Re: JSH: Now the fun part

On Aug 6, 9:29 pm, JSH <jst...@xxxxxxxxx> wrote:
On Aug 6, 5:13 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:



On Aug 6, 8:56 pm, JSH <jst...@xxxxxxxxx> wrote:

On Aug 6, 4:51 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Aug 6, 8:07 pm, JSH <jst...@xxxxxxxxx> wrote:

On Aug 6, 3:30 am, hagman <goo...@xxxxxxxxxxxxx> wrote:

On 6 Aug., 06:24, JSH <jst...@xxxxxxxxx> wrote:

Assume I'm correct in that from the ring of objects you can prove that
units are being used in the ring of algebraic integers where they are
not units to wrap up constants like 7 so that you can divide them off
and remain in that ring.

Then you find a consistent argument as I've stated.

For those who don't know I have the following definition for the ring
of objects:

The object ring is defined by two conditions, and includes all numbers
such that these conditions are true:

1. 1 and -1 are the only rationals that are units in the ring.

2. Given a member m of the ring there must exist a non-zero member n
such that mn is an integer, and if mn is not a factor of m, then n
cannot be a unit in the ring.

Hm, to make that comprehendable, let us call a subring A of the
complex numbers "object-like" if
(1) A intersect Q = Z (Q=the rationals, Z=the rational integers)
(2) if m in A and m*A intersect Z is {0}, then m=0.
(The last condition about factors and units is trivial under the
prerequisite that we are working inside the integral domain C.)

Posters keep making that claim

because it is trivially true

Think so? I think you don't understand how the conditions work.

but I think you all keep missing
something, especially since you include pi below when pi is excluded
from the object ring by those rules as is e.

Nope. Clearly what you say is not what you mean. Why should the
rules
exclude pi (the rules do not exclude sqrt(10) = 3.1622...
why should they exclude pi = 3.14159...)?

- William Hughes

How do you define pi in the ring?

James Harris

E.g. a period of any solution to f''(x) = - f(x).

[Note the rules say nothing about how members
of the ring can or cannot be defined]

- William Hughes

How do you define the second derivative in the object ring?

You don't. There is no need to.
[Note the rules say nothing about how members
of the ring can or cannot be defined. In particular
they do not say that members of the ring can only be
definied by use of things in the ring]


- William Hughes


.

Relevant Pages

  • Re: JSH: The "Published" paper he dosent what you to know about.
    ... I'm not sure what he wants from his "object ring", ... In "The Theory of Algebraic Integers, ... Note the "theory of ideals" in the last sentence. ...
    (sci.math)
  • Re: JSH "problem"
    ... (Or in fact with any ring? ... I started following he was already fixated on the algebraic integers. ... The one he's going on forever about in these threads is the delusion ... as a factor in the object ring". ...
    (sci.math)
  • Re: JSH "problem"
    ... two, non-integral roots, one of the roots is in the object ring and the other ... which is an algebraic integer. ... So the ring of integers of Qcontains the other root, ...
    (sci.math)
  • James Harris object ring
    ... James Harris considers the following properties of a commutative ring R: ... Every non-unit of Z remains a non-unit in R, ... Lets call a ring satisfiying 1-4 an>object ring<. ...
    (sci.math)
  • Re: JSH: Now the fun part
    ... and remain in that ring. ... A is called "object ring" if it is maximal object-like. ... The point of the rules is to exclude numbers like 1/2, ...
    (sci.math)
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