3
$\begingroup$

Does a mathematical operation 'op' exist so that there is a unique result for every expression? so:

a op bb op a
a op bc op a
a op bc op d
a op b = a op b

Thanks in advance.

  • 2
    Your property list doesn't stop nonunique results. For example, let $x\circ y := x$. This trivial operation discards the second input, and it satisfies all the listed properties whenever $a,b,c,d$ are distinct, yet $x\circ y$ and $x\circ z$ have the same result even if $y\ne z$. You should add after the second property these two: $$a\circ b\ne a\circ c $$ $$a\circ b\ne c\circ b$$ At any rate, if the set of elements under consideration is $X$ (and is infinite), any [injection](http://en.wikipedia.org/wiki/Injective_function) $X\times X\to X$ will determine such an operation.2011-09-08
  • 1
    Are we to take a, b, c, d distinct from each other?2011-09-08
  • 0
    @Mark Bennet: Unless stated, some of a,b,c,d could match and the requirements must still be met.2011-09-12
  • 0
    @Ross Millikan: If a=b the two sides of the first expression are equal, so my thought was that there was some unstated condition.2011-09-12
  • 0
    @Mark Bennet: good point. Looks like they have to be distinct2011-09-12

4 Answers 4