also bigger than or equal to 2, then it will be less than or equal to 2 or which is or other way to saying is that we should show that this inequality cannot be strict. We have already shown that alpha square is less than or equal to 2. Now, suppose alpha square is strictly less than 2. Alpha square is strictly less than 2. We have to show that this cannot happen. Our idea is to show that this cannot happen. Alpha square is strictly less than 2. Then we will do something similar again. Here we have taken alpha minus epsilon. We will now consider alpha plus epsilon. We will choose an epsilon in appropriate way and get some contradiction. So, let us say consider epsilon bigger than 0 and then of course, alpha plus epsilon is also bigger than 0. And then see our idea is basically this. If alpha square is less than 2, we will find some epsilon such that alpha plus epsilon whole square is also less than 2. Is that possible? In fact, that is what we shall verify. Those are the calculations that we shall verify. So, consider a sum epsilon bigger than 0 and alpha plus epsilon is also bigger than 2. Then alpha plus epsilon whole square is equal to alpha square plus 2 alpha epsilon plus epsilon square. Now, alpha square is less than 2. Alpha square is less than 2, which is same as saying that if I take this number 2 minus alpha square, that is a positive number. Now, the question is suppose I choose an epsilon in such a way that this part 2 alpha epsilon plus epsilon square, suppose this is less than this. Then what will be the meaning of this? Then that whole thing will be less than 2. So, let me just say that thing here first. Choose epsilon bigger than 0 such that 2 alpha epsilon plus epsilon square is less than 2 minus alpha square. How this is to be done or whether this can be done or not? That we shall see little later. But suppose this can be done. Suppose this can be done. Then what does that mean? It will mean that then alpha plus epsilon whole square that is equal to that is less than alpha square plus 2 minus alpha square and that is less than 2. That is less than 2 and it means that for this epsilon, what does that mean? That alpha plus epsilon is bigger than 0 and its square is less than 2. That means this means alpha plus epsilon belongs to A and this means alpha plus epsilon belongs to A. Now, with epsilon strictly bigger than 0. Now, can that happen? Alpha is an upper bound of A. So, no number bigger than alpha can belong to A. So, this is the contradiction. So, now only thing remains is that whether we can choose such an epsilon. Now, for this to happen, what must be the case? Anyway, see if we choose any particular epsilon and suppose this inequality that is what we need is this inequality. If that inequality works for any particular epsilon, it will also work for any other epsilon smaller than that. That is clear. Suppose I let us say it works for epsilon equal to half. It will also work for epsilon equal to 1 by 4, 1 by 8 or anything. So, it is because of this that I can assume right from the beginning that epsilon is less than 1. I can assume right in the beginning that I can choose epsilon smaller than 1. So, I can say that we may choose 0 less than epsilon less than 1. Then, the point of choosing this epsilon is less than 1 is that this will also imply that epsilon square is less than epsilon. If epsilon is less than 1, then epsilon square is less than epsilon. Then, now the choice becomes very easy. That is the whole idea here. Then, 2 alpha epsilon plus epsilon square, this is less than it will be less than 2 alpha epsilon plus epsilon, which is nothing but epsilon into 2 alpha plus 1 epsilon into 2 alpha plus 1. Now, can we choose epsilon is such that this is less than 2 minus alpha square 2 alpha plus 1 is a positive number. So, epsilon into 2 alpha plus 1 less than 2 minus alpha square is this is same as saying that epsilon should be less than 2 minus alpha square divided by 2 alpha plus 1. That is and of course epsilon. Now, 2 minus alpha square divided by 2 alpha plus 1 this is a positive number. Can we always choose some positive number with some positive number which is strictly less than that? We can always do that. We also wanted it should also be smaller than 1. So, we can say that choose epsilon which is smaller than minimum of these two. Take 1 and 2 minus alpha square divided by 2 alpha plus 1. These are two positive numbers. Their minimum is also a positive number. For example, you can choose epsilon to be half of that minimum. Then it will be less than 1 as well as less than this. This is the kind of technique that you use in many proofs. That is why I have done this in detail here. Is this clear? Whatever we have done. So, now what is the final condition? So, this shows that alpha square less than 2. This is not possible because we have got a connection to this. So, that means alpha square has to be equal to 2. We already shown that alpha square is less than or equal to 2. But now we have shown that it is not strictly less than 2. So, we must have alpha square is equal to 2. So, let us again take a look. What did we prove that there exists of course, uniqueness is trivial. There exists a unique x which is strictly that x is alpha. That alpha is bigger than 0 and alpha square is 2 and that is unique. All those things are proved. Now, let me give you an exercise. Exercise is the following. There is nothing particular about this 2 and that 2. You can take any number here instead of 2 and similarly you can take any natural number here instead of 2. By the way after having proved this then only this notation is meaning proved 2 or 2 power half or whatever it is. So, I will simply state this theorem and proof will be left to you as it is. So, for every x in r with x bigger than let me instead of x let me take it as a for every a in r with a bigger than 0 and a natural number n. There exists unique x in r with which is positive with x bigger than 0 such that x to the power n is equal to a. Proof left to you as an exercise. Do it on your own. Whatever we have done for 2, you do a similar thing for n. Again uniqueness will be straight forward to prove. You similarly define the set a with appropriate modifications here. Whatever set here we have defined, you define the set with appropriate modification. It will be a set of all x bigger than 0 with x to the power n less than or equal to a. Then show that that set is non-empty bounded above etcetera. Take its supremum and there in a similar way show that that x to the power n is equal to a. So, whatever we have done for this square, you will have to take the nth power everywhere and come across the appropriate inequalities, prove the appropriate inequalities. Now, let us go to one more very important property of real numbers. See till now we have seen how natural numbers and real numbers are related to each other. Now, we look at how rational numbers and real numbers are related to each other. This important property is sometimes expressed by saying that