Originally posted by ChronicLeakyWhat about non-Cantorian set theory which adds the negation of the 'axiom of choice ' to restricted set theory?
This thread is not about physics. Bertrand Russell didn't "debunk" any of this -- it's all consistent with Zermelo-Fraenkel set theory and none of the usual fiddly bits of the foundations of mathematics have shown up yet, although they will (this discussion is likely to invoke the axiom of choice, for instance).
While Russell did, quixotically, atte ...[text shortened]...
EDIT I decline your suggestion that I play some chess, since I have no interest in it.
Originally posted by Thales2What's your point? I'm constructing numbers for BDN within the context of one particular formalisation of set theory. Besides, at best your question addresses the future -- the only sets I've dealt with so far are well-ordered, so despite their infinitude, there's no need for the A of C.
What about non-Cantorian set theory which adds the negation of the 'axiom of choice ' to restricted set theory?
The A of C is independent of the other axioms of Zermelo-Fraenkel set theory, so while you could perhaps perform constructions analogous to (but radically different from) the constructions I'm doing here, that would not address BDN's original question -- to define numbers in a way that is consistent with his intuition -- in the spirit in which it was asked.
If you're interested, DoctorScribbles and I had some discussions about constructivism a while back which deal, among other things, with some of the issues raised by the A of C, which I can find for you if you like. For the purposes of this discussion, though, we'll be using ZFC for the reasons explained above, although I'll be explicit about when I am using the A of C if that makes you feel better.
Perhaps you'd like to start yet another spin-off thread (you haven't started the field equation one yet, I see), in which you construct some objects without the A of C which agree with BDN's intuition about numbers.
Originally posted by ChronicLeakyLoL
I'll stop posting in this thread if you can continue it in the spirit it is going in with no serious errors, in a way that ultimately characterises the reals as a field with the same cardinality as the power set of N, on one hand, as a complete metric space on the other. Since this thread is for BDN's edification, you also have to include some discussi ...[text shortened]... continue this thread in this manner, by all means do so. If not, then f|_|ck off, troglodyte.
Hey rCL... I like it. You about made me choke! I am laughing good on this response to said trog.
I am enjoying this a lot. I am able to follow you so far, but still lack any intuition for it. I have to generalize it and try to visualize the ashtray as zero, red cigarettes as the original iteration to the right of said marker and green cigs to the left. Which tells you a lot about my poor old noodle.
Originally posted by StarValleyWyHey Mike, glad you like and thanks for reading. I think any sort of sensible visualisation like the one you describe is at least a good start to putting this stuff in your intuitive mindset.
LoL
Hey rCL... I like it. You about made me choke! I am laughing good on this response to said trog.
I am enjoying this a lot. I am able to follow you so far, but still lack any intuition for it. I have to generalize it and try to visualize the ashtray as zero, red cigarettes as the original iteration to the right of said marker and green cigs to the left. Which tells you a lot about my poor old noodle.
If the trog starts that spin-off thread in a way that BDN finds helpful, I'll quit smoking 😉.
Originally posted by ChronicLeakyhttp://www.blitzgeist.com/maths_notebooks.html
Hey Mike, glad you like and thanks for reading. I think any sort of sensible visualisation like the one you describe is at least a good start to putting this stuff in your intuitive mindset.
If the trog starts that spin-off thread in a way that BDN finds helpful, I'll quit smoking 😉.
This is what comes up when one searches for 'BDN' on the internet.
Are you the author?
Originally posted by ChronicLeakyWhy not just quit anyway?
Hey Mike, glad you like and thanks for reading. I think any sort of sensible visualisation like the one you describe is at least a good start to putting this stuff in your intuitive mindset.
If the trog starts that spin-off thread in a way that BDN finds helpful, I'll quit smoking 😉.
Originally posted by ChronicLeakyThen why not just quit this internet forum and smoking together, thereby avoiding the need to make bold and pointless wagers as well as improving your life expetancy?
Because then I'd have fewer things to stake in bold and pointless wagers on internet forums, obviously.
Originally posted by Nordlyslol
It's a whole new mathematical discipline. Fag maths.
My son works training various people from around the world in chemical warfare anti-terrorism methodology.
He had a group of GB troops in a week long exercise and he was quite proud that at the end of the week he had most of them convinced that the phrase "I need to bum a fag" was not a thing you want to say to an american.
Originally posted by StarValleyWyExactly. You're a programmer, you'll understand the recursive construction of the natural numbers noe problem.
Hmmm... if one quit f(quit)... f(ff(quit,fersurequit))... f(fff(((quit,ferdamnsurequit)))...
Naw. It can't be that simple. I better quit while i'm beheaded here.
Right, after this amusing little interlude, here's the next bit.
Having constructed the integers, we'll concisely list all the properties in a way that justifies the use of some more familiar notation, and then we'll expand things.
The set Z of integers is a countable set equipped with a total ordering relation > ("greater than or equal to"😉 such that for all n, m and p in Z:
1. m < m
2. m < n and n < p implies m < p
3. m < n and n < m imples m = n
4. m < n or n < m
Z is also equipped with two operations (functions from Z x Z to Z), called "+"
and "*" (we just write mn instead of m*n) such that:
1. m + (n + p) = (m + n) + p
2. There is an element 0 of Z such that 0 + m = m for all m
3. For all m, there is an element, -m, such that -m + m = 0
4. m + n = n + m
(Recalle these properties mean that Z and + together constitute an "abelian group"😉. Also,
1. m(np) = (mn)p
2. mn = nm
3. m(n + p) = mn + mp
4. There is an element, 1, of Z such that 1m = m
The preceding 8 properties define Z with + and * as something called a
"commutative ring". This is an algebraic structure -- a set together
with some functions relating its elements. If we deduce things about the
integers without invoking the rules about ordering, then our results will be true for any other structure obeying the same relationship-axioms, ie for all commutative rings.
However, we can prove some other things about the integers, eventually using the ordering. First, note that if n is an integer, 0n = (m - m)n for any integer m, and this is equal to -mn + mn = -(mn) + (mn) = 0. Also, the ordering implies that 0 and 1 are different integers (a fact not contained in the algebraic properties of commutative rings -- there are commutative rings in which this is not the case). Finally, if we have two integers, m and n, and we specify that n is not zero, and that mn = 0, then:
mn = mn - mn,
so, using the rules above:
mn + mn = mn
m(n + n) = mn
First, suppose m > 0 and m is not 0. Then if n > 0, we have that m(n + n) > mn and they are not equal. Likewise, if n < 0, we have the reverse situation.
Similar considerations show the problem is the same if m is less than 0. Thus m must be 0. We codify this in two additional algebraic rules obeyed by the integers:
1.) 1 =!= 0
2.) mn = 0 implies either m = 0 or n = 0 or both
Formally speaking, this makes the integers something called an "integral domain" (you guessed it -- the name "integral domain" actually comes from the word "integer" because the formal structure "integral domain" is designed to abstract the algebraic properties of the integers). It's worth noting that when one speaks of an integral domain abstractly, one does not assume an ordering of any kind. Instead, one adds to the commutative ring axioms the two things we just proved, as additional axioms.
The main attraction of the two results we just proved is that, under certain conditions, we can "invert" the operation of multiplication. Formally, if a, b and c are integers and ab = cb and b is not 0, then if ab = 0, we know a and c must both be 0. If not, then ab - cb = 0 = (a - c)b. But we know that either b or a - c is 0, and we assumed b isn't, so a - c = 0, ie a = c. This theorem, the "law of cancellation", shows that under some conditions we can, in the integers, do what you intuitively recognise at division.