finish ch1 corrections

Author: oscarlevin

source/exercises/logic-conc.ptx

@@ -83,7 +83,7 @@
         can you conclude about Peter and Robert if you know that Quincy is indeed fat?
         Explain
         (you may reference
-        <xref ref="tt">Problem</xref>).
+        <xref ref="tt">Question</xref>).
       </p>
     </statement>
     <solution>
@@ -279,7 +279,7 @@
 
           <li>
             <p>
-              For every integer <m>x</m> and every integer <m>y</m> there is an integer <m>n</m> such that if
+              For every integer <m>x</m> and every integer <m>y</m>, there is an integer <m>n</m> such that if
               <m>x &gt; 0</m> then <m>nx &gt; y</m>.
             </p>
           </li>
@@ -452,8 +452,8 @@
   <exercise>
     <statement>
       <p>
-        Consider the statement: for all integers <m>n</m>,
-        if <m>n</m> is even and <m>n \le 7</m> then <m>n</m> is negative or <m>n \in \{0,2,4,6\}</m>.
+        Consider the statement, <q>For all integers <m>n</m>,
+        if <m>n</m> is even and <m>n \le 7</m>, then <m>n</m> is negative or <m>n \in \{0,2,4,6\}</m>.</q>
 
         <ol>
           <li>
@@ -661,8 +661,8 @@
   <exercise>
     <statement>
       <p>
-        Consider the statement: for all integers <m>n</m>,
-        if <m>n</m> is odd, then <m>7n</m> is odd.
+        Consider the statement, <q>For all integers <m>n</m>,
+        if <m>n</m> is odd, then <m>7n</m> is odd.</q>
 
         <ol>
           <li>

source/exercises/logic-proofs.ptx

@@ -5,7 +5,7 @@
   <exercise>
     <statement>
 
-      <p> For a given predicate <m>P(x)</m>, you might believe that the statements <m>\forall x P(x)</m> or <m>\exists x P(x)</m> are either true or false. How would you decide if you were correct in each case? You have four choices: you could give an example of an element <m>n</m> in the domain for which <m>P(n)</m> is true or for which <m>P(n)</m> if false, or you could argue that no matter what <m>n</m> is, <m>P(n)</m> is true or is false. <ol>
+      <p> For a given predicate <m>P(x)</m>, you might believe that the statements <m>\forall x P(x)</m> or <m>\exists x P(x)</m> are either true or false. How would you decide if you were correct in each case? You have four choices: You could give an example of an element <m>n</m> in the domain for which <m>P(n)</m> is true or for which <m>P(n)</m> is false, or you could argue that no matter what <m>n</m> is, <m>P(n)</m> is true or is false. <ol>
           <li>
             <p> What would you need to do to prove <m>\forall x P(x)</m> is true? </p>
 
@@ -52,8 +52,8 @@
   <exercise>
     <statement>
       <p>
-        Consider the statement <q>for all integers <m>a</m> and <m>b</m>,
-        if <m>a + b</m> is even, then <m>a</m> and <m>b</m> are even</q>
+        Consider the statement, <q>For all integers <m>a</m> and <m>b</m>,
+        if <m>a + b</m> is even, then <m>a</m> and <m>b</m> are even.</q>
 
         <ol>
           <li>
@@ -233,8 +233,8 @@
   <exercise>
     <statement>
       <p>
-        Consider the statement: for all integers <m>n</m>,
-        if <m>n</m> is even then <m>8n</m> is even.
+        Consider the statement, <q>For all integers <m>n</m>,
+        if <m>n</m> is even then <m>8n</m> is even.</q>
 
         <ol>
           <li>
@@ -356,7 +356,7 @@
   <exercise>
     <statement>
       <p>
-        Prove that for all integers <m>n</m>, it is the case that <m>n</m> is  even if and only if <m>3n</m> is even.  That is, prove both implications: if <m>n</m> is even, then <m>3n</m> is even, and if <m>3n</m> is even, then <m>n</m> is even. 
+        Prove that for all integers <m>n</m>, it is the case that <m>n</m> is  even if and only if <m>3n</m> is even.  That is, prove both implications: If <m>n</m> is even, then <m>3n</m> is even, and if <m>3n</m> is even, then <m>n</m> is even. 
       </p>
     </statement>
     <hint>
@@ -385,8 +385,8 @@
   <exercise>
     <statement>
       <p>
-        Consider the statement: for all integers <m>a</m> and <m>b</m>,
-        if <m>a</m> is even and <m>b</m> is a multiple of 3, then <m>ab</m> is a multiple of 6.
+        Consider the statement, <q>For all integers <m>a</m> and <m>b</m>,
+        if <m>a</m> is even and <m>b</m> is a multiple of 3, then <m>ab</m> is a multiple of 6.</q>
 
