auto-update(nvim): 2025-01-07 01:00:54

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

@@ -286,14 +286,23 @@
 #let example = example-style("item", "Example")
 
 #let proof(body, name: none) = {
-  thmtitle[Proof]
+  [_Proof_]
   if name != none {
     [ #thmname[#name]]
   }
-  thmtitle[.]
+  [.]
   body
   h(1fr)
-  $square$
+
+  // Add a word-joiner so that the proof square and the last word before the
+  // 1fr spacing are kept together.
+  sym.wj
+
+  // Add a non-breaking space to ensure a minimum amount of space between the
+  // text and the proof square.
+  sym.space.nobreak
+
+  $square.stroked$
 }
 
 #let fact = thmplain(

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

@@ -103,7 +103,258 @@ Problems:
       [F], [F], [T], [F], [T], [T],
       [F], [F], [F], [F], [F], [F],
     )
+  ] \
+  h. $not P and not Q$
+  #table(
+    align: center,
+    columns: 5,
+    [$P$], [$Q$], [$not P$], [$not Q$], [$not P and not Q$],
+    [T], [T], [F], [F], [F],
+    [T], [F], [F], [T], [F],
+    [F], [T], [T], [F], [F],
+    [F], [F], [T], [T], [T],
+  ) \
+  j. $(P and Q) or (P and R)$
+  #table(
+    align: center,
+    columns: 6,
+    [$P$], [$Q$], [$R$], [$P and Q$], [$P and R$], [$(P and Q) or (P and R)$],
+    [T], [T], [T], [T], [T], [T],
+    [T], [F], [T], [F], [T], [T],
+    [T], [T], [F], [T], [F], [T],
+    [T], [F], [F], [F], [F], [F],
+    [F], [T], [T], [F], [F], [F],
+    [F], [F], [T], [F], [F], [F],
+    [F], [T], [F], [F], [F], [F],
+    [F], [F], [F], [F], [F], [F],
+  ) \
+  L. $(P and Q) or (P and not S)$
+  #table(
+    align: center,
+    columns: 7,
+    [$P$],
+    [$Q$],
+    [$S$],
+    [$not S$],
+    [$P and Q$],
+    [$P and not S$],
+    [$(P and Q) or (P and not S)$],
+
+    [T], [T], [T], [F], [T], [F], [T],
+    [T], [F], [T], [F], [F], [F], [F],
+    [T], [T], [F], [T], [T], [T], [T],
+    [T], [F], [F], [T], [F], [T], [T],
+    [F], [T], [T], [F], [F], [F], [F],
+    [F], [F], [T], [F], [F], [F], [F],
+    [F], [T], [F], [T], [F], [F], [F],
+    [F], [F], [F], [T], [F], [F], [F],
+  )
+
+4. \
+  c. $(P or Q) and (R or S)$ is true. \
+  d. $(not P or not Q) or (not R or not S)$ is true. \
+  e. $not P or not Q$ is false. \
+  f. $(not Q or S) and (Q or S)$ is false. \
+  h. $K and not (S or Q)$ is false.
+
+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.
+
+  #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
+
+    $
+      (P and Q) or R &= (P or R) and (Q or R) \
+      P and (Q or R) &= (P or Q) and (P or R)
+    $
+
+    Clearly the choices of $P$ and $Q$ are false, $R$ is true again yields a contradictory result where proposition 1 is true while proposition 2 is false.
+  ]
+
+  d. $not (P and Q), not P and not Q$ \
+
+  The statements are not equivalent.
+
+  #proof[
+    By DeMorgan's,
+    $ not (P and Q) = not P or not Q $
+    Therefore, choosing $P,Q$ with $P != Q$ gives a contradictory result between the two propositions.
+  ]
+
+  e. $(P and Q) or R, P or (Q and R)$
+
+  They are not equivalent.
+
+  #proof[
+    Symbolic manipulation gives
+    $
+      (P and Q) or R &= (P or R) and (Q or R) \
+      P or (Q and R) &= (P or Q) and (P or R)
+    $
+    Then select the conjunctive statement in parentheses that occurs
+    exclusively in one of the statements and set both of its propositions to
+    false. In this case we take $Q$ and $R$ false while $P$ is true, which
+    makes proposition 1 false while proposition 2 is true.
+  ]
+
+  f. $(P and Q) or P, P$
+
+  They are equivalent, trivially. Clearly $Q$ is independent of the outcome's truth value which only depends on $P$.
+
+7. \
+
+  c. Julius Caesar was born in 1492 or 1493 and died in 1776. \
+
+  This proposition is of the form $(P or Q) and R$, where
+
+  $
+    P &= #[Caesar was born in 1492] \
+    Q &= #[Caesar was born in 1493] \
+    R &= #[Caesar died in 1776] \
+  $
+
+  It is false, clearly, as Caesar certainly could not have died in 1776.
+
+  g. It is not the case that $-5$ and $13$ are elements of $NN$, but $4$ is in the set of rational numbers.
+
+  This proposition takes the form $not (P and Q) and R$, where
+
+  $
+    P &= -5 in NN \
+    Q &= 13 in NN \
+    R &= 4 in QQ
+  $
+
+  It is true, since $P$ is false, $Q$ is true, $R$ is true, which implies $not
+  (P and Q)$ is true, thus making the proposition as a whole true.
+
+10. \
+  c. $(P and Q) or (not P or not Q)$ \
+
+  This is a tautology that is true $forall P,Q$.
+
+  #proof[
+    #table(
+      align: center,
+      columns: 5,
+      [$P$],
+      [$Q$],
+      [$P and Q$],
+      [$not P or not Q$],
+      [$(P and Q) or (not P or not Q)$],
+
+      [T], [T], [T], [F], [T],
+      [T], [F], [F], [T], [T],
+      [F], [T], [F], [T], [T],
+      [F], [F], [T], [T], [T],
+    )
+  ]
+
+  e. $(Q and not P) and not (P and R)$ \
+
+  This is neither a tautology nor a contradiction.
+
+  #proof[
+    #table(
+      align: center,
+      columns: 5,
+      [$P$],
+      [$Q$],
+      [$Q and not P$],
+      [$not (P and R)$],
+      [$(Q and not P) and not (P and R)$],
+
+      [T], [T], [F], [F], [F],
+      [F], [T], [F], [T], [F],
+      [T], [F], [T], [T], [T],
+      [F], [F], [F], [T], [F],
+    )
+  ]
+
+11. \
+
+  b. Cleveland will win the first game or the second game.
+
+  Cleveland loses both the first and second games.
+
+  e. Roses are red and violets are blue.
+
+  Roses are green.
+
+  i. The function $g$ has a relative maximum at $x = 2$ or $x = 4$ and a relative minimum at $x = 3$.
+
+  $ g(x) "is given by" integral^x_0 (t - 5)(t-6) dif t $
+
+12a.
+
+Restore parentheses to $not not P or not Q and not S$.
+
+$ not (not P) or (not Q and not S) $
+
+13. \
+
+  a. Make a truth table for the exclusive or ($xor$)
+
+  #table(
+    align: center,
+    columns: 3,
+    [$P$], [$Q$], [$P xor Q$],
+    [T], [T], [F],
+    [F], [T], [T],
+    [T], [F], [T],
+    [F], [F], [F],
+  )
+
+  b. Show that $A xor B = (A or B) and not (A and B)$.
+
+  Intuitively this fact makes sense. Either $A$ or $B$ are true, but $A$ and $B$ can't _both_ be true.
+
+  #proof[
+    By direct computation,
+
+    #table(
+      align: center,
+      columns: 6,
+      [$P$],
+      [$Q$],
+      [$P xor Q$],
+      [$A or B$],
+      [$not (A and B)$],
+      [$(A or B) and not (A and B)$],
+
+      [T], [T], [F], [T], [F], [F],
+      [F], [T], [T], [T], [T], [T],
+      [T], [F], [T], [T], [T], [T],
+      [F], [F], [F], [F], [T], [F],
+    )
+  ]
 
