This motivates yet another definition: if we can write an integer p as p = mn, then we say that "m divides p" and write m|p (we would then also say n|p). There are huge and fruitful applications of this principle, both in the integers specifically (this is called multiplicative number theory) and in more abstract settings. For the moment, though, we have more numbers to construct, so we're just going to use some consequences of this notion to expand our universe of discourse again.
You may find it annoying that the "division" hinted at above is not something that can be readily performed in the integers, because "most" of the time, not only will a given integer not divide another one, but we won't even be able to find an integer (other than 1 and -1) which divides both members of any given pair of integers*. There are several ways out of this situation, and most of them involve changing the integers into something called a "field".
You'll notice that, regarded as an integral domain, the integers contain two elements, 1 and -1, which divide 1. In general, we call such an element a "unit". If we had some hypothetical integral domain in which every element other than 0 were a unit, what would happen? If m and n are elements of this hypothetical integral domain, then they are units, so there exist p and q in this domain such that mp = 1 and nq = 1. This means that (mq)n = m(qn) = m. In other words, there is an element, mq, which, when multiplied by n, yields m -- this is intuitively the essence of division.
From this we can speculate that if we were to embed the integers in a larger context -- an integral domain in which every element is a unit (such an integral domain is called a "field"😉 -- which preserved the properties of the integers already laid down, then we'd have a useful new domain of numbers in which we can do more arithmetic. We'll proceed how we did before, by naming a set, Q, and starting to put things in it.
First, we stipulate that every integer is also an element of Q. Next, for each nonzero integer m, we add an element m' to Q, subject to some constraints. First, we state that 1' = 1 and -1' = -1. In other words, we don't add anything new for 1 and -1. In every other case, we do add to Q a new element m' for each m.
Next, we define a function *: QxQ --> Q such that if m and n are in Q and also integers, then m*n gives the same integer that it did when we were multiplying in Z, and also that, for all m in Q, m*m' = 1. We then fully define * by requiring that Q\{0} (the set of nonzero elements of Q) be, with respect to this multiplication, an abelian group.
Notice that this requirement automatically adds elements to Q that weren't there before: it's implicit in the definition of an abelian group that the image of the operation-function is contained in the original set. However, when m and n are different inegers, Q as defined so far does not contain mn'. However, the requirement that Q\{0} be an abelian group with this multiplication automatically adds all such expressions to Q. They're well-defined, too, as far as multiplying goes: mn' is the element of Q which, when multiplied by n, gives m(nn'😉 = m1 = m, because our * must have the property that (mn'😉n = m(nn'😉, and m and 1 are just integers which we know how to multiply. Note that every element of Q is now a unit.
For the rest, we define multiplication by (mn'😉*(pq'😉 = (mp)(nq)' = mpn'q'.
Furthermore, we can define an addition operation on Q: if m and n are in Q and both integers, we add them as before. However, the above discussion reveals that we can write every element of Q in the form mn'. We thus DEFINE our addition function by:
For integers m, n, p, q (n and q are not 0), mn' + pq' = (mq + np)(nq)'. We can check that this reduces to integer addition when mn' and pq' are integers, and that in fact all of the axioms of abelian-groupness are satisfied on the whole of Q.
You may have already picked up on a little problem. If mn' is in Q, and there is some integer s which divides both m and n, the we can write m = as and n = bs, so mn' = (as)(bs)' = asb's' = ass'b' = a1b' = ab'. This means that each element of Q may be defined in several essentially different ways. We get around this by separating the elements of Q into classes in the following way:
If mn' and pq' are in Q, then we say mn' = pq' if and only if mq = np. This allowas us to cleanly split Q into subsets, each of which contains elements which are all equal according to the above definition. We then choose a representative element of each of these subsets in an unabiguous way: we choose the element mn' which n positive and as small as possible (in other words, we invoke the ordering of the integers again). This is familiarly known as reducing fractions to the lowest common denominator. We now define Q to be the set of these representatives. Notice that this certainly doesn't take any integers out, since they all take the form n*1', and 1 is as small as positive nonzero integers get. Also, it doesn't affect the operations we've defined in an essential way; all we have to do, after performing an operation and getting some result, is to look up our result in the division of Q into subsets, and replace it with the representative of the subset in which it appears.
We've got enough intuition now to get rid of the silly ' business, and write "1/m" for m' and "n/m" for nm'.
