0:39

7 is actually not a member of this interval.

Â For the same reason as 5 wasn't an element of this interval.

Â But nevertheless, 7 is still the least upper bound.

Â There is no smaller upper bound of that interval than 7.

Â So it is a least upper bound. So least upper bounds are not the same as

Â maximums, and in this case, 0 is a member of that set.

Â That interval is defined remember as a set of reals, such as 0 less than or

Â equal to x less than or equal to 1. And in this case, the endpoints are 0 and

Â 1 are elements of the interval. So 0 is in there, and it's clearly the

Â minimum elements of that interval. Well, the answer is the first one is

Â correct. This is what it means to say that the

Â rational line is dense. Between any two rationals, you can find a

Â third one. The second one is actually true, but it

Â doesn't express density. It doesn't express density, because

Â whether or not there is an irrational number between two rationals is, is sort

Â of irrelevant. itâ€™s, itâ€™s the question is about the

Â rational line being dense. And this is actually making a statements

Â about the real line. So it's true, but actually irrelevant to

Â the notion of density of the rational line.

Â So that one's not true. I mean, that one's true, but it's not the

Â answer to the question. And this one that's actually expressing

Â the notion of completeness. Now, if by a least upper bound, we mean

Â least upper bound in the rationals and the rationals on Q, then that would say

Â that the rational line is complete. Which is false, if however we interpreted

Â this to main, every set of rationals that is bounded above has at least upper bound

Â in the real numbers, then that would be an instance of the completeness of the

Â real line. And this is sure about the existence of

Â least upper bounds is what distinguishes the reals from the rationals and it's

Â what makes the reals a very powerful system.

Â For doing advanced mathematics and calculus in particular, and and

Â demonstrate the, the fact that the rationals is not complete is what

Â demonstrates the [INAUDIBLE] the impoverished niche of the rationals in

Â terms of mathematics and doing things like calculus.

Â Okay, how did you get on? What I want to do now is well actually,

Â what I really want to do is introduce the beginnings of the subject known as real

Â analysis. Now, this isn't real anaylsis as opposed

Â to fake analysis. Real here is essential short for real

Â numbers. Well, for the real number system if you

Â like. It's the analysis of the real numbers.

Â And I'm going to began with a theorem, the rational line is not complete.

Â Now, if you've done that assignment, assignment 10.1 that I've asked you to

Â do,you should be familiar with what that means.

Â But let me remind you in case you are decided to play, if plate was in and go

Â ahead without doing that assignment. Well, let me just remind you that

Â completeness means, if A subset of reals has an upper bound, then it has a least

Â upper bound in the set of reals. That is that was completeness as a

Â product of reals. But as I mentioned at the time, these

Â notions also apply to any set. So in terms of the rationals,

Â completeness would mean if A is a set of rationals having, an upper bound then it

Â has at least upper bound in the rationals.

Â What this theorem says is that this property does not hold for the rational

Â numbers. Remember, [INAUDIBLE] for real numbers

Â here. But if I replaced r by q and talked about

Â the rational numbers then this property would not hold.

Â It does however hold for the real numbers.

Â Now, there's the completeness property for the real line.

Â we won't be able to prove that but, I'll be able to indicate how it's possible to

Â construct the reals in order to make it possible to prove that.

Â 5:54

Okay, here is the proof of the theorem. Let A be the set of all rationals r, so

Â set r is now negative and r squared is less than two.

Â Now, already you can probably sense what's going on.

Â This is going to hinder on the property of that square root of 2 if it's a

Â rational. So let me draw a picture, is 0, is 2 a is

Â going to be a set well everything is going to be greater than or equal to 0.

Â And A if going to up to some point less than 2.

Â Well, A only contains rationals less than 2, now, whose square is less than 2, so

Â those rationals themselves less than 2, so it's going to something like this.

Â And we all know, that's just lurking in there somewhere is the square root of 2.

Â I should stress that throughout this argument, the argument, I'm about, I'm

Â about to give, we're talking purely about the rationals.

Â So I'm not going to be talking about any reals, her's why I sort of put this down

Â here somewhat faintly. This is to help guide our intuition.

Â This is just to motivate what's going to go on.

Â But the entire argument I give is going to be in terms of rational numbers, not

Â real numbers. I deliberately did not write r less than

Â the square root of 2, because there is no such thing as the square root of 2 in the

Â rationals. I'm using sets of rationals in this

Â argument. It's an arguments about the rational

Â numbers, not about the real numbers. Okay, well, A is bounded above, for

Â example 2 is an upper bound, you want to need to find 1 and 2 will do just fine.

Â I will show that A has no least upper bound.

