1
$\begingroup$

I want to find a function $L:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ satisfies: $ L(a,b)=L(b,a) $

and

$ L(a,L(b,c))=L(L(a,b),c) $

I tried several functions, but only two trivial solution works, i.e. $L(m,n)=m+n+C$ or $L(m,n)=Cmn$.

So I wonder are they the only solution? If not, please show me an example..

2 Answers 2

4

Consider $L(m,n) = mn + m+ n$.

  • 0
    @Amr...opps..I forget that distributive law is in the definition of ring...ok .your new structure is quite good .thank you!2012-11-11
0

Another function L(a,b)=(ab)mod 2

let $p:Z→Z$ be a bijection, then define $A(m,n)=p^{−1}(p(m)+p(n))$ and define $M(m,n)=p^{−1}(p(m)p(n))$. The structure (Z,A,M) is a ring. In fact, by Letting p(n)=n+1 we find that $M(m,n)=p^{−1}(p(m)p(n))=mn+m+n$ and $A(m,n)=m+n+1$