We've now got a set Q with two operations, + and *, such that:
1. Q with + is an abelian group
2. Q\{0} with * is an abelian group
(Algebraic properties)
3. Q has a totally ordered subset (the integers)
Given everything we've done thus far, we can extend this ordering to all of Q: if we have m/n and p/q in Q, we say p/q < m/n under exactly one of the following circumstances:
-- nq > 0 and mq > np
-- nq < 0 and mq < np
(Note that if we're scrupulously following the "each element of Q is represented by an element with a positive denominator" convention we introduced above, the second case won't arise.)
This ordering is, by definition, derived from the well-understood ordering of the integers, and reduces to that ordering when the elements of Q in question are integers.
Now things get counterintuitive and fun. First, suppose A and B are elements of Q and A < B, but not equal. Then (1/2)(A + B) is in Q, since it is an abelian group under multiplication and addition. But if we write A = m/n and B = p/q, then the above expression is m/2n + p/2q = (2mq + 2np)/4nq. Applying the above definition shows that this element of Q is both greater than A and less than B. In other words, between any two elements of Q, no matter how close, there is another element of Q. This is in stark contrast to the integers, where for any integer n, there is a unique "closest" greater integer, namely n+1 and a uniques "closest" smaller integer, namely n-1. The ordering on Q is an extension of the ordering on Z, and yet we've just seen that between any two integers there is an infinite set of elements of Q. We say Q is "densely ordered".
You may not find this problematic until you consider the following. Remember that elements of Q are, by construction, uniquely expressible in the form m/n, where m and n are in Z. Thus there is a subset, which we'll call S, of ZxZ which such that there is a bijection between Q and S: simply associate m/n with (m,n). Note that S is not the whole of ZxZ, because, for instance, (2,4) is not in S, since the element 2/4 of Q is equal to 1/2 by our construction.
Now, for each integer N at least 0, consider the subset of ZxZ, S(N) = {(x,y) : x,y are in Z, and xx + yy < NN}. For instance, S(2) contains (0,0), (-1,0), (1,0), (0,-1), (0,1), (0,2), (2,0), (-1,1), (1,-1), (1,1) and (-1,-1), (-2,0), (0,-2) (remember I'm using < to mean "less than or equal to"😉. Note that S(N) is a finite set for all N, and that S(N+1) contains S(N) for all N. If you'd like to picture this informally, think of a sheet of graph paper with (0,0) marked off at some point. Then mark the 4 line-intersections at a distance of N from this point, and connect your points with a circle. S(N) consists of the line-intersections on or in this circle.
Now S(N) is finite, and contains k(N) elements, so we can number the elements of S(N) any way we like. Let's call them a(1), a(2), ... a(k(N)). Then S(N+1) contains a(1) ... a(k(N)) and, in addition, the finite number of pairs of integers (x,y) such that xx + yy = N+1. There are at least 4 of these and rather less than N+1 of them. Let us call their number l(N+1). Then we see that ZxZ is the union of the sets S(0), S(1)\S(0), ... S(N+1)\S(N) .... These sets are all disjoint, and the Nth one has a finite number, l(N), of elements. There is thus a clear bijection between ZxZ and the natural numbers. Since S is an (infinite) subset of ZxZ, there is clearly a bijection between Z and the natural numbers. Since there is a bijection between Q and S, there is also a bijection between Q and the natural numbers.
You'll remember that Z is in a bijection with the natural numbers. Thus, despite the fact that Q is densely ordered and stretches to infinity in both directions, there is exactly one element of Q for each integer. We say Q is "countable" and readjust our intuitions to deal with this seemingly strange confluence of facts.
Thus we have completely characterised a new set, Q, of numbers -- the "rational numbers". The rationals constitutes THE totally ordered field which can be put into a bijection with the integers and which is densely ordered. Any other structure with these properties is formally equivalent to the rationals.
The obvious thing to do next is to construct the real numbers. This can be done in many ways. The most obvious way to do it from here (ie the way best motivated by our construction of the rationals) is to construct the so-called "topological completion" of the rational numbers and correspondingly expand our arithmetic. From there, I can define a few more things and satisfy the demands I made to Thales2. This will require a bit of technical jargon and the rearrangement of a few stops on your intuition organ. This is probably good; we haven't thought much about the size and measurement implications of our ordering, and this would accomplish this.
