2
$\begingroup$

I need to find a and b with whom this expression is associative: $$x\bullet y=(a+x)(b+y)$$ Also known that $$x,y\in Z$$ So firstly I write: $$(x\bullet y)\bullet z=x\bullet (y\bullet z)$$ Then I express and express them and after small algebra I get: $$ab^2+b^2x+az+abz+bxz=a^2b+a^2z+bx+abx+axz$$ And I don't know what to do now, I don't know if I should do it, but I tried to assign to $x=0$ and $z=1$ but nothing better happened.

Could you please share some ideas?

EDIT: Sorry, but I forgot to mention that I need to find such a and b with whom it should be associative

  • 5
    Unless there's something special about $a$ and $b$, your operator isn't associative...2011-10-16
  • 0
    @J.M. sorry, forgot to add that I need to find such a and b with whom it should be associative2011-10-16
  • 0
    Obviously, $a=b=0$ works.2011-10-16
  • 0
    @Templar: What are $a,b,x,y$ supposed to be? Real numbers? Integers?2011-10-16
  • 0
    @ZevChonoles x,y is integers, nothing said about a, b2011-10-16
  • 0
    I feel lost. How can we write x*y for (a+x)(b+y) when (a+x)(b+y) has four variables, and x*y only has two?2011-10-18

4 Answers 4