fix typos

Author: oscarlevin

source/bookinfo.ptx

@@ -19,6 +19,10 @@
     \def\R{\mathbb R} 
     \def\C{\mathbb C} 
     \def\U{\mathcal U} 
+    \def\x{\mathbf{x}}
+    \def\y{\mathbf{y}}
+    \def\X{\mathcal{X}}
+    \def\Y{\mathcal{Y}}
     \def\pow{\mathcal P} 
     \def\inv{^{-1}} 
     \def\st{:} 

source/ch_counting.ptx

@@ -15,7 +15,7 @@
           <p>
             In a group of 10 people,
             if everyone shakes hands with everyone else exactly once,
-            how many handshakes took place?
+            how many handshakes take place?
           </p>
         </li>
 
@@ -38,7 +38,7 @@
       let's look at the basics of counting.
     </p>
   </introduction>
-
+  <clearpage/>
   <xi:include href="sec_counting-pascal.ptx"/>
   <xi:include href="sec_counting-combine-outcomes.ptx"/>
   <xi:include href="sec_counting-non-disjoint.ptx" />

source/ch_logic.ptx

@@ -4,7 +4,7 @@
   <title>Logic and Proofs</title>
   <introduction>
     <p>
-      Logic is the study of consequence. Given a few mathematical statements or facts, we would like to be able to draw some conclusions. For example, if I told you that a particular real-valued function was continuous on the interval <m>[0,1]</m>, and <m>f(0) = -1</m> and <m>f(1) = 5</m>, can we conclude that there is some point between <m>[0,1]</m> where the graph of the function crosses the <m>x</m>-axis? Yes, we can, thanks to the Intermediate Value Theorem from Calculus. Can we conclude that there is exactly one point? No. Whenever we find an <q>answer</q> in math, we really have a (perhaps hidden) argument. 
+      Logic is the study of consequence. Given a few mathematical statements or facts, we would like to be able to draw some conclusions. For example, if I told you that a particular real-valued function was continuous on the interval <m>[0,1]</m>, and <m>f(0) = -1</m> and <m>f(1) = 5</m>, can we conclude that there is some point between <m>[0,1]</m> where the graph of the function crosses the <m>x</m>-axis? Yes, we can, thanks to the Intermediate Value Theorem from calculus. Can we conclude that there is exactly one point? No. Whenever we find an <q>answer</q> in math, we really have a (perhaps hidden) argument. 
     </p>
 
     <p>

source/exercises/seq-basics.ptx

@@ -373,7 +373,7 @@
 
           <li>
             <p>
-              Describe what you have found above in terms of Pascal's Triangle.
+              Describe what you have found above in terms of <xref ref="fig-pascal-large" text="custom">Pascal's triangle</xref>.
               What pattern have you discovered?<idx><h>Pascal's triangle</h></idx>
             </p>
           </li>

source/exercises/seq-induction.ptx

@@ -575,13 +575,13 @@
     <statement>
       <p>
         Use induction to prove that <m>\d\sum_{k=0}^n {n \choose k} = 2^n</m>.
-        That is, the sum of the <m>n</m>th row of Pascal's Triangle is <m>2^n</m>.
+        That is, the sum of the <m>n</m>th row of Pascal's triangle is <m>2^n</m>.
       </p>
     </statement>
     <hint>
       <p>
         Here's the idea:
-        since every entry in Pascal's Triangle is the sum of the two entries above it,
+        since every entry in Pascal's triangle is the sum of the two entries above it,
         we can get the <m>k+1</m>st row by adding up all the pairs of entry from the <m>k</m>th row.
         But doing this uses each entry on the <m>k</m>th row twice.
         Thus each time we drop to the next row, we double the total.

