Originally posted by StarrmanAgreed! And because I haven't had anything to wreck in ages, I just wrecked the whole set (that's the set of posts written by ChronicLeaky in this thread until now containing more than one line). The first page of the wreck list will be flooded with maths posts - sweet!
More people shoul rec this thread, this is the sort of thing that needs to be at the top of the recommendation list.
I don't have time to digest the last few posts properly right now--hope to have absorbed them by Monday--but it would be interesting to see what kind of comments the already mathematically clued-up might make.
A sub-thread to this entire thread, which as yet remains implicit and may never manifest, is, What is the deep connection if any between H.P. Lovecraft and Riemannian numbers?
Originally posted by Bosse de NageDepends on how complex an analysis you require. At the end of the day the association may not be real but just the product of an imaginary relationship.
I don't have time to digest the last few posts properly right now--hope to have absorbed them by Monday--but it would be interesting to see what kind of comments the already mathematically clued-up might make.
A sub-thread to this entire thread, which as yet remains implicit and may never manifest, is, What is the deep connection if any between H.P. Lovecraft and Riemannian numbers?
Originally posted by kmax87Its always difficult to integrate the scant sum total of knowledge any mere mortal could opine on Riemann into a debate anyway.
Depends on how complex an analysis you require. At the end of the day the association may not be real but just the product of an imaginary relationship.
The following construction of the integers is motivated by the fact that counting gives rise to arithmetic on the natural numbers, but this arithmetic has certain counterintuitive limitations which necessitate expanding things a bit. It will be helpful, at the beginning, to refer back to previous posts.
Remember that we constructed the natural numbers as sets of sets. Suppose we have two natural numbers, called m and n. Suppose that, in the languag of the ordering we defined earlier, m < n. Now let's build a set, m' which is any set we like, subject to two conditions:
1. m' and m are disjoint. This means that anything in the set m is not in the set m'.
2. There is a bijection from m to m'.
Now consider the union of the sets m' and n. This is the set consisting of everything in m', together with everything in n. Note that since m < n, the set m is contained in n. Since m' and m are disjoint, nothing in m' is in n, so every element we add to n by taking its union with m' is new. Let p be the natural number which for which there is a bijection between p and the union of m' and n. A little reflection will show that this p is unique, and we define the sum of the natural numbers m and n by this third natural number, p. As a shorthand, we write m + n = p.
Shout if it isn't clear why this definition formalises the notion of addition of natural number with which you are familiar.
There's another way to construct addition on the natural numbers which is equivalent to this, and I'll introduce it because it is an example of the phenomenon of recursion, which is sexy and which we'll meet again later.
Recall that N' is the set of natural numbers. Define a function, called "+", from N' x N' to N', in the following manner:
First rule: +(1, 1) = 2
Second rule: for any m and n in N', +(m, n+1) = +(m, n) + 1.
Third rule: +(m, n) = +(n, m) for all m and n in N'
In the second rule, we've taken advantage of the fact that we showed before what "+ 1" means, in terms of bijections. Thus, formally, the "+" symbol of our function means something slightly different from the "+" symbol in "+ 1". The function is defined recursively: if we'd like to add m and n + 1, we take +(m, n) and add 1 in the way we demonstrated earlier. Of course, this requires that we know what +(m, n) is. But we do. We find the unique k, if it exists, such that k + 1 = n and add 1 to +(m, k). If such a k does not exist, then we're down to n = 1, and we invoke the third rule: +(m, 1) = +(1, m). Then, if m does not equal 1, we repeat the descent, until our answer is +(m, n) = +(1, 1) + 1 + 1 + 1 + ... + 1. We know what +(1, 1) is, by the first rule, so we're done. As a convention, we'll write "m + n" instead of "+(m, n)".
(This might be a nice place to introduce a real formal proof: we might want to show that these definitions really are the same, by showing that for all pairs (m, n), they give the same answer. I'd be happy to talk through this, but it would probably be more beneficial if you thought of some ideas about how to go about it, first, and then we talked it through together. If you'd like to do this, I'll start you off by suggesting you consider the smallest n such that +(n, m) by the first definition gives a different answer that the second definition for some m. From there, using anything mentioned thus far, try to deduce some conclusion which contradicts something that's already been said. This will allow you to conclude that there can be no such n.)
If we spent some time adding natural numbers, we'd notice some disconcerting things. Addition as we've defined it generalises the counting process; before, to jump around the set of natural numbers, we had to proceed by steps of 1, and this could only take us from a smalle number to a larger one. We then learnt how to jump from small numbers to large in bigger steps, by addition. However, you may have noticed in the second definition of addition that we mentioned that to each natural number n different from 1 there corresponds a natural number k such that k + 1 = n. This was unproblematic at the time, since it was clear what adding 1 meant. However, we may wonder how to find the natural number k such that, for some given m, k + m = n.
We could try to construct a definition of subtract like the ones above, but it won't be very tidy, because whether such a k exists depends on m and n. Instead, we are going to expand our number system and extend the definition of addition so that, within this new system, we can move with equal ease in both directions.
We'll start by abstracting some properties of addition on the natural numbers from the above definition. In what follows, m, n and k can be any natural numbers.
Property 1 ("commutativity"😉: m + n = n + m (this follows directly from the third rule of the second definitiong)
Property 2 ("associativity"😉: (m + n) + k = m + (n + k) (this follows directly from the second definition; brackets mean "do the addition in the brackets first, then take the result and add it to the natural number outside the brackets"😉
Another result which is open to proof (a bit easier than the preceding one) is that m < n if and only if there is a natural number k such that m + k = n.
