From e9a0ae083a509af70da43d9f60e62745558ac189 Mon Sep 17 00:00:00 2001
From: Kevin C <mathapptician@gmail.com>
Date: Thu, 5 Jun 2025 11:59:43 -0400
Subject: [PATCH] Updated induction hypothesis in 6.6.2

---
 html/06_Induction.html | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/html/06_Induction.html b/html/06_Induction.html
index 5abbfc38..c5f572f6 100644
--- a/html/06_Induction.html
+++ b/html/06_Induction.html
@@ -1922,7 +1922,7 @@ <h3><span class="section-number">6.5.4. </span>Exercises<a class="headerlink" hr
 <p class="admonition-title">Proof</p>
 <p>We prove this by strong induction relative to the expression <span class="math notranslate nohighlight">\(2n - d\)</span>.  Suppose that for all
 integers <span class="math notranslate nohighlight">\(m\)</span> and <span class="math notranslate nohighlight">\(c\)</span> with <span class="math notranslate nohighlight">\(|2m - c|&lt;|2n-d|\)</span>, it is true that
-<span class="math notranslate nohighlight">\(\operatorname{mod}(n, d) + d \cdot \operatorname{div}(n, d) = n\)</span>.</p>
+<span class="math notranslate nohighlight">\(\operatorname{mod}(m, c) + c \cdot \operatorname{div}(m, c) = m\)</span>.</p>
 <p><strong>Case 1</strong> (<span class="math notranslate nohighlight">\(nd&lt;0\)</span>): Then by the inductive hypothesis</p>
 <div class="math notranslate nohighlight">
 \[\operatorname{mod}(n+d, d) + d \cdot \operatorname{div}(n+d, d) = n+d,\]</div>
@@ -2830,4 +2830,4 @@ <h3><span class="section-number">6.7.7. </span>Exercises<a class="headerlink" hr
   </script> 
 
 </body>
-</html>
\ No newline at end of file
+</html>