Welcome back as they say in the media. Where the correct chances number two and
number three. Five is not the maximum element of this
interval. In fact five is not an element of this
interval at all. This remember is the set of all x in the
reals, switch the zero, strictly less than x, strictly less than five.
So five itself is not a member of that. So not only is it not the maximum element,
it's not an element at all. Seven is actually not a member of this
interval, for the same reason as five wasn't an element of this interval.
But nevertheless seven is still the least upper bound.
There is no smaller upper bound of that interval than seven so it is a least upper
bound. So a least upper bounds are not the same
as maximums. And in this case, zero is a member of that
set. That interval is defined remember as a set
of reals so which the zero <= x <= one And in this case, the end point zero and one
are elements of the interval, so zero is in there and it's clearly the minimum
element 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, but the question is about the
rational line being dense. This is actually making a statement 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 least upper bound, we mean least
upper bound in the rationals, in the rational IQ.
Than that would say that the rational line is complete, which is false.
If, however we interpreted this to mean 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 true about the existence of least upper bounds is what distinguishes
the reals from the rationals. And this is what makes the reals a very
powerful system for doing advanced mathematics and calculus in particular and
demonstrate, and the, the, the fact that the rationals is not complete is, is what
demonstrates the, the, the improvised nature 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, actually, what
I really want do is introduce the beginnings of the subject known as real
analysis. Now this isn't real analysis as opposed to
fake analysis. Real here is, is, is essentially short for
real numbers, well for the real number system, if you like.
It's the analysis of the real numbers. We're going begin with a theorem, that
rational nine, is not complete. Now, if you've done that assignment,
assignment 10.1 that I asked you do to, you should be familiar with what that
means, but let me remind you, in case you decided to play it, play it, risky and,
and go ahead without doing that assignments.
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 applied to 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 is such a rationals, having an upper bound, then
this has a least upper bound in the rationals.
What this theorem says is, is that this property does not hold for the rational
numbers. Remember R stands for real numbers here.
But if I replaced R by Q and talk about the rational numbers, then this property
would not hold. It does, however, hold for the real
numbers. At least the realness property for the
real line we won't be able to prove that, but I will be able to indicate how it's
possible to construct the reals, in order to make it possible to prove that.
Okay. Here's the proof of the theorem.
Let A be the set of all rationals R, such that R is non negative and R^2 < two.
Now already you can probably sense whats going on, this is going to hinge around
that property that the square root of two is rational.
So let me draw a picture. There's zero, there's two.
A is going to be a set, while everything is going to be greater than or equal to
zero. And A is going to go up to some point less
than two. Well, A only contains rationals less than
two whose square is less than two. So those rationals, themselves, are less
than two. So, it's going to be something like this.
And we all know that it's lurking in there somewhere is the square root of two.
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. Here's why I've sort of put this down here
somewhat faintly. This is to help guide our intuition, this
is just to sort of motivate what's going to go on but the entire argument I'll give
is going to be in terms of rational numbers, not real numbers.
I deliberately did not write how less than the square root of two because there is no
such thing as the square root of two in the rationals.
I'm using sets of rationals, in this argument.
It's an argument about the rational numbers, not about the real numbers.
Okay. Well, A is bounded above.
For example, two is an upper bound. We only need to find one, and, two will do
just fine. I will show that A has no least upper
bound. That would mean the day is a set of
rationals which has a upper bound but not a least upper bound, and hence a rational
line is not complete. Because completeness would say that any
set of rationals with an upper bound in rationals has a least upper bound in
rationals. Well, how would I show that there's no
least upper bound? Well, that X and Q be any of a bound of A.
And sure there is a smaller one. Again let me stress smaller one in the
rationals. Remember we use as of Q to denote
rationals because Q stands for quotient and rational numbers are numbers that the
quotients have introduced. We can't use the letter R for rational
because R is used for real numbers, unfortunate I know but there we are.
Well, since we're talking about the rationals, that upper bound X is going to
be of the form P / Q. Where P and Q, we're introduced.
