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

  • 2
    And they said studying for the Math GRE was a waste of time...2012-11-11
  • 0
    In fact, I'm considering how to give a different ring structure on $\mathbb{N}$. Using your $L(m,n)$ as addition, is there a suitable multiplication?..Or, do you know an different ring structure on $N$?2012-11-11
  • 0
    @hxhxhx88: $\mathbb{N}$ with the usual operations isn't a ring; and it also isn't a ring with $m \cdot n = L(m,n)$ as its addition (or multiplication).2012-11-11
  • 0
    He probably meant Z2012-11-11
  • 0
    Let p:Z→Z be a bijection, then define A(m,n)=p(m)+p(n) and define M(m,n)=p(m)p(n). The structure (Z,A,M) is a ring2012-11-11
  • 0
    @CliveNewstead..yeah..I mean $Z$..and I got one..addition using $P(m,n)=m+n+1$ and multiplication using $L(m,n)=mn+m+n$.2012-11-11
  • 0
    @Amr.thx..one more question..what if further more I want the addition and multiplication satisfying distributive law?2012-11-11
  • 0
    Opps I made a mistake by assuming that (Z,A,M) is a ring in previous comments2012-11-11
  • 2
    Again 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 ring2012-11-11
  • 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 $A(m,n)=m+n+1$