         <ol>
           <li>
@@ -417,8 +417,8 @@
   <exercise>
     <statement>
       <p>
-        Prove the statement: For all integers <m>n</m>,
-        if <m>5n</m> is odd, then <m>n</m> is odd.
+        Prove the statement, <q>For all integers <m>n</m>,
+        if <m>5n</m> is odd, then <m>n</m> is odd.</q>
         Clearly state the style of proof you are using.
       </p>
     </statement>
@@ -427,9 +427,9 @@
   <exercise>
     <statement>
       <p>
-        Prove the statement: For all integers <m>a</m>, <m>b</m>,
+        Prove the statement, <q>For all integers <m>a</m>, <m>b</m>,
         and <m>c</m>,
-        if <m>a^2 + b^2 = c^2</m>, then <m>a</m> or <m>b</m> is even.
+        if <m>a^2 + b^2 = c^2</m>, then <m>a</m> or <m>b</m> is even.</q>
       </p>
     </statement>
     <hint>
@@ -490,7 +490,7 @@
     <statement>
       <p>
         Suppose you have a collection of rare 5-cent stamps and 8-cent stamps.
-        You desperately need to mail a letter and having no other stamps available, decide to dip into your collection.  The question is, what amounts of postage can you make?
+        You desperately need to mail a letter and, having no other stamps available, decide to dip into your collection.  The question is, what amounts of postage can you make?
 
         <ol>
           <li>
@@ -530,7 +530,7 @@
       <p>
         Prove: <m>x=y</m> if and only if <m>xy=\dfrac{(x+y)^2}{4}</m>.
         Note, you will need to prove two
-        <q>directions</q> here:
+        <q>directions</q> here,
         the <q>if</q> and the <q>only if</q> part.
       </p>
     </statement>
@@ -624,10 +624,10 @@
     <idx><h>playing cards</h></idx>
     <statement>
       <p>
-        A standard deck of 52 cards consists of 4 suites (hearts,
+        A standard deck of 52 cards consists of 4 suits (hearts,
         diamonds,
         spades, and clubs) each containing 13 different values (Ace, 2, 3, <ellipsis/>, 10, J, Q, K).
-        If you draw some number of cards at random you might or might not have a pair
+        If you draw some number of cards at random, you might or might not have a pair
         (two cards with the same value)
         or three cards all of the same suit.
         However, if you draw enough cards,
@@ -702,7 +702,7 @@
     </idx>
     <statement>
       <p>
-        Suppose you have an <m>n\times n</m> chessboard but your dog has eaten one of the corner squares.  You have dominoes that each cover exactly two squares of the board.  Can you cover the remaining squares on the board with non-overlapping dominoes?
+        Suppose you have an <m>n\times n</m> chessboard, but your dog has eaten one of the corner squares.  You have dominoes that each cover exactly two squares of the board.  Can you cover the remaining squares on the board with non-overlapping dominoes?
         What needs to be true about <m>n</m>?
         Give necessary and sufficient conditions
         (that is, say exactly which values of <m>n</m> work and which do not work).

source/exercises/logic-rules.ptx

@@ -17,7 +17,7 @@
         </p>
 
         <p>
-          Troll 2: We are cousins or we are both knaves.
+          Troll 2: We are cousins, or we are both knaves.
         </p>
       </blockquote>
 
@@ -49,7 +49,7 @@
           <li>
             <title>Quinn</title>
             <p>
-              Ryan is a knight and if Pat is a knight, then so am I.
+              Ryan is a knight, and if Pat is a knight, then so am I.
             </p>
           </li>
 
@@ -86,15 +86,15 @@
             <p>
         Assuming the statement is true, what
               (if anything)
-              can you conclude if there will be cake?
+              can you conclude if you know there will be cake?
             </p>
           </li>
 
           <li>
             <p>
         Assuming the statement is true, what
               (if anything)
-              can you conclude if there will not be cake?
+              can you conclude if you know there will not be cake?
             </p>
           </li>
 
