Closed, finite set under *.

Originally posted by Palynka
But don't you have to multiply by 1/a[k] to get there, which is not guaranteed to be in the set?

Since the a[ i] aren't defined (we don't know if they are numbers, matrices, differential operators, integral operators or whatever) we can think like this:

a[1]*...*a[k]*...*a[n]=a[k] now if * is commutative between them, that was what I implicitly admitted before, we can write a[1]*...*a[k-1]*a[k+1]*...*a[n]*a[k]=a[k]. Let us define A=a[1]*...*a[k-1]*a[k+1]*...*a[n].

Thus we have A.a[k]=a[k] for every a[k] and since we have made no special hypothesis about the nature of a[k] it follows that A=1. Even if a[k]=0 A=1 is still valid. If we were talking about a particular a[k] then a[k] must be different from 0 like you say, but the thing is that we are talking about every a[k] so we never multiply by 1/a[k] we look at equation and come to the conclusion that A=1 is the only way out.

Another thing is that A=1 doesn't mean that A is the number 1 it should be read as "A is the unit operator".

Edit: Just seen your edit. 🙂

Originally posted by genius
I have more of a sketch proof (I know, I know...but i'm not sure how correct it is). I came across a similar problem in one of my modules-Finite Maths-where we were asked to prove that in a finite field F there exists an integer n such that 1+1+...+1=0 (where there are n 1s). I couldn't think how to tackle it so i showed it (roughly) for a general fini ...[text shortened]... t my logic when i made an elementary mistake the other day, so I could be completly wrong...

hmmmmmmmmm.
I think I know what your reasoning was for the 1+1+1+...+1=0 but... I have to refrain myself to say more because I think it is very related to the hijack I made to your thread.

Thought: In your proof did you assume that S was associative under *?

Also, as the the 1+1+...+1=0, that is what you essentially proved - that and closed and abelian set under an operation * (here, +) has an identity (here, 0).

Originally posted by genius
Thought: In your proof did you assume that S was associative under *?

I think I only assumed that it is commutative... But let me get a closer look on my proof.

Edit: After a closer look I think it's safe to say that I only assumed S was commutative under *. Could be wrong though. Haven't lunched yet.

Originally posted by adam warlock
I think I only assumed that it is commutative... But let me get a closer look on my proof.

If you did then it is a group.

(Which would imply that the number of S'-subgroups-must divide the order of S-that is, the number of elements in S).

Originally posted by genius
If you did then it is a group.

(Which would imply that the number of S'-subgroups-must divide the order of S-that is, the number of elements in S).

((((((...(a[1]*a[2])a[3])....)a[k-1])a[k])a[k+1])...)a[n]) and then you just go passing a[k] to the right. I think that the only thing that's being assumed is commutability.

Argh! I hate it when habit forces itself upon a mathematical rigorous proof.

If no one else posts an answer please post yours just so we can discuss it.

When I talk about S' I'm not talking about subgroups. Sorry for the sloppy notation. I'm thinking about different S-like sets that you can form. I hope this time I'm being understandable.

Originally posted by adam warlock
When I talk about S' I'm not talking about subgroups. Sorry for the sloppy notation. I'm thinking about different S-like sets that you can form. I hope this time I'm being understandable.

I was assuming S was thus a group, so all S' would be groups too...

However, I feel that Lagranges theorem must still hold. I'm not sure, however, if the mapping f😕->Sa defined by af=a*x is well defined if * is not associative. I'll look into it later though-I think I shall turn my laptop off and do some work now...

Originally posted by genius
I was assuming S was thus a group, so all S' would be groups too...

However, I feel that Lagranges theorem must still hold. I'm not sure, however, if the mapping f😕->Sa defined by af=a*x is well defined if * is not associative. I'll look into it later though-I think I shall turn my laptop off and do some work now...

Ok.See you later then. You probem got me thinking a lot you know? Thanks! 🙂

Well here we got the proof without resorting to commutative or associative properties.
S={a[1],a[2],...,a[n]}

Let us form the product a[1]*...*a[k-1]*a[k+1]*...*a[n]*a[k]=a[k] and the rest was previously seen here.
Since S is closed to * we can have our product in any order that we wish and this is one that makes our job easy.

And now my question, propperly asked, is: Is S unique.

Originally posted by adam warlock
Let us form the product a[1]*...*a[k-1]*a[k+1]*...*a[n]*a[k]=a[k] and the rest was previously seen here.
Since S is closed to * we can have our product in any order that we wish and this is one that makes our job easy.

Is it obvious you can do this? I could believe it, but I'm not sure it's that straightforward.

Originally posted by adam warlock
And now my question, propperly asked, is: Is S unique.

No. Here's a counter-example.

* is multiplication modulo 3.
then:
{1} is closed
{1, 2} is also closed

{1, 2} is not closed.

Edit: I realized just now what modulus 3 means but I think we were sticking to regular multiplication. And the question of unicity is more fun for nontrivial sets.

Originally posted by adam warlock
{1, 2} is not closed.

Edit: I realized just now what modulus 3 means but I think we were sticking to regular multiplication. And the question of unicity is more fun for nontrivial sets.

Well, the question - and your proof - made no assumptions about the nature of the operation. You even tried to avoid using commutativity - if we were talking about regular multiplication that wouldn't be a problem.

If you're considering regular multiplication on the real numbers then I think it's fairly clear [earlier version said obvious, but then I missed a few details, so it obviously wasn't!] there's no finite set with members other than 0 and +-1. So {0}, {1}, {1, -1}, {0, 1}, {0, 1, -1}

I think using things other than real numbers is no less artificial than using other types of multiplication. But here's a counter example with regular multiplication on complex numbers.

{1}
{1, -1}
{1, i, -1, -i}

or, more generally

{exp[2*n*pi/m]} for n = 0 to (m-1)}

is closed.

Originally posted by mtthw
Well, the question - and your proof - made no assumptions about the nature of the operation. You even tried to avoid using commutativity - if we were talking about regular multiplication that wouldn't be a problem.

If you're considering regular multiplication on the real numbers then I think it's pretty obvious there's no finite set with that property other than {1} [Edit: and {-1, 1}]

At first I didn't try to avoid commutability but since it was a point I tried to show that commutability isn't even required for the result to hold.
But they are other finite real sets. And since you are so advanced here you go {1,a,1/a} with a different from 0,1, and -1.

Assuming that the elements of S are all different of course.