With these properties in mind, we'll ask a very simple question: given a natural number n, what natural number k is such that n + k = n? A little reflection shows there isn't one. Let's make one up. Define a set, N'', consisting of N' together with an element we'll call 0, such that for all n in N'', n + 0 = n. When we want to add two elements of N'', we apply the preceding rule if at least one of them is 0, and the original addition definition if they are both natural numbers. You may wish to check that properties 1 and 2 still hold for all of N''. 0 is called the "additive identity", and it is unique because we said so -- no natural number has the property which defines it, and it is defined only by that property, so anything else with that property IS it.
This helps a bit, but now we ask our nagging little question of all of N'' -- given some k, for what m in N'' does k + m = 0? A little more reflection shows that if k = 0, then m = 0. Otherwise, we're out of luck, since addition of any two natural numbers gives another natural number, of which 0 is not a member.
We get rid of this problem by expanding our universe yet again. Let's define a set, called M', such that there is a bijection f:N' --> M and M and N' are disjoint, and 0 is not in M. Then we consider a set called Z which consists of everything in M together with everything in N' and the element 0.
We now define a function + : Z --> Z in steps. First of all, if m and n are in N'', + is just as it was before. In addition, we make use of our bijection f to make a new rule: for all n in N', n + f(n) = f(n) + n = 0. Also, we stipulate that 0 + n = n for all n in M. If m and n are both in M, then by the bijectiveness of f, m = f(x) and n = f(y) for some x and y in N', where we know how to add, so we define m + n = f(x + y). If n is in N' and m is in M, then we note that m = f(x) for some x in N' and consider two possibilities. If x < n, then m + n is the unique natural number such that x + m + n = n. If x > n, then m + n is the unique natural number such that x + m + n = x.
By convention, we write f(x) as "-x" and "x + -y" as "x - y".
If we went through and abstracted everything we've done thus far, we'd find that in fact everything we've said about Z is encapsulated in the following rules, which we'll take to define Z from now on:
Z is a set, together with a function "+" from Z x Z to Z such that:
A1. For all x, y in Z, x + y = y + x
A2. For all x, y, w in Z, (x + y) + w = x + (y + w)
A3. There exists an element 0 in Z such that x + 0 = x for all x in Z
A4. For all x in Z, there exists -x in Z such that -x + x = 0
F1. There exists a bijection from N' to Z (to see where this comes from, let your bijection pair 0 with 1, 1 with 2, -1 with 3, 2 with 4, -2 with 5, etc)
There is an ordering relation, " < ", on Z such that:
O1. For all x in Z, x < x + 1
O2. If x < y and y < z, x < z
I've organised these rules, which completely determine everything we've said about Z, according to the type of property they generalise. The ones marked "A" are "algebraic" properties -- they deal with the structure that the addition operation imposes on the set. Mathematicians often consider sets with operations defined ONLY by A1 - A4. Such a structure is called an "abelian group", and is probably the most basic structured object found in mathematics. Different situations give rise to loads of different abelian groups, each defined by adding in other sorts of properties, as we've done. (The word "abelian" means that property A1 holds. If it doesn't, we just have a group. Many definitions will include the addition rule, for all groups, that if x and y are in the group, x + y must also be, but this is superfluous because we already said at the beginning that the image of the + function is Z.)
The F property states that the integers are a group in a bijection with the natural numbers. This tells us that the integers must be an infinite set, but we can count them (by a method like the one I sketched). The F comes from the term "free abelian group". This refers to any abelian group which can be put into a bijection with the integers in such a way that the addition on the group mirrors that of the integers.
The O properties refer to ordering. They specify that given two integers, we can always tell which is the larger, unless they are equal.
(BDN is a better linguist than I, so he should be able to work out why we use the letter Z to denote the integers.)
What have we accomplished? We built natural numbers and defined addition on them as a way to generalise counting. In order to make our addition work in every possible way (at least according to the group axioms, which are just generalisations of our additiontuitions), we expanded the realm in which we can add to a larger set, the integers, while carefully preserving every property of the natural numbers in an expanded, analogous property (algebra, countable infinity, ordering).
To construct more numbers, it's clear that we'll have to find something new to do in the integers that doesn't quite allow them to have nicely intuitive algebraic properties anymore, requiring us to build new numbers. Next time, we'll learn to multiply in the integers, decide that there are multiplicative things we want to do that the integers won't allow, and so we'll expand things again and build the rational numbers.
Originally posted by ChronicLeakyHow about playing some chess and cutting the cackle on a subject about which you seem to be more confused than most.
It's time for a small digression on sets. A full discussion of set theory is not really necessary here, but we'll need a few ideas. It's hard to find sources on sets which don't draw most of their examples from other maths, which doesn't seem to be what you're looking for, but I've stuck a few links on the bottom of this post.
One of the first peop ...[text shortened]... l-written exposition so far, so ask for clarification wherever required.
Originally posted by Thales2Hahaha, is this going to be another of your ridiculous claims that you can't substantiate? You do realise this is a forum, designed for posting in? It's a shame you don't have anything to contribute to a conversation other than racist cackle and yet I imagine you expect people to read what you write.
How about playing some chess and cutting the cackle on a subject about which you seem to be more confused than most.