Hello!
It is interesting that I am also from Russia just as like K.A. Rubcov, and since 2004 me and one my friend, we have often been thinking and talking about the zeration. We seriosely tried to resolve this problem rationally. Our discussion on the matter sometimes stopped, or sometimes continued along all these years. This friend had even found on the web an article of K.A. Rubcov in Russain language, and we made some notes on it.
But we use another symbol for the operation: "@", i. e. a@a = a+2; and we call it, in Russian, with a word "набирание" which is close to English words "making a set", "making a collection". We have chosen this word because of it's psycological sense: before additing some numbers to each other, we must imagine them together, in a collection, in a set.
We two tried to satisfy all the restrictions presented here:
1) a@a = a+2 like a+a = a*2
2) a+(b@c) = a+b@a+c like a*(b+c) = a*b+a*c
3) a@b = b@a just we want it be so
4) a@b < a+b < a*b for all a not equal b (probably)
And I think that we achieved a good solution. Now I am preparing an article about it and thinking of where to publish it.
If you are still interested in it or just do not deny the possibility of a good solution, then I could hurry up with an article.
It is interesting that I am also from Russia just as like K.A. Rubcov, and since 2004 me and one my friend, we have often been thinking and talking about the zeration. We seriosely tried to resolve this problem rationally. Our discussion on the matter sometimes stopped, or sometimes continued along all these years. This friend had even found on the web an article of K.A. Rubcov in Russain language, and we made some notes on it.
But we use another symbol for the operation: "@", i. e. a@a = a+2; and we call it, in Russian, with a word "набирание" which is close to English words "making a set", "making a collection". We have chosen this word because of it's psycological sense: before additing some numbers to each other, we must imagine them together, in a collection, in a set.
We two tried to satisfy all the restrictions presented here:
1) a@a = a+2 like a+a = a*2
2) a+(b@c) = a+b@a+c like a*(b+c) = a*b+a*c
3) a@b = b@a just we want it be so
4) a@b < a+b < a*b for all a not equal b (probably)
And I think that we achieved a good solution. Now I am preparing an article about it and thinking of where to publish it.
If you are still interested in it or just do not deny the possibility of a good solution, then I could hurry up with an article.