A Few Math Questions

[Grog]

Hmm... now that I've re-examined it more closely, it looks valid. I wonder, does this work if instead of
i x a = a
a' x a = i

we take

i x a = a
a x a' = i
and do not assume that (a')' = a?
then we can't put together line 3... instead, we get

c = a x a' = a x i x a' = a x (a x a') x a' = a x (a' x (a')') x a' -> not very useful

Interesting...

I think c = a' x a is the way to start if you have a x a' = i

[drachefly]

c = a' x a = a' x a x a' x a = c x c
i = c' x c = blah

yeah.

Okay. Kuroneko, what was it you were getting at, then?

Are one-sided inverses and one-sided identities only the province of non-associative algebras?

[Kuroneko]

I should have clarified. Your identity was defined as i×a = a and inverse as a×a' = i. Group structure does not follow from that, but Grog is mostly correct in that it follows from one-sidedness from the same side. Thus, either {i×a = a, a'×a = i} or {a×i = a, a×a' = i} are sufficient for group structure, but not {i×a = a, a×a' = i} and not {a×i, a'×a = i}. The latter axioms only give a group if i is assumed to be unique, which it was not in this case.

[drachefly]

That's what I was getting at with my suggestion. (making i x a and a x a' be different-sided)

And this morning while trying to get out of bed I realized what the problem was!

The problem with Grog's proof is at the very end, but to make the error clear I need to reproduce the entire proof, more clearly. E = there exists. A = for all. | = such that.

E i | A a,
i x a = a

A b, E b' |
b x b' = i

A c, E d |
d = c' x c

(use the middle line of Grog's proof to establish that
d = d x d

c x d = d x c = c
but that's not showing that
A a, d x a = a!
d was defined specifically with c in mind, so just showing that d is an identity for c does not show that it's an identity for the entire set!

Phew. There we go.

[Kuroneko]

c x d = d x c = c but that's not showing that A a, d x a = a! d was defined specifically with c in mind, so just showing that d is an identity for c does not show that it's an identity for the entire set!

Right. That's why uniqueness of identity should be assumed to force group structure. The problem does not occur if the identity and inverse are same-sided. For example, if they are left-sided, then by liberal use of the associative property, a×a' = i×(a×a') = (a"×a')×(a×a') = a"×((a'×a)×a') = a"×(i×a') = a"×a' = i, so right-sidedness for inverse follows. Using this, right-sidedness of identity becomes trivial, and so uniqueness of identity follows from the standard group structure.