An Alternative to Standard Arithmetic
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From: jkcender on 5 Aug 2010 19:32 On Aug 5, 12:33 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > As usual, let me see whether there's a way to make this > more rigorous. > > After Little points out that associativity of addition > fails in jkcender's system, jkcender decides to use polynomials in z to define his system I have already agreed with Tim Little that the associative property of addition with the new zero numbers does not work the same way as with 0. The associative property fails here in the same way that the associative property of addition fails when adding Reals and imaginaries. However, the associative property applies to the new zero numbers in the same way that the associative property applies to the addition of Reals and imaginaries. Perhaps this is a better example than polynomials, although both have their shortcomings.
From: jkcender on 5 Aug 2010 20:13 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. > Notice that although the coefficients are positive, zero > coefficients are allowed, since a polynomial 4z+0z^2+5z^3 > is really just the polynomial 4z+5z^3. The number symbolized by "0" does not exist in the system under question. "z" replaces 0. So 4z+0z^2+5z^3 is undefined, at least 0z^2 is undefined. If we write zz^2, then z^3 is final form. In this sense, there are no zero coefficients. Any z coefficient simply combines, so 1z1z + 5z^3 = 1z^2 + 5z^3. > 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. > But what we can't have is (0,0), since this would be the forbidden zero polynomial. (nz,nz) would not be forbidden. Again as best I understand, barring some intricacy of Golden's of which I'm unaware. > So what we need is (R+[z])^2\{(0,0)}. Notice that this > set is closed under component wise addition since R+ itself > is closed under addition. I am not at all sure what happens here (if anything) when nz is substituted for 0. (R+[z])^2\{(nz,nz)}
From: jkcender on 5 Aug 2010 20:14 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. > Notice that although the coefficients are positive, zero > coefficients are allowed, since a polynomial 4z+0z^2+5z^3 > is really just the polynomial 4z+5z^3. The number symbolized by "0" does not exist in the system under question. "z" replaces 0. So 4z+0z^2+5z^3 is undefined, at least 0z^2 is undefined. If we write zz^2, then z^3 is final form. In this sense, there are no zero coefficients. Any z coefficient simply combines, so 1z1z + 5z^3 = 1z^2 + 5z^3. > 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. > But what we can't have is (0,0), since this would be the forbidden zero polynomial. (nz,nz) would not be forbidden. Again as best I understand, barring some intricacy of Golden's of which I'm unaware. > So what we need is (R+[z])^2\{(0,0)}. Notice that this > set is closed under component wise addition since R+ itself > is closed under addition. I am not at all sure what happens here (if anything) when nz is substituted for 0. (R+[z])^2\{(nz,nz)}
From: jkcender on 5 Aug 2010 20:14 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. > Notice that although the coefficients are positive, zero > coefficients are allowed, since a polynomial 4z+0z^2+5z^3 > is really just the polynomial 4z+5z^3. The number symbolized by "0" does not exist in the system under question. "z" replaces 0. So 4z+0z^2+5z^3 is undefined, at least 0z^2 is undefined. If we write zz^2, then z^3 is final form. In this sense, there are no zero coefficients. Any z coefficient simply combines, so 1z1z + 5z^3 = 1z^2 + 5z^3. > 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. > But what we can't have is (0,0), since this would be the forbidden zero polynomial. (nz,nz) would not be forbidden. Again as best I understand, barring some intricacy of Golden's of which I'm unaware. > So what we need is (R+[z])^2\{(0,0)}. Notice that this > set is closed under component wise addition since R+ itself > is closed under addition. I am not at all sure what happens here (if anything) when nz is substituted for 0. (R+[z])^2\{(nz,nz)}
From: Jonathan Cender on 6 Aug 2010 18:10
On Aug 5, 12:33 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > Before I leave, let me point out that I'm not sure whether > jkcender's addition is commutative, either. For we know > that 1+(-1) = z, but what is (-1)+1? We see that: > > (-1)+1 = -1z^0 + 1z^0 > = -1(1z^0 + -1z^0) > = -1z^1 > = -z > > a problem which doesn't go away by using polynomials. Addition of a Real and its additive inverse is not commutative with nz unlike with 0. However, note what perseveres: ordered pairs of numbers and their additive inverses.* This means that the sums a-b, -(a-b) resulting in (c, -c), where c is any Real and -c is its additive inverse, still holds when a=b. So when a=b we have a-a = az and -(a-a)= -az The relationship is preserved into the z numbers. And it continues on so az-az = az^2, -(az-az) = -az^2 and so forth. *Additive inverse means the sum of two numbers resulting in some zero number nz^r rather than the zero number, 0. |