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Updated induction hypothesis in 6.6.2 #35

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Updated induction hypothesis in 6.6.2 #35

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4 changes: 2 additions & 2 deletions html/06_Induction.html
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Expand Up @@ -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>
Expand Down Expand Up @@ -2830,4 +2830,4 @@ <h3><span class="section-number">6.7.7. </span>Exercises<a class="headerlink" hr
</script>

</body>
</html>
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