We could also satisfy my demands by constructing by so-called "Dedekind Cuts". This approach has the advantage of building directly on the idea of the rationals as an ordered field, but has the disadvantage of requiring a bit more background material, as well as missing out important and interesting ideas from analysis which the first approach uses.
Finally, there are slightly less rigorous approaches to constructing the reals. One of my first-year lecturers, a fellow called Martin Liebeck, wrote an excellent book called "A Concise Introduction to Pure Mathematics", which constructs the reals in terms of their decimal expansions, as well as introducing the geometric idea of the "number line". I'd recommend giving this or a similar book a look, but unless it's desparately desired, I'm not going to use these approaches for the construction of the reals in this thread, because they don't depend on much of what we've done so far. I will, however, explain these ideas in terms of the construction I do use.
I'm leaning toward the first method at the moment, but it may be several weeks before the next installment, so I have plenty of time to make up my mind.
*(The first serious result I ever proved, a little before I became RC, made precise the definition of "most" in this sentence -- given two integers n and m, the probability that there is no integer other than 1 and -1 which divides both of them is 6/(pi^2). This is pretty standard and well-known, but I dug it at the time.)
Originally posted by Bosse de NageThat's a good question. I have often wondered about how one could improve maths lessons in school. When I went to school, I kept having the feeling that I didn't really have an idea what I was doing. I was one of the best students, I could follow all the steps and solve the problems, but I kept feeling there was something missing. I couldn't pinpoint the problem, but in retrospect I believe it was the missing groundwork. It was only during the last years of highschool (where I specialised in maths) and in an analysis class I took at university just for fun that I got some of that groundwork, and it was deeply satisfying. I think I would have profited from hearing about that much earlier. But on the other hand, some people seemed to have trouble understanding the relevance of it even at university, so it would probably be quite a challenge to present it in a way that would be both interesting and comprehensible for younger students, most of whom don't have a particular interest in maths (but maybe they don't have an interest because they have never been confronted with real maths?).
Fascinating thread...If you were teaching this stuff, what would the youngest appropriate learner age group be?
Originally posted by Bosse de NageI'll echo Noodles' comments here, to some extent. I think that some experience with arithmetic, and maybe some intuitive appreciation for its contextual limitations, is needed to see the point of this. As for being able to handle the abstractions inherent in some of the definitions above (and a treatment without explicitly talking about, say, groups, would be less abstract at the expense of being more technically detailed and less conceptually portable), I'd say anyone who can understand the rules of chess* and deduce some of their consequences is more than capable of the level of abstraction required by this discussion (I say "more than", because there are good intuitive concepts motivating the formalisation of arithmetic, while the rules of chess are pretty arbitrary).
Fascinating thread...If you were teaching this stuff, what would the youngest appropriate learner age group be?
The ability to follow this discussed comes from interest and motivating experience, which are probably mildly positively correlated with age. Actually, that's probably not even true. I think I could present a MORE abstract version of this to a mildly bright** child who didn't yet know how to add, conventionally, as simply the rules of a game, and xyr understanding would depend only on xyr interest level. I could also present it to, say, an accountant with no interest in it at all, and xe would probably not make the effort needed to understand.
I'm highly in favour of teaching set theory and some algebra in school before the "procedural" maths that one generally meets is even touched at all, because it tends to hold interest better, and "procedures" and algorithms can be learnt more easily if motivated by theory (and, strictly speaking, needn't be taught at all if the students have understood the theory, although I've never met anyone who could deduce every algorithm likely to be useful from first principles, without the benefit of knowing what is likely to be useful, which requires a lot of experience).
This sort of question is very immediate to me; I'm likely going to be teaching some undergraduate mathematics starting in September, and depending on how much curricular freedom I will have as a teaching assistant, I may take an "abstraction first, procedure later" approach to whatever it is I'm teaching.
*Thales2, I have now talked about chess in this thread.
** By which I just mean "linguistically adept enough to handle the clearest explanation I could muster".
Originally posted by ChronicLeakyWe actually did some set theory in grade 1 (and again later, I think). But this wasn't used as groundworks for anything else we did (or I missed the connection).
