Refined the epsilon lemma/dfn

main.tex

@@ -1032,6 +1032,10 @@ Where $\epsilon_q^1$ and $\epsilon_q^2$ are defined as follows:
 	
 \end{lemmadfn}
 
+It's worth noting that $\epsilon_q^2$ is potentially larger than $\epsilon_q^2$
+but calculating it involves a $\gcd$, a modulo reduction, and a modulo $n$
+inverse, for each $q$ considered.
+
 \begin{proof}
 
 Consider the following tautology:
@@ -1070,20 +1074,16 @@ Consider the following tautology:
 	\label{eqn:better_eps_problem_k_mod_n}
 \end{align}
 
-In our situation, we want to find a gap between the right-hand-side of eqn
-\ref{eqn:positive_rad_condition_in_terms_of_q_beta},
-and the least element of $\frac{1}{m}\ZZ$ which is strictly greater.
-This amounts to finding the least $k \in \ZZ_{>0}$ for which 
-eqn \ref{eqn:finding_better_eps_problem} holds.
+In our situation, we want to find the least $k$ satisfying 
+eqn \ref{eqn:finding_better_eps_problem}.
 Since such a $k$ must also satisfy eqn \ref{eqn:better_eps_problem_k_mod_n},
-we can pick the smallest $k \in \ZZ_{>0}$ which satisfies this new condition
-(a computation only depending on $q$ and $\beta$, but not $r$)
-and we are then guaranteed that the gap is at least $\frac{k}{2mn^2}$.
-
-A potentially larger gap can also be guaranteed if we choose the least
-$k \in \ZZ_{>0}$ satisfying eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
-instead, but at the cost of computing several $\gcd$'s and modulo reductions
-for each $q$ considered.
+we can pick the smallest $k_q^1 \in \ZZ_{>0}$ which satisfies this new condition
+(a computation only depending on $q$ and $\beta$, but not $r$).
+We are then guaranteed that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_q^1$.
+Furthermore, $k$ also satisfies
+eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
+so we can also pick the smallest $k_q^2 \in \ZZ_{>0}$ satisfying this condition,
+which also guarantees that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_q^2$.
 
 \end{proof}