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+++ b/documents/by-course/math-8/course-notes/main.typ
@@ -284,3 +284,62 @@ A conditional may be true even when the antedecent and consequent are unrelated.
 
   It is true because $P(x,y)$ is injective (one-to-one).
 ]
+
+= Missed a bunch of lecture :(
+
+Probably not any important content, though.
+
+= Lecture #datetime(day: 17, month: 1, year: 2025).display()
+
+== Proof of a biconditional statement
+
+To prove a biconditional statement of the form $P <=> Q$, we need to show
+$ P => Q and Q => P $
+
+#theorem("Fundamental Theorem of Arithmetic")[
+  $forall x in ZZ, x > 1$, $x$ can be written as a product of prime factors.
+]
+
+#example[
+  Assume $p$ is prime. Then $p | b$ iff $p | b^2$.
+
+  #proof[
+
+  ]
+]
+
+== Proof by contradiction
+
+#definition[
+  A proof by contradiction of the statement $P$ proceeds by assuming $not P$,
+  then showing that this fact leads to a contradiction.
+
+  A proof by contradiction of $P => Q$ proceeds by assuming $P and not Q$, then
+  showing a contradiction, implying that $P$ indeed implies $Q$.
+]
+
+#definition[
+  A real number $x in RR$ is called rational iff
+  $ exists p,q in ZZ, x = p / q $
+
+  $x$ is irrational if it is not rational.
+]
+
+#example[
+  Prove that if $x$ is rational and $y$ is irrational, then $2x - y$ is irrational.
+
+  #proof[
+    Suppose that $x$ is rational and $y$ is irrational but $2x - y$ is
+    rational. Then $x = a / b$ and $2x - y = c/d$ where $a,b,c,d in ZZ$ and $b
+    != 0, d != 0$.
+
+    Then
+    $
+      2a / b - y &= c / d \
+      y &= (2a) / b - c / d \
+      y = (2a d - b c) / (b d) &= m / n
+    $
+    which implies $y = m/n$ and therefore we have a contradiction. So $2x - y$
+    is irrational.
+  ]
+]