Welcome back as they say in the media. Where the correct answers number two and number three. 5 is not the maximum element of this interval, in fact, 5 is not an element of this interval at all. This, remember, is a set of all x in the reals, so which the 0 strictly less than x, strictly less than 5. So 5 itself is not a member there. So not only is it not the maximum element, it's not an element at all. 7 is actually not a member of this interval. For the same reason as 5 wasn't an element of this interval. But nevertheless, 7 is still the least upper bound. There is no smaller upper bound of that interval than 7. So it is a least upper bound. So least upper bounds are not the same as maximums, and in this case, 0 is a member of that set. That interval is defined remember as a set of reals, such as 0 less than or equal to x less than or equal to 1. And in this case, the endpoints are 0 and 1 are elements of the interval. So 0 is in there, and it's clearly the minimum elements of that interval. Well, the answer is the first one is correct. This is what it means to say that the rational line is dense. Between any two rationals, you can find a third one. The second one is actually true, but it doesn't express density. It doesn't express density, because whether or not there is an irrational number between two rationals is, is sort of irrelevant. it’s, it’s the question is about the rational line being dense. And this is actually making a statements about the real line. So it's true, but actually irrelevant to the notion of density of the rational line. So that one's not true. I mean, that one's true, but it's not the answer to the question. And this one that's actually expressing the notion of completeness. Now, if by a least upper bound, we mean least upper bound in the rationals and the rationals on Q, then that would say that the rational line is complete. Which is false, if however we interpreted this to main, every set of rationals that is bounded above has at least upper bound in the real numbers, then that would be an instance of the completeness of the real line. And this is sure about the existence of least upper bounds is what distinguishes the reals from the rationals and it's what makes the reals a very powerful system. For doing advanced mathematics and calculus in particular, and and demonstrate the, the fact that the rationals is not complete is what demonstrates the [INAUDIBLE] the impoverished niche of the rationals in terms of mathematics and doing things like calculus. Okay, how did you get on? What I want to do now is well actually, what I really want to do is introduce the beginnings of the subject known as real analysis. Now, this isn't real anaylsis as opposed to fake analysis. Real here is essential short for real numbers. Well, for the real number system if you like. It's the analysis of the real numbers. And I'm going to began with a theorem, the rational line is not complete. Now, if you've done that assignment, assignment 10.1 that I've asked you to do,you should be familiar with what that means. But let me remind you in case you are decided to play, if plate was in and go ahead without doing that assignment. Well, let me just remind you that completeness means, if A subset of reals has an upper bound, then it has a least upper bound in the set of reals. That is that was completeness as a product of reals. But as I mentioned at the time, these notions also apply to any set. So in terms of the rationals, completeness would mean if A is a set of rationals having, an upper bound then it has at least upper bound in the rationals. What this theorem says is that this property does not hold for the rational numbers. Remember, [INAUDIBLE] for real numbers here. But if I replaced r by q and talked about the rational numbers then this property would not hold. It does however hold for the real numbers. Now, there's the completeness property for the real line. we won't be able to prove that but, I'll be able to indicate how it's possible to construct the reals in order to make it possible to prove that. Okay, here is the proof of the theorem. Let A be the set of all rationals r, so set r is now negative and r squared is less than two. Now, already you can probably sense what's going on. This is going to hinder on the property of that square root of 2 if it's a rational. So let me draw a picture, is 0, is 2 a is going to be a set well everything is going to be greater than or equal to 0. And A if going to up to some point less than 2. Well, A only contains rationals less than 2, now, whose square is less than 2, so those rationals themselves less than 2, so it's going to something like this. And we all know, that's just lurking in there somewhere is the square root of 2. I should stress that throughout this argument, the argument, I'm about, I'm about to give, we're talking purely about the rationals. So I'm not going to be talking about any reals, her's why I sort of put this down here somewhat faintly. This is to help guide our intuition. This is just to motivate what's going to go on. But the entire argument I give is going to be in terms of rational numbers, not real numbers. I deliberately did not write r less than the square root of 2, because there is no such thing as the square root of 2 in the rationals. I'm using sets of rationals in this argument. It's an