From ab73cfe5c786d90f6e64ea0faacda2168ef3c337 Mon Sep 17 00:00:00 2001
From: Youwen Wu <youwenw@gmail.com>
Date: Fri, 17 Jan 2025 15:09:08 -0800
Subject: [PATCH] auto-update(nvim): 2025-01-17 15:09:08

---
 .../by-course/math-8/course-notes/main.typ    | 88 +++++++++++++++++++
 1 file changed, 88 insertions(+)

diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ
index d142614..ff1b9ad 100644
--- a/documents/by-course/math-8/course-notes/main.typ
+++ b/documents/by-course/math-8/course-notes/main.typ
@@ -343,3 +343,91 @@ $ P => Q and Q => P $
     is irrational.
   ]
 ]
+
+#fact[$sqrt(2)$ is irrational.]
+
+The proof of this fact generalizes nicely to show that the square root of any
+non-perfect square is irrational.
+
+#example[
+  Prove that $sqrt(6)$ is irrational.
+
+  #proof[
+    Seeking a contradiction, suppose $sqrt(6)$ is rational. Then $sqrt(6) =
+    a/b, exists a,b in ZZ$ with $b != 0$.
+
+
+    Then
+    $ a^2 = 6b^2 = 3 (2b^2) $
+    Since $a^2 = a dot a$, $a$ has $j = 0,1,2, ...$ factors of $3$ in its
+    unique prime factorization, then $a^2$ has $2j$ factors of $3$, which is to
+    say that $a^2$ has an even $hash$ factors of 3.
+
+    Similarly, $b^2$ has an even $hash$ of 3, say $2k$, where $k = 0, 1, 2,
+    ...$. Then $3(2b^2)$ has $2k + 1$ $hash$ factors of 3, an odd amount.
+
+    But $a^2$ = $3(b^2)$ so they must have the same factors, a contradiction.
+    Therefore $sqrt(6)$ must be irrational.
+  ]
+]
+
+#exercise[
+  Prove that $sqrt(2)$ is irrational using the method above.
+]
+
+#exercise[
+  Show that $sqrt(15)$ is irrational.
+]
+
+#exercise("Euclid's Theorem")[
+  Show that there are an infinite amount of prime numbers.
+]
+
+== Proofs involving quantifiers
+
+Many of our proofs up to this point have been of the form
+$ forall x in U, P(x) => Q(x) $
+
+To prove a statement of this form,
+
++ Let $x in U$.
++ Assume $P(x)$.
++ Show $Q(x)$.
++ Conclude $forall x in U, P(x) => Q(x)$.
+
+#example[
+  Prove that $forall x,y in ZZ$, $2x + 14y != 3$.
+
+  #proof[
+    Seeking a contradiction, suppose instead $2x + 14y = 3$. Then
+    $2x $ is even, and $14y = 2(7y)$ is even, and therefore $2x + 14y$ is even.
+    But $3$ is odd, so they cannot be equal.
+
+    Alternatively simply write
+    $ x + 7y = 1.5 $
+    but the ring of $ZZ$ is closed under addition so we have a contradiction.
+  ]
+]
+
+To prove a statement of the form
+
+$ exists x in U, P(x) $
+
++ Find at least one $x in U$ that makes $P(x)$ true.
++ Conclude $exists x in U, P(x)$.
+
+#example[
+  Prove that there is a natural number $N$ such that for all natural numbers $n > N$,
+  $ 1 / n < 0.02 $
+  We want $1/n < 0.02$, so the idea is to play around with this statement. Taking reciprocals,
+  $ n > 50 $
+  #proof[
+    Let $N = 50$. Then $n in NN$ and $n > N = 50$,
+    $ 1 / n < 1 / 50 = 0.02 $
+  ]
+]
+
+#exercise[
+  Prove that between any two rational numbers $x$ and $y$ there is another
+  rational number $z$.
+]