# Doubt in computing $g^\frac{1}{\delta+x}$ where $x \in \mathbb{Z}$

I was going through Zero Knowledge Set Membership and came across the following:

Given $$x \in \mathbb{Z}$$ and $$g$$ is the generator of a multiplicative group $$\mathbb{G}$$ how do we compute $$g^\frac{1}{\delta+x}$$.

My doubt is, since $$\mathbb{G}$$ is a multiplicative group, doesn't that require $$\frac{1}{\delta + x} \in \mathbb{Z}$$. But since $$x \in \mathbb{Z}$$ is that possible?

• $g^\frac{1}{\delta+x} = g^{{(\delta+x)}^{-1}}$. – kelalaka Apr 22 at 10:19
• To be more explicit, in this context $\frac{1}{\delta+x}$ stands for that number $y$ such that $y \times (\delta + x) = 1$ modulo the order of $g$ – poncho Apr 22 at 11:30