arguments about the rational numbers, not about the real numbers. Okay, well, A is bounded above, for example 2 is an upper bound, you want to need to find 1 and 2 will do just fine. I will show that A has no least upper bound. That would mean that A is a set of rationals, which has an upper bound but no least upper bound, and hence, the rational line is not complete. Because completeness would say that any set of rationals with an upper bound in the rationals has a least upper bound in the rationals. Well, how would I show that there's no least upper bound? Well, let x and Q be any upper bound of A and show there's a smaller one. Again, let me stress, smaller 1 in the rationals. Remember, we use the letter Q to denote rationals, because Q stands for quotient and rational numbers are numbers that are quotients of integers. We can't use the letter R for rational, because R is used for real numbers. Unfortunate, I know, but there we are. And since we're talking about the rationals, that upper bound x, you're going to be able to form p over, where p and q are integers. In fact, they can be natural numbers. Because this set A is is, is positive integer and it's not negative number, it is, it is set A is not negative rationals, it's everything to the right of the origin. So everything is positive, so I don't have to worry about negative numbers here, so these two integers can be chosen positive And I want to show that there's a smaller upper bound. Well, lets suppose x squared is less than 2. It's either less than 2 or it's greater or equal than 2. It's one of the two. Let's just see what happens if x is less than 2. In that case, looking at this equation, 2Q squared is bigger than P squared. Now, as n gets larger, n squared divided by 2n plus 1 increases without bound. So we can pick an ne N so large that n squared over 2n plus 1 is bigger than p squared divided by 2q squared minus p squared. Now, you might not see where I'm going with this. But hopefully, you can believe everything I've said. Okay, we're assuming x squared is less than 2. Well actually, in a moment, we'll arrive at a contradiction, so the, the conclusion I'm going to get out of this, is that x squared is in fact not less than 2, but this is where we're starting. If x squared is less than two and because of that definition 2q squared is greater than p squared. Okay, so 2q squared minus p squared is positive, that means this number is a positive number. And what I'm seeing is because we've got an n squared here and a, and a, and a linear term involving n here, the bigger N gets, this gets increasingly large, it gets as this large it wanted to be. So I can pick in big enough so that this number is bigger than that one, and if you rearrange that, you'll find that 2n squared q squared is bigger than n plus 1 squared P squared. Okay, I'll leave you to do the algebra for getting from there to there. Hence, n plus 1 over n squared times p squared over q squared is less than 2. Just rearranging that, taking those terms to the other side. Now, let y be n plus 1 over n times p over q. Now, notice that y is a rational number. It's a quotient of integers and y squared is less than 2. Because this says that y squared is less than 2. By the way, this, by now, you should have begin to smell why I, I I, I started looking at this term. I was trying to get this number y. Remember, I started with an x as, as an upper bound, and I wanted to show that there's a smaller one. And I'm going to work towards that and I've got I've introduced this y. So y is in q and y squared less than 2. So y is an element of that set A, but wait a minute, y is equal to a number slightly bigger than 1 times x. So that means that y is actually bigger than x. So the number y that I've constructed is in the set A, and yet is bigger than X. Well, that's a contradiction. Since x is an upper bound of A, that supposition must be false. So x squared has to be greater than or equal to 2, okay? So what I've done is I've taken an upper bound of A. I'm going to show there's a smaller one. And as a first step towards doing that, I've shown by contradiction. That, that upper bound has to have its square greater than or equal to 2. Now, I'm going to go ahead, using this extra information, to show that there's a smaller upper bound, enhances no chance of any x being a least upper bound. Let me recap where we've got to. Let A be the set of all rationals as a nonzero and for which r squared is less than 2. We let x be an upper bound of A and we had x in the form p over q, where p and q are integers. Okay, so we, we have that. And our goal is to show that A has an upper bound smaller than x, hence, that cannot be at least upper bound, which would show that the rationals are not complete and we just showed that x squared is greater than or equal to 2. Hence, since the square root of 2 is irrational, x squared is strictly bigger than 2. x is irrational, x squared can't be equal to 2, so it's strictly greater than 2. Thus, since x equals p over q, p squared is bigger than 2 q squared. I'm going to use this fact to find an upper bound of A smaller than x. To do that, I'm going to pick n, an integer so large that the following is true. n squared divided by 2 n plus 1 is bigger than 2q squared over p squared minus 2q squared i,e., rearranging that, p squared n squared greater than 2q