diff --git a/documents/by-course/math-8/pset-1/2vd.typ b/documents/by-course/math-8/pset-1/2vd.typ
deleted file mode 100644
index 06deb27..0000000
--- a/documents/by-course/math-8/pset-1/2vd.typ
+++ /dev/null
@@ -1,341 +0,0 @@
-#import "@preview/ctheorems:1.1.3": *
-#import "@preview/showybox:2.0.3": showybox
-
-#let colors = (
-  rgb("#9E9E9E"),
-  rgb("#F44336"),
-  rgb("#E91E63"),
-  rgb("#9C27B0"),
-  rgb("#673AB7"),
-  rgb("#3F51B5"),
-  rgb("#2196F3"),
-  rgb("#03A9F4"),
-  rgb("#00BCD4"),
-  rgb("#009688"),
-  rgb("#4CAF50"),
-  rgb("#8BC34A"),
-  rgb("#CDDC39"),
-  rgb("#FFEB3B"),
-  rgb("#FFC107"),
-  rgb("#FF9800"),
-  rgb("#FF5722"),
-  rgb("#795548"),
-  rgb("#9E9E9E"),
-)
-
-#let dvdtyp(
-  title: "",
-  subtitle: "",
-  author: "",
-  abstract: none,
-  bibliography: none,
-  paper-size: "a4",
-  date: "today",
-  body,
-) = {
-  set document(title: title, author: author)
-
-  set std.bibliography(style: "springer-mathphys", title: [References])
-
-  show: thmrules
-
-  set page(
-    numbering: "1",
-    number-align: center,
-    header: locate(loc => {
-      if loc.page() == 1 {
-        return
-      }
-      box(stroke: (bottom: 0.7pt), inset: 0.4em)[#text(
-          font: "New Computer Modern",
-        )[
-          *#author* --- #datetime.today().display("[day] [month repr:long] [year]")
-          #h(1fr)
-          *#title*
-        ]]
-    }),
-    paper: paper-size,
-    // The margins depend on the paper size.
-    margin: (
-      left: (86pt / 216mm) * 100%,
-      right: (86pt / 216mm) * 100%,
-    ),
-  )
-
-
-  set heading(numbering: "1.")
-  show heading: it => {
-    set text(font: "Libertinus Serif")
-
-    block[
-      #if it.numbering != none {
-        text(rgb("#2196F3"), weight: 500)[#sym.section]
-
-        text(rgb("#2196F3"))[#counter(heading).display() ]
-      }
-      #it.body
-      #v(0.5em)
-    ]
-  }
-
-  set text(font: "New Computer Modern", lang: "en")
-
-  show math.equation: set text(weight: 400)
-
-
-  // Title row.
-  align(center)[
-    #set text(font: "Libertinus Serif")
-    #block(text(weight: 700, 26pt, title))
-
-
-    #if subtitle != none [#text(12pt, weight: 500)[#(
-          subtitle
-        )]]
-
-    #if author != none [#text(16pt)[#smallcaps(author)]]
-    #v(1.2em, weak: true)
-
-    #if date == "today" {
-      datetime.today().display("[day] [month repr:long] [year]")
-    } else {
-      date
-    }
-
-  ]
-
-  if abstract != none [
-    #v(2.2em)
-    #set text(font: "Libertinus Serif")
-    #pad(x: 14%, abstract)
-    #v(1em)
-  ]
-
-  set outline(fill: repeat[~.], indent: 1em)
-
-  show outline: set heading(numbering: none)
-  show outline: set par(first-line-indent: 0em)
-
-  show outline.entry.where(level: 1): it => {
-    text(font: "Libertinus Serif", rgb("#2196F3"))[#strong[#it]]
-  }
-  show outline.entry: it => {
-    h(1em)
-    text(font: "Libertinus Serif", rgb("#2196F3"))[#it]
-  }
-
-
-  // Main body.
-  set par(
-    justify: true,
-    spacing: 0.65em,
-    first-line-indent: 2em,
-  )
-
-  body
-
-  // Display the bibliography, if any is given.
