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On the interval $(-1, 1)$, consider the binary operation

$$x*y=\dfrac{2xy+3(x+y)+2}{3xy+2(x+y)+3}$$

with $x, y \in (-1, 1)$. I have to find the number of solutions for the equation:

$$\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$$

Finding the left-hand side would be incredibly painful, so I didn't try that. I looked at the previous sub-point of this exercise and it looked like it might help, but I don't know how to use it exactly. In this previous sub-point I showed that for the function:

$$f:(-1, 1) \rightarrow (0, \infty) \hspace{2cm} f(x) = \dfrac{1}{5} \cdot \dfrac{1-x}{1+x}$$

it is true that:

$$f(x * y) = f(x)f(y)$$

$\forall x,y \in (-1, 1)$.

Still I don't know how to find the number of solutions for:

$$\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$$

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    Hint: Apply $f$ both sides of the equation. – Peter Foreman Mar 18 '20 at 19:18
  • Is the star operation associative ? This could be checked in about 10 minutes of suffering. If so, then it would combine well with Peter Foreman's hint. – Simon Mar 18 '20 at 19:41
  • Actually, you don't need general associativity, you only need it in the case where every operand is equal, i.e. to $x$. This "restricted associativity" follows from commutativity of the star operation, which can be seen immediately to hold. – Simon Mar 18 '20 at 20:03
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    You have associativity in general. $f((xy)z) = f(xy)f(z) = (f(x)f(y))f(z) = f(x)(f(y)f(z)) = f(x)f(yz) = f(x(yz))$ and $f$ is a bijection, so $(xy)z = x(yz)$. – Jair Taylor Mar 18 '20 at 20:54
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    This is an example of a formal group law – Jair Taylor Mar 18 '20 at 20:56
  • That's really cool @Jair Taylor ! Thank you for the interesting link. Reading it now. – Simon Mar 18 '20 at 23:03

1 Answers1

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What a strange operation! The trick is to find an isomorphism, and they have given you the hint: because $f$ is a bijection, we can say that the operation of $*$ on $(-1, 1)$ is isomorphic to the operation of usual multiplication on $(0, \infty)$. Thus, in terms of $y = f(x) \in (0, +\infty)$, the problem is simply to solve $y^{10} = f(1/10) \in (0, +\infty)$, which has exactly one solution.

(One should check, quite painlessly, that having an isomorphism is "really this good"; for example, that we also get $x * ... * x = f^{-1}(f(x)^n)$ and so on. Indeed, a bijection satisfying the homomorphism law is the correct notion of an isomorphism between magmas.)

Christopher Gadzinski
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  • @CristopherGadzinski How did you find that $y^{10} = f(1/10)$ has only one solution? –  Mar 19 '20 at 21:44