auto-update(nvim): 2025-01-18 01:37:05

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

@@ -465,7 +465,7 @@ $ exists x in U, P(x) $
 #exercise[
   Prove that between any two rational numbers $x$ and $y$ there is another
   rational number $z$.
-]
+]<rational-between>
 
 == Solutions
 
@@ -484,3 +484,78 @@ the original list $p_1, p_2, ..., p_n$.
 If $p_k$ were in the original list, then since $P$ is divisible by $p_k$, and $P
 + 1$ is also divisible by $p_k$, 1 must be divisible by $p_k$ which is
 impossible. So $p_k$ is a new prime.
+
+For completeness, let's finish the proof explicitly. Start with primes $p_1$,
+$p_2$. The method above implies the existence of another prime, which we denote
+$p_3$. Repeat this to find additional primes $p_(k+1)$.
+
+*@perfectsquare.* This is a generalization of the proof $sqrt(6)$ is
+irrational. Seeking a contradiction, suppose $sqrt(a)$ is irrational.
+
+$
+  exists p,q in ZZ, sqrt(a) = p / q \
+  p^2 = a q^2
+$
+
+Then by the fundamental theorem of arithmetic,
+
+$ a = b_1 dot b_2 dot ... dot b_n $
+
+where $b_i$ is prime.
+
+$ p^2 = (Pi^n_(i=1) b_1) dot q^2 $
+
+Notice all $b_i$ are unique (again by the same theorem) and without loss of
+generality, choose a $b_k$, $1 <= k <= n$.
+
+Then $p$ has $j = 1,2,...$ $hash$ of $b_k$ in its factors. Then $p^2$ has $2j$
+$hash$ of $b_k$. Similarly, $q$ has $L = 1,2,...$ $hash$ of $b_k$, and $q^2$
+has $2L$. Then $(Pi ^n _(i=1)) q^2$ has $2L + 1$ $hash$ of $b_k$. But
+
+$ p^2 = (Pi^n_(i=1)) q^2 $
+
+and by unique factorization they must have the same $hash$ of the prime factor $b_k$, so $sqrt(a)$ is irrational.
+
+Note that if $a$ was a perfect square, TODO
+
+*@rational-between.* Effectively we are asked to show that given $x,y in QQ$,
+where $x < y$, $exists z in QQ$ such that $x < z < y$.
+
+First, let us take the difference between $x$ and $y$.
+
+$
+  exists a,b in ZZ \
+  x = a / b \
+  y = c / d \
+  y - x = (b c - a d) / (b d)
+$
+
+So we know that
+
+$
+  y = x + (b c - a d) / (b d)
+$
+
+We see that if we can find any nonzero integer less than $(b c - a d)/(b d)$,
+adding it to $x$ gives us the number between $x$ and $y$ we desire. One option is
+
+$
+  (b c - a d) / (b d + 1) < (b c - a d) / (b d) \
+  x < x + (b c - a d) / (b d + 1) < x + (b c - a d) / (b d) = y \
+$
+
+So one such $z$ is
+
+$ z = x + (b c - a d) / (b d + 1) $
+
+#remark[
+  There is a minor hole in this proof, if $b d + 1 = 0$, $z$ is undefined. We
+  can easily avoid this by replacing all instances of $b d + 1$ with, say, $b d
+  + 2$, when $b d + 1 = 0$.
+]
+
+#remark[
+  A much easier and less roundabout proof is to take
+  $ z = (x + y) / 2 $
+  Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
+]

documents/by-course/pstat-120a/hw1/main.typ

@@ -0,0 +1,37 @@
+#import "@youwen/zen:0.1.0": *
+#import "@preview/ctheorems:1.1.3": *
+
+#show: zen.with(
+  title: "Homework 1",
+  author: "Youwen Wu",
+  date: "Winter 2025",
+)
+
+1. #[
+    #set enum(numbering: "a)", spacing: 2em)
+
+    + #[
+        We know that $B$ and $B'$ are disjoint. That is, $B sect B' =
+        emptyset$. Additionally,
+        $
+          E = (A sect B) subset B \
+          F = (A sect B') subset B' \
+        $
+        Then we note
+        $
+          forall x in E, x in B, x in.not B' \
+          forall y in F, y in B, y in.not B'
+        $
+        So clearly $E$ and $F$ have no common elements, and
+        $
+          E sect F = emptyset
+        $
+      ]
+
+    + #[
+        $
+          E union F &= (A sect B) union (A sect B') \
+          &=
+        $
+      ]
+  ]

documents/by-course/pstat-120a/hw1/package.nix

@@ -0,0 +1,37 @@
+{
+  pkgs,
+  typstPackagesCache,
+  typixLib,
+  cleanTypstSource,
+  flakeSelf,
+  ...
+}:
+let
+  src = cleanTypstSource ./.;
+  commonArgs = {
+    typstSource = "main.typ";
+
+    fontPaths = [
+      # Add paths to fonts here
+      # "${pkgs.roboto}/share/fonts/truetype"
+    ];
+
+    virtualPaths = [
+      # Add paths that must be locally accessible to typst here
+      # {
+      #   dest = "icons";
+      #   src = "${inputs.font-awesome}/svgs/regular";
+      # }
+    ];
+
+    XDG_CACHE_HOME = typstPackagesCache;
+    SOURCE_DATE_EPOCH = builtins.toString flakeSelf.lastModified;
+  };
+
+in
+typixLib.buildTypstProject (
+  commonArgs
+  // {
+    inherit src;
+  }
+)