#### Comment Preferences

• ##### Another way to do it(2+ / 0-)
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Orinoco, jessical

Permit me to suggest a couple more efficient ways to demonstrate associativity.

First way: We need to check that (aPb)Pc equals aP(bPc) for every possible combination of variables a, b, and c. Since each variable can assume any of three values, there are 3x3x3 or 27 cases to check.

But when any of the variables is the identity G, the desired result is trivial because G drops out of any operation. Take a=G as an example: On the left side, (aPb) = b regardless of b's value, so the result is bPc; on the right side, G P (bPc) is necessarily bPc, and we're done. An identical argument applies if b=G or if c=G.

There are 2x2x2 = 8 remaining cases to check, those where none of a, b, or c is equal to G. Two of these fall instantly to commutativity: (aPa)Pa = aP(aPa) no matter what, given that P is commutative. This leaves only 6 cases where we actually must look at the table.

Now here's an even faster way: This week's operation P is essentially identical to last week's B, except that last week's Y has been renamed as this week's G, G as R, and R as Y. (That is, if you take last week's table and change every Y to a G, every G to an R, and every R to a Y, you get exactly this week's table, though perhaps with whole rows or columns rearranged, which doesn't matter.) Technically we say that the two operations are isomorphic. Since B is known to be associative, so must be P.

• ##### Oh, very cool. Thank you.(1+ / 0-)
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jessical

The goats have been ramming through the first method.

So I could explain the second method to the goats, and they could explain it to everyone else. I may wait, though, until I introduce a five by five Caley table, so the brute force method breaks down and the goats are forced to find a better way.

The isomorphism bothers me a bit, though, I had hoped to use Butt as an addition operation, and Push as a multiplication operation, to show distributivity. But if they are both isomorphic to addition, will this work?

"The problems of incompetent, corrupt, corporatist government are incompetence, corruption and corporatism, not government." Jerome a Paris

[ Parent ]

• ##### It will not . . .(1+ / 0-)
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Orinoco

. . . and the reason is only indirectly related to isomorphism.

That it doesn't work is easily demonstrated. Take, for example, y P (g B g).  This is y P r which is g. On the other hand, y P g is of course y (g being P's identity) so (y P g) B (y P g) equals y B y which is y.

The problem is that a typical multiplication operator won't have the nice properties of an addition operator. Most blatantly the additive identity (at least) won't have an inverse! Even the definition of field says that only the nonzero elements need multiplicative inverses. So Push, as an operator, is too strong to function well as multiplication.

I'm not saying that there's no such thing as a pair of operators where one distributes over the other and both provide inverses of every element; such a thing may well exist. I just don't know of an example.

Note the connection to my earlier statement that associativity is a more basic property. The simplest structure with two operators is a ring, in which addition must be commutative and associative and have an identity and inverses. But multiplication needn't be commutative and needn't have inverses or even an identity! All that's required is that it be associative and distribute over addition. Everything else is extra.

• ##### So I need another operator(0+ / 0-)

that is associative and distributes over addition. Hmmm. I guess I have some work to do before next Saturday.

"The problems of incompetent, corrupt, corporatist government are incompetence, corruption and corporatism, not government." Jerome a Paris

[ Parent ]

• ##### Oh, so it's an operator you need?(1+ / 0-)
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Orinoco

Got one handy right here, fresh today.

After all, both P and B are really just addition mod 3. So make your new operator multiplication mod 3 and you're done. May as well keep Y as the additive identity, i.e. zero, and make G the multiplicative identity, i.e. one. So Y times anything is Y and G times x is x for any x. The rest is easy.

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