I'm highly in favour of teaching set theory and some algebra in school before the "procedural" maths that one generally meets is even touched at all, because it tends to hold interest better, and "procedures" and algorithms can be learnt more easily if motivated by theory (and, strictly speaking, needn't be taught at all if the students have understood ...[text shortened]... the benefit of knowing what is likely to be useful, which requires a lot of experience).
In grade 5 (first year of grammar school) we did some formal logic, which apparently wasn't part of the curriculum, but which our teacher thought was important. I agree with her, I think it should be part of the curriculum. I can see again and again that even adults have problems with basic logic (you only need to look at the discussions in Debates and Spirituality).
This sort of question is very immediate to me; I'm likely going to be teaching some undergraduate mathematics starting in September, and depending on how much curricular freedom I will have as a teaching assistant, I may take an "abstraction first, procedure later" approach to whatever it is I'm teaching.
That would be quite interesting. I hope you'll be allowed to try it. If you do, let us know how it goes! I am quite sure that approach would have worked well for me; but I tend to learn things "upside down" compared to others (of course in my view, it's the others who have it the wrong way 😉).
Originally posted by ChronicLeakyI had a flash back here. I remember reading about Bohr, Dirac Heisenberg and Schrodinger when they met severally and in pairs to try and work out the Quantum Physics. In 1926 Bohr was still the grand pooba of the new movement and he tended to drive everyone nuts with his meandering "philosophical questions". As Schrodinger said..."Soon you no longer know whether you really take the position he is attacking, or whether you really must attack the position he is defending."
** By which I just mean "linguistically adept enough to handle the clearest explanation I could muster".
Dirac was more blunt. He said "Bohr's arguments were mainly of a qualitative nature, and I was not able to really pinpoint the facts behind them. What I wanted was statements which could be expressed in terms of equations, and Bohr's work very seldom provided such statements." But then Dirac was a mathamatician and Bohr was ... "limited"? Perhaps.
But Heisenberg was determined to understand and hammer out a definition of "uncertainty" that would either preserve or destroy forever "determinism". And if determinism were to suffer then most great physicists thought that the end of physics would be close behind. Einstein was the leader in the effort to preserve classical physics in his backing of Schrodingers waves over the "probablility" option.
To make a long and interesting story short, Heisenberg and Bohr basically spent every waking hour for seven months seeking "words" or "a word" that would convey meaning to quantum mechanics. The first candidate they considered was "Ungenauigkeit" or "inexactness". They then published a paper "Uber den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik".
The key word root "auschaulich" they thought a good one.
So they were completely shocked to see three different journals translate this as:
1 - On The Perceptual Content of Quantum Theoretical Kinematics and Mechanics
2 - On The Physical Content of Quantum Theoretical Kinematics and Mechanics
3 - On The Intuitive Content of Quantum Theoretical Kinematics and Mechanics
They struggled until Heisenberg actually broke down in tears and they had to separate and seek out sanity each alone for a few months... Bohr skiing and Heisenberg beach combing and sunning on the rocks.
And when all was said and done? They agreed to issue a statement that the new concept was not definable currently. Bohr's exact words were... "Our words don't fit."
I wonder if this is an accident or if our intuitions which comprise our linguistics isn't limited by the sheer unknowability of "uncertainty"?
Anyway... a long little story with the punch line... "I would long to learn it, but my mind doesn't allow for it. I don't have the ability to match the concept to any known words. And my mind deals only with words and images." Poor me. Anyway, very interesting still. Carry on.
Originally posted by StarValleyWyIf I don't get back to this post properly within the next couple of months, harass me about it, because I'd be interested to talk more about QM, about which I don't know a whole lot.
I had a flash back here. I remember reading about Bohr, Dirac Heisenberg and Schrodinger when they met severally and in pairs to try and work out the Quantum Physics. In 1926 Bohr was still the grand pooba of the new movement and he tended to drive everyone nuts with his meandering "philosophical questions". As Schrodinger said..."Soon you no longer kno ...[text shortened]... ages." Poor me. Anyway, very interesting still. Carry on.
I read about a certain thought experiment recently, and it inspired me to write a bad short story called "God Plays Dice With Billiard Balls", which can be found here:
blog.myspace.com/d_in_t
if you are interested. Mad propz if you can recognise the experiment!
Originally posted by NordlysI very much agree that some instruction in logic is very important. Actually, given how quickly people learn things which interest them regardless of whether they are formally taught, and how poorly people learn things which bore them, no matter how well-taught they are, it would probably be better if schools focused more on sound methods of thinking than on specific subject matter.