source/practice/counting-probability.ptx

@@ -311,7 +311,7 @@
             Compare the probability of getting tails when flipping a coin once to the probability of drawing a single red card.
           </p>
           <p>
-            <m>P(\text{tails}) =</m><var name="'1/2'" width="10" />; <m>P(\text{Red}) = </m> <var name="'1/2'" width="10" />.  Difference: <var name="'0'" width="10" />
+            <m>P(\text{tails}) =</m><var name="'1/2'" width="10" />; <m>P(\text{red}) = </m> <var name="'1/2'" width="10" />.  Difference: <var name="'0'" width="10" />
           </p>
         </statement>
       </task>
@@ -321,7 +321,7 @@
             Compare the probability of getting two tails when flipping a coin twice to the probability of drawing a two red cards (without replacement).
           </p>
           <p>
-            <m>P(\text{2 tails}) =</m><var name="'1/4'" width="10" />; <m>P(\text{2 Reds}) = </m> <var name="'(4*3)/(8*7)'" width="10" />.  Difference: <var name="'1/28'" width="10" />
+            <m>P(\text{2 tails}) =</m><var name="'1/4'" width="10" />; <m>P(\text{2 reds}) = </m> <var name="'(4*3)/(8*7)'" width="10" />.  Difference: <var name="'1/28'" width="10" />
           </p>
         </statement>
       </task>
@@ -331,7 +331,7 @@
             Compare the probability of getting three tails when flipping a coin thrice to the probability of drawing a three red cards (without replacement).
           </p>
           <p>
-            <m>P(\text{3 tails}) =</m><var name="'1/8'" width="10" />; <m>P(\text{3 Reds}) = </m> <var name="'(4*3*2)/(8*7*6)'" width="10" />.  Difference: <var name="'3/56'" width="10" />
+            <m>P(\text{3 tails}) =</m><var name="'1/8'" width="10" />; <m>P(\text{3 reds}) = </m> <var name="'(4*3*2)/(8*7*6)'" width="10" />.  Difference: <var name="'3/56'" width="10" />
           </p>
         </statement>
       </task>
@@ -341,7 +341,7 @@
             Compare the probability of getting four tails when flipping a coin four times to the probability of drawing a four red cards (without replacement).
           </p>
           <p>
-            <m>P(\text{4 tails}) =</m><var name="'1/16'" width="10" />; <m>P(\text{4 Reds}) = </m> <var name="'(4*3*2*1)/(8*7*6*5)'" width="10" />.  Difference: <var name="'27/560'" width="10" />
+            <m>P(\text{4 tails}) =</m><var name="'1/16'" width="10" />; <m>P(\text{4 reds}) = </m> <var name="'(4*3*2*1)/(8*7*6*5)'" width="10" />.  Difference: <var name="'27/560'" width="10" />
           </p>
         </statement>
       </task>
@@ -351,7 +351,7 @@
             Compare the probability of getting five tails when flipping a coin five times to the probability of drawing a five red cards (without replacement).
           </p>
           <p>
-            <m>P(\text{5 tails}) =</m><var name="'1/32'" width="10" />; <m>P(\text{5 Reds}) = </m> <var name="'0'" width="10" />.  Difference: <var name="'1/32'" width="10" />
+            <m>P(\text{5 tails}) =</m><var name="'1/32'" width="10" />; <m>P(\text{5 reds}) = </m> <var name="'0'" width="10" />.  Difference: <var name="'1/32'" width="10" />
           </p>
         </statement>
       </task>

source/sec_counting-combine-outcomes.ptx

@@ -288,7 +288,7 @@
 			</statement>
 			<solution>
 				<p>
-					We learned how to count bit strings with a fixed length and weight in <xref ref="sec_counting-pascal"/>.  The total number of <m>10</m>-bit strings of weight <m>6</m> is <m>\binom{10}{6} = 210</m> (where that value was found using <xref ref="fig-pascal-large">Pascal's Triangle</xref>).  However, not all of these are ones we want to count: we want to count just those that start with <m>11</m> or <m>00</m> (and not those that start with <m>10</m> or <m>01</m>).
+					We learned how to count bit strings with a fixed length and weight in <xref ref="sec_counting-pascal"/>.  The total number of <m>10</m>-bit strings of weight <m>6</m> is <m>\binom{10}{6} = 210</m> (where that value was found using <xref ref="fig-pascal-large">Pascal's triangle</xref>).  However, not all of these are ones we want to count: we want to count just those that start with <m>11</m> or <m>00</m> (and not those that start with <m>10</m> or <m>01</m>).
 				</p>
 