In fact, they can be natural numbers because this set A is a, is, is positive
integers. There's no negative number.
It is set A is not negative rationals. It's everything's to the right of the
origin, so everything's 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^2 < two, its either
less than two or its greater or equal to two, its one of the two, lets just see
what happens if X < two In that case, looking at this equation, 2q^2 > p^2.
Now, is n gets larger, n^2 / 2n + one increases without bound.
So we can pick n in n so large that n^2 / 2n + one > p^2 / 2q^2 - p^2.
Now you might not see where I'm going with this.
But hopefully you can believe everything I've said.
Okay, well assuming x^2 < two. We'll actually, in a moment, we'll have a
contradiction. So the, the conclusion I'm going to get
out of this is that x^2 is, in fact, not less than two, but this is where we're
starting. If x^2 < two, then because of that
definition 2q^2 > p^2. Okay?
So 2q^2 - p^2 is positive. That means this number is a positive
number. And what I'm saying is that because we got
an n^2 here in the linear term involving n here, a bigger n gets this gets
increasingly large, it gets as now as you wanted to be, so I can pick an n big
enough so that this number is bigger than that one.
And if you rearrange that, you find that 2n^2q^2 > (n+1)^2p^2.
Okay, I'll leave you to do the algebra for getting from there to there.
Hence, (n+1/n)^2 p^2/q^2 < two. Just rearranging that taking those terms
to the other side. Now let y = (n+1/n) p / q And notice, that
y is a rational number, it's a quotient of integers.
And y^2 < two. Because they say y ^ two < two.
By the way this, by now should of began to smell why I, I d-, started looking at this
tab. I was trying to get this number y.
Remember I started with an x as an upper bound and I wanted to show that the
smaller one, and I'm going to work towards that, now I've got to, I've introduce as
y, okay. So y is in q, and y^2 < two So y is an
element of that set A. Wait a minute.
Y is equal to a number slightly bigger than one 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^2 has to be greater than or equal to two.
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 it's square
greater than or equal to two. Now I am going to go ahead using this
extra information to show that there is a smaller upper bound and hence there is no
chance of any x being the least upper bound.
Let me recap where we've got to. We let A be the set of all rationals.
For the non zero, and for which r^2 < two, we'll let x be an upper bound of A.
And we have x in the form, p / q Where p and q are integers.
Okay. So we have that.
And our goal is to show that A has an upper bound smaller than X.
Hence, there cannot be at least upper bound which would show that the rationals
are not complete. And we just showed that x^2 >= two Hence,
since the square root of two is irrational, x^2 is strictly bigger than
two. X is irrational, x^2 can't be equal to
two, so it's strictly greater than two. Thus since x = p / q.
P^2 > 2q^2. I'm going to use this fact to find an
upper bound of a < x To do that, I'm going to pick n, an integer so large that the
following is true n^2 / 2n + one > 2q^2 / p^2 - 2q^2 I.e, rearranging that, p^2n^2 >
2q^2(n + one)^2. So then we just rearrange this, do a
little bit of algebra, and you get this. I.e.
P^2 / q^2 (n / n + one)^2 > two. Again, you just rearrange that and do a
little bit of algebra to get that. Let y = n / (n + one)p / 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^2 > two. More over, since n / n + one < one, y < x.
Because p / q = x and y is just this guy times x so it's, see, it's less than x.
But, for any a and a, a^2 < two < y^2 so a < 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 handwriting.
This proves the therm. 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, your, the very gateway to
modern real analysis. To our final topic in this course.
I'd like to say 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, a technique or a concept for doing an awful
lot of work in real analysis. To put it another way, sequences of real
numbers, they're 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 cut it sequences are a big deal.
Now, what is a sequence? Well, in everyday terms it's a list.
1,2,3. 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, this infinite extent here
is by writing it A n 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'll, 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, one, two, three, and so forth.
That's an infinite sequence. In terms of our notation, I would just
write that as n, for n goes from one to infinity.
Or I could have n the sequence that, consists simply of an infinite sequence of
symbols. Seven going on forever.