We actually did some set theory in grade 1 (and again later, I think). But this wasn't used as groundworks for anything else we did (or I missed the connection).
In grade 5 (first year of grammar school) we did some formal logic, which apparently wasn't part of the curriculum, but which our teacher thought was important. I agree with her, I think it shoul ...[text shortened]... mpared to others (of course in my view, it's the others who have it the wrong way 😉).
I'm generally pretty upside-down myself.
Originally posted by ChronicLeakyI enjoyed the story a lot. Didn't we once have a discussion about "space billiards" during a chess game or two? (edit) Or maybe it was "spaced out monkeys and chimp credits"! I kept seeing Pauli sleeping in and leisurly driving everyone nuts as he set meetings to discuss exclusion vs. bose-einstein spin, missing said meetings and explaining that he couldn't be in two places at once and that his bed was much prefered in early morning than hard chairs in stale meeting rooms. lol
If I don't get back to this post properly within the next couple of months, harass me about it, because I'd be interested to talk more about QM, about which I don't know a whole lot.
I read about a certain thought experiment recently, and it inspired me to write a bad short story called "God Plays Dice With Billiard Balls", which can be found here:
...[text shortened]... yspace.com/d_in_t
if you are interested. Mad propz if you can recognise the experiment!
I have no idea of the EXACT EXPERIMENT. (pun intended)😉
Right, BDN, I've got the time and the inclination today and with the rest of my finals approaching, I might not for a while, so I'm going to build the reals for you. There's no need to respond to this or even read it right away; I'm just exercising a little Cauchion*.
Recall that we've thus far constructed the countable ordered field of rational numbers. I'm going to take the first approach to constructing the reals which I alluded to earlier, except I thought of a simple way to introduce the necessary concepts about metric spaces in the limited context of the rationals. The full truth shall come out later, once we've got the reals.
As we've done before, we're going to motivate our expansion of our number-world with an algebraic question. In fact, we're going to go back to the Pythagoreans to do it. Our question is:
Given a rational number a, what rational numbers x exist such that x*x = a? Must such rational numbers always exist? The answer, as a student of Pythagoras pointed out, is no, and the proof is such a classic that I'm going to reproduce it here in a potentially patronising amount of detail, in the language we've used thus far.
Theorem: There does not exist a rational number x such that x*x = 2 (a rational number).
Proof: Suppose that there does exist such a rational number x such that x*x = 2. Then, by definition, there are integers m and n such that m/n = x. Furthermore, recall that each rational is formally represented as the quotient of two integers such that no integer larger than 1 divides both of them. Colloquially, say we've chosen m/n in lowest terms.
By our assumption, m*m/n*n = 2. Thus m*m = 2n*n. This means that 2 divides m*m, so we can conlude that 2 divides m, for if not, then we'd be able to write m = 2k + 1 for some integer k (any integer can be written either in the form 2k or 2k+1 -- ie each integer is even or odd). Having done this, we find m*m = 4*k*k + 4*k + 1 = 2*(2*k*k + 2*k) + 1, which is a contradiction, since we know 2 divides m*m. Thus our assumption that m = 2*k+1 is false, and we can write m = 2*k for some integer k. Substituting this back into the equation m*m = 2n*n gives:
4*k*k = 2*n*n
or equivalently,
2*k*k = n*n. Thus 2 divides n*n, so by an argument identical to the one above, 2 divides n. Thus the
situation is:
1. 2 divides m
2. 2 divides n
3. No integer larger than 1 divides both m and n
Thus our options are to conclude that 2 is not larger than 1, which is absurd, or abandon our assumption that such an x exists. We have to opt for the latter or abandon the ordering of the integers, so we conclude that no rational x exists such that x*x = 2. QED**
By now, we're pretty comfortable with encountering limitations of this type in our number systems, and looking for ways to expand them. We might, at this point, suggest defining some object to be such that multiplying it by itself gives 2, and then tossing it into a set with our rationals, and finally defining some operations to extend the arithmetic of rationals to this new set. In fact, we could do that, and it gives an example of what's called an "algebraic number field of dimension 2", but it isn't really what we're looking for. For one thing, in this algebraic number field (take my word for this, I'm afraid, although the proof isn't formally too different from the one above), there is no x such that x*x =3. We might, undaunted by this fact, make up some object, call it "the square root of 3", and toss it into the mix. There is a field of study which generalises this process and leads to numerous interesting results, algebraic number theory***. However, this doesn't really give us what we want.