-  if bibliography != none {
-    show std.bibliography: set text(footnote-size)
-    show std.bibliography: set block(above: 11pt)
-    show std.bibliography: pad.with(x: 0.5pt)
-    bibliography
-  }
-}
-
-#let thmtitle(t, color: rgb("#000000")) = {
-  return text(
-    font: "Libertinus Serif",
-    weight: "semibold",
-    fill: color,
-  )[#t]
-}
-#let thmname(t, color: rgb("#000000")) = {
-  return text(font: "Libertinus Serif", fill: color)[(#t)]
-}
-
-#let thmtext(t, color: rgb("#000000")) = {
-  let a = t.children
-  if (a.at(0) == [ ]) {
-    a.remove(0)
-  }
-  t = a.join()
-
-  return text(font: "New Computer Modern", fill: color)[#t]
-}
-
-#let thmbase(
-  identifier,
-  head,
-  ..blockargs,
-  supplement: auto,
-  padding: (top: 0.5em, bottom: 0.5em),
-  namefmt: x => [(#x)],
-  titlefmt: strong,
-  bodyfmt: x => x,
-  separator: [. \ ],
-  base: "heading",
-  base_level: none,
-) = {
-  if supplement == auto {
-    supplement = head
-  }
-  let boxfmt(name, number, body, title: auto, ..blockargs_individual) = {
-    if not name == none {
-      name = [ #namefmt(name)]
-    } else {
-      name = []
-    }
-    if title == auto {
-      title = head
-    }
-    if not number == none {
-      title += " " + number
-    }
-    title = titlefmt(title)
-    body = [#pad(top: 2pt, bodyfmt(body))]
-    pad(
-      ..padding,
-      showybox(
-        width: 100%,
-        radius: 0.3em,
-        breakable: true,
-        padding: (top: 0em, bottom: 0em),
-        ..blockargs.named(),
-        ..blockargs_individual.named(),
-        [
-          #title#name#titlefmt(separator)#body
-        ],
-      ),
-    )
-  }
-
-  let auxthmenv = thmenv(
-    identifier,
-    base,
-    base_level,
-    boxfmt,
-  ).with(supplement: supplement)
-
-  return auxthmenv.with(numbering: "1.1")
-}
-
-#let styled-thmbase = thmbase.with(
-  titlefmt: thmtitle,
-  namefmt: thmname,
-  bodyfmt: thmtext,
-)
-
-#let builder-thmbox(color: rgb("#000000"), ..builderargs) = styled-thmbase.with(
-  titlefmt: thmtitle.with(color: color.darken(30%)),
-  bodyfmt: thmtext.with(color: color.darken(70%)),
-  namefmt: thmname.with(color: color.darken(30%)),
-  frame: (
-    body-color: color.lighten(92%),
-    border-color: color.darken(10%),
-    thickness: 1.5pt,
-    inset: 1.2em,
-    radius: 0.3em,
-  ),
-  ..builderargs,
-)
-
-#let builder-thmline(
-  color: rgb("#000000"),
-  ..builderargs,
-) = styled-thmbase.with(
-  titlefmt: thmtitle.with(color: color.darken(30%)),
-  bodyfmt: thmtext.with(color: color.darken(70%)),
-  namefmt: thmname.with(color: color.darken(30%)),
-  frame: (
-    body-color: color.lighten(92%),
-    border-color: color.darken(10%),
-    thickness: (left: 2pt),
-    inset: 1.2em,
-    radius: 0em,
-  ),
-  ..builderargs,
-)
-
-#let problem-style = builder-thmbox(
-  color: colors.at(11),
-  shadow: (offset: (x: 2pt, y: 2pt), color: luma(70%)),
-)
-
-#let exercise = problem-style("item", "Exercise")
-#let problem = exercise
-
-#let theorem-style = builder-thmbox(
-  color: colors.at(6),
-  shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
-)
-
-#let example-style = builder-thmbox(
-  color: colors.at(16),
-  shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
-)
-
-#let theorem = theorem-style("item", "Theorem")
-#let lemma = theorem-style("item", "Lemma")
-#let corollary = theorem-style("item", "Corollary")
-
-#let definition-style = builder-thmline(color: colors.at(8))
-
-// #let definition = definition-style("definition", "Definition")
-#let proposition = definition-style("item", "Proposition")
-#let remark = definition-style("item", "Remark")
-#let observation = definition-style("item", "Observation")
-
-// #let example-style = builder-thmline(color: colors.at(16))
-
-#let example = example-style("item", "Example")
-
-#let proof(body, name: none) = {
-  v(0.5em)
-  [_Proof_]
-  if name != none {
-    [ #thmname[#name]]
-  }
-  [.]
