An Alternative to Standard Arithmetic
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From: Tim Little on 6 Aug 2010 23:18 On 2010-08-05, jkcender <jkcmsal(a)yahoo.com> wrote: > On Aug 4, 7:26pm, Tim Little <t...(a)little-possums.net> wrote: >> 1 + ((-1) + z) >> = 1 + (-1) ... by (3) >> = z ... by (4). You haven't addressed this derivation. Which step is incorrect? It should be easy, there are only two and I've labelled both of them with the numbering of rules you provided. > 1 + (-1)= z therefore 1 + (-1) + z = 1z + 1z = 2z also 1 + ((-1) + > z) = (1 + (-1)) + z = 2z > Associativity holds. You're assuming that associativity holds to show that associativity holds. If you're stating that associativity holds as an *axiom*, you should say so and accept that your system is then inconsistent. - Tim
From: Tim Little on 6 Aug 2010 23:20 On 2010-08-05, jkcender <jkcmsal(a)yahoo.com> wrote: > rewritten with all zero numbers we can see that it does not equal z. > 1z^0 + ((-1z^0) + z^1) =/= z. I hope it is clear here that, just like > in any polynomial, terms with different powers cannot be added. So your rule (3) is actually false? It is not true that x + nz = x? - Tim
From: Tim Little on 6 Aug 2010 23:22 On 2010-08-05, jkcender <jkcmsal(a)yahoo.com> wrote: > The associative property fails here in the same way that the > associative property of addition fails when adding Reals and > imaginaries. The associative property of addition works perfectly well when adding reals and imaginaries. The difference is that nobody working in complex arithmetic claims that imaginaries are additive identities for reals. - Tim
From: Transfer Principle on 6 Aug 2010 23:31 On Aug 5, 5:13 pm, jkcender <jkcm...(a)yahoo.com> wrote: > On Aug 5, 12:33 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > I point out that there is a similar trick to make Tim Golden's polysigned > > numbers work, so we may try the same thing here. > I am not familiar with Tim Golden's polysigned numbers work. Having > said that I will point out a couple of possible problems with the > following as applies to the new zero numbers based upon my > understanding of the symbols used here. My apologies if there is some > intricacy of Golden's of which I'm unaware. I apologize for being a bit confusing here. I refer to several concepts in this post, so let me address them one at a time. First of all, Cender, are you familiar with the standard construction of the set R of real numbers via Cauchy sequences or Dedekind cuts? Posters like Little and Tonio definitely are familiar with these, and so it might helpful to know about this when discussing the z-numbers with them. Basically, what I'm trying to do here is proceed with the usual construction, but as soon as the number 0 appears, replace it with the new number z. Typically, one starts with the natural numbers N, then the integers Z, forms the rational numbers Q, and finally the real numbers R. The step from Z to Q is probably the step that is best-known. We start with fractions, which are pairs of integers -- for example, one-half is 1/2. Then we have equivalence classes of fractions, where some fractions are declared to be equivalent, such as 1/2 and 2/4. The rule (equivalence relation) which determines whether two fractions are equivalent is as follows: a/b = c/d iff ad = bc. Then we add rules for how to multiply, add, and compare two fractions, and voila! We have the set Q of rationals. Where do rationals come from? They come from _dividing_ two integers, since it was realized that some division problems, such as 1 divided by 2, are impossible unless we have fractions. Similarly, integers come from _subtracting_ two naturals, since it was realized that some subtraction problems, such as 1-2, are impossible unless we have negative integers. And so the construction from N to Z is similar to that from Z to Q, except now we have ordered pairs of naturals that we're trying to subtract, not divide. So just as a/b means a divided by b, at this step of the construction, (a,b) means a-b. Similarly, we have equivalence classes of these ordered pairs, except that the equivalence relation is now that (a,b) = (c,d) iff we have a+d = b+c. Thus (2,1), (3,2), (4,3), and so on are equivalent. All of these equal the integer +1. Similarly, (1,2), (2,3), (3,4), and so on equal the integer -1. But what about the ordered pair (1,1)? Aha -- this is exactly where the number 0 first appears in our construction, and this is exactly where we want to take 0 out and put our z in. The key to constructing Cender's system therefore is to avoid 1-1 = 0 and have 1-1 = z instead. But 0 appears fairly early in our construction from N to Z to Q to R. What I'd like to do is go as far as we can before worrying about 0. As it turns out, the poster David Ullrich recently posted how we can go from N to the _positive_ rationals Q+ and the _positive_ reals R+ without worrying about zero. The positive reals work exactly the same in both R+ and Cender's system. And so we can use R+ as a starting point. (Try a Google search for David Ullrich algebra question in 2010 for more information on the construction from N to R+.) > > Indeed, the polynomial 0 is allowed -- (P,0) is a polynomial with all positive > > coefficients, and (0,P) is a polynomial with all negative coefficients. > So what happens in Golden's system when we substitute z for 0 in the > above? For example, (P,nz) is a polynomial with all positive > coefficients, and (nz,P) is a polynomial with all negative > coefficients? As best I understand, I agree with this. Again, barring > some intricacy of Golden's of which I'm unaware. Ah, Golden's system. As I wrote earlier, the key to constructing Cender's system is avoiding 1-1 = 0. As it turns out, Golden wants to avoid 1-1 = 0 as well, but in a different manner. In Golden's system, instead of having a