That would be expressed in this way. Or I could have the following sequence,
three, one, four, one, five, nine etc, enumerating the decimal digits apply.
Well, there's no simple formula like this to capture this one.
I'll have to use some expression like this.
Or let me give you another example. That could have the sequence consisting of
(-one)^n + one from n = one to infinity. That's a sequence that consists of +one,
-one, +one, -one, +one, -one, etc. 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 this little sequence is a, just infinitely some numbers.
Now lets look at the following example. Look at the sequence consisting of the
numbers one / n, from n = one to infinity. Okay.
That's the sequence one, one-half, one-third, a quarter and so on.
And the thing to notice about this, is it the numbers are closer and closer to zero.
In fact, they get arbitrarily close to zero, or this one.
One + one / two^n from n = one to infinity.
That encompasses numbers one and a half, one and a quarter, one and one / eight,
one and one / sixteen etc. And these numbers.
It's arbitrarily close to one. I'm going back to this example, here.
The sequence three, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1415926.
That sequence gets arbitrarily close to pie that's as far as I know the decimal
expansion by the way. So these sequences have the property that
as you go along them they get arbitrarily close to a fixed number, zero in the first
case, one in the second case, and pi in the third case.
There's a, there's a general property here that we're going to capture in the, by way
of a definition. If the numbers in the sequence n, n from
one to infinity, gets arbitrarily closer to some fixed number a, we say that
sequence tends to the limit A and write n as our A, is N as our infinity.
An alternative notation is we sometimes write it this way.
The limit as n goes to infinity, of a sub n = a.
Not all sequences tend to a limit. Look at this one for example, this
alternating sequence. +One, -one, +one, -one.
That doesn't approach any particular number.
It bounces back and forth between + one and -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.
And we would sometimes say that the sequence tends to infinity but that's
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.
Okay, let's move on. Well, so far everything's been very
intuitive. Let's get a little bit more formal now.
I've got a sequence An, n from one to infinity.
The intuitive explanation, or the intuitive, description that I gave, of
this, that the sequence or the members of the sequence, tend to a limit a and tends
to infinity, that sort of corresponds, fairly roughly, to, the fact that the
absolute value of an - a becomes arbitrarily close to zero.
Boy, my writing really is a problem. Isn't it close.
Now let me give you the formal definition, and the formal definition.
Involves being precise about what that means.
A n turns to A is N turns to infinity, if and only if for all real numbers epsilon,
greater than zero there is a natural number n So except for all n >= n Absolute
value of A n - A < epsilon. And now, perhaps for the first time you
see why we spent so much effort understanding quantifiers and being
particular about the order in which quantifiers appeared.
This definition is absolutely crucial in real analysis.
It's absolutely crucial to our definition and our understanding of the real numbers.
And that means it's absolutely crucial to calculus.
And hence, it's absolutely crucial to physics, science, engineering, technology.
Etcetera, etcetera, etcetera. This is the real stuff folks.
Okay, let's try and understand how that captures this intuition.
The geometric intuition we have I think it's fairly clear, that you're going along
a sequence, and the numbers get closer and closer and closer to some fixed number.
How does this somewhat complicated-looking expression capture that in a precise,
formal way. Well.
Let's peel away this part for a minute. By the way it's, it's traditional in
mathematics to use epsilon in this context.
And at the back of your mind, you should think that epsilon is not just positive
but it's very small and positive. Like a tenth, or a hundredth, or a
millionth, or a zillionth, or whatever. As, as well see when I, as you'll see when
I go through this, it's, it's when epsilon becomes small, but remains positive that
this thing kicks into power. Okay, so let's forget now a bit.
Let's look at the other part. Exist an m, such that for all m >= n,
absolute value A m - a = epsilon. And I've avoided mentioning expressly the
set n now. Notice that in the first place, I didn't
mention that m was a natural number, because the context makes that clear n is
a natural number and which [inaudible] and when we discuss quantified areas when I, I
pointed out that mathematicians do this kind of simplification all the time.
So it makes it easy to read that's all. Okay.