-  body
-  h(1fr)
-
-  // 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
-
-  $square.stroked$
-  v(0.5em)
-}
-
-#let fact = thmplain(
-  "item",
-  "Fact",
-  titlefmt: content => [*#content.*],
-  namefmt: content => [_(#content)._],
-  separator: [],
-  inset: 0pt,
-  padding: (bottom: 0.5em, top: 0.5em),
-)
-#let abuse = thmplain(
-  "item",
-  "Abuse of Notation",
-  titlefmt: content => [*#content.*],
-  namefmt: content => [_(#content)._],
-  separator: [],
-  inset: 0pt,
-  padding: (bottom: 0.5em, top: 0.5em),
-)
-#let definition = thmplain(
-  "item",
-  "Definition",
-  titlefmt: content => [*#content.*],
-  namefmt: content => [_(#content)._],
-  separator: [],
-  inset: 0pt,
-  padding: (bottom: 0.5em, top: 0.5em),
-)
diff --git a/documents/by-course/math-8/pset-1/main.typ b/documents/by-course/math-8/pset-1/main.typ
index f1b07bf..25132d5 100644
--- a/documents/by-course/math-8/pset-1/main.typ
+++ b/documents/by-course/math-8/pset-1/main.typ
@@ -1,6 +1,6 @@
-#import "./dvd.typ": *
+#import "@youwen/zen:0.1.0": *
 
-#show: dvdtyp.with(
+#show: zen.with(
   title: "Homework 1",
   author: "Youwen Wu",
 )
diff --git a/documents/by-course/math-8/pset-3/main.typ b/documents/by-course/math-8/pset-3/main.typ
new file mode 100644
index 0000000..c810b99
--- /dev/null
+++ b/documents/by-course/math-8/pset-3/main.typ
@@ -0,0 +1,278 @@
+#import "@youwen/zen:0.1.0": *
+#import "@preview/mitex:0.2.5": *
+
+#show: zen.with(
+  title: "Homework 3",
+  author: "Youwen Wu",
+)
+
+#set heading(numbering: none)
+
+#set par(first-line-indent: 0pt, spacing: 1em)
+
+#let nonzero = $ZZ_(!=0)$
+
+Problems:
+
+1.5: $hash$ 3cdefgh, 4de, 6de, 7ab, 9, 10, 11, 12a
+
+1.6: $hash$ 1bd, 4abcd, 6abefik
+
+2.1: $hash$ 4, 5, 6abcd, 8, 11ab, 14ab, 15abcd
+
+= 1.5
+
+*3c.*
+
+If $x^2$ is not divisible by 4, then $x$ is odd.
+
+#proof[
+  Suppose $x$ is not odd. We seek to show that $x^2$ is divisible by $4$.
+
+  Then $x$ is even and $exists k in ZZ_(!=0), x = 2k$. Thus
+  $x^2 = 4k^2$, which implies $4 | x^2$. So $x$ is not odd implies $x^2$ is
+  divisible by 4, and therefore the contrapositive (which is our original
+  statement) is also true.
+]
+
+*3d.*
+
+If $x y$ is even, then either $x$ or $y$ is even.
+
+#proof[
+  Suppose that $not (x "or" y" is even")$. In other words, $x$ and $y$ are both
+  odd. We seek to show that $x y$ is odd.
+
+  Then $exists j,k in ZZ_(n!=0), x = 2j + 1, y = 2k + 1$. So $x y = 4 j k + 2j +
+  2k + 1 = 2 (2 j k + j + k) + 1$ where $2 j k + j + k$ is an integer so $x y$
+  is odd. This is the contrapositive, so the original statement is also true.
+]
+
+*3e.*
+
+If $x + y$ is even, then $x$ and $y$ have the same parity.
+
+#proof[
+  Suppose $x$ and $y$ had different parities. We seek to show that $x + y$ is
+  odd.
+
+  Without loss of generality, let us inspect the case where $x$ is odd. Then
+  $y$ must be even and
+  $
+    exists j,k in ZZ_(!=0), x = 2j + 1, y = 2k \
+    x + y = 2j + 1 + 2k = 2(j + k) + 1 = 2n + 1, n in ZZ_(n!=0)
+  $
+  Repeat the same reasoning for when $x$ is even and $y$ is odd. We've shown
+  that $x$ and $y$ having different parities implies that $x + y$ is odd.
+  Therefore the contrapositive is also true.
+]
+
+*3f.*
+
+If $x y$ is odd, then both $x$ and $y$ are odd.