@@ -150,7 +150,7 @@
         Geoff Poshingten is out at a fancy pizza joint and decides to order a calzone. When the
         waiter asks what he would like in it, he replies, <q>I want either pepperoni or sausage.
           Also, if I have sausage, then I must also include quail.
-          Oh, and if I have pepperoni or quail then I must also have ricotta cheese.</q>
+          Oh, and if I have pepperoni or quail, then I must also have ricotta cheese.</q>
 
         <ol>
           <li>
@@ -259,7 +259,7 @@
   <exercise>
     <statement>
       <p>
-        Use De Morgan's Laws, and any other logical equivalence facts you know to simplify the
+        Use De Morgan's Laws and any other logical equivalence facts you know to simplify the
         following statements. Show all your steps. Your final statements should have negations only
         appear directly next to the sentence variables or predicates (<m>P</m>, <m>Q</m>, <m>E(x)</m>, etc.),
         and no double negations. It would be a good idea to use only conjunctions, disjunctions, and
@@ -551,7 +551,7 @@
 
           <li>
             <p> There is a number <m>
-        n</m> for which no other number is either less <m>n</m> than or equal to <m>n</m>. </p>
+        n</m> for which no other number is less than or equal to <m>n</m>. </p>
           </li>
 
           <li>

source/exercises/logic-structures.ptx

@@ -86,7 +86,7 @@
     <exercise>
       <introduction>
         <p>
-          Let <m>f:X \to Y</m> be a function and let <m>A</m> and <m>B</m> be subsets of <m>X</m>. 
+          Let <m>f:X \to Y</m> be a function, and let <m>A</m> and <m>B</m> be subsets of <m>X</m>. 
         </p>
       </introduction>
       <task>
@@ -108,7 +108,7 @@
     <exercise>
       <introduction>
         <p>
-          Let <m>f:X \to Y</m> be a function and let <m>A</m> and <m>B</m> be subsets of <m>X</m>.
+          Let <m>f:X \to Y</m> be a function, and let <m>A</m> and <m>B</m> be subsets of <m>X</m>.
         </p>
       </introduction>
       <task>
@@ -198,15 +198,15 @@
     <exercise>
       <statement>
         <p>
-          Let <m>f:X \to Y</m> be a function and let <m>A</m> and <m>B</m> be subsets of <m>Y</m>.  Prove that <m>f\inv(A \cap B) = f\inv(A) \cap f\inv(B)</m>.
+          Let <m>f:X \to Y</m> be a function, and let <m>A</m> and <m>B</m> be subsets of <m>Y</m>.  Prove that <m>f\inv(A \cap B) = f\inv(A) \cap f\inv(B)</m>.
         </p>
       </statement>
     </exercise>
 
     <exercise>
       <statement>
         <p>
-          Let <m>f:X \to Y</m> be a function and let <m>A</m> and <m>B</m> be subsets of <m>Y</m>.  Prove that <m>f\inv(A \cup B) = f\inv(A) \cup f\inv(B)</m>.
+          Let <m>f:X \to Y</m> be a function, and let <m>A</m> and <m>B</m> be subsets of <m>Y</m>.  Prove that <m>f\inv(A \cup B) = f\inv(A) \cup f\inv(B)</m>.
         </p>
       </statement>
     </exercise>

