My Math Forum Expanding the set of Integers to include more than pos/negs

 Number Theory Number Theory Math Forum

January 5th, 2010, 05:26 PM   #2
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Re: Expanding the set of Integers to include more than pos/negs

Quote:
 Originally Posted by jamesuminator It is very difficult to imagine this idea because it defies the traditional economic layout of mathematics we created.
This time, instead of referencing abstract algebra (still slightly relevant), I'll reference foundations of mathematics/formal logic/set theory.

[Here I originally had a point-by-point description of Peano arithmetic; I decided to delete it in favor of a less formal discussion.]

Let's start with the theory of addition on the natural numbers N = {0, 1, 2, ...}. Then define a structure (a, b) where a in in N and b is in {+, -}. Let (a, b) = (c, d) if a = b and c = d, but also let (0, b) = (0, d) regardless of b and d. (Otherwise they are unequal.) Then define (a, b) + (c, d) as (a + c, b) if b = d and (a, b) + (a, d) = (0, a) if b ? d.

This should be enough to give addition on the integers. The ordinary relations <, >, etc. can be defined in the obvious way. And of course we can then treat the whole structure (a, b) as a single number. Call the set of such numbers Z, as usual.

So now you're asking about extending this to a structure (n, a) with n in Z and a in {*, /}. I imagine you want equality to be (m, a) = (n, b) iff [(m = n and a = b) or m = n = 1], and multiplication to be (m, a) * (n, b) = (mn, a) if a = b and (m, a) * (m, b) = (1, a) if a ? b. Is that right?

 January 6th, 2010, 02:43 PM #3 Senior Member   Joined: Aug 2008 Posts: 133 Thanks: 0 Re: Expanding the set of Integers to include more than pos/negs Yes, this is exactly what I mean; you express it far better than I do. Edit: Actually, I'm willing to argue another set if this one proves to have faults, such that: m is in the set of N, and c is {/,*}, m = n and c = d; or as you described it. (m, c) = (n, d); only if m = n, and c = d. (m, c) * (n, d) = (m+n, c) such that c = d. And the main conjecture: (m, c) * (m, d) = (0,--) such that d =/= d. This is clearly less complex, but I will let you continue with the first manner of description and consider the latter afterwards.
January 7th, 2010, 05:32 AM   #4
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Re: Expanding the set of Integers to include more than pos/negs

Quote:
 Originally Posted by jamesuminator Actually, I'm willing to argue another set if this one proves to have faults, such that:
OK, let's go!

Quote:
 Originally Posted by jamesuminator (m, c) = (n, d); only if m = n, and c = d.
In particular, you have (1, /) ? (1, *).

Quote:
 Originally Posted by jamesuminator (m, c) * (n, d) = (m+n, c) such that c = d.
This is quite odd; is it what you intended? I would have expected
(m, c) * (n, d) = (m*n, c) if c = d
which is different in two places.

Quote:
 Originally Posted by jamesuminator And the main conjecture: (m, c) * (m, d) = (0,--) such that d =/= d.
??

(0, --) is not a number in your system (call it J), since -- is not in {/, *}.
d ? d is always true.
You haven't defined what (m, c) * (n, d) is for c ? d, so nothing can be determined about that case.

January 7th, 2010, 12:44 PM   #5
Senior Member

Joined: Aug 2008

Posts: 133
Thanks: 0

Re: Expanding the set of Integers to include more than pos/negs

Quote:
 In particular, you have (1, /) ? (1, *).
Yes, in the same way -1 =/= 1.

Quote:
 This is quite odd; is it what you intended? I would have expected (m, c) * (n, d) = (m*n, c) if c = d which is different in two places.
This is an alternative method of developing the set; and may prove to be the better way given occam's razor.
It's easy to invision it as: when a number is divisive or multiplicative {*, /}, multiplication is to it, as addition is to integers {+, -}.

Quote:
 d ? d is always true.
Sorry, it was c =/= d; typo.

Quote:
 (0, --) is not a number in your system (call it J), since -- is not in {/, *}.
By that I just meant the number 0; and since 0 can have either/or properties I used that blank thing. It should be assumed to be in the system and its property is simply arbitrary: * or / works just as fine.

Quote:
 You haven't defined what (m, c) * (n, d) is for c ? d, so nothing can be determined about that case.
(m, c) * (n, d) = (m-n, d) where c =/= d.

