Ring theory, binomial theorem in a general ring

[Meital]

Hello everyone,

I am trying to find the form of the binomial theorem in general ring; that is, find an expression for ( a+b)^n, where n is a positive integer. I tried to solve it by induction, but usually we use induction when we know the answer and we try to show it is true. But here, I don't think here this will work. So I tried to write few terms, for n =1 a +b for n = 2 we have a^2 + 2ab + b^2 = a^2 + ab + ba + b^2, and for n = 3 it is a^3 + aab + aba + abb + baa+ bab + bba + b^3, so it is like the terms of (a+b)^n is consisted of all the possible arrangements between a and b in a sum, and every term of the sum has n entries either all a, all b, or a combination of a and b. I know we can do that since if a and b are elements of a general ring then we have addition is associative, so is multiplication, and both distributive laws hold. but the thing is I don't know how to write that in a fomal way or say, have it down as a formula or expression. Can someone help me put all that together? Thanks in advance.

[matt grime]

There is nothing to stop you inventing your own notation, as long as you explain it - that after is all the usual binomial expression is.

 

Here's a naive way of writing it:

 

[math](a+b)^n = \sum_{s \in S(a,b,n)} s[/math]

 

where S(a,b,n) is the set of all words in a and b of total length n.

 

You can refine that since S is the disjoint union of, say S(a,b,n,p) which is the words in a and b of length n with exactly p a's.

 

Can you see how to prove that all this is true? Think about showing how every element in the expansion is some word of length n and how every word of length n can be gotten from the expansion.