fix typos

Author: oscarlevin

source/activities/coloring-vertices.ptx

@@ -16,7 +16,7 @@
 
 				<li>
 					<p>
-						Find the <em>fewest</em> number of colors you need to properly coloring the vertices of the graph.
+						Find the <em>fewest</em> number of colors you need to properly color the vertices of the graph.
 						This is called the <term>chromatic number</term> of the graph.
 						Think about how you know your answer is correct.
 					</p>

source/exercises/gt-paths.ptx

@@ -8,7 +8,7 @@
         You and your friends want to tour the southwest by car.
         You will visit the nine states below,
         with the following rather odd rule:
-        you must cross each border between neighboring states exactly once
+        You must cross each border between neighboring states exactly once
         (so, for example,
         you must cross the Colorado-Utah border exactly once).
         Can you do it?
@@ -118,7 +118,7 @@
         <ol>
           <li>
             <p>
-              Edward wants to give a tour of his new pad to a lady-mouse-friend.
+              Edward wants to give a tour of his new pad to a lady-mouse friend.
               Is it possible for them to walk through every doorway exactly once?
               If so, in which rooms must they begin and end the tour?
               Explain.

source/exercises/gt-planar.ptx

@@ -7,7 +7,7 @@
   <exercise>
     <statement>
       <p>
-        Is it possible for a planar graph to have 6 vertices, 10 edges and 5 faces?
+        Is it possible for a planar graph to have 6 vertices, 10 edges, and 5 faces?
         Explain.
       </p>
     </statement>

source/sec_gt-coloring.ptx

@@ -290,13 +290,13 @@
       It appears that there is no limit to how large chromatic numbers can get.
       It should not come as a surprise that <m>K_n</m> has chromatic number <m>n</m>.
       So how could there possibly be an answer to the original map coloring question?
-      If the chromatic number of graph can be arbitrarily large,
+      If the chromatic number of a graph can be arbitrarily large,
       then it seems like there would be no upper bound to the number of colors needed for any map.
       But there is.
     </p>
 
     <p>
-      The key observation is that while it is true that for any number <m>n</m>,
+      The key observation is that while it is true that for any number <m>n</m>
       there is a graph with chromatic number <m>n</m>,
       only some graphs arrive as representations of maps.
       If you convert a map to a graph,
@@ -363,7 +363,7 @@
         <p>
           Radio stations broadcast their signal at certain frequencies.
           However, there are a limited number of frequencies to choose from,
-          so nationwide many stations use the same frequency.
+          so nationwide, many stations use the same frequency.
           This works because the stations are far enough apart that their signals will not interfere;
           no one radio could pick them up at the same time.
         </p>
@@ -410,7 +410,7 @@
         <p>
           This graph has chromatic number 5.
           A proper 5-coloring is shown on the right.
-          Notice that the graph contains a copy of the complete graph <m>K_5</m> so no fewer than 5 colors can be used.
+          Notice that the graph contains a copy of the complete graph <m>K_5</m>, so no fewer than 5 colors can be used.
         </p>
 
         <sidebyside widths="35% 30%" margins="auto" valign="bottom">
@@ -482,7 +482,7 @@
       While we might not be able to find the exact chromatic number of a graph easily,
       we can often give a reasonable range for the chromatic number.
       In other words,
-      we can give upper and lower bounds for chromatic number.
+      we can give upper and lower bounds for the chromatic number.
     </p>
 
     <p>
@@ -594,7 +594,7 @@
       The proof of this theorem is <em>just</em>
       complicated enough that we will not present it here
       (although you are asked to prove a special case in the exercises).
-      The adventurous reader is encouraged to find a book on graph theory for suggestions on how to prove the theorem.
+      The adventurous reader is encouraged to find a book on graph theory to find suggestions for how to prove the theorem.
     </p>
   </subsection>
 
@@ -624,7 +624,7 @@
         <p>
           Six friends decide to spend the afternoon playing chess.
           Everyone will play everyone else once.
-          They have plenty of chess sets but nobody wants to play more than one game at a time.
+          They have plenty of chess sets, but nobody wants to play more than one game at a time.
           Games will last an hour
           (thanks to their handy chess clocks).
           How many hours will the tournament last?
@@ -664,7 +664,7 @@
         <p>
           Notice that for sure <m>\chi'(K_6) \ge 5</m>,
           since there is a vertex of degree 5.
-          It turns out 5 colors is enough
+          It turns out, 5 colors is enough
           (go find such a coloring).
           Therefore the friends will play for 5 hours.
         </p>
@@ -740,7 +740,7 @@
         For example,
         we know that you need to go up to <m>K_{17}</m> in order to force a monochromatic triangle using three colors,
         but nobody knows how big you need to go with more colors.
-        Similarly, we know that using two colors <m>K_{18}</m> is the smallest graph that forces a monochromatic copy of <m>K_4</m>,
+        Similarly, we know that using two colors, <m>K_{18}</m> is the smallest graph that forces a monochromatic copy of <m>K_4</m>,
         but the best we have to force a monochromatic <m>K_{5}</m> is a range,
         somewhere from <m>K_{43}</m> to <m>K_{49}</m>.
         If you are interested in these sorts of questions,

source/sec_gt-paths.ptx

@@ -404,7 +404,7 @@
   </subsection>
 
     <subsection>
-      <title>Conditions for Euler Trials</title>
+      <title>Conditions for Euler Trails</title>
 
     <p>
       One way to guarantee that a graph does <em>not</em>
@@ -431,7 +431,7 @@
     </p>
 
