stoutfellow (![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
stoutfellow) wrote2004-12-21 03:10 pm
stoutfellow) wrote2004-12-21 03:10 pm
Entry tags:
Some More Arithmetic
Cranking it up a notch...
What I'm about to describe is one of an enormous family of examples. I'll say a few words about how they work at the end, but I'm not going into details.
The objects we're considering are called "Z2-sets". A Z2-set is a set, equipped with a notion of - call it partnership. Each element of a Z2-set has a partner, which is also an element of that set. The rule is that the partner of the partner of an element is the original element - if Laura is Rob's partner, then Rob is Laura's partner. That's the only rule; in particular, an element can be its own partner.
Let's look at some examples. If a Z2-set has only one element, that element is a singleton - it is its own partner. For practical purposes, such a Z2-set acts just like the number 1. A Z2-set with two members might consist of two singletons - {Sally, Mel} - or of a doubleton, a pair of elements which are each other's partners - {Laura, Rob}. The first one acts like the number 2; the second one is something new, which we'll call D. A Z2-set with three elements either contains three singletons (3) or a singleton and a doubleton (D+1). In general, a Z2-set consists of a certain number of singletons and a certain number of doubletons, and two Z2-sets "look alike" (the way two ordinary sets with the same number of elements do) if they have the same number of each.
Addition first. To describe a Z2-set, we need a set, and we need a notion of partnership. It'll be easiest if we look at an example; the disjoint union of a singleton {Sally} and a doubleton {Buddy, Pickles} is the disjoint union of the sets, with the obvious notion of partnership: it's {Sally, Buddy, Pickles}, with Sally a singleton and Buddy and Pickles partners. (This justifies calling a set like this D+1, as I did above.)
Multiplication next. Let's take the Cartesian product of {Rob, Laura} with {Buddy, Pickles}. As a set, it's {(Rob, Buddy), (Rob, Pickles), (Laura, Buddy), (Laura, Pickles)}. To take the partner of a pair, just take the partner of each entry: the partner of (Rob, Buddy) is (Laura, Pickles), and the partner of (Rob, Pickles) is (Laura, Buddy). We just took the product of two doubletons, and it turned out to consist of two doubletons: DxD=2D. That's all you need to know about multiplication here; if you want to multiply 5+3D by 7+D, just use the distributive law and the fact that DxD=2D.
Exponentiation is the tricky one. Once again, if I want to construct {Rob, Laura}^{Buddy, Pickles}, I start by listing the functions from {Buddy, Pickles} to {Rob, Laura}; there are four of them.
f1(Buddy)=f1(Pickles)=Rob;
f2(Buddy)=Rob, f2(Pickles)=Laura;
f3(Buddy)=Laura, f3(Pickles)=Rob;
f4(Buddy)=f4(Pickles)=Laura.
How does the partnership work? If two functions f and g are partners, and if f takes element 1 to element 2, g takes the partner of element 1 to the partner of element 2. Let's check this out. f1 takes Buddy to Rob, so the partner of f1 takes Pickles to Laura. f1 also takes Pickles to Rob, so the partner of f1 takes Buddy to Laura. That is, the partner of f1 is f4. f2 takes Buddy to Rob, so the partner of f2 takes Pickles to Laura; f2 takes Pickles to Laura, so its partner takes Buddy to Rob. So f2 is its own partner. f3 is also its own partner. {Rob, Laura}^{Buddy, Pickles} consists of two singletons and a doubleton; D^D=D+2.
If you carried out the same procedure to evaluate D^(D+1), you'd get 4D; (D+1)^D is 2D+3. The problem is that, unlike with multiplication, there's no obvious way to relate (A+B)^C to A^C, B^C, and other combinations. (Compare, in ordinary arithmetic, (a+b)^2 = a^2 + 2ab + b^2, and the like.) That's the question I asked my Master's student to study.
As I mentioned above, Z2-sets are one of a large - in fact, infinite - family of examples. Each consists of selecting a "group" - which is a technical term, which I'm not going to define; in essence, each set has some transformations associated with it, which interact with each other according to rules set by the group. In the Z2 case, the transformation is the one that takes each element to its partner, and the requirement is that if you do it twice, everything's back where it started. You could work with Z3-sets; the requirement then is that the transformation goes back where it started the third time, instead of the second. Likewise you have Z4, Z5, and so on. V-sets have two notions of partner - call them "first partner" and "second partner"; your first partner's first partner is you, your second partner's second partner is you, and your first partner's second partner is your second partner's first partner. And so on; the possibilities are endless, and each one gives rise to a different arithmetic, including a different notion of exponentiation.