positive and negative number combine to give zero, he wants to have a third sign, *, in addition to the two standard signs - and +. Then one has to combine all three signs to obtain zero, as in -1+1*1 = 0. To construct the polysigned numbers, we use equivalence classes of ordered _triples_, so that (1,1,1) = 0. To learn more about polysigned numbers, one can refer to the following link: http://www.bandtechnology.com/PolySigned/PolySigned.html Leaving Golden's numbers aside, let's begin the construction of Cender's numbers from R+. As Cender points out, many of his numbers can be written as a polynomial in z. Because of this, we start out by consider the set R+[z] of polynomials in z with positive real coefficients. Now -- and this is where I confused Cender earlier -- we notice that in a polynomial such as 4z+5z^3, there is no z^2 term. The z^2 term is _missing_. Indeed, so are the constant, z^4 term, z^5 term, etc. -- in fact, every polynomial must have all but finitely many terms missing. And so, R+[z] must contain a polynomial such that _every_ term is missing. In standard theory, we would normally refer to this as the "zero polynomial," but in Cender's theory, there is no 0. So let's instead call it the "vanishing" polynomial. But what about negative coefficients, since after all, the coefficients of the polys come from R+? Let's go back to the construction of Z from N, where we had ordered pairs such as (1,2) to denote -1. In a way, the first natural in the ordered pair means +1, and the second natural in the ordered pair means -2, and +1-2 = -1. And so (4z,5z^3) denotes 4z-5z^3. But what if all the coefficients have the same sign? This is where the vanishing poly comes in. We have that (4z+5z^3,vanishing) denotes 4z+5z^3, while if we want -4z-5z^3, we use (vanishing,4z+5z^3). But what we can't have is (vanishing,vanishing) -- because this would correspond to traditional 0, which doesn't exist. And so we have the set of all ordered pairs of polys in z, with positive coefficients, such that at least one of the polys doesn't vanish. It is the Cartesian product of R+[z] with itself, with one element missing. All we need now is an equivalence relation which tells us which expressions are equal. For example, we want to make the pair (1,1) equivalent to (z,vanishing), since this tells us that 1-1 = z, the fundamental equation. This is difficult. I will need some more time to think about this for a while. But thank you, Cender, for providing us with such an interesting system!
From: Jonathan Cender on 10 Aug 2010 21:50
On Aug 6, 5:18 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-08-05, jkcender <jkcm...(a)yahoo.com> wrote: > > > On Aug 4, 7:26pm, Tim Little <t...(a)little-possums.net> wrote: > >> 1 + ((-1) + z) > >> = 1 + (-1) ... by (3) > >> = z ... by (4). > > You haven't addressed this derivation. Which step is incorrect? It > should be easy, there are only two and I've labelled both of them with > the numbering of rules you provided. Tim, First let me say I appreciate your patience and the time youve spent on thinking about this. Your questions are helping me examine basic assumptions that in some cases I havent even known Ive made about the consequences of changing the Peano axioms in this way. Im quite excited by your questions because, if this alternate arithmetic is to be useful, they need a good answer. Specifically about the above, I now understand what you meant. The ellipses threw me and I jumped to the wrong conclusion and answered a question you didnt actually ask. Heres my attempt to answer. Reals and zero numbers are not associative as Ive said before. So parentheses matter. As long as those parentheses are there, z is the answer, not 2z. To repeat, what you have written above is correct. It is equal to z by 4 and 5 if (-1) + z is not in final form. The z just drops off. Simple. But if the parentheses change the answer changes; in this case to 2z. What you may have teased out here is that it is possible to have more than one arithmetic with the new zero. Rule 4 mushes too many things together. Lets separate them out. Rule 4a. Additive identity. x + nz = x. Rule 4b. x + nz is in final form Rule 4a would be for one arithmetic and 4b for a separate one. 4b would be a lot more complicated and harder to show whether it would work or not. At least for me right now. Before looking just at rule 4a math remember that Rule 5, since it is not commutative as Transfer Principle showed at the end of his first post, is now n-n= nz and (n-n)= -nz OK, for arithmetic with just rule 4a, Ive begun to wonder if separating it out like this is a key to a number of issues Ive struggled with. By following the steps you have questioned, nz and nz can always be reduced to z and z because nz = z + (n-1)z. Then set z = 1 + (-1) as youve done and apply rules 4a and 5 again. Then the need for rule 6 goes away as long as a suitable order of operations is set up. [Rule 6 was there mainly to resolve one of the less common reasons for the lack of definition for division by zero anyway. But this resolves it, too.] The result of this is that z is a lot more like 0 except that division is defined for z. For example, Rule 2 just becomes Zero times any Real number is z. x * nz = xnz = z Looking forward to your reply, especially if there's something I'm overlooking. Jonathan > > > 1 + (-1)= z therefore 1 + (-1) + z = 1z + 1z = 2z also 1 + ((-1) + > > z) = (1 + (-1)) + z = 2z > > Associativity holds. > > You're assuming that associativity holds to show that associativity > holds. If you're stating that associativity holds as an *axiom*, you > should say so and accept that your system is then inconsistent. > > - Tim I hope the above response in this post also answers this part. Associativity doesnt hold. Let me know if theres something still not clear or Im missing something. - Jonathan |