 = Exercises

documents/by-course/pstat-120a/course-notes/main.typ

@@ -4,7 +4,7 @@
 #show: dvdtyp.with(
   title: "PSTAT120A Course Notes",
   author: "Youwen Wu",
-  date: "Winter 2024",
+  date: "Winter 2025",
   subtitle: "Taught by Brian Wainwright",
 )
 
@@ -27,11 +27,11 @@ upon them.
 #remark[
   Keep in mind that without $cal(Z F C)$ or another model of set theory that
   resolves fundamental issues, our set theory is subject to paradoxes like
-  Russell's.
+  Russell's. Whoops, the universe doesn't exist.
 ]
 
 #definition[
-  A Set is a collection of elements.
+  A *Set* is a collection of elements.
 ]
 
 #example[Examples of sets][
@@ -101,9 +101,9 @@ as the notation for $n$ dimensional spaces in $RR$?
   + $(A sect B)' = A' union B'$
 ]
 
-#remark[Generalized DeMorgan's][
-  + $(union_i A_i)' = sect_i A_i'$
-  + $(sect_i A_i)' = union_i A_i'$
+#fact[Generalized DeMorgan's][
+  + $(union_i A_i)' = sect_i A_i '$
+  + $(sect_i A_i)' = union_i A_i '$
 ]
 
 == Sizes of infinity
@@ -125,14 +125,26 @@ When a set is uncountably infinite, its cardinality is greater than $aleph_0$.
   + The natural numbers $NN$.
   + The rationals $QQ$.
   + The natural numbers $ZZ$.
+  + The set of all logical tautologies.
 ]
 
 #example("Uncountable sets")[
   + The real numbers $RR$.
   + The real numbers in the interval $[0,1]$.
+  + The _power set_ of $ZZ$, which is the set of all subsets of $ZZ$.
+]
+
+#remark[
+  All the uncountable sets above have cardinality $2^(aleph_0)$ or $aleph_1$ or
+  $frak(c)$ or $beth_1$. This is the _cardinality of the continuum_, also
+  called "aleph 1" or "beth 1".
+
+  However, in general uncountably infinite sets do not have the same
+  cardinality.
 ]
 
 #fact[
   If a set is countably infinite, then it has a bijection with $ZZ$. This means
-  every set with cardinality $aleph_0$ has a bijection to $ZZ$.
+  every set with cardinality $aleph_0$ has a bijection to $ZZ$. More generally,
+  any sets with the same cardinality have a bijection between them.
 ]