Prev: Dear Friend! You Nigerian Lottery ticket won 1,255,000 USD. I'm authorized to transfer this amount directly to your banking account.
Next: I'm gonna try this one more time CANTOR DISPROOF
From: Tegiri Nenashi on 7 Jun 2010 20:24
I'm aware of two (partial)) order definitions (and looking forward to
expand my perspective):
1. (Semi)lattice: x < y <-> x ^ y = y.
2. N: x < y <-> exists z: x + z = y.
(There is no pun intended with the name "natural order" applied in
second case)

Now it comes with a bit of surprise that there is no natural order
definition for Z. Or there is? Perhaps, some algebras (other than Z
introduce order in some other way?
From: Ostap Bender on 7 Jun 2010 22:32
On Jun 7, 5:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote:
> I'm aware of two (partial)) order definitions (and looking forward to
> expand my perspective):
> 1. (Semi)lattice: x < y <-> x ^ y = y.
> 2. N: x < y <-> exists z: x + z = y.
> (There is no pun intended with the name "natural order" applied in
> second case)
>
> Now it comes with a bit of surprise that there is no natural order
> definition for Z. Or there is? Perhaps, some algebras (other than Z
> introduce order in some other way?

What do you mean by Z? The set of integers?


From: Arturo Magidin on 7 Jun 2010 22:53
On Jun 7, 7:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote:
> I'm aware of two (partial)) order definitions (and looking forward to
> expand my perspective):
> 1. (Semi)lattice: x < y <-> x ^ y = y.
> 2. N: x < y <-> exists z: x + z = y.
> (There is no pun intended with the name "natural order" applied in
> second case)
>
> Now it comes with a bit of surprise that there is no natural order
> definition for Z.

What do you mean by "natural order"? A definition like 2? Yes, 2 does
not define an order relation on Z.

However, when people talk about the "natural order" of Z, they don't
refer to (2).

Generally, given a set S, a partial order on S is a binary relation #
such that:

(i) For all s in S, s#s; [reflexivity]
(ii) For all s and t in S, if s#t and t#s, then t=s [anti-simmetry];
(iii) For all s, t, u in S, if s#t and t#u, then s#u. [transitivity].

If in addition we have
(iv) For all s and t in S, either s#t or t#s [totality]
then we say the order is a "total order".

A "strict partial order" on a set S is a binary relation < such that
(1) For all s in S, s<s does not hold [areflexivity];
(2) For all s, t, u in S, if s<t and t<u, then s<u [transitivity]

If in addition we have
(3) For all s and t in S, exactly one of
s<t; s=t; t<s
holds [trichotomy]
then we have a "total strict order".

If # is a partial order, then we can define a strict order by letting
s<t if and only if s#t and s=/=t.

If < is a strict partial order, then we can define a partial order by
letting s#t if and only if s=t or s<t.

Lattices are special kinds of partially ordered sets, in which every
pair of elements has a least upper bound and a greatest lower bound.

The integers have "natural" (or canonical) total order, which can be
either derived from the order of the natural numbers, or that can be
defined "algebraically" by defining the non-zero natural numbers to be
the "positive class". Details available on request.

--
Arturo Magidin
From: Ostap Bender on 7 Jun 2010 23:02
On Jun 7, 5:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote:

Te giri ne nashi? Govorite po-russki?
From: William Elliot on 8 Jun 2010 01:48
On Mon, 7 Jun 2010, Arturo Magidin wrote:
> On Jun 7, 7:24�pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote:

>> I'm aware of two (partial)) order definitions (and looking forward to
>> expand my perspective):
>> 1. (Semi)lattice: x < y <-> x ^ y = y.
>> 2. N: x < y <-> exists z: x + z = y.
>> (There is no pun intended with the name "natural order" applied in
>> second case)
>>
>> Now it comes with a bit of surprise that there is no natural order
>> definition for Z.
>
> The integers have "natural" (or canonical) total order, which can be
> either derived from the order of the natural numbers, or that can be
> defined "algebraically" by defining the non-zero natural numbers to be
> the "positive class". Details available on request.
>
Let N = /\{ A subset Z | 1 in A, for all x in A, x + 1 in A }.
Define x < y as y - x in N.

--
Embed Z into R. Let R+ = { x^2 | x in R }.
Define x < y as y - x in R+\0.

----
 |  Next  |  Last
Pages: 1 2 3 4 5 6
Prev: Dear Friend! You Nigerian Lottery ticket won 1,255,000 USD. I'm authorized to transfer this amount directly to your banking account.
Next: I'm gonna try this one more time CANTOR DISPROOF