Notice that, if we adopt the notational convention of writing "x^n" to mean "multiply n copies of a rational number x together, where n is a positive integer", every expression we can make in a finite number of steps using our rational operations can be written as a sum of finitely many rational numbers, each of which is the product of finitely many rational numbers. In particular, given some integer n and some rationals, called a(0), a(1), a(2), a(3),...a(n), we can ask a more general version of the question which gave us the square root of 2:
"Is there a rational number x such that:
a(n)*x^n + a(n-1)*x^(n-1) + ... +a(1)*x = a(0) ?"
Such and expression is termed a "rational polynomial". In the special case where n = 2, and a(2) = 1, a(1) = 0 and a(0) = 2, we proved that the answer is no. If, however, we built a set which we stipulated to consist of all the rationals plus a set of objects which we defined, through extension of our arithmetic, to consist of numbers which, between the lot of them, satisfy every possible rational polynomial, we would still not have constructed the real numbers as we intuitively understand them. This can't be formally clarified until we've constructed the real numbers, unfortunately, but this set is a very important one, called the set of "algebraic numbers over the rationals", and we shall return to it later.
What other properties of the rationals can we take advantage of? Recall that we showed before that between any two rationals, there is another rational. We'll say that, by virtue of having this property, the rationals are a "dense set" (although, since we've yet to build the reals, we're abusing the standard definition of the term a little bit).
With this in mind, we introduce a function d:QxQ --> Q. "d" stands for "distance", for reasons which are about to become clear, and our definition is going to make Q into something formally very similar to one of those elusive metric spaces -- similar enough, in fact, to allow us to build the real numbers we need to be able to say what a metric space actually is****.
For any two rationals x and y, we say d(x, y) = |x - y|. This notation means:
1. Subtract y from x
2. If the result is at least 0, then d(x, y) = x - y
3. If the result is less than 0, then d(x, y) = y - x.
We'll prove four essential properties of this function d, which we'll call the "rational metric".
Theorem:
1. For all rationals x, d(x, x) = 0
2. If d(x, y) = 0, then x = y
3. For all rationals x and y, d(x, y) = d(y, x)
4. For all rationals x, y and z, d(x, z) < d(x, y) + d(y, z).
(Remember that < means "less than or equal to".)
Proof:
1. This follows directly from the definition, since x - x = 0
2. If d(x, y) = 0, then |x - y| = 0, so by definition x - y = 0, so x = y by the rules of our rational arithmetic.
3. d(x, y) = |x - y| = |y - x| by definition, and this is just d(y, x).
4. d(x, y) + d(y, z) = |x - y| + |y - z|. If both of the expressions within bars are positive, we have x - y +
y - z = x - z = |x - z| = d(x, z). If the first is positive and the second negative, we have x - y + z - y = x
- z + 2(z - y) > x - z = d(x, z). Finally, if both are negative, we have y - x + z - y = z - x = d(x, z).
We're finished.
Part 4. of the theorem is very important, but the general name for it makes no sense in this context, so we'll call it "part 4" when invoking it.
Now we can rephrase our statement about the rationals being dense in a more formal setting:
"For any two rationals x and y, there exists a rational z such that d(x, z) < d(x, y)."
This is actually a bit weaker than what we said before, since z merely has to be closer to x than y is; it needn't be between x and y, but this is as much as we need.
Next, observe that 0 is rational, and for any integer n, 1/n is rational. Also, observe that if m and n are integers, and n < m, then 1/n > 1/m. In particular, this means that no matter what distance you specify, I can tell you an integer n such that d(0, 1/n) is smaller than that specified distance.
Now suppose you have a rational x. Then, by the density property, if you tell me an integer n, I can tell you a rational y such that d(x, y) < 1/n. If you keep specifying bigger and bigger values of n, I'll keep specifying values of y closer and closer to x, ad infinitum.
This motivates a definition:
Convergence of Rational Sequences
We call an ordered subset of the rationals (ie a function from N to Q), a "rational sequence". If the nth term in the sequence is, say, x(n), we denote the whole sequence by {x(n)}.