     <p>
-      You run into a similar problem whenever you have a vertex of any odd degree. If you start at such a vertex, you will not be able to end there (after traversing every edge exactly once). After using one edge to leave the starting vertex, you will be left with an even number of edges emanating from the vertex. Half of these could be used for returning to the vertex, the other half for leaving. So you return, then leave. Return, then leave. The only way to use up all the edges is to use the last one by leaving the vertex. On the other hand, if you have a vertex with odd degree that you do not start a trail at, then you will eventually get stuck at that vertex. The trail will use pairs of edges incident to the vertex to arrive and leave again. Eventually all but one of these edges will be used up, leaving only an edge to arrive by, and none to leave again.
+      You run into a similar problem whenever you have a vertex of any odd degree. If you start at such a vertex, you will not be able to end there (after traversing every edge exactly once). After using one edge to leave the starting vertex, you will be left with an even number of edges emanating from the vertex. Half of these could be used for returning to the vertex, the other half for leaving. So you return, then leave. Return, then leave. The only way to use up all the edges is to use the last one by leaving the vertex. On the other hand, if you have a vertex with odd degree at which you do not start a trail, then you will eventually get stuck at that vertex. The trail will use pairs of edges incident to the vertex to arrive and leave again. Eventually all but one of these edges will be used up, leaving only an edge to arrive by, and none to leave again.
     </p>
 
     <p>

source/sec_gt-planar.ptx

@@ -128,7 +128,7 @@
 
     <p>
       <idx><h>planar representation</h></idx>
-      The graphs are the same, so if one is planar, the other must be too.
+      The graphs are the same, so if one is planar, the other must be, too.
       However, the original drawing of the graph was not a
       <term>planar representation</term> of the graph.
     </p>
@@ -359,7 +359,7 @@
   </subsection>
 
     <subsection>
-      <title>Euler's formula for planar graphs</title>
+      <title>Euler's Formula for Planar Graphs</title>
 
     <p>
       There is a connection between the number of vertices (<m>v</m>),
@@ -455,11 +455,11 @@
       and doing so never changes the quantity <m>v - e + f</m>,
       that quantity will be the same for all graphs.
       But notice that our starting graph <m>P_2</m> has <m>v = 2</m>,
-      <m>e = 1</m> and <m>f = 1</m>, so <m>v - e + f = 2</m>.
+      <m>e = 1</m>, and <m>f = 1</m>, so <m>v - e + f = 2</m>.
     </p>
 
     <p>
-      The argument we have outlined above is not quite correct, since we made the unjustified assumption that all graphs can be built up from <m>P_2</m> using only the two moves we described.  To avoid this issue, we can use a minimal criminal argument.  You are asked to do this in the exercises, but the idea is essentially the same as we have here, except that start with a minimal connected planar graph that does not satisfy the formula, then <em>remove</em> either an edge or a vertex (and its edge) to get a smaller connected planar graph that does satisfy the formula.  But just like the adding moves we have described above, removing an edge or a vertex does not change the quantity <m>v - e + f</m>.
+      The argument we have outlined above is not quite correct, since we made the unjustified assumption that all graphs can be built up from <m>P_2</m> using only the two moves we described.  To avoid this issue, we can use a minimal criminal argument.  You are asked to do this in the exercises, but the idea is essentially the same as we have here, except that we start with a minimal connected planar graph that does not satisfy the formula, then <em>remove</em> either an edge or a vertex (and its edge) to get a smaller connected planar graph that does satisfy the formula.  But just like the adding moves we have described above, removing an edge or a vertex does not change the quantity <m>v - e + f</m>.
     </p>
   </subsection>
 
@@ -576,7 +576,7 @@
         But this is impossible,
         since we have already determined that <m>f = 7</m> and <m>e = 10</m>,
         and <m>21 \not\le 20</m>.
-        This is a contradiction so in fact <m>K_5</m> is not planar.
+        This is a contradiction, so in fact <m>K_5</m> is not planar.
       </p>
     </proof>
 
@@ -657,7 +657,7 @@
       Then we find a relationship between the number of faces and the number of edges based on how many edges surround each face.
       This is the only difference.
       In the proof for <m>K_5</m>,
-      we got <m>3f \le 2e</m> and for <m>K_{3,3}</m> we go <m>4f \le 2e</m>.
+      we got <m>3f \le 2e</m> and for <m>K_{3,3}</m> we had <m>4f \le 2e</m>.
       The coefficient of <m>f</m> is the key.
       It is the smallest number of edges that could surround any face.
       If some number of edges surround a face,
@@ -695,7 +695,7 @@
       <p>
         There are exactly four other regular polyhedra:
         the tetrahedron, octahedron, dodecahedron,
-        and icosahedron with 4, 8, 12, and 20 faces respectively.
+        and icosahedron, with 4, 8, 12, and 20 faces respectively.
         How many vertices and edges do each of these have?
       </p>
     </investigation>
@@ -822,7 +822,7 @@
         <b>Case 1</b>: Each face is a triangle.
         Let <m>f</m> be the number of faces.
         There are then <m>3f/2</m> edges.
-        Using Euler's formula we have
+        Using Euler's formula, we have
         <m>v - 3f/2 + f = 2</m> so <m>v = 2 + f/2</m>.
         Now each vertex has the same degree, say <m>k</m>.
         So the number of edges is also <m>kv/2</m>.
@@ -857,7 +857,7 @@
       <p>
         <b>Case 2</b>: Each face is a square.
         Now we have <m>e = 4f/2 = 2f</m>.
-        Using Euler's formula we get <m>v = 2 + f</m>,
+        Using Euler's formula, we get <m>v = 2 + f</m>,
         and counting edges using the degree <m>k</m> of each vertex gives us
         <me>
           e = 2f = \frac{k(2+f)}{2}