Start formatting result about new eps into theorem/dfn

main.tex

@@ -24,6 +24,7 @@
 \newcommand{\minorheading}[1]{{\noindent\normalfont\normalsize\bfseries #1}}
 
 \newtheorem{theorem}{Theorem}[section]
+\newtheorem{lemmadfn}{Lemma/Definition}[section]
 
 \begin{document}
 
@@ -985,7 +986,36 @@ rhs_numerator = (
 \noindent
 Let $\aa^{'}$ be an integer representative of $\aa^{-1}$ in $\ZZ/n\ZZ$.
 
-Considering the following tautology:
+Next, we seek to find a larger $\epsilon$ to use in place of $\epsilon_F$ in the
+proof of theorem \ref{thm:rmax_with_uniform_eps}:
+
+\begin{lemmadfn}[
+	Finding better alternatives to $\epsilon_F$:
+	$\epsilon_q^1$ and $\epsilon_q^1$
+]
+Suppose $d \in \frac{1}{m}\ZZ$ is satisfies the condition in
+eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}.
+That is:
+
+\begin{equation*}
+	\sage{positive_radius_condition.subs([q_value_expr,beta_value_expr]).factor()}
+\end{equation*}
+
+\noindent
+Then we have:
+
+\begin{equation*}
+	d - \frac{(\aa r + 2\bb)\aa}{2n^2}
+	\geq \epsilon_q^2 \geq \epsilon_q^1 > 0
+\end{equation*}
+
+Where $\epsilon_q^1$ and $\epsilon_q^2$ are defined as follows:
+	
+\end{lemmadfn}
+
+\begin{proof}
+
+Consider the following tautology:
 
 \begin{align}
 	&\frac{ x }{ m }
@@ -1021,12 +1051,12 @@ Considering the following tautology:
 	\label{eqn:better_eps_problem_k_mod_n}
 \end{align}
 
-In our situation, we want to find the gap between the right-hand-side of eqn
+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.
-Since such a $k$ satisfies eqn \ref{eqn:better_eps_problem_k_mod_n},
+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 $\frac{k}{2mn^2}$.
@@ -1036,6 +1066,7 @@ $k \in \ZZ_{>0}$ 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.
 
+\end{proof}