# Closed, finite set under *.

genius
Posers and Puzzles 10 Oct '07 08:51
1. genius
Wayward Soul
10 Oct '07 08:513 edits
Prove that a set S={a[0], a[1], a[2], ... ,a[n]} that is closed under the operation * contains an element a[ i] such that a[ i]*a[p]=a[p]*a[ i]=a[p] (that is to say, an identity element).

If this is not true, given a counter-example.
2. 10 Oct '07 09:38
Counterexample: Define a[p]*a[q]=a[p].
Baby Gauss
10 Oct '07 10:321 edit
Originally posted by genius
Prove that a set S={a[0], a[1], a[2], ... ,a[n]} that is closed under the operation * contains an element a[ i] such that a[ i]*a[p]=a[p]*a[ i]=a[p] (that is to say, an identity element).

If this is not true, given a counter-example.
Is S is closed under the operation we have that a[p]*a[q] is still an element of S. So we can define define a[p]*a[q]=a[s] where a[s] is still an element of S.

Then we notice that a[1]*a[2]*a[3]*...*a[n-1]*a[n]=a[k] is still an element of S by continuosly applying the previous argument. But since we exausted all the elements of S we can be sure that a[k]=a[p] for some p
Baby Gauss
10 Oct '07 10:333 edits
Is S is closed under the operation we have that a[p]*a[q] is still an element of S. So we can define define a[p]*a[q]=a[s] where a[s] is still an element of S.
Baby Gauss
10 Oct '07 10:341 edit
This keeps on cutting my full post. I'll send you the answer via PM.
Baby Gauss
10 Oct '07 10:402 edits
Meh, this is my last try...
http://tinypic.com/view.php?pic=2gwhoav&s=2
Baby Gauss
10 Oct '07 11:372 edits
I think this will do the trick and since I can't edit one of my previous posts I'll have to spam this thread one more time. Sorry for that one. Anyway what was causing the problem was a less than sign.

If S is closed under the operation * we have that a[p]*a[q] is still an element of S. So we
can define a[p]*a[q]=a[s] where a[s] is still an element of S.

Then we notice that a[1]*a[2]*a[3]*...*a[n-1]*a[n]=a[k] is still an element of S by
continuously applying the previous argument. But since we exhausted all the elements
of S we can be sure that a[k]=a[p] for some p in {1,2,...,n}. So without loss of generality let's take for instance k=2. Then we have a[1]*a[2]*a[3]*...*a[n-1]*a[n]=a[2] and in which case it'll be a[1]*a[3]*...*a[n-1]*a[n]=1.

Note that the particular value of k that we took here doesn't really matter since with
every other value we could make the following conclusion:
a[1]*a[2]*a[3]*...*a[n-1]*a[n]=a[k] for some k in {1,2,...,n} then we would have
a[1]*a[2]*...a[k-1]*a[k]*a[k+1]*...*a[n-1]*a[n]=a[k] which would imply
a[1]*a[2]*...a[k-1]*a[k+1]*...*a[n-1]*a[n]=1
8. Palynka
Upward Spiral
10 Oct '07 11:431 edit
I think this will do the trick and since I can't edit one of my previous posts I'll have to spam this thread one more time. Sorry for that one. Anyway what was causing the problem was a less than sign.

