Since you have gone down this route I think I will throw the Banach-Tarski paradox into the mix. A rather nice video on it can be found here on YouTube:If you divide an object into pieces, you can't just assume that you keep all the previous axioms without introducing contradictions, you have to show it. Your axioms ought to be written in terms of the most basic building blocks of the system.
Again we have confusion. I can see what you might be trying to reach for but you still aren't defining the "major players" all that well.Chiro, Dan, Archie
I think I finally understand the mistake I have been making. I have spent the last several hours trying to find a solution to Chiro and Dan's last few post. I believe I might have done so...
A = any number
S = any set
Let the ordered pair (x,y) be described as follows...
((z1,z2),(z1,z2)) → (x,y) : ∀A in S
∀A ≠ 0: (z1,z2) = (A,A) = A
∀A = 0: (z1,z2) = (0,1) = 0
combined with the axiom
"Any A in binary operation of multiplication or division is only representing z1, or z2."
Sure! We always like good feedback. But please put the thread in the Feedback Forum. (And you don't need my permission to post anything so long as it conforms to the Forum Rules below.)Also I would ask permission from Dan, to post a new topic extolling the virtue's of the members of this forum, regarding this EXCELLENT peer review.
But if (2, 2) is what we would typically call the number "2" why all the extra work? Not counting the 0 stuff it would seem that you are just adding complexity where none is needed. You could simply say that the number 2 has the properties of space and value and not worry about the whole formalism. Is there ever a situation where we would have, say (2, 3)? What would that mean?Dan
(z1,z2), are not members of any set...(they be considered pieces of the members of the set)
"z1 and z2 for A, (any number) is "really" just the number given put into what appears to be an order pair"
z1,z2 for A = (A,A)
z1,z2 for x = (x,x)
z1,z2 for y = (y,y)
z1,z2 for 2 = (2,2)
z1,z2 for 3 = (3,3)
Any number has the same quantity of space, as it has value......except zero.
(_)= one ACTUAL space, no value = z2
(1)= one value, no space = z1
If I put these things together I get a number
(1) = 1
Listen, if there is ever a problem with your posts rest assured someone will mention it to you. Just go ahead and post and don't worry about any apologies.If you "think" you may be seeing what I am reaching for... I will continue....please let me know if otherwise... so that I may adhere to my aforementioned promise directly!
It is true that elements of a set are ALWAYS numbers?´/quote]
No. The set of all Disney elephants contains Dumbo, some other elephants, and no numbers.
Your parts are, mathematically, numbers. You might want them to represent something else, but unless you introduce some rules to the contrary, they appear to function exactly as numbers.I am talking about pieces of numbers... I am not aware of any set theory that currently "breaks" numbers up into their parts.....parts that are NOT numbers.
What you have thus far described (not in these terms exactly, although you were closing in on it) is that you have a set of ordered pairs \(\displaystyle (z_1,z_2) \in \mathbb N \times \mathbb N\). You call \(\displaystyle z_1\) the "space" and \(\displaystyle z_2\) the "value" (although these terms mean nothing mathematically: their behaviour is defined by other statements). You all have implicitly defined a bijection \(\displaystyle \mathbb N \times \mathbb N \mapsto \mathbb N\) defined by \(\displaystyle (a,a) \mapsto a\).
That's all well and good, but very thin. What happens next needs some formalisation. It reads as a very ad hoc construction that looks unlikely to stand up to any rigorous interrogation. You say that, for multiplication \(\displaystyle (a,a) \times (b,b) = (a, ab)\): taking the space from \(\displaystyle (a,a)\) and then putting \(\displaystyle a\) copies of value \(\displaystyle b\) into those spaces. (I note here that you have absolutely used the fact that both the space and the value are functioning exactly as numbers). However, as noted before, we now have something that isn't in the definition of the set. You also claim that we have a map \(\displaystyle (a,ab) \mapsto ab\) which stops the mapping from being a bijection because we have \(\displaystyle (ab,ab) \mapsto ab\) already. Essentially, we seem to be heading to a place where the "space" is completely immaterial to the mapping. You have also stated that, due to commutativity of multiplication we have \(\displaystyle (b,b) \times (a,a) = (b, ab)\) and that this also maps \(\displaystyle (b,ab) \mapsto b\) further damaging the bijection idea and further illustrating that (up to now) the "space" is irrelevant.
None of this stops you from continuing to define this system, although it does make me question the point to it all.
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