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 eddybob123 October 25th, 2013 06:30 PM

The set of primes as an algebraic structure

Let $P$ denote the set of all prime numbers, and let $K_P$ be an algebraic structure over $P$ (so for example, let $K_P\$ be the group $(P\,,\,\cdot\,)$ where $\cdot$ is a (currently unknown) binary operation). We can do a lot with this algebraic structure, and the more structure it has, the more theorems we can endow on the prime numbers.
However, finding such an algebra is not easy. For it to be useful, it must have a lot of structure, and a binary operation that is closed over the primes is rarely associative, let alone commutative, distributive, or idempotent.
So what do you guys think? Is such a binary operation within our grasp? Would an algebraic structure of this type be applicable in any way? Please share your thoughts on the subject.

 CRGreathouse October 25th, 2013 08:55 PM

Re: The set of primes as an algebraic structure

Sure, here's an example. Let S(p) denote the prime succeeding p and P(p) denote the prime preceding p, if such a number exists. So P(5) = 3, P(11) = 7, and so forth. Then let $p\cdot 2=p=2\cdot p$ and $p\cdot S(q)=S(p\cdot q)=S(q)\cdot p.$ So for example 3*7 = S(2*7) = S(7) = 11.

 mathbalarka October 25th, 2013 10:54 PM

Re: The set of primes as an algebraic structure

Quote:
 Originally Posted by eddybob123 Is such a binary operation within our grasp?
Actually, I believe lots of such are.

Quote:
 Originally Posted by eddybob123 Would an algebraic structure of this type be applicable in any way?
That'd be a hard question. Perhaps it can be developed in theoretical grounds.

 eddybob123 October 26th, 2013 09:40 AM

Re: The set of primes as an algebraic structure

Quote:
 Originally Posted by CRGreathouse Sure, here's an example. Let S(p) denote the prime succeeding p and P(p) denote the prime preceding p, if such a number exists. So P(5) = 3, P(11) = 7, and so forth. Then let $p\cdot 2=p=2\cdot p$ and $p\cdot S(q)=S(p\cdot q)=S(q)\cdot p.$ So for example 3*7 = S(2*7) = S(7) = 11.
I don't understand. Why does the 3 change into a 2?

 CRGreathouse October 26th, 2013 09:45 AM

Re: The set of primes as an algebraic structure

Quote:

Originally Posted by eddybob123
Quote:
 Originally Posted by CRGreathouse Sure, here's an example. Let S(p) denote the prime succeeding p and P(p) denote the prime preceding p, if such a number exists. So P(5) = 3, P(11) = 7, and so forth. Then let $p\cdot 2=p=2\cdot p$ and $p\cdot S(q)=S(p\cdot q)=S(q)\cdot p.$ So for example 3*7 = S(2*7) = S(7) = 11.
I don't understand. Why does the 3 change into a 2?

3 = S(2), and you're converting S(p) * q to S(p*q) by the rules.

 eddybob123 October 26th, 2013 11:23 AM

Re: The set of primes as an algebraic structure

The rules you stated were
1. 2*p=p*2=p
2. p*S(q)=S(q)*p=S(p*q)

From these rules, 3*7=S(3*5)=S(15)=13

 CRGreathouse October 26th, 2013 11:24 AM

Re: The set of primes as an algebraic structure

Quote:
 Originally Posted by eddybob123 The rules you stated were 1. 2*p=p*2=p 2. p*S(q)=S(q)*p=S(p*q) From these rules, 3*7=S(3*5)=S(15)=13
No. The * is the group operator, not the usual multiplication operator in Z. So 3*7 = S(3*5) = S(S(3*3)) = S(S(S(3*2))) = S(S(S(3))) = S(S(5)) = S(7) = 11.

As it happens the operation is commutative, so it's faster to decrease the smaller number: 3*7 = 7*3 = S(7*2) = S(7) = 11.

 eddybob123 October 26th, 2013 11:27 AM

Re: The set of primes as an algebraic structure

Quote:
 Originally Posted by CRGreathouse No. The * is the group operator, not the usual multiplication operator in Z.
Ah, I see, so the operation is recursively defined. What structure does it have? It's closed and has identity. Is it associative or does an inverse exist for each element?

 CRGreathouse October 26th, 2013 11:40 AM

Re: The set of primes as an algebraic structure

Quote:
 Originally Posted by eddybob123 Ah, I see, so the operation is recursively defined.
Yes.

Quote:
 Originally Posted by eddybob123 What structure does it have?
I'll choose not to answer for now, lest I spoil the secret.

Quote:
 Originally Posted by eddybob123 It's closed by definition, and it wouldn't be too hard to find if it's associative or not.
Yes, and it is.

Quote:
 Originally Posted by eddybob123 Any others?
Sure. Here's a dumb one, which gives infinitely many variations on the one above.

To compute p*q, first replace 11 with 13 if p or q is 11, and vice versa (replace 13 with 11 if p or q is 13). Then compute the operator described above. Finally, swap 11s and 13s as in the first step.

Clearly this has the same structure as the first one, but it orders the primes 2, 3, 5, 7, [color=#800000]13, 11[/color], 17, 19, ... giving essentially the same result but with different names.

I have a more complicated operation in mind but I'm not sure how to describe it. Here are some values:
13*3 = 11
7*7 = 2
17*23 = 47
3*13 = 11
2*5 = 5
5*2 = 5
19*3 = 17
3*7 = 5
13*5 = 19
7*29 = 31
2*13 = 13
11*17 = 5
2*47 = 47
5*7 = 3
59*43 = 113

 CRGreathouse October 26th, 2013 11:44 AM

Re: The set of primes as an algebraic structure

Quote:
 Originally Posted by eddybob123 Is it associative or does an inverse exist for each element?
My first example is an abelian semigroup, so it is associative but has no inverses (well, except 2 which is by definition the identity). In fact it's monotonic, in the sense that p*q >= max(p, q). My second example is an abelian group so it does have inverses -- in fact each element is its own inverse.

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