Having trouble finding it, but there was a question posted last week regarding a binary operation $x,y \in \mathbb R, x\oplus y=\frac{xy}{x+y}$. It also addressed Resistors in parallel and worked towards a corresponding product rule for derivatives. The question seemed to be begging for a Group Theory analysis to answer the OP's question, at least to start. I hit a hurdle looking into it. Inverses and identity I came up with seemed off and in particular, I found what seems to be an inverse operation, but doesn't correspond to the inverse elements I found. Is there away to get the inverse operator below from inverse elements?
Can you get a group using that operator?
Operation is obviously commutative.
It's closed.
Associative?
$(x \oplus y ) \oplus z = \frac{\frac{xyz}{x+y}}{\frac{xy}{x+y}+z}=\frac{xyz}{xy+xz+yz}$
$x\oplus (y \oplus z) = \frac{\frac{yzx}{y+z}}{\frac{yz}{y+z}+x}=\frac{xyz}{yz+xy+xz}$
So it is associative.
- Identity?
$x \oplus e=x \implies \frac{xe}{x+e}=x\implies xe=x^2+xe$
There doesn't seem to be an identity element. You do get the result if you take e to be $\pm \infty$. But I have misgivings about that.
- Inverses. Here's where I'm stuck.
Let's allow for now $e \in \{\infty, -\infty\}$
Then solve for y so that $\frac{xy}{x+y}=e$
This doesn't seem to lend itself to inverses. The denominator needs to go to zero so $y=-x$. This makes the numerator negative for all values but the denominator more approaches zero to yield the identity rather than equaling zero. Further if $y$ approaches x from the left or the right changes the sign, so that's an additional complication. So does this mean for certain there are no inverses?
The paper linked to that question gave an inverse operator $x\ominus y = \frac{xy}{y-x}$
Then $(x\oplus y) \ominus y = \frac{\frac{xy^2}{x+y}}{y-\frac{xy}{x+y}}=x$
So it works. The candidates for inverse elements above suggested a sign change for one of the operands in both the numerator and the denominator, but the inverse operator just changes the sign of one operand in the denominator.
There seems to be an algebraic structure almost like a group here. Could there be a way to slightly modify $\oplus$ to get an actual group?