 				<p>
@@ -476,7 +476,7 @@
 			</statement>
 			<solution>
 				<p>
-					We can count lattice paths with <xref ref="fig-pascal-large" text="custom">Pascal's Triangle</xref>, but how do we ensure that the path passes through a particular point?  Well, any path that passes through <m>(4,7)</m> must first go from <m>(0,0)</m> to <m>(4,7)</m>, and then complete its journey with a path from <m>(4,7)</m> to <m>(10,10)</m>. 
+					We can count lattice paths with <xref ref="fig-pascal-large" text="custom">Pascal's triangle</xref>, but how do we ensure that the path passes through a particular point?  Well, any path that passes through <m>(4,7)</m> must first go from <m>(0,0)</m> to <m>(4,7)</m>, and then complete its journey with a path from <m>(4,7)</m> to <m>(10,10)</m>. 
 				</p>
 				<p>
 					The number of paths from <m>(0,0)</m> to <m>(4,7)</m> is <m>\binom{11}{4} = 330</m>.

source/sec_counting-combperm.ptx

@@ -544,7 +544,7 @@
 				This question is actually too vague.  Do we mean how many ways can we select five players?  Or do we mean how many ways can we pick five players to fill the five positions?<fn>I'm told the five positions are called point guard, shooting guard, small forward, power forward, and center.  Who knew?</fn>  Let's answer both of these questions.
 			</p>
 			<p>
-				First, if we just want to select five out of the 12 players, that is just like picking five out of 12 pizza toppings (although less delicious).  We know that there are <m>\binom{12}{5}</m> ways to do this, and from <xref ref="fig-pascal-large" text="custom">Pascal's Triangle</xref> we know that this is 792.
+				First, if we just want to select five out of the 12 players, that is just like picking five out of 12 pizza toppings (although less delicious).  We know that there are <m>\binom{12}{5}</m> ways to do this, and from <xref ref="fig-pascal-large" text="custom">Pascal's triangle</xref> we know that this is 792.
 			</p>
 			<p>
 				On the other hand, if we wanted to pick five players for the five different positions,... well, we could start by picking one of the 792 different <em>sets</em> of five players, and then permute them into the five positions.  Of the five players on the court, we pick one of the five to be the point guard, then one of the remaining four to be the shooting guard, and so on.  This gives us <m>5\cdot 4\cdot 3\cdot 2\cdot 1 = 5!</m> ways to arrange the players.  So the total number of ways to pick five players for the five positions is <me>\binom{12}{5}\cdot 5! = 792\cdot 120 = 95,040</me>.

source/sec_counting-non-disjoint.ptx

@@ -671,7 +671,7 @@
       </solution>
     </example>
     <p>
-      We will see how to answer the counting problem above in the case that letters can be repeated in <xref ref="sec_counting-multisets"/>.  If we restricted the question further and required that the letters are distininct (and in alphabetical order) we actually know how to answer the question already.  Since for any <em>set</em> of three letters, there is exactly one 3-letter word that has those letters in alphabetical order, we can simply count the number of sets of three letters.  This is <m>\binom{26}{3}</m>.  Then we just look at Pascal's Triangle to find the value... oh shoot.  Our triangle doesn't go down that far.  I guess we should think of a way to compute the value without the triangle.  Read on!
+      We will see how to answer the counting problem above in the case that letters can be repeated in <xref ref="sec_counting-multisets"/>.  If we restricted the question further and required that the letters are distinct (and in alphabetical order) we actually know how to answer the question already.  Since for any <em>set</em> of three letters, there is exactly one 3-letter word that has those letters in alphabetical order, we can simply count the number of sets of three letters.  This is <m>\binom{26}{3}</m>.  Then we just look at <xref ref="fig-pascal-large" text="custom">Pascal's triangle</xref> to find the value... oh shoot.  Our triangle doesn't go down that far.  I guess we should think of a way to compute the value without the triangle.  Read on!
     </p>
   </subsection>
   <reading-questions xml:id="rqs-counting-non-disjoint">