Fix typo in binary conversion -- fixes #1

Author: agmath

Slides/D2_BinaryAndFloatingPointNumbers.html

@@ -11,7 +11,7 @@
   <meta name="generator" content="quarto-1.5.43">
 
   <meta name="author" content="Dr.&nbsp;Gilbert">
-  <meta name="dcterms.date" content="2026-01-08">
+  <meta name="dcterms.date" content="2026-01-09">
   <title>MAT 370: Binary and Floating Point Arithmetic</title>
   <meta name="apple-mobile-web-app-capable" content="yes">
   <meta name="apple-mobile-web-app-status-bar-style" content="black-translucent">
@@ -337,7 +337,7 @@ <h1 class="title">MAT 370: Binary and Floating Point Arithmetic</h1>
 </div>
 </div>
 
-  <p class="date">January 8, 2026</p>
+  <p class="date">January 9, 2026</p>
 </section>
 <section class="slide level2">
 
@@ -535,7 +535,7 @@ <h2>Binary Representations of Integers</h2>
 <p><span class="math display">\[\begin{align*} 84 &amp;= 2\left(42\right) + 0\\
 42 &amp;= 2\left(21\right) + 0\\
 21 &amp;= 2\left(10\right) + 1\\
-10 &amp;= 2\left(5\right) + 1
+10 &amp;= 2\left(5\right) + 0
 \end{align*}\]</span></p>
 </div><div class="column" style="width:60%;">
 
@@ -566,7 +566,7 @@ <h2>Binary Representations of Integers</h2>
 <p><span class="math display">\[\begin{align*} 84 &amp;= 2\left(42\right) + 0\\
 42 &amp;= 2\left(21\right) + 0\\
 21 &amp;= 2\left(10\right) + 1\\
-10 &amp;= 2\left(5\right) + 1\\
+10 &amp;= 2\left(5\right) + 0\\
 5 &amp;= 2\left(2\right) + 1
 \end{align*}\]</span></p>
 </div><div class="column" style="width:60%;">
@@ -598,7 +598,7 @@ <h2>Binary Representations of Integers</h2>
 <p><span class="math display">\[\begin{align*} 84 &amp;= 2\left(42\right) + 0\\
 42 &amp;= 2\left(21\right) + 0\\
 21 &amp;= 2\left(10\right) + 1\\
-10 &amp;= 2\left(5\right) + 1\\
+10 &amp;= 2\left(5\right) + 0\\
 5 &amp;= 2\left(2\right) + 1\\
 2 &amp;= 2\left(1\right) + 0
 \end{align*}\]</span></p>
@@ -631,20 +631,20 @@ <h2>Binary Representations of Integers</h2>
 <p><span class="math display">\[\begin{align*} 84 &amp;= 2\left(42\right) + 0\\
 42 &amp;= 2\left(21\right) + 0\\
 21 &amp;= 2\left(10\right) + 1\\
-10 &amp;= 2\left(5\right) + 1\\
+10 &amp;= 2\left(5\right) + 0\\
 5 &amp;= 2\left(2\right) + 1\\
 2 &amp;= 2\left(1\right) + 0\\
 1 &amp;= 2\left(0\right) + 1
 \end{align*}\]</span></p>
 </div><div class="column" style="width:60%;">
 <div class="fragment">
-<p>Reading the remainders from bottom to top, we find that <span class="math inline">\(\left(84\right)_{10} = \left(1011100\right)_{2}\)</span>.</p>
+<p>Reading the remainders from bottom to top, we find that <span class="math inline">\(\left(84\right)_{10} = \left(1010100\right)_{2}\)</span>.</p>
 </div>
 <div class="fragment">
 <p>Each <em>bi</em>nary dig<em>it</em> is referred to as a <em>bit</em>, and it is common to consider collections of <span class="math inline">\(8\)</span> bits together – known as a <em>byte</em>.</p>
 </div>
 <div class="fragment">
-<p>That is, we would more commonly write <span class="math inline">\(\left(01011100\right)_{2}\)</span> as the binary representation of <span class="math inline">\(84\)</span>.</p>
+<p>That is, we would more commonly write <span class="math inline">\(\left(01010100\right)_{2}\)</span> as the binary representation of <span class="math inline">\(84\)</span>.</p>
 </div>
 </div>
 </div>

Slides/D2_BinaryAndFloatingPointNumbers.qmd

@@ -259,7 +259,7 @@ Below we show how we can use the *Euclidean Algorithm* to convert an integer ($8
 \begin{align*} 84 &= 2\left(42\right) + 0\\
 42 &= 2\left(21\right) + 0\\
 21 &= 2\left(10\right) + 1\\
-10 &= 2\left(5\right) + 1
+10 &= 2\left(5\right) + 0
 \end{align*}
 
 :::
@@ -300,7 +300,7 @@ Below we show how we can use the *Euclidean Algorithm* to convert an integer ($8
 \begin{align*} 84 &= 2\left(42\right) + 0\\
 42 &= 2\left(21\right) + 0\\
 21 &= 2\left(10\right) + 1\\
-10 &= 2\left(5\right) + 1\\
+10 &= 2\left(5\right) + 0\\
 5 &= 2\left(2\right) + 1
 \end{align*}
 
@@ -342,7 +342,7 @@ Below we show how we can use the *Euclidean Algorithm* to convert an integer ($8
 \begin{align*} 84 &= 2\left(42\right) + 0\\
 42 &= 2\left(21\right) + 0\\
 21 &= 2\left(10\right) + 1\\
-10 &= 2\left(5\right) + 1\\
+10 &= 2\left(5\right) + 0\\
 5 &= 2\left(2\right) + 1\\
 2 &= 2\left(1\right) + 0
 \end{align*}
@@ -385,7 +385,7 @@ Below we show how we can use the *Euclidean Algorithm* to convert an integer ($8
 \begin{align*} 84 &= 2\left(42\right) + 0\\
 42 &= 2\left(21\right) + 0\\
 21 &= 2\left(10\right) + 1\\
-10 &= 2\left(5\right) + 1\\
+10 &= 2\left(5\right) + 0\\
 5 &= 2\left(2\right) + 1\\
 2 &= 2\left(1\right) + 0\\
 1 &= 2\left(0\right) + 1
@@ -397,7 +397,7 @@ Below we show how we can use the *Euclidean Algorithm* to convert an integer ($8
 
 :::{.fragment}
 
-Reading the remainders from bottom to top, we find that $\left(84\right)_{10} = \left(1011100\right)_{2}$. 
+Reading the remainders from bottom to top, we find that $\left(84\right)_{10} = \left(1010100\right)_{2}$. 
 
 :::
 
@@ -409,7 +409,7 @@ Each *bi*nary dig*it* is referred to as a *bit*, and it is common to consider co
 
 :::{.fragment}
 
-That is, we would more commonly write $\left(01011100\right)_{2}$ as the binary representation of $84$.
+That is, we would more commonly write $\left(01010100\right)_{2}$ as the binary representation of $84$.
 
 :::