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In elementary concepts, mostly scalar exponents shows up in group operations:

$g^x$

As one may encounter in more advanced papers, there are rational exponents over generators. Simply seems like:

$g^{1/x}$

For example, it can be seen at the fouth definition in the paper: q-weak Diffie Hellman assumption

  • Q1: How should we interpret rational exponents on generators?
  • Q2: What is elliptic curve notation of the rational multiplicators? $(1/x)G$ ?
  • Q3: What is relationship between $g^{x}$ and $g^{1/x}$?

Thank you.

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If $$x \times \frac{1}{x} = 1$$

then $$g^{x\frac{1}{x}} = g^1 = g$$

Alternatively, you could denote $\frac{1}{x}$ as $x^{-1}$.

In the context of cryptography, $\frac{1}{x}$ will usually mean the modular multiplicative inverse of $x$, because cryptography usually works with modular arithmetic.

With modular arithmetic, the way to evaluate the division operation is to multiply by the inverse. For example: $$3 \times 7 \equiv 10 \bmod 11\\10 \times 4 \equiv 7 \bmod 11$$

You wouldn't be able to use the traditional division operation to divide $10$ by $3$ to obtain $7$. But you can use multiplication by the inverse of $3$ (which happens to be $4$ here) to obtain the correct result. So we could also denote the above by: $$10 \times 3^{-1} \equiv 7 \bmod 11$$ or equivalently: $$10 \times \frac{1}{3} \equiv 7 \bmod 11$$

So in general, $\frac{y}{x}$ means $y \times x^{-1}$.

Q2: What is elliptic curve notation of the rational multiplicators? $(1/x)$G ?

What you've written looks plausible, but it wouldn't surprise me if they tended to use $x^{-1}G$ instead.*

Either way you denote it, people will probably understand the intent - especially if you provide helpful definitions for your notation.

* I'm not well-versed in literature on elliptic curves, so I'll leave it to someone else to confirm/deny that.

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$g^{1/x}$ (or $g^{x^{-1}}$) is any element of the ambient group $G$ which, when raised to the $x$th power, yields $g$.

If, as is typical, $G$ is a cyclic group of prime order $p$, $g$ is a generator, and $x \not\equiv 0 \pmod p$, then it can be shown that $g^{1/x}$ is unique and equals $g^{x^{-1} \bmod p}$.

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