 January 7th, 2010, 02:31 PM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Expanding the set of Integers to include more than pos/negs OK, let me see if we're communicating. (I'm not entirely sure I understand you, or that you've communicated all you need.) J is the set of ordered pairs (m, c) with m in N = {0, 1, 2, ...} and c = {/, *}. For (m, c) and (n, d) in J, (m, c) = (n, d) iff m = n and c = d. (m, c) * (n, c) = (m+n, c) (m, c) * (n, d) = (m-n, d) where c ? d. Is this right? (Please check carefully, because I'll make conclusions based on exactly what is written.)
 January 7th, 2010, 02:45 PM #7 Senior Member   Joined: Aug 2008 Posts: 133 Thanks: 0 Re: Expanding the set of Integers to include more than pos/negs Yes exactly. It is also worth noting, (m, c) / (n, d) = (m-n, c) if c = d (m, c) / (n, d) = (m+n, d) if c =/= d
 January 7th, 2010, 03:03 PM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Expanding the set of Integers to include more than pos/negs My first observation is that "=" is not an equivalence relation on J, since (1, *) * (1, /) = (0, /) by Axiom 3 (1, /) * (1. *) = (0, *) by Axiom 3 (0, /) ? (0, *) by Axiom 1
 January 9th, 2010, 11:16 AM #9 Senior Member   Joined: Aug 2008 Posts: 133 Thanks: 0 Re: Expanding the set of Integers to include more than pos/negs Hmm, I see what is wrong now. I seem to have misappropriated something. Let me reanalyze: (m, c) * (n, d) = (m-n, c) iff (c =/= d, and m > n) (m, c) * (n, d) = (m-n, d) iff (c =/= d, and m < n) such that: (5, *) * (3, /) = (5-3, *) = (2, *) (5, *) * (7, /) = (5-7, *) = (-2, *) = (2, /) So therefore it would instill that c & d are dependent on the values of m & n. You really have to imagine it as a whole set, so let me explain this using a general grammar. M(0) is successorship, M(1) is addition, and M(-1) is subtraction, etc etc... Let m & n belong to the set of N; and let c & d belong to the same set of indexes as M(n), such that {...-2 : /, -1 : -, 0 is successorship property, 1 : +... etc etc}; and let x belong to the set of Z. (m, c) M(x) (n, d) = (m+n, c) iff (c = d = x) eg; (4, 1:+) M(1) (3, 1) = (7, 1) = 4 + 3 = 7 (m, c) M(x) (n, d) = (m-n, c) iff (c =/= d, (c = |d| = x) and m>n) eg; (6, 2) M(2) (3, -2) = (6-3, 2) = (3, 2) = (6, *) * (3, /) = (6-3, *) = (3, *) (m, c) M(x) (n, d) = (m-n, d) iff (c =/= d, (c = |d| = x) and m= (c + 2) >= (d + 2). But let's not get into that yet, let's try to understand the most basic rules.
January 9th, 2010, 08:25 PM   #10
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Re: Expanding the set of Integers to include more than pos/negs

So now you have:

J is the set of ordered pairs (m, c) with m in N = {0, 1, 2, ...} and c = {/, *}.
1. (m, c) = (n, d) iff m = n and c = d.
2. (m, c) * (n, c) = (m+n, c)
3. (m, c) * (n, d) = (m-n, c) iff (c ? d, and m > n)
4. (m, c) * (n, d) = (m-n, d) iff (c ? d, and m < n)

Comment 1: I notice that (m, *) * (m, /) is undefined. Is this intentional?

Quote:
 Originally Posted by jamesuminator (5, *) * (7, /) = (5-7, *) = (-2, *) = (2, /)
Comment 2: The first equality does not follow from your axioms; it should be
(5, *) * (7, /) = (-2, /)

Comment 3: (-2, /) is not a member of J, and thus * as you have defined it is not a map from J x J to J. Is this what you intended?

Comment 4: (-2, *) ? (2, /) by Axiom 1. You wrote the opposite; what did you intend?

 Tags expanding, include, integers, pos or negs, set

### hyper operators of negative order

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post matqkks Number Theory 2 June 17th, 2013 10:40 AM sophieT66 Algebra 1 October 26th, 2010 12:36 PM MarkFL New Users 2 July 1st, 2010 12:11 PM everettjsj2 Calculus 2 February 28th, 2010 01:49 PM sophieT66 Number Theory 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top