An Alternative to Standard Arithmetic
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From: jkcender on 4 Aug 2010 22:30 Here is an alternative to Standard Arithmetic. Unlike Nonstandard Arithmetics which all take Standard Arithmetic as given, the alternative results from changing one of its axioms. New number zeros replace 0 and a new zero axiom is given. Foundationally, alternatives to the empty set and relevant parts of first order logic are provided. These changes are based on the other notion of nothing in mathematics, the notion exemplified by the placeholder zero. The remainder of this post consists of a very brief overview of the new zero numbers and some of the rules for them. A link to a draft paper is at the end. The new zero numbers are multiplicative inverses (not reciprocals). Here z stands for the zero number, z(inv) for its inverse; n, x, y are Real numbers. Like the imaginary number i, z is always paired with a Real number. z represents the absence of a given array of numbers; its inverse their presence. Following John A Wheeler and Roger Penrose, the array used here is the Real numbers. [note 1] Unlike their array, this one doesnt contain zero since it is no longer a Real. [note 2] 1. nz * nz(inv) = 1, and xz * yz(inv) = xy [note 3] 2. Zero times anything else is still a zero number. x * nz = xnz, and xz * nz = xn(z)^2, 3. Division defined. x/nz = x/n * z(inv) = (x/n)z(inv), and xz/yz = x/ y, and xz(inv)/yz(inv) = x/y 4. Additive identity. x + nz = x or x + nz may be considered in final form, but z + z = 2z. 5. Additive inverse. n n = nz, and nz nz = (nz)^2 6. xz is not equal to yz, rather xz + yz = (x+y)z 7. (nz)^0 = 1 (other zeros such as the exponent zero still use the symbol 0) Rules for z(inv) are such that apply to any monomial. Also x + nz(inv) is in final form. As mentioned, Penrose and Wheeler had a different notation for z(inv), its purpose being to lay out an n-dimensional- real space. The arithmetic rules for z(inv) turn their notation into a way to construct a space operationally. For more on the new numbers, see my draft paper What If There Were A Different Zero at http://www.docstoc.com/docs/48156633/What-If-There-A-Different-Zero [note 1] The Road to Reality, pp. 379-80, etc. [note 2] Penrose and Wheeler do not change zero and they use a different symbol than z(inv). I use the same symbol they do in my paper; z is used here for clarity. [note 3] As is the case with the Riemann Sphere, there is an exception. Here, z does not = 1/z(inv). The exception for the sphere at complex infinity is the opposite, 1 does not = 0 * inf as mentioned in Mathematica and mathworld.wolfram.com/DivisionbyZero.html This is the one exception Ive found to the usual rules of arithmetic. As with Riemanns work, it does not seem to be significant.
From: Tim Little on 5 Aug 2010 00:50 On 2010-08-05, jkcender <jkcmsal(a)yahoo.com> wrote: > 4. Additive identity. x + nz = x or x + nz may be considered in > final form, but z + z = 2z. > 5. Additive inverse. n – n = nz, and nz – nz = (nz)^2 Just to clarify, (1 + (-1)) + z = 2z, but 1 + ((-1) + z) = z right? So this form of addition does not obey the associative law? - Tim
From: jkcender on 5 Aug 2010 01:15 On Aug 4, 6:50 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-08-05, jkcender <jkcm...(a)yahoo.com> wrote: > > > 4. Additive identity. x + nz = x or x + nz may be considered in > > final form, but z + z = 2z. > Just to clarify, (1 + (-1)) + z = 2z, but 1 + ((-1) + z) = z right? > So this form of addition does not obey the associative law? > > - Tim I think you are referring to 4. If so, no. 1 + ((-1) + z) = (1 + (-1)) + z = 2z. The associative law is obeyed here. If wondering about "x + nz may be considered in final form," this is only after consolidation of like terms.
From: jkcender on 5 Aug 2010 01:19 > 5. Additive inverse. n n = nz, and nz nz = (nz)^2 Tim called my attention to 5. To avoid possible confusion, only the z part is squared. The n is not.
From: Tim Little on 5 Aug 2010 01:26
On 2010-08-05, jkcender <jkcmsal(a)yahoo.com> wrote: > On Aug 4, 6:50 pm, Tim Little <t...(a)little-possums.net> wrote: >> Just to clarify, (1 + (-1)) + z = 2z, but 1 + ((-1) + z) = z right? >> So this form of addition does not obey the associative law? > > I think you are referring to 4. If so, no. 1 + ((-1) + z) = (1 + (-1)) > + z = 2z. I'm referring to both 3 and 4. 1 + ((-1) + z) = 1 + (-1) ... by (3) = z ... by (4). If associativity holds, then 1 + ((-1) + z) = (1 + (-1)) + z = z + z ... by (4) = 2z ... by (3). Since you already stated that 2z =/= z, associativity cannot hold. - Tim |