+
+#proof[
+  Suppose that both $x$ and $y$ are odd was not true, in other words $x$ or $y$
+  are even (here or is the logical $or$). We seek to show that this implies $x y
+$ is even. Then we have two cases: either $x$ or
+  $y$, but not both, are even, and the case where $x$ and $y$ are both even.
+
+  We look at the first case, without loss of generality, assume $x$ is even and
+  $y$ is odd. Then
+
+  $
+    exists j,k in ZZ_(!= 0), x = 2j, y = 2k + 1 \
+    x y = 2j(2k + 1) = 4 j k + 2j = 2(2 j k + j) = 2n, n in ZZ_(!=0)
+  $
+
+  So $x$ and $y$ having different parities implies $x y$ is even. The argument holds identically when instead $y$ is even and $x$ is odd. Now we turn our attention to the case when $x$ and $y$ are both even. Then
+
+  $
+    exists j,k in ZZ_(!=0), x = 2j, y = 2k \
+    x y = 2j dot 2k = 4 j k = 2 (2 j k) = 2n, n in ZZ_(!=0)
+  $
+
+  So this also implies that $x y$ is even. Therefore $x$ or $y$ being even indeed
+  implies $x y$ is also even, and the contrapositive is also true.
+]
+
+*3g.*
+
+If 8 does not divide $x^2 - 1$, then $x$ is even.
+
+#proof[
+  Suppose that $x$ is odd. We seek to show that 8 does in fact divide $x^2 - 1$.
+
+  $
+    exists k in nonzero, x = 2k + 1 \
+    x^2 - 1 = 4k^2 + 4k = 4(k^2 + k)
+  $
+  Now we need to see if it's divisible by 8. First consider the unique case $k = 0$, where $x^2 - 1 = 0$. Clearly $8 | 0$ so 8 does divide $x^2-1$.
+
+  Now we consider all other values. Notice that for all other possible values
+  of $k$, $k^2 + k$ is greater than 1. We see this by noting that $k^2 + k$ is
+  a quadratic with its absolute minima at $k = 1/2$, therefore we can check the
+  two non-zero integers closest to this value. For $k = 1$, $k^2 + k = 2$.
+  Since we already checked $k=-1$, let's check $k=-2$, which gives $k^2 + k =
+  2$. For all values of $k$ greater than 1 or less than $-2$, $k^2 + k$ must be
+  greater than 2 (because it's a quadratic).
+
+  Therefore $4(k^2 + k)$ can be written as $8n$ for some integer $n$, so 8
+  indeed divides $x^2 - 1$ for all possible $k$, and the contrapositive is also
+  true.
+]
+
+*3h.*
+
+If $x$ does not divide $y z$, then $x$ does not divide $z$.
+
+#proof[
+  Assume $x$ does divide $z$. We seek to show that $x$ does divide $y z$. If $x
+  | z$ then $exists k in nonzero, k x = z$. So $y z$ can be written as $y k x$.
+  But this shows $exists j in nonzero, j x = y z$, namely $j = y k$, which
+  means $x | y z$, and the contrapositive is also true.
+]
+
+*4d.*
+
+If $(x+1)(x-1) < 0$, then $x < 1$.
+
+#proof[
+  Suppose that $x >= 1$. We seek to show that $(x+1)(x-1) >= 0$. Expanding out
+  factors,
+  $
+    (x+1)(x-1) = x^2 - 1>= 0
+  $
+  This quadratic is zero at exactly $x = 1$ and positive for all $x > 1$. So
+  it's true for all possible values of $x$. Therefore our original statement is
+  also true.
+]
+
+*4e.*
+
+If $x(x-4) > -3$, then $x < 1$ or $x > 3$.
+
+#proof[
+  Suppose that $x >= 1$ and $x <= 3$. We seek to show $x(x-4) <= -3$. Expanding
+  out factors,
+
+  $
+    x(x-4) = x^2 - 4x > -3
+  $
+
+  This quadratic has its stationary point at $x = 2$. Let's check its value at $x
+= 1$ and $x = 3$.
+
+  At $x = 1$, $x(x-4) = -3$. So for all values greater than 1 until $x = 2$, $x^2 - 4x$ is less than $-3$. Our inequality is satisfied.
+
+  At $x = 3$, $x(x-4) = -3$ again. So for all values less than 3 until $x = 2$,
+  $x^2 - 4x$ is less than $-3$. Our inequality is satisfied for both $x >=1$
+  and $x <=3$, so $x^2 -4x > -3$ is true for all possible $x$ and the
+  contrapositive is also true.