Â That would mean that A is a set of rationals, which has an upper bound but

Â no least upper bound, and hence, the rational line is not complete.

Â Because completeness would say that any set of rationals with an upper bound in

Â the rationals has a least upper bound in the rationals.

Â Well, how would I show that there's no least upper bound?

Â Well, let x and Q be any upper bound of A and show there's a smaller one.

Â 8:27

Again, let me stress, smaller 1 in the rationals.

Â Remember, we use the letter Q to denote rationals, because Q stands for quotient

Â and rational numbers are numbers that are quotients of integers.

Â We can't use the letter R for rational, because R is used for real numbers.

Â Unfortunate, I know, but there we are. And since we're talking about the

Â rationals, that upper bound x, you're going to be able to form p over, where p

Â and q are integers. In fact, they can be natural numbers.

Â Because this set A is is, is positive integer and it's not negative number, it

Â is, it is set A is not negative rationals, it's everything to the right

Â of the origin. So everything is positive, so I don't

Â have to worry about negative numbers here, so these two integers can be chosen

Â positive And I want to show that there's a smaller upper bound.

Â 9:37

Well, lets suppose x squared is less than 2.

Â It's either less than 2 or it's greater or equal than 2.

Â It's one of the two. Let's just see what happens if x is less

Â than 2. In that case, looking at this equation,

Â 2Q squared is bigger than P squared. Now, as n gets larger, n squared divided

Â by 2n plus 1 increases without bound. So we can pick an ne N so large that n

Â squared over 2n plus 1 is bigger than p squared divided by 2q squared minus p

Â squared. Now, you might not see where I'm going

Â with this. But hopefully, you can believe everything

Â I've said. Okay, we're assuming x squared is less

Â than 2. Well actually, in a moment, we'll arrive

Â at a contradiction, so the, the conclusion I'm going to get out of this,

Â is that x squared is in fact not less than 2, but this is where we're starting.

Â If x squared is less than two and because of that definition 2q squared is greater

Â than p squared. Okay, so 2q squared minus p squared is

Â positive, that means this number is a positive number.

Â And what I'm seeing is because we've got an n squared here and a, and a, and a

Â linear term involving n here, the bigger N gets, this gets increasingly large, it

Â gets as this large it wanted to be. So I can pick in big enough so that this

Â number is bigger than that one, and if you rearrange that, you'll find that 2n

Â squared q squared is bigger than n plus 1 squared P squared.

Â Okay, I'll leave you to do the algebra for getting from there to there.

Â Hence, n plus 1 over n squared times p squared over q squared is less than 2.

Â Just rearranging that, taking those terms to the other side.

Â 12:00

Now, let y be n plus 1 over n times p over q.

Â Now, notice that y is a rational number. It's a quotient of integers and y squared

Â is less than 2. Because this says that y squared is less

Â than 2. By the way, this, by now, you should have

Â begin to smell why I, I I, I started looking at this term.

Â I was trying to get this number y. Remember, I started with an x as, as an

Â upper bound, and I wanted to show that there's a smaller one.

Â And I'm going to work towards that and I've got I've introduced this y.

Â So y is in q and y squared less than 2. So y is an element of that set A, but

Â wait a minute, y is equal to a number slightly bigger than 1 times x.

Â So that means that y is actually bigger than x.

Â So the number y that I've constructed is in the set A, and yet is bigger than X.

Â Well, that's a contradiction. Since x is an upper bound of A, that

Â supposition must be false. So x squared has to be greater than or

Â equal to 2, okay? So what I've done is I've taken an upper

Â bound of A. I'm going to show there's a smaller one.

Â And as a first step towards doing that, I've shown by contradiction.

Â That, that upper bound has to have its square greater than or equal to 2.

Â Now, I'm going to go ahead, using this extra information, to show that there's a

Â smaller upper bound, enhances no chance of any x being a least upper bound.

Â Let me recap where we've got to. Let A be the set of all rationals as a

Â nonzero and for which r squared is less than 2.

Â We let x be an upper bound of A and we had x in the form p over q, where p and q

Â are integers. Okay, so we, we have that.

Â And our goal is to show that A has an upper bound smaller than x, hence, that

Â cannot be at least upper bound, which would show that the rationals are not

Â complete and we just showed that x squared is greater than or equal to 2.

Â Hence, since the square root of 2 is irrational, x squared is strictly bigger

Â than 2. x is irrational, x squared can't be equal

Â to 2, so it's strictly greater than 2. Thus, since x equals p over q, p squared

Â is bigger than 2 q squared. I'm going to use this fact to find an

Â upper bound of A smaller than x. To do that, I'm going to pick n, an

Â integer so large that the following is true.

