auto-update(nvim): 2025-01-13 02:08:30

documents/by-course/math-8/pset-1/main.typ

@@ -13,6 +13,8 @@
   ),
 )
 
+#set par(first-line-indent: 0pt, spacing: 1em)
+
 Problems:
 
 1.1: \#1ceij, 2c, 3cdeghjL, 4cdefh, 6, 7cg, 10ce, 11bei, 12a, 13
@@ -27,16 +29,19 @@ Problems:
 
   True.
   ] \
+
   e. #[
   Either $pi$ is rational and $17$ is a prime, or $7 < 13$ and $81$ is a perfect square.
 
   True.
   ] \
+
   i. #[
     It is not the case that $39$ is prime, or that 64 is a power of 2.
 
     False.
   ] \
+
   j. #[
   There are more than three false statements in this book, and this statement is one of them.
 
@@ -160,35 +165,9 @@ Problems:
 6. \
   a. $not P and not Q$, $not (P and not Q)$ are not equivalent because we can choose $P$ and $Q$ to both be true which gives false for proposition 1 and true for proposition 2. \
   b. $(not P) or (not Q), not (P or not Q)$ are not equivalent because we can choose $P$ to be true and $Q$ to be false, and proposition 1 is true while proposition 2 is false. \
-  c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent, by a truth table.
+  c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent
 
   #proof[
-    Direct computation, by a truth table. Hopefully this suffices to show the absurdity of computing nontrivial boolean results by truth table.
-    #table(
-      align: center,
-      columns: (1fr, 1fr, 1fr, 1fr, 1fr, 2fr, 2fr),
-      [$P$],
-      [$Q$],
-      [$R$],
-      [$P and Q$],
-      [$Q or R$],
-      [$(P and Q) or R$],
-      [$P and (Q or R)$],
-
-      [T], [T], [T], [T], [T], [T], [T],
-      [T], [F], [T], [F], [T], [T], [T],
-      [T], [T], [F], [F], [T], [T], [T],
-      [T], [F], [F], [F], [F], [F], [F],
-      [F], [T], [T], [F], [T], [T], [F],
-      [F], [F], [T], [], [], [], [],
-      [F], [T], [F], [], [], [], [],
-      [F], [F], [F], [], [], [], [],
-    )
-
-    We need not complete the table as we have already identified a contradiction for when $P$ and $Q$ are false, $R$ true.
-  ]
-
-  #remark(numbering: none)[
     We can also show this result directly by
 
     $
@@ -358,3 +337,183 @@ $ not (not P) or (not Q and not S) $
   ]
 
 = Exercises
+
+1. \
+  b. If #math.underbrace("the moon is made of cheese", "antedecent"), then #math.underbrace("triangles have four sides", "consequent").
+
+  d. #math.underbrace([The differentiability of $f$], "antedecent") is sufficient for #math.underbrace[$f$ to be continuous][consequent].
+
+2. \
+  b.
+
+  Converse: If triangles have four sides, then squares have three sides.
+
+  Contrapositive: If triangles don't have four sides, then squares don't have three sides.
+
+  #linebreak()
+
+  d.
+
+  Converse: The continuity of $f$ is sufficient for $f$ to be differentiable.
+
+  Contrapositive: If $f$ is not continuous, then $f$ is not differentiable.
+
+3. \
+  a. $Q$ must be true.
+
+  #linebreak()
+
+  b. $Q$ must be true.
+
+  #linebreak()
+
+  c. $Q$ is false.
+
+  #linebreak()
+
+  d. $Q$ is false.
+
+  #linebreak()
+
+  e. $Q$ is true.
+
+5. \
+  c. True.
+
+  #linebreak()
+
+  f. True.
+
+  #linebreak()
+
+  g. True.
+
+6. \
+  b. False.
+
+  #linebreak()
+
+  c. False
+
+  #linebreak()
+
+  g. False.
+
+  #linebreak()
+
+7. \
+  b.
+
+  #table(
+    columns: 6,
+    align: center,
+    [$P$],
+    [$Q$],
+    [$not P$],
+    [$not P => Q$],
+    [$Q <=> P$],
+    [$(not P => Q) or (Q <=> P)$],
+
+    [T], [T], [F], [T], [T], [T],
+    [F], [T], [T], [T], [F], [T],
+    [T], [F], [F], [T], [F], [T],
+    [F], [F], [T], [F], [T], [T],
+  )
+
+  #linebreak()
+
+  e.
+  #table(
+    columns: 5,
+    align: center,
+    [$P$], [$Q$], [$not Q$], [$Q <=> P$], [$not Q => (Q <=> P)$],
+    [$T$], [$T$], [F], [T], [T],
+    [$F$], [$T$], [F], [F], [T],
+    [$T$], [$F$], [T], [F], [F],
+    [$F$], [$F$], [T], [T], [T],
+  )
+
+10. \
+  b. If $n$ is prime, then $n = 2$ or $n$ is odd.
+
+  $
+    n "is prime" => (n=2) or (n mod 2 = 1)
+  $
+
+  f.
+
+  $
+    2 < n - 6 => 2n < 4 or n > 4
+  $
+
+  g.
+
+  $
+    6 >= n - 3 <=> n > 4 or n > 10
+  $
+
+12. \
+  b.
+
+  #table(
+    columns: 9,
+    align: center,
+    [$P$],
+    [$Q$],
+    [$R$],
+    [$not R$],
+    [$not Q$],
+    [$P and Q$],
+    [$P and not R$],
+    [$(P and Q) => R$],
+    [$(P and not R) => not Q$],
+
+    [T], [T], [T], [F], [F], [T], [F], [T], [T],
+    [T], [F], [T], [F], [T], [F], [F], [T], [T],
+    [T], [T], [F], [T], [F], [T], [T], [F], [F],
+    [T], [F], [F], [T], [T], [F], [T], [T], [T],
+    [F], [T], [T], [F], [F], [F], [F], [T], [T],
+    [F], [F], [T], [F], [T], [F], [F], [T], [T],
+    [F], [T], [F], [T], [F], [F], [F], [T], [T],
+    [F], [F], [F], [T], [T], [F], [F], [T], [T],
+  )
+
+  So they are equivalent.
+
+  #linebreak()
+
+  c.
+
+  #proof[
+    $
+      P => (Q and R) &<=> not (Q and R) => not P &&("contrapositive") \
+      not (Q and R) => not P &<=> (not Q or not R) => not P space &&("DeMorgan's") \
+    $
+  ]
+
+13. \
+  a. Continuity implies differentiability.
+
+  #linebreak()
+
+  b. Differentiability implies continuity.
+
+  #linebreak()
+
+  c. It is not possible.
+
+  #linebreak()
+
+  d. $P => Q$, where $P$ is true and $Q$ is true.
+
+16. \
+  c. Contradiction.
+
+  #linebreak()
+
+  d. Neither. Statement is equivalent to $P => Q$.
+
+  #linebreak()
+
+  e. Tautology. $P and (Q or not Q)$ is independent of $Q$ so this reduces to
+  $P <=> P$.