auto-update(nvim): 2025-01-08 13:39:53

documents/by-course/math-8/course-notes/main.typ

@@ -40,7 +40,7 @@ The lowest homework score will be dropped.
 
 == Exams
 
-Each exam is 20% of the grade. The final exam wil replace the lowest of the
+Each exam is 20% of the grade. The final exam will replace the lowest of the
 first two exam scores if it is higher.
 
 = Meeting #datetime(year: 2025, month: 1, day: 6).display()
@@ -132,3 +132,150 @@ Q$ is the proposition "If $P$, then $Q$."
 ]
 
 A conditional may be true even when the antedecent and consequent are unrelated.
+
+= Lecture #datetime(day: 8, month: 1, year: 2025).display()
+
+== More propositional forms
+
+#definition[
+  Let $P$ and $Q$ be propositions. The *biconditional sentence*
+  $ P <=> Q $
+  is true exactly when $P$ and $Q$ are both true or both false.
+]
+
+#example[Ways of stating $P <=> Q$][
+  - $P$ if and only if $Q$
+  - $P$ iff. $Q$
+  - $P$ is equivalent to $Q$
+]
+
+#exercise[
+  Translate each statement into symbols, where $a$ is a fixed real number.
+
+  + $a > 5$ is sufficient for $a > 3$
+  + $a > 3$ is necessary for $a > 5$
+  + $a > 5$ only if $a > 3$
+  + $|a| = -a$ whenever $a < 0$
+  + $|a| = 2$ is necessary and sufficient for $a^2 = 4$
+]
+
+#definition[
+  #set enum(numbering: "a.")
+  Let $P$ and $Q$ be propositions.
+
+  + The converse of $P => Q$ is $Q => P$
+  + The contrapositive of $P => Q$ is $not Q => not P$
+]
+
+#theorem[
+  Let $P$ and $Q$ be propositions. Then:
+]
+
+#example[
+  If $f(x)$ is differentiable at $x = a$, then $f(x)$ is continuous at $x = a$.
+
+  $ P => Q $
+
+  + $not Q => not P$: if $f$ is not continuous at $x=a$, then $f$ is not differentiable at $x = a$.
+
+  + $Q => P$: if $f$ is continuous at $x = a$, then $f$ is diffferentiable at $x = a$.
+]
+
+#fact[
+  We apply our new logical connectives in the following order: $not, and, or, => , <=>$
+]
+
+#example[
+  Include parentheses to clarify the expression.
+
+  $ P or Q => not R <=> S and T $
+]
+
+#theorem[
+  #set enum(numbering: "a.")
+  For propositions $P$, $Q$, and $R$, the following are equivalent:
+
+  + $P => Q "and" not P or Q$
+  + $P <=> Q "and" (P => Q) and (Q => P)$
+  + $not (P => Q) "and" P and not Q$
+]
+
+== Quantified statements
+
+#definition[
+  A *predicate* or *open sentence* is a sentence involving one or more variables.
+]
+
+#example[
+  Consider the open sentence $P(x,y): x^2 + y^2 = 25$. Write a true and a false proposition.
+
+  $
+    P(3,-4) &: 3^2 + (-4)^2 = 25 &"(true)" \
+    P(2,0) &: 2^2 + 0^2 = 25 &"(false)" \
+  $
+]
+
+
+#definition[
+  The *universe* is the set of all objects available for substitution into an open sentence. Denoted $U$.
+]
+
+#definition[
+  A *truth set* is all objects in $U$ that make an open sentence true.
+]
+
+#example[
+  Let the universe be the set of all real numbers for the open sentence $P(x) : x^2 + x = 6$. Find the truth set.
+
+  $
+    U = RR \
+    "truth set:" {2,-3}
+  $
+]
+
+#definition[
+  Let $P(x)$ be an open sentence with variable $x$.
+
+  The *universal quantifier* is the sentence
+
+  $ forall x in U, P(x) $
+
+  The *existential quantifier* is the sentence:
+  $ exists x in U, P(x) $
+
+  The *unique existence quantifier* is the sentence
+  $ exists! x in U, P(x) $
+]
+
+#example[
+  Let the universe be the set of all real numbers and consider the open sentence
+  $ P(x) : x^2 + 1 >= 0 $
+  Consider the quantified sentence $forall x in U,P(x)$. Then
+  $ forall x in RR, x^2 + 1 >= 0 $
+  is a true statement.
+
+  However, if instead $U = CC$, then the sentence is false.
+]
+
+#example[
+  Let the universe be the set of all real numbers and consider the open sentence
+
+  $
+    Q(x) "where" x in ZZ \
+    R(x) "is a perfect square" \
+  $
+  Consider the quantified sentence
+  $ exists Q(x), R(x) $
+]
+
+#example[
+  Let the universe be the set of all real numbers and consider the open sentence
+
+  $ P(x,y) : y = x^3 + 4 $
+
+  Consider the quantified sentence
+
+  $ forall y in U, exists! x in U, P(x,y) $
+
+  It is true because $P(x,y)$ is injective (one-to-one).
+]