rationales are dense in real numbers. What we want to say is that given any two real numbers, there exists a rational number between those two real numbers. So, given any two real numbers, there exists a rational number between the two means between those two real numbers. So, suppose x and y are real numbers and we are saying the two real numbers, it means they are two different real numbers. So, let me say a and b. Suppose a and b are two real numbers and a not equal to b. So, the first step I want to say that it is enough to consider the case when both a and b are positive and both a and b are positive. We can also assume that a is less than if both if they are different means either a is less than b or b is less than a. So, a and b are after all notations. So, we can assume that a is less than b. So, what I want to say is that it is enough to prove it is enough to consider the case. Let me say 0 less than a less than b. In other words, I want to say it is enough to consider the case when both are positive. Why it is enough? What are the other possible cases? One is negative and one is positive. One is negative and one is 0 is there. So, that is nothing to prove. If both are negative, see suppose we have proved this case, let us say q is a rational number. If a is less than q less than b, because in that is same as say that minus b less than minus q less than minus a. So, suppose you take two negative numbers, you take their minus of those two numbers, they will be positive. Find the rational between those two positive numbers, minus of that will be rational between the two negative numbers. So, it is enough to consider this case. That is clear? Now, let us see how we go about this. See, after all what is our idea? We want to find the rational numbers. That means we want to find numbers p by q. We want to find number, we want to find natural numbers p and q or integers p and q such that q not equal to 0. This should happen a less than p by q less than b. Maybe one of these inequalities may be less than or equal to also. That does not matter. Of course, if a and b both are positive, is it clear that p and q also must be positive? So, we have to basically choose p and q in such a way that this happens. To do that, we have to say how to choose q and how to choose p. Now, first consider the case. Let us consider the number 1 by b minus a and 1 by b. Both of these are positive numbers. You can take maximum of these. They are also positive. You can take maximum of these. There is also positive. Now, whatever is positive real number, since we have already proved that the set of all natural numbers is not bounded above, I can always find a natural number which is bigger than the maximum of these two. Is it clear to you? This is a real number. Maximum of these two numbers is a real number. I can always find a natural number which is bigger than this real number. That number I will call q. You can say there exists q in n such that this maximum is less than or equal to q. That is 1 by b less than or equal to q and 1 by b minus a is also less than or equal to q. That is also less than or equal to q. This is something we want to prove. Let us just record what we got here. Then we will have these two inequalities. Then 1 by b less than or equal to q and 1 by b minus a. This is also less than or equal to q. Now, let me take this set. I will again call this set as a. I will take these are the set of all natural numbers in n such that m by q is less than or equal to b. m by q is less than or equal to b. That is same as saying m less than or equal to b times q. Is this set non-empty? This is an obvious number that belongs to it. What is it? 1. 1 belongs to it. It is obviously bounded above. Of course, this b times q may not be a natural number. It is just a real number. We are taking all those natural numbers which are less than or equal to this real number. Since this set is bounded above, can we see that there must be a maximum in this? It is a subset of natural numbers. It is not a subset of real numbers. It is a subset of natural numbers and that is bounded above. There must be maximum among this. That maximum I will call p. Let p be equal to maximum of a. What does it mean? That p belongs to a. Since p is a maximum, that means p plus 1 does not belong to a. That is not only that. Fine. This is what we will require that is. Then p belongs to a and p plus 1 does not belong to a. We shall use both these facts. Now, p belongs to a means what? Of course, p is a maximum of a. That means, remember a is the subset of natural numbers. We are choosing maximum of that. This is also a natural number. This is a natural number. What does p belong to a mean? p is less than or equal to b q. This means p is less than or equal to b q. Remember q is also a natural number. So, this implies p by q less than or equal to b. This is one of the things that we want. We also wanted that p by q is bigger than or equal to a. That will follow from this part. p plus 1 does not belong to a means what? p plus 1 must be strictly bigger than b q. p plus 1 must be strictly bigger than b q. So, this means p plus 1 strictly bigger than b q. This also means p by q plus 1 by q is bigger than b. p by q plus 1 by q is bigger than b. Let me write it here. So, p by q is bigger than b minus 1 by q. Now, what do we know about 1 by q? It is here. 1 by b is less than or equal to q. So, this means 1 by q is less than or equal to b. This is not something that is useful now. This part is useful now. 1 by b minus a is less than or equal to q. So, this means 1 by q is less than or equal to b minus a. Hence, what can we say about minus 1 by q? Minus 1 by q is bigger than or equal to minus of b minus a. So, this is bigger than or equal to b minus b minus a. That is same as a. So, here we have proved that p by q is less than or equal to b. Here, we have proved that p by q is bigger than a. That is what we wanted. Remember, we wanted to show this. That is a less than or equal to p by q less than or equal to b. We have proved that. So, we have shown that between any two real numbers, there exists a rational numbers. Let me again remind you that this is what is expressed by saying that rationals are dense in reals. Now, remember in this proof, we directly use the l u b action in the proof itself. In this proof, we directly use l u b action in the proof itself. In this proof, we did not use it directly, but did we use it indirectly? For example, here we said that there exists q in n such that this is less than or equal to q. There we use the fact that n is not bounded above. This we can get because the set of all rational numbers is not bounded above. That was the property that we got from l u b action. So, what I wanted to say again is that practically everything that you will prove about real numbers, it will follow either directly or indirectly from l u b action. We shall see the instances of this in the next few lectures when we discuss the real number system. We will stop with that for now.