If S is closed under the operation * we have that a[p]*a[q] is still an element of S. So we
can define a[p]*a[q]=a[s] where a[s] is still an e a[k+1]*...*a[n-1]*a[n]=a[k] which would imply
a[1]*a[2]*...a[k-1]*a[k+1]*...*a[n-1]*a[n]=1
But how do you know that 1 (the RHS) is a part of the set? (that is, if we don't, how does that imply that an identity element exists?)
Baby Gauss
10 Oct '07 11:56
Originally posted by Palynka
But how do you know that 1 (the RHS) is a part of the set? (that is, if we don't, how does that imply that an identity element exists?)
Because the set is closed. That means that if you multiply elements of the set you will ALWAYS get elements of the set. a[1]*...*a[k-1]*a[k+1]*...a[n] belongs to the set with no doubt and by the equality a[1]*...*a[k-1]*a[k]*a[k+1]*...a[n]=a[k] it just equals 1. That is of course if the multiplication operation on the set S is commutative.
Baby Gauss
10 Oct '07 12:222 edits
Originally posted by David113
Counterexample: Define a[p]*a[q]=a[p].
I don't know if you're joking or not, but the counter-example would be a set S closed under the multilication operation that doesn't have the unit element in it.
Just changing the letters in the indexes doesn't cut it cause they are dummy indexes.
11. genius
Wayward Soul
10 Oct '07 12:572 edits
I think this will do the trick and since I can't edit one of my previous posts I'll have to spam this thread one more time. Sorry for that one. Anyway what was causing the problem was a less than sign.

If S is closed under the operation * we have that a[p]*a[q] is still an element of S. So we
can define a[p]*a[q]=a[s] where a[s] is still an e a[k+1]*...*a[n-1]*a[n]=a[k] which would imply
a[1]*a[2]*...a[k-1]*a[k+1]*...*a[n-1]*a[n]=1
As I said in the PM, i think you are implying that the set is commutative. I'll have a proper look at it tonight, but i don't think a[1]*a[2]*...a[k-1]*a[k]*a[k+1]*...*a[n-1]*a[n]=a[k] would imply a[1]*a[2]*...a[k-1]*a[k+1]*...*a[n-1]*a[n]=1 unless S was commutative.

EDIT: but the identity is commutative with all S...hmm...

EDIT2: spent so long contemplating your answer that i forgot to plunge my coffee. Its strong...
Baby Gauss
10 Oct '07 13:11
Originally posted by genius
As I said in the PM, i think you are implying that the set is commutative. I'll have a proper look at it tonight, but i don't think a[1]*a[2]*...a[k-1]*a[k]*a[k+1]*...*a[n-1]*a[n]=a[k] would imply a[1]*a[2]*...a[k-1]*a[k+1]*...*a[n-1]*a[n]=1 unless S was commutative.

EDIT: but the identity is commutative with all S...hmm...

EDIT2: spent so long contemplating your answer that i forgot to plunge my coffee. Its strong...
Yes, yes I'm assuming that the * defined in the set is commutative but I think it is a fair assumption.

a[1]*a[2]*...a[k-1]*a[k]*a[k+1]*...*a[n-1]*a[n]=a[k] the things is if we can swap a[k] all the way through the end of the product, or front, to have a[1]*a[2]*...a[k-1]*a[k+1]*...*a[n-1]*a[n]*a[k]=a[k]. If so the result follows. So what really at stake here is not the fact of the identity being commutative but if the a[k] are commutative with eachother.

By the way were did you get this problem from? And do you an answer that doesn't use the assumption of commutability of a[k]. If so just tell me and I'll think harder.
Baby Gauss
10 Oct '07 13:19
{0,0} is a counter example if the elements need not to be different.

But if we assume that * is commutative then the previous result still holds.

Let's assume that * is indeed commutative and that the elements are different. Then how many elements have the set S. How many set S' can you find?

If the elements need not to be different which is the form of the set S? And how many set S' exist now?
14. Palynka
Upward Spiral
10 Oct '07 13:272 edits
Because the set is closed. That means that if you multiply elements of the set you will ALWAYS get elements of the set. a[1]*...*a[k-1]*a[k+1]*...a[n] belongs to the set with no doubt and by the equality a[1]*...*a[k-1]*a[k]*a[k+1]*...a[n]=a[k] it just equals 1. That is of course if the multiplication operation on the set S is commutative.
But don't you have to multiply by 1/a[k] to get there, which is not guaranteed to be in the set?

Edit - Ah, ok, I see it. You just call a[1]*...*a[k-1]*a[k]*a[k+1]*...a[n] as a[ i] and then a[ i]*a[k]=a[k]. Nice.
15. genius
Wayward Soul
10 Oct '07 13:371 edit