If {x(n)} is a rational sequence, and x another rational number, we say "{x(n)} converges to x", and write "x(n) --> x", if the following holds:
For any integer m, there exists another integer M such that, for all integers n > M, d(x(n), x) < 1/m.
The best way to illustrate this is with an example. Suppose, for each integer n, x(n) = (n+1)/4n. Then, if you tell me an integer m, I'll say:
Let M = m/4. Then if n > M, we have d(x(n), 1/4) = |(n+1)/4n - 1/4| = 1/(4n). But n > m/4, so 4n > m, so
1/(4n) < 1/m. Thus x(n) --> 1/4.
Clearly not every rational sequence converges to something, but we notice that, of those rational sequences that do converge to something, they all share a certain property:
Cauchion
If {x(n)} is a rational sequence, then for any integer m, there exists an integer M such that if n and k are
integers greater than M, then d(x(k), x(n)) < 1/m.
The definition of convergence tells us that the terms of a convergent sequence, as we enumerate them, become arbitrarily close to some fixed (rational) value. The "Cauchion" condition tells us that the terms, as we enumerate them, become arbitrarily close to one another.
("Cauchion" comes from the fact that the standard name for such a sequence in a slightly more developed context is a "Cauchy sequence".)
We can now prove a theorem which will effectively give us a reason to build the real numbers.
Theorem:
If {x(n)} is a sequence of rational numbers converging to some (rational) number x, then {x(n)} has the Cauchion property.
Proof:
We've assumed x(n) --> x. Let's fix an integer m. Then, by the definition of convergence, there is an integer M such that if n and k are integers larger than M, the following are both true:
d(x(n), x) < 1/(2m)
d(x(k), x) < 1/(2m)
We can get away with using "1/(2m)" because, by the definition of convergence, we can choose any integer we like in the denominator of the right-hand side of those inequalities, no matter how small, and they will still hold, provided we choose M to be large enough. Now we invoke part 4 of the theorem on properties of d:
d(x(n), x(k)) < d(x(n), x) + d(x(k), x) < 1/(2m) + 1/(2m) = 1/m
Thus for any integer m, we can find and integer M so that, if n and k > M, d(x(n), x(k)) < 1/m. This is exactly what the Cauchion property states, so we're done. // or something
Finally, we mention a few facts which we'll state here and prove in more generality later. They're pretty intuitive, though:
1. If y(n) --> y and z(n) --> z, and for all n, we let w(n) = y(n) + z(n), then w(n) --> y + z
2. If y(n) --> y and z(n) --> z, and for all n, we let w(n) = y(n)*z(n), then w(n) --> yz
3. If x(n) --> x and x(n) --> y, then x = y.
Intuitively, this means that if a sequence of rationals grows arbitrarily close to a single rational value, then the terms in the sequence must also become mutually closer together. This seems pretty sensible; there's no real intuitive way to visualise things being otherwise. In fact, if we're visualising, it would seem the converse should be true too, and we ask whether every rational sequence with the Cauchion property converges.
To answer this, consider the sequence of rationals defined as follows.
First, x(0) = 1. Next, for each integer n at least 0, we define x(n+1) = x(n)/2 + 1/x(n). This first few terms of this sequence are 1, 3/2, 17/12, ... This suggests that this sequence has the Cauchion property, and indeed it's not hard to show it does.
On the other hand, suppose that there is some x in our set of rationals such that x(n) --> x. Then by what we stated above, x(n)*x(n) --> x^2. If we rearrange the definition above a bit, we see:
x(n+1)*x(n) = x(n)^2/2 + 1
Now, letting n grow, we know x(n+1) --> x, x(n) --> x, and x(n)^2 --> x, and, since two sequences (the left and the right side) which are equal in every term must, if they converge, converge to the same thing. This shows
that:
x^2 = x^2/2 + 1
In other words, x^2 = 2. Recapitulating: {x(n)} is a sequence with the Cauchion property, and converges to x, where x^2 = 2. But the first fact we proved in this part of the discussion is that no rational x exists which has this property!
Thus not every rational sequence with the Cauchion property converges to anything we know about -- under the process of sequence convergence, the rational numbers are incomplete, and this is what will motivate us to build the reals.
