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I'm trying to gain some intuition about modular multiplicative inverses. Suppose I have a prime number $n$ and the group $(\mathbb{Z} / n \mathbb{Z})^\times$, and I pick some constant $\alpha$ much smaller than 1. Let $A = \{ 1, 2, 3, \ldots , \alpha n \}$, and let $S = A^{-1}$. Is it true that as $n$ approaches infinity, $S$ is uniform? (More precisely, if you look at any interval $I$ of the set $ \{ 1, 2, 3, \ldots, n-1 \}$, the number of elements of $S$ falling in this interval is proportional to $\displaystyle{\frac{|I|}{n - 1}}$?) Is there a rigorous proof of this?

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  • $\begingroup$ I think there are some misunderstandings about the sets at hand. What $n$ is varying? Is $\alpha$ a constant of $\mathbb{Z}$ (and what do you mean by $\alpha <<1$ but the set $A$ is listed in increasing order)? If by $S$ you mean the, say, sequence of inverses for the corresponding elements (like, $1$'s inverse, $2$'s inverse,...) then for arbitrary $n$ (in $\mathbb{Z}/n\mathbb{Z}^\times$) the elements in $A$ may not have inverses. $\endgroup$ – Eoin Jun 19 '15 at 0:24
  • $\begingroup$ @Eoin $\alpha << 1$ means $\alpha$ is much smaller than 1. The set $A$ varies with $n$. Also, $n$ has to be prime as it approaches infinity. $\endgroup$ – John Doe Jun 19 '15 at 0:26
  • $\begingroup$ If $\alpha<<1$ is supposed to mean much smaller than $1$, but still an integer, then $\alpha=0$ or is a negative integer so the questions I would have are i) Are the elements of $A$ integers (where the increasing sequence doesn't make sense with $\alpha$ stated so the elements of $A$ are ambiguous), or elements of the group $\mathbb{Z}/n\mathbb{Z}^\times$(where $n$ doesn't make sense because $\alpha n=0$ is not in $\Bbb{Z}/n\Bbb{Z}^\times$)? ii) Are the $n$ that varies in $A$ and the $n$ that is in $\mathbb{Z}/n\mathbb{Z}^\times$ the same? If not, you should relabel one to avoid confusion. $\endgroup$ – Eoin Jun 19 '15 at 0:34
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    $\begingroup$ A related issue involving the distribution of inverses modulo $p$: Prove or disprove that $\lim_{p\to \infty}\frac{1}{p^3}\sum_{i=1}^{p-1}ii^{-1}=\frac{1}4$ $\endgroup$ – punctured dusk Jun 19 '15 at 10:12
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    $\begingroup$ You've asked the same question twice, right? math.stackexchange.com/questions/1330981/… $\endgroup$ – Gerry Myerson Jun 19 '15 at 10:26

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