What does this part mean? So we we've been given an epsilon ratio
and we've been given some positive number epsilon and at the back of your mind think
of it as a small positive number. What this means, is that, from some point
onwards, from some point, n onwards, all the members of the sequence, are within a
distance, epsilon of A. So, all the numbers in a sequence a n, n =
one to infinity within a distance of epsilon from A.
Still with me? I want to see what this thing means, when
we have the whole expression. This says that for any epsilon greater
than zero, this things holds. So for any epsilon zero, there is a point
such that from that point onwards, all the numbers are within a distance epsilon from
A. As I mentioned a moment ago, the intuition
here is that we can [inaudible] greater than zero as small as we want.
So let me draw a picture now. You got some number A and as a sequence
bouncing around here somewhere. You know maybe it's coming in from the
left, from right, or maybe it's bouncing around but whether these numbers A1, A2
scattered along this line here. Now were given an epsilon greater than
zero, so let's say suppose it's down here, so here's let's say negative is A -
epsilon, and here is A + epsilon. Okay.
I've just got a distance epsilon to the left, and a distance epsilon to the right.
And the argument I'm about to give, will hold for any epsilon.
At least that's what it says. So I'm given an epsilon, I look at its
interval. And what the rest of this formula says, is
that from some point on, all the elements of the sequence, they're in here.
If I take a small of epsilon, let's do it say, here and here, let's call it epsilon
prime. So it will be A - epsilon prime, and A
would be A + epsilon prime. So I take a smaller epsilon, still
positive, so the formula will hold for the epsilon prime.
There will be some point from which all the elements of the sequence, beyond that
point, are within this region. And then I could take an even smaller one,
and then even smaller one. Notice that the n depends on the epsilon.
Remember that example about the American melanoma foundation, wasn't a problem for
them getting the quantifiers the wrong way around.
If we have the quantifiers the wrong way around here, we'd be in big trouble, this
wouldn't work. The point is, given an epsilon you could
find an n. For each epsilon you may have to go
further out in the sequence until you come to a point where all the elements are
within that distance. But what this says is, is that you can
always, by going sufficiently far out in the sequence, reach a stage where all of
the numbers in the sequence from then onwards are within a given distance of A.
And you can do it out smaller, and smaller, and smaller A.
And that's capturing this intuition. And it does it beautifully and elegantly
and with enormous power. Okay, let me give you a couple of
examples. First example.
Let's look at the sequence, one / n, n goes from one to infinity.
Now we know, that one over n tends to zero as n tends to infinity.
Let's prove this rigorously, in terms of the definition of limits that I just gave.
Okay? So I want to prove that fact.
What I have to show, is that for all that epsilon greater than zero, all real
numbers that are greater than zero, there's an n in the natural numbers such
that all for all natural numbers m >= n absolute value, one / m - zero < epsilon.
Well let's simplify that. For all real numbers epsilon > zero, there
is an n, and let me just drop that explicit mention of the natural numbers.
For all m >= n that just says absolute value of one / m < epsilon which actually
just means one / m < epsilon Because m is a positive integer.
Okay. So I have to verify this in order to prove
that. Okay, so how do I verify that?
Well let epsilon > zero be given. Another way of saying it would be to say,
let epsilon > zero be arbitrary. Different way of saying the same thing
essentially, at least in this circumstance.
What I need to do is find an n. Such that For all m >= n, one / m <
epsilon. Well, Pick any n such that n is bigger
than one / epsilon. Epsilon, if epsilon is a very small
number, this will be a very large number, so I might have to pick a large n.
Don't pick such an N. Now intuitively you know that since the
natural numbers go on forever as it were, getting bigger and bigger, that's always
possible. That actually uses a principal of
mathematics, call it Archimedean property, so named after Archimedes.
See assignments 10.2 for a little bit more about that.
So pick an n big enough so that n > one / epsilon.
Then, If m >=n one / m <= one / n < epsilon.
And we're done. Notice that the choice of n depended on
epsilon. Given an epsilon, I picked my n, in terms
of the epsilon. Please note, American Melanoma Foundation.