+]
+
+*6d.*
+
+If $a - b$ is odd, then $a + b$ is odd.
+
+#proof[
+  Suppose, seeking a contradiction, that if $a - b$ is even, then $a + b$ is odd.
+
+  Then $exists k in nonzero,  a - b = 2k$. Which means we can write $a + b = 2k
+  + 2b$. But we can factor this as $2(k + b)$ and so $a + b = 2n, n in
+  nonzero$, implying it is even. However we assumed that $a + b$ should be odd,
+  a contradiction. Therefore $a - b$ must be odd.
+]
+
+*6e.*
+
+If $a < b$ and $a b < 3$, then $a = 1$.
+
+#proof[
+  Suppose, seeking a contradiction, that $a >= b$ and $a b >= 3$ implies $a =
+  1$. Consider the specific cases $a = b$, $a b = 3$. Then $a = 3/a$, and $a =
+  plus.minus sqrt(3)$. However we assumed $a = 1$ is implied, a contradiction.
+  Therefore we must have $a < b$ and $a b < 3$.
+]
+
+*7a.*
+
+$a c$ divides $b$ and $b$ divides $b + 3$ if and only if $a = 2$ and $b = 3$.
+
+We first show the result left to right, namely, $a c | b c => a | b$.
+
+$
+  exists k in nonzero, k a c = b c \
+  k a = b
+$
+
+which is the definition of $a | b$.
+
+Now we show the right to left direction, namely, $a | b => a c | b c$.
+
+$
+  exists k in nonzero, k a = b \
+  k a dot c = b dot c
+$
+
+So $a c | b c$ by the definition of divisibility. Therefore the biconditional
+is true, as we have shown both directions.
+
+*7b.*
+
+The right to left direction is very easy in this case. By directly plugging in $a = 2$, $b = 3$, we see that $2 + 1 | 3$ and $3 | 3 + 3$. To show the left to right case,
+
+*9.*
+
+#proof[
+  Suppose that instead $n/(n+1) <= n/(n+2)$. We can be assured the following
+  operations do not flip the inequality as $n$ cannot be negative.
+  $
+    n(n+2) <= n(n+1) \
+    n + 2 <= n + 1 \
+    2 <= 1
+  $
+  So $n/(n+1) > n/(n+2)$.
+]
+
+*10.*
+
+#proof[
+  Suppose that $sqrt(5)$ was rational. That is, $sqrt(5) = p/q$ for nonzero
+  integers $p$ and $q$. Additionally, assume that $p/q$ is in its most reduced
+  form, that is, $p$ and $q$ share no common factors besides 1. Then
+
+  $
+    p^2 = 5q \
+  $
+  implies that
+  $ 5 | p^2 $
+  We need to show that $5 | p^2 => 5 | p$.
+
+  By the fundamental theorem of arithmetic, $p^2$ has 5 as one of its unique
+  prime factors. If 5 was not a factor of $p$, then $p^2 = p dot p$ would not
+  have 5 in its factors either. So 5 is a factor of $p$ and thus $5 | p$. Note
+  that 5 appears at least twice amongst the prime factors of $p^2$. Then $5q$
+  should also have at least two 5s in its prime factorization. Then $q$ has at
+  least one 5 in its prime factorization. However we assumed that $p$ and $q$
+  share no common factors besides 1, so this is a contradiction. Therefore
+  $sqrt(5)$ is not rational.
+]
+
+*11.*
+
+#proof[
+  We say that two numbers $x$ and $y$ are within $1/2$ unit from one another if
+  $|x - y| < 1/2$. Consider the distance between $z$, and $y$, if it is within
+  $1/2$, then we are done. Otherwise $z - y >= 1/2$.
+
+  We know that
+  $ (1 - z) + x + (y-x) + (z-y) = 1 $
+  because this is the length of all the line segments partitioned by $x,y,z$,
+  which is the interval 1. If $(z - y) >= 1/2$, then everything else must be
+  less than $1/2$. So the maximum value of $y-x$, the distance between $x$ and
+  $y$, is less than $1/2$. Therefore either $y$ and $z$ are within $1/2$ unit
+  of each other or $y$ and $x$ are.
+]
+
+*12a.*
diff --git a/documents/by-course/math-8/pset-3/package.nix b/documents/by-course/math-8/pset-3/package.nix
new file mode 100644
index 0000000..7879e7e
--- /dev/null
+++ b/documents/by-course/math-8/pset-3/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;
+  }
+)