In a manner very analogous to our definition of the Cauchion property, we'll say two rational sequences {x(n)} and {y(n)} are "mutually Cauchious" if the rational sequence d(x(n), y(n)) --> 0. Intuitively, this means that the terms of {x(n)} and {y(n)} get as close together as we like, and it's not hard to see that if they each converge to a rational number, they converge to the same rational number; if x(n) --> y and y(n) --> y, then for any integer m, we can find an integer M such that n > M implies
d(x(n), x) < 1/(2m)
d(y(n), y) < 1/(2m)
so
d(x,y) < d(y(n), y) + d(x(n), x) < 1/m.
Thus d(x,y) is at most 1/m, for every integer m. Thus d(x,y) = 0, and so x = y.
Given a rational sequence with the Cauchion property, we define its "mutual Cauchion class" to be the set of all rational sequences to which it is mutually Cauchious. Clearly, every sequence in a mutual Cauchion class is mutually Cauchious to every other sequence in the class, and if two sequences are in different mutual Cauchion classes, they are not mutually Cauchious. Finally, each Cauchious sequence is in some mutual Cauchion class, because each sequence is clearly mutually Cauchious to itself.
We define a new set, R, which is the set of all mutual Cauchion classes which contain sequences with the Cauchion property. We call a mutual Cauchion class a "real number". Now given a rational number x, the sequence whose every term is x both converges to x and is mutually Cauchious to every other sequence which converges to x, and not mutually Cauchious to any other sequence. Thus each rational number is associated uniquely with a real number (a Cauchion class), and we may as well consider R to contain the rational numbers.
Let's define some arithmetic.
1. If X and Y are real numbers, then choose a rational sequence {x(n)} in the Cauchion class of X and a rational sequence in the Cauchion class of {y(n)}. Then the sequence {x(n) + y(n)} defines a Cauchion class and we call this Cauchion class X + Y (this is real addition).
2. If X and Y are real numbers in the Cauchion classes which respectively contain {x(n)} and {y(n)}, we find the Cauchion class of the sequence {x(n)y(n)} and call it XY (this is real multiplication).
It's not immediately clear that these operations are well-defined. We have to show that it doesn't matter which sequences we choose from the Cauchion classes of X and Y to define X + Y and XY.
Suppose {x(n)} and {z(n)} are both in the Cauchion class of X and {y(n)} is in the Cauchion class of Y. Then from the definition of d and mutual Cauchion, it's not hard to show that {x(n) + y(n)} defines the same Cauchion class as {z(n) + y(n)} and {x(n)y(n)} defines the same Cauchion class as {z(n)y(n)}. I'd recommend trying to prove this formally, and am happy to talk about it in detail if you get stuck. This proves that our real addition and real multiplication are well-defined; any two real numbers have a unique sum and product, which is also a real number.
Also, we can prove that for real numbers X, Y and Z, (X+Y)+Z = X+(Y+Z) and (XY)Z = X(YZ) and X+Y = Y+X and XY =
YX. These all basically follow from the analogous properties of rational addition and multiplication.
Now consider the Cauchion class of 0. This is represented most simply by the rational sequence {0}, and indeed this shows immediately that for all real numbers X, X + 0 = X, and that for any X, we can define a real number -X: if {x(n)} is in the Cauchion class of X, -X is the Cauchion class containing {-x(n)}. All of this tells us that R is an abelian group under our addition!
Similarly, the Cauchion class, called 1, which contains the sequence {1}, is a multiplicative identity in the reals, and that if X is a Cauchion class different from 0, containing a sequence {x(n)} (which does not converge to 0), then the sequence {1/x(n)} has the Cauchion property and its Cauchion class, called X^-1, is such that XX^-1 = 1.
The reals are thus a field with respect to these two operations, and we can associate each rational number, as we saw, with a real, and if we take the subset of R containing Cauchion classes of rational sequences which converge to rational limits, then adding or multiplying them as real numbers gives other Cauchion classes containing rational sequences which converge to rational numbers, and these rational numbers are the same ones we'd get if we added or multiplied our original rational number with rational arithmetic. This is all justified by the things we have proved so far, and I've expressed it non-technically because it's a bit of a digression. Formally, it shows that the set of real numbers which are Cauchion classes of rationally convergent sequences is "isomorphic" to the set of all rationals. This means that, for structural purposes, we may regard these sets as the same set, and conclude that, at least in this sense, the real numbers contain the rationals.