# Rational exponents on group generators

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.

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