What I'm about to describe is one of an enormous family of examples. I'll say a few words about how they work at the end, but I'm not going into details.
The objects we're considering are called "Z2-sets". A Z2-set is a set, equipped with a notion of - call it partnership. Each element of a Z2-set has a partner, which is also an element of that set. The rule is that the partner of the partner of an element is the original element - if Laura is Rob's partner, then Rob is Laura's partner. That's the only rule; in particular, an element can be its own partner.
Let's look at some examples. If a Z2-set has only one element, that element is a singleton - it is its own partner. For practical purposes, such a Z2-set acts just like the number 1. A Z2-set with two members might consist of two singletons - {Sally, Mel} - or of a doubleton, a pair of elements which are each other's partners - {Laura, Rob}. The first one acts like the number 2; the second one is something new, which we'll call D. A Z2-set with three elements either contains three singletons (3) or a singleton and a doubleton (D+1). In general, a Z2-set consists of a certain number of singletons and a certain number of doubletons, and two Z2-sets "look alike" (the way two ordinary sets with the same number of elements do) if they have the same number of each.
Addition first. To describe a Z2-set, we need a set, and we need a notion of partnership. It'll be easiest if we look at an example; the disjoint union of a singleton {Sally} and a doubleton {Buddy, Pickles} is the disjoint union of the sets, with the obvious notion of partnership: it's {Sally, Buddy, Pickles}, with Sally a singleton and Buddy and Pickles partners. (This justifies calling a set like this D+1, as I did above.)
Multiplication next. Let's take the Cartesian product of {Rob, Laura} with {Buddy, Pickles}. As a set, it's {(Rob, Buddy), (Rob, Pickles), (Laura, Buddy), (Laura, Pickles)}. To take the partner of a pair, just take the partner of each entry: the partner of (Rob, Buddy) is (Laura, Pickles), and the partner of (Rob, Pickles) is (Laura, Buddy). We just took the product of two doubletons, and it turned out to consist of two doubletons: DxD=2D. That's all you need to know about multiplication here; if you want to multiply 5+3D by 7+D, just use the distributive law and the fact that DxD=2D.
Exponentiation is the tricky one. Once again, if I want to construct {Rob, Laura}^{Buddy, Pickles}, I start by listing the functions from {Buddy, Pickles} to {Rob, Laura}; there are four of them.
f1(Buddy)=f1(Pickles)=Rob;
f2(Buddy)=Rob, f2(Pickles)=Laura;
f3(Buddy)=Laura, f3(Pickles)=Rob;
f4(Buddy)=f4(Pickles)=Laura.
How does the partnership work? If two functions f and g are partners, and if f takes element 1 to element 2, g takes the partner of element 1 to the partner of element 2. Let's check this out. f1 takes Buddy to Rob, so the partner of f1 takes Pickles to Laura. f1 also takes Pickles to Rob, so the partner of f1 takes Buddy to Laura. That is, the partner of f1 is f4. f2 takes Buddy to Rob, so the partner of f2 takes Pickles to Laura; f2 takes Pickles to Laura, so its partner takes Buddy to Rob. So f2 is its own partner. f3 is also its own partner. {Rob, Laura}^{Buddy, Pickles} consists of two singletons and a doubleton; D^D=D+2.
If you carried out the same procedure to evaluate D^(D+1), you'd get 4D; (D+1)^D is 2D+3. The problem is that, unlike with multiplication, there's no obvious way to relate (A+B)^C to A^C, B^C, and other combinations. (Compare, in ordinary arithmetic, (a+b)^2 = a^2 + 2ab + b^2, and the like.) That's the question I asked my Master's student to study.
As I mentioned above, Z2-sets are one of a large - in fact, infinite - family of examples. Each consists of selecting a "group" - which is a technical term, which I'm not going to define; in essence, each set has some transformations associated with it, which interact with each other according to rules set by the group. In the Z2 case, the transformation is the one that takes each element to its partner, and the requirement is that if you do it twice, everything's back where it started. You could work with Z3-sets; the requirement then is that the transformation goes back where it started the third time, instead of the second. Likewise you have Z4, Z5, and so on. V-sets have two notions of partner - call them "first partner" and "second partner"; your first partner's first partner is you, your second partner's second partner is you, and your first partner's second partner is your second partner's first partner. And so on; the possibilities are endless, and each one gives rise to a different arithmetic, including a different notion of exponentiation.