littlebabyjesus
one of Maxwell's demons
Would it be possible to give a brief outline of how the proof goes?
Epistemological Holism and The Incompleteness Theorem
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Would it be possible to give a brief outline of how the proof goes?
Can you remember the proof?
Over a distance of 14 years and never having used field theory since? 
I think it was to do with breaking down what operators ("+" in this instance) mean in terms of bijections of ordered pairs within a set that obeys the rules associated with a field. As such, you needed various results from set theory. But recreate the actual proof? I wish.
I think it was to do with breaking down what operators ("+" in this instance) mean in terms of bijections of ordered pairs within a set that obeys the rules associated with a field. As such, you needed various results from set theory. But recreate the actual proof? I wish.
littlebabyjesus
one of Maxwell's demons
I think that 1 + 1 = 0 is only valid in a ring with two elements, in which case 2 is identical to 0.
I'm not going to start debating field theory at this point though. I know when I'm over my head!
Not quite it could be the direct product of rings with two elements. Why is all this stuff still in my head?
littlebabyjesus
one of Maxwell's demons
I don't quite get this. If you add something that isn't additive identity to something else, you can't come up with additive identity unless the thing you're adding to the initial quantity is such that it reduces that initial quantity to the additive identity. And that thing that you're adding to the initial quantity cannot be the quantity itself unless the initial quantity is itself the additive identity. In other words, 0 + 0 = 0 but you cannot add any other quantity to itself and produce 0. You can only add its 'anti-self' to it to produce zero.
Sorry, probably not expressing that very well - the thought seems clear enough until I start trying to write it down.
Extend that thought: that's why it only works if 1 is its own inverse, i.e. the ring only has two elements (1 and 0). In other words, modulo 2 arithmetic.
I think you're going to have to explain that one further.
In abstract algebra you can do what you like (as long as it is consistent).
Here's a way to think of this one. Think Boolean algebra, think instead of "1" think "true" think instead of "0" think "false" think instead of "+" think "exclusive or".
The mathematical object we are talking about is a Boolean ring. Its very similar to a Boolean algebra except that it deals with "exclusive or" not "inclusive or".
littlebabyjesus
one of Maxwell's demons
Yes, I see that. However, looking at the statement 1 + 1 = 0 (modulo 2) from the outside, as it were, isn't the complete statement something like 1 + 1 = 0 and one loop has been completed.
Still only has two elements though.
It's just a multidimensional version of what you said.
An addition might look like:
(1,0,0,1,1)+(0,0,1,1,0)=(1,0,1,0,1)
No, less of the "as it were", please, this is maths. There is no "outside". There is no counting of loops. 1 + 1 = 0 (mod 2) and that's the end of it. 1 is its own inverse in this ring.
Right, I see. I already alluded to the fact that I was in one-dimensional space, though, when I said it was about ordered pairs. I can't remotely remember the specifics, but I would imagine that any proof of 1 + 1 = 2 would rely on that. So yes, restricted in scope. But still relevent to the 1-D world.
littlebabyjesus
one of Maxwell's demons
Ok, no as it were. However, imposing the limit mod 2, you are limiting the amount of information you're providing. You are simply saying whether or not the result is divisible by 2 - 0 for yes, and 1 for no.
So, that 1 + 1 = 0 (mod 2) does not mean that it is possible that 1 + 1 = 0 .
So, that 1 + 1 = 0 (mod 2) does not mean that it is possible that 1 + 1 = 0 .
This is ordered pairs as well, in my example it's ordered pairs of ordered qunituples. Tell me if I'm annoying you...
littlebabyjesus
one of Maxwell's demons
EVERYTHING is maths.
Do you have a problem with the idea that 1 + 1 = 2 can be proved within a one-dimensional vector space? Bear in mind that I'm trying to remember the details of a 14 year-old problem, here. That's half a lifetime of maths ago.
TruXta
tired
Fucking platonist you!
1+1 is distinct from both 1 and 0 in monogenic fields of more than two elements (and you can label 1+1 with this symbol "2"). I will accept that. However I'm not sure that's what your proof was about.
TruXta
tired
Yeah, but you don't need to talk about field theory in order to understand the incomp theorem. Anyway, play on kids. I'll get off your lawn.
littlebabyjesus
one of Maxwell's demons
I'll give you this: I regret bringing up field theory
I don't. It's not something I know much about and it's made me think.
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