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


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:

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

  • 4
    $\begingroup$ Hint: Apply $f$ both sides of the equation. $\endgroup$ – Peter Foreman Mar 18 at 19:18
  • $\begingroup$ 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. $\endgroup$ – Simon Mar 18 at 19:41
  • $\begingroup$ 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. $\endgroup$ – Simon Mar 18 at 20:03
  • 1
    $\begingroup$ You have associativity in general. $f((x*y)*z) = f(x*y)f(z) = (f(x)f(y))f(z) = f(x)(f(y)f(z)) = f(x)f(y*z) = f(x*(y*z))$ and $f$ is a bijection, so $(x*y)*z = x*(y*z)$. $\endgroup$ – Jair Taylor Mar 18 at 20:54
  • 1
    $\begingroup$ This is an example of a formal group law $\endgroup$ – Jair Taylor Mar 18 at 20:56

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.)

  • $\begingroup$ @CristopherGadzinski How did you find that $y^{10} = f(1/10)$ has only one solution? $\endgroup$ – user1502 Mar 19 at 21:44

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