At least when you apply quantifiers in mathematics, it's a big deal if you get
them in the wrong order. The n depends upon the epsilon.
The smaller the epsilon is, the bigger the n has to be.
Intuitively, for this sequence. The smaller the tolerance you impose on
the number being close to zero the further out in the sequence you have to go, the
bigger the n is before your within that tolerance.
So quantify order matters. The choice of n depended on epsilon.
Different epsilon. Different n.
Okay. Let me give you one more of example.
And then let me look at this sequence. N / n +one of n = one to infinity, that's
a sequence a half, two thirds, three quarters, four fifths and so on.
And let me prove, in terms of the definition, that's n / n + one tends to
one as n tends to infinity. I mean intuitively it obviously does,
these numbers get closer and closer to one.
But to prove that, what I have to show is the following.
For all epsilon > zero, here is a natural number n such that for all natural numbers
m >= n Absolute value m / m + one - one < epsilon.
Okay. I am going prove it the same way as
before. I'm going to assume I'm given an epsilon,
then I am going find an n, dependent on epsilon to make this thing true.
The epsilon > zero will be given. We need to find an end such that for all m
> = n m / m +one - one in absolute value is < epsilon.
Well I'm going to pick in so large that n > one / epsilon.
That's actually the same choice as in the previous example.
It just works out that way because I am using very simple examples, in more
complicated examples, the choice is perhaps no quite so simple as straight
forward as this. Okay.
Well lets see why that one works in this case.
Stand for any m >= n Of the following. M / m + one - one in absolute value,
equals - one of n + one in absolute value, just work that out as a single fraction,
which is just one / m + one < one / m <= one / n because m >= n which is less than
epsilon. And we're done.
Okay, you should now be in a position to attempt the questions on assignment 10.2.
Bear in mind that the two examples I've given are particularly simple ones where
the choice of the n was very simple. You won't always be so lucky, you might
have to work a little bit harder to find the n.
But my focus here, as it's been throughout the course, wasn't so much on the details
and on the procedures. And if you graduated from high school, I
assume you can do mathematical procedures. You can do the algebra.
The focus is on the reasoning. In particular, in the context of
sequences, the key thing is how, given an epsilon, you have to choose an end that
depends upon it. It's this quantifier switch that's really
crucial. For all epsilon, there is an n.
If you go on to further studies in, in mathematics, and in, particular, real
analysis, you're going to see variance of this kind of definition, not just for, for
limits of sequences but for the definition of continuity and, and then when you come
to look at the formal definitions of integration and so forth.
So you're going to come across this kind of a definition a lot.
Coming up with this formulation was one of the jewels in the crown of late nineteenth
century mathematics. Absolutely brilliant piece of work and it
gave rise to pretty well all of modern mathematics, at least all of modern
mathematics that's to do with, with real numbers.
Okay? Well, good luck with assignment 10.2.
Well, you made it. All you have to do now is complete the
final assignment, take the final exam, participate in the peer review process.
And you should be well set for taking college level mathematics courses.
Or for simply reasoning more effectively in your job or in your everyday life.
Since you got this far, I'm sure you'll complete these last steps, so let me offer
you my congratulations now. These massive online courses are still
hugely experimental, the software platform is still being developed and debugged.
And the peer review process is very much a work in progress.
Moreover, this is the first time I or any of my TAs have designed or given such a
course. So simply sticking with us and finishing
the course is an accomplishment all by itself.
Future generations of students will surely have it easier in some ways.
What they won't find easier however is the material itself.
Learning facts, is relatively easy. Learning to think a whole new way is
extremely difficult. However well you think you mastered the
material, you should feel pleased with completing the entire process under your
own motivation. As a life-long teacher, much of my
motivation and reward has been getting to know each generation of students and
working with them, helping them overcome difficulties and then seeing them succeed.
With a mook, that regular close contact's absent, and I miss it.
On the other hand, being able to reach tens of thousands of students, as opposed
to 25, brings its own reward. Thank you for sharing the past five weeks
with me. And good mathematical thinking.