fix(6.7.5): state correct inductive hypothesis

html/06_Induction.html

@@ -2663,8 +2663,9 @@ <h3><span class="section-number">6.6.6. </span>Exercises<a class="headerlink" hr
 <div class="admonition-proof admonition">
 <p class="admonition-title">Proof</p>
 <p>We prove this by strong induction on <span class="math notranslate nohighlight">\(b\)</span>.   Suppose that for all
-integers <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(y\)</span> with <span class="math notranslate nohighlight">\(|y|&lt;|b|\)</span>, it is true that
-<span class="math notranslate nohighlight">\(0 \le  \operatorname{gcd}(x, y)\)</span>.</p>
+integers <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(y\)</span> with <span class="math notranslate nohighlight">\(|y|&lt;|b|\)</span>, it is true that</p>
+<div class="math notranslate nohighlight">
+\[L(x,y)x+R(x,y)y=\operatorname{gcd}(x,y).\]</div>
 <p><strong>Case 1</strong> (<span class="math notranslate nohighlight">\(0&lt;b\)</span>): Let <span class="math notranslate nohighlight">\(q=\operatorname{div}(a,b)\)</span> and let
 <span class="math notranslate nohighlight">\(r=\operatorname{mod}(a,b)\)</span>, so that <span class="math notranslate nohighlight">\(a=r+bq\)</span> (by <a class="reference internal" href="#mod-add-div"><span class="std std-numref">Example 6.6.2</span></a>).</p>
 <p>Then by the recurrence definitions,</p>