squared times n plus 1 squared. So you just rearrange this to a little bit of algebra and you get this, i.e., p squared of a q squared times n or of n plus 1 squared is greater than 2. Again, you just rearrange that and do a little bit of algebra to get that. Let y be n over n plus 1 times p over Q. Then, y is an element of Q, y is a rational number. It's a quotient of integers. So it's in Q and more over. y squared is bigger than 2, moreover, since n divided by n plus 1 is less than 1, y is less than x, because p over q is x and y is just this guy, times x. So it's, it's 6 less than x. But, for any a in A, a squared is less than 2 is less than y squared, so a is less than y. Hence, y is an upper bound of a, which is smaller than x. Thus, a does not have a least upper bound. And this proves the theorem. I guess my mathematics is better than my handwritting. This proves the theorem. Final remark. The construction of r from Q, can be done in several different ways, but in all cases the aim is to prevent an argument like the above going through for r. And with that, you're at the very gateway to modern real analysis. For our final topic in this course, I'd like to say a little bit about real number sequences. these are connected with one of the ways of constructing the real numbers from the rationals. And they, also give us a technique or a concept for doing an awful lot of work in real analysis. To put it another way, sequences of real numbers are a big deal in modern real analysis, which means they're a big deal in calculus. And anything that's a big deal in calculus is a big deal in science and engineering and technology so whichever way you put it sequences are a big deal. Now, what is a sequence? Well, in everyday terms it's a list, a1, a2, a3, and let's put some commas in here, of numbers. So, we have a number, a number, a number going on to infinity. The way we normally express this and try to capture this is infinite extent here is by writing it an and where n goes from one to infinity, and this is what's called an infinite sequence. If you look in textbooks, you'll find a more formal definition to the sequences of function from the set of natural numbers into the real numbers but, for the purposes of what I, the kind of things I want to talk about here, it's enough to think of it simply as an infinite list of real numbers. For example, the sequence of natural numbers 1, 2, 3, and so forth. That's an infinite sequence. In terms of our notation, I would just write that as n, for n goes from 1 to infinity. Or I could have the sequence that consists simply of an infinite sequence of 7s. 7 going on forever, and that would be expressed in this way. Or, I could have the following sequence. 3, 1, 4, 1, 5, 9, et cetera, anywhere that's in the decimal digits of pi. But there's no simple formula like this to capture this one. I have to use some expression like this, or, let me give you another example. I could have the sequence consisting of negative 1 to the n plus 1 from n equals 1 to infinity. That's a sequence that consists of plus 1, negative 1 plus 1, negative 1 plus 1, negative 1, et cetera. That's an example of what's known as an alternating sequence, meaning that the sign alternates as you go through the sequence. Okay, so that's what sequences are, just infinite list of numbers. Now, let's look at the following example, look at the sequence consisting of the numbers 1 over n from n if its 1 to infinity. Again, that's the sequence 1, a half, a third, a quarter, and so on. And the things to notice about this is that the numbers get closer and closer to 0. In fact, to get arbitrarily close to 0 or this one. 1 plus 1 over 2 to the n, from n equals 1 to infinity. That consists of the numbers 1 and a half, 1 and a quarter, and 1 and an eighth, 1 and a sixteenth, and these numbers is arbitrarily close to 1. And going back to this example, here, the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1415926. That sequence gets arbitrarily close to pi. That's as far as I know the decimal expansion, by the way. So these sequences of the property as you go along them, it gets arbitrarily close to a fixed number. 0 in the first case, 1 in the second case, and pi in the third case. And there's there's a general property here that we are going to capture in, by a way of a definition. If the numbers in a sequence n, n from 1 to infinity, get arbitrarily closer to some fixed number a, we say that sequence tends to the limit a, and write n arrows a as n arrows infinity. An alternative for notation is we sometimes write it this way, limit as n goes to infinity of a sub n equals a. Not all sequences tend to a limit. look at this one for example, this alternating sequence, plus one, negative one, plus one, negative one. That doesn't approach any particular number, it bounces back and forth between plus one and negative one. this one in a trivial sense doesn't tend to a limit, this one tends to the limit seven. It doesn't just tend to it, it, never gets away from it. this one doesn't tend to a limit at all. These numbers get bigger and bigger and bigger. we would sometimes say that the sequence tends to infinity. but thats, beyond the scope of, the limited amount I want to talk about sequences in, in this course. The point is, some sequences don't tend to a limit. Other sequences do tend to a limit.