Â n squared divided by 2 n plus 1 is bigger than 2q squared over p squared minus 2q

Â squared i,e., rearranging that, p squared n squared greater than 2q squared times n

Â plus 1 squared. So you just rearrange this to a little

Â bit of algebra and you get this, i.e., p squared of a q squared times n or of n

Â plus 1 squared is greater than 2. Again, you just rearrange that and do a

Â little bit of algebra to get that. Let y be n over n plus 1 times p over Q.

Â Then, y is an element of Q, y is a rational number.

Â It's a quotient of integers. So it's in Q and more over.

Â 17:53

Hence, y is an upper bound of a, which is smaller than x.

Â Thus, a does not have a least upper bound.

Â And this proves the theorem. I guess my mathematics is better than my

Â handwritting. This proves the theorem.

Â Final remark. The construction of r from Q, can be done

Â in several different ways, but in all cases the aim is to prevent an argument

Â like the above going through for r. And with that, you're at the very gateway

Â to modern real analysis. For our final topic in this course, I'd

Â like to say a little bit about real number sequences.

Â these are connected with one of the ways of constructing the real numbers from the

Â rationals. And they, also give us a technique or a

Â concept for doing an awful lot of work in real analysis.

Â To put it another way, sequences of real numbers are a big deal in modern real

Â analysis, which means they're a big deal in calculus.

Â And anything that's a big deal in calculus is a big deal in science and

Â engineering and technology so whichever way you put it sequences are a big deal.

Â Now, what is a sequence? Well, in everyday terms it's a list, a1,

Â a2, a3, and let's put some commas in here, of numbers.

Â So, we have a number, a number, a number going on to infinity.

Â 19:46

The way we normally express this and try to capture this is infinite extent here

Â is by writing it an and where n goes from one to infinity, and this is what's

Â called an infinite sequence. If you look in textbooks, you'll find a

Â more formal definition to the sequences of function from the set of natural

Â numbers into the real numbers but, for the purposes of what I, the kind of

Â things I want to talk about here, it's enough to think of it simply as an

Â infinite list of real numbers. For example, the sequence of natural

Â numbers 1, 2, 3, and so forth. That's an infinite sequence.

Â 20:41

In terms of our notation, I would just write that as n, for n goes from 1 to

Â infinity. Or I could have the sequence that

Â consists simply of an infinite sequence of 7s.

Â 7 going on forever, and that would be expressed in this way.

Â Or, I could have the following sequence. 3, 1, 4, 1, 5, 9, et cetera, anywhere

Â that's in the decimal digits of pi. But there's no simple formula like this

Â to capture this one. I have to use some expression like this,

Â or, let me give you another example. I could have the sequence consisting of

Â negative 1 to the n plus 1 from n equals 1 to infinity.

Â That's a sequence that consists of plus 1, negative 1 plus 1, negative 1 plus 1,

Â negative 1, et cetera. That's an example of what's known as an

Â alternating sequence, meaning that the sign alternates as you go through the

Â sequence. Okay, so that's what sequences are, just

Â infinite list of numbers. Now, let's look at the following example,

Â look at the sequence consisting of the numbers 1 over n from n if its 1 to

Â infinity. Again, that's the sequence 1, a half, a

Â third, a quarter, and so on. And the things to notice about this is

Â that the numbers get closer and closer to 0.

Â In fact, to get arbitrarily close to 0 or this one.

Â 1 plus 1 over 2 to the n, from n equals 1 to infinity.

Â That consists of the numbers 1 and a half, 1 and a quarter, and 1 and an

Â eighth, 1 and a sixteenth, and these numbers is arbitrarily close to 1.

Â And going back to this example, here, the sequence 3, 3.1, 3.14, 3.141, 3.1415,

Â 3.14159, 3.141592, 3.1415926. That sequence gets arbitrarily close to

Â pi. That's as far as I know the decimal

Â expansion, by the way. So these sequences of the property as you

Â go along them, it gets arbitrarily close to a fixed number.

Â 0 in the first case, 1 in the second case, and pi in the third case.

Â And there's there's a general property here that we are going to capture in, by

Â a way of a definition. If the numbers in a sequence n, n from 1

Â to infinity, get arbitrarily closer to some fixed number a, we say that sequence

Â tends to the limit a, and write n arrows a as n arrows infinity.

Â An alternative for notation is we sometimes write it this way, limit as n

Â goes to infinity of a sub n equals a. Not all sequences tend to a limit.

Â look at this one for example, this alternating sequence, plus one, negative

Â one, plus one, negative one. That doesn't approach any particular

Â number, it bounces back and forth between plus one and negative one.

Â