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


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


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.


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.


$g^{1/x}$ (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 $g^{x^{-1} \bmod p}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.