source/practice/logic-proofs.ptx

@@ -13,7 +13,7 @@
     <blocks>
       <block order="1">
         <p>
-          Let <m>n</m> be an arbitrary integer and assume <m>n</m> is even.
+          Let <m>n</m> be an arbitrary integer, and assume <m>n</m> is even.
         </p>
       </block>
       <block order="2">
@@ -28,12 +28,12 @@
       </block>
       <block correct="no">
         <p>
-          Let <m>n</m> be an arbitrary integer and assume <m>7n</m> is even.
+          Let <m>n</m> be an arbitrary integer, and assume <m>7n</m> is even.
         </p>
       </block>
       <block correct="no">
         <p>
-          Let <m>n</m> be an arbitrary integer and assume <m>7n</m> is odd.
+          Let <m>n</m> be an arbitrary integer, and assume <m>7n</m> is odd.
         </p>
       </block>
       <block correct="no">
@@ -68,7 +68,7 @@
     </statement>
     <blocks>
       <block correct="no">
-        <p> Let <m>n</m> be an arbitrary integer and assume <m>n</m> is even. </p>
+        <p> Let <m>n</m> be an arbitrary integer, and assume <m>n</m> is even. </p>
       </block>
       <block correct="no">
         <p>
@@ -80,10 +80,10 @@
           <m>7n</m> must be even. </p>
       </block>
       <block correct="no">
-        <p> Let <m>n</m> be an arbitrary integer and assume <m>7n</m> is even. </p>
+        <p> Let <m>n</m> be an arbitrary integer, and assume <m>7n</m> is even. </p>
       </block>
       <block order="1">
-        <p> Let <m>n</m> be an arbitrary integer and assume <m>n</m> is odd. </p>
+        <p> Let <m>n</m> be an arbitrary integer, and assume <m>n</m> is odd. </p>
       </block>
       <block order="2">
         <p> Since <m>7</m> is odd and the product of an odd number and an odd number is odd, </p>
@@ -113,17 +113,17 @@
     <blocks>
       <block correct="no">
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume that <m>a+b</m> is odd.
+          Let <m>a</m> and <m>b</m> be integers, and assume that <m>a+b</m> is odd.
         </p>
       </block>
       <block correct="no">
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume that if <m>a+b</m> is odd, then either <m>a</m> or <m>b</m> is odd.
+          Let <m>a</m> and <m>b</m> be integers, and assume that if <m>a+b</m> is odd, then either <m>a</m> or <m>b</m> is odd.
         </p>
       </block>
       <block>
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume both are even.
+          Let <m>a</m> and <m>b</m> be integers, and assume both are even.
         </p>
       </block>
       <block>
@@ -156,7 +156,7 @@
   <exercise label="proofs-odd-contradiction" adaptive="yes">
     <statement>
       <p>
-        Consider the same statement, <q>For any numbers <m>a</m> and <m>b</m>, if <m>a+b</m> is odd, then either <m>a</m> or <m>b</m> is odd</q>.
+        Consider the same statement, <q>For any numbers <m>a</m> and <m>b</m>, if <m>a+b</m> is odd, then either <m>a</m> or <m>b</m> is odd.</q>
       </p>
       <p>
         Give a valid proof of the statement, this time using a <em>proof by contradiction</em> using some of the statements below.
@@ -165,22 +165,22 @@
     <blocks>
       <block correct="no">
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume that <m>a+b</m> is odd.
+          Let <m>a</m> and <m>b</m> be integers, and assume that <m>a+b</m> is odd.
         </p>
       </block>
       <block correct="no">
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume that if <m>a+b</m> is odd, then either <m>a</m> or <m>b</m> is odd.
+          Let <m>a</m> and <m>b</m> be integers, and assume that if <m>a+b</m> is odd, then either <m>a</m> or <m>b</m> is odd.
         </p>
       </block>
       <block correct="no">
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume both are even.
+          Let <m>a</m> and <m>b</m> be integers, and assume both are even.
         </p>
       </block>
       <block>
         <p>
-          Let <m>a</m> and <m>b</m> be integers and assume that <m>a+b</m> is odd but <m>a</m> and <m>b</m> are both even.
+          Let <m>a</m> and <m>b</m> be integers, and assume that <m>a+b</m> is odd but <m>a</m> and <m>b</m> are both even.
         </p>
       </block>
       <block correct="no">
@@ -218,7 +218,7 @@
     <matches>
       <match order="1">
         <premise>Assume <m>f: A \to B</m> is a bijection</premise>
-        <response>Direct Proof</response>
+        <response>Direct proof</response>
       </match>
       <match order="3">
         <premise>Assume <m>|A| \ne |B|</m></premise>

source/practice/logic-structures.ptx

@@ -105,7 +105,7 @@
     <blocks>
       <block order="4">
         <p>
-          Suppose <m>B \subseteq A \cap B</m> and let <m>b</m> be an element of <m>B</m>.
+          Suppose <m>B \subseteq A \cap B</m>, and let <m>b</m> be an element of <m>B</m>.
         </p>
       </block>
       <block order="2">
@@ -174,7 +174,7 @@
       </block>
       <block order="7">
         <p>
-          Then <m>x</m> is an element of <m>A \cap B</m> or <m>x</m> is an element of <m>A</m>.
+          Then <m>x</m> is an element of <m>A \cap B</m>, or <m>x</m> is an element of <m>A</m>.
         </p>
       </block>
       <block order="1">