Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

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

share|improve this answer
2  
And they said studying for the Math GRE was a waste of time... –  Isaac Solomon Nov 11 '12 at 9:28
    
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 $N$? –  hxhxhx88 Nov 11 '12 at 9:34
    
@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). –  Clive Newstead Nov 11 '12 at 9:35
    
He probably meant Z –  Amr Nov 11 '12 at 9:36
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 ring –  Amr Nov 11 '12 at 9:54
show 5 more comments

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 $A(m,n)=m+n+1$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.