homework-template:0.1.0

packages/preview/homework-template/0.1.0/LICENSE

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+MIT License
+
+Copyright (c) 2026 dhayer200
+
+Permission is hereby granted, free of charge, to any person obtaining a copy
+of this software and associated documentation files (the "Software"), to deal
+in the Software without restriction, including without limitation the rights
+to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
+copies of the Software, and to permit persons to whom the Software is
+furnished to do so, subject to the following conditions:
+
+The above copyright notice and this permission notice shall be included in all
+copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
+LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
+OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
+SOFTWARE.

packages/preview/homework-template/0.1.0/README.md

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+# homework-template
+
+A minimal Typst package for math homework with styled boxes for questions, parts, answers, proofs, definitions, and theorems.
+
+## Usage
+
+```typst
+#import "@preview/homework-template:0.1.0": *
+
+#header(
+  name: "Your Name",
+  course: "Math 110 — Linear Algebra",
+  hw: "3",
+  date: "March 4, 2026",
+  professor: "Prof. Smith",   // optional
+)
+
+#qs(title: [Prove that the additive identity is unique.])[
+  #pt(title: [Uniqueness of zero])[
+    #prf[
+      Suppose $0$ and $0'$ are both identities. Then $0 = 0 + 0' = 0'$.
+    ]
+  ]
+]
+```
+
+See [`example.typ`](example.typ) for a full working document.
+
+## Functions
+
+| Function | Description |
+|----------|-------------|
+| `header(name, course, hw, date, professor?, topic?)` | Page header with rule |
+| `qs(title?)[ ]` | Numbered question box |
+| `pt(title?)[ ]` | Lettered part (a., b., …); nests to i., ii., … |
+| `ans[ ]` | Answer/solution block |
+| `prf[ ]` | Proof block with flush-right QED mark |
+| `defn(title?)[ ]` | Definition box |
+| `thm(title?)[ ]` | Theorem box |
+| `eg(title?)[ ]` | Example box |
+| `notn(title?)[ ]` | Notation box |
+| `note[ ]` | Left-ruled remark |
+| `vc(sym)` | Vector arrow shorthand: `vc(v)` → $\vec{v}$ |
+
+## License
+
+MIT

packages/preview/homework-template/0.1.0/example.typ

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+#import "@local/homework-template:0.1.0": *
+
+#header(
+  name: "Deep Hayer",
+  course: "Math 110 — Linear Algebra",
+  hw: "3",
+  date: "March 4, 2026",
+  professor: "Prof. Sheldon Axler",
+)
+
+// ─────────────────────────────────────────────────────────────────────────────
+// Background definitions the homework builds on
+// ─────────────────────────────────────────────────────────────────────────────
+
+#defn(title: [vector space])[
+  A _vector space_ over $FF$ is a set $V$ together with addition and scalar
+  multiplication satisfying commutativity, associativity, additive identity,
+  additive inverses, multiplicative identity, and distributive properties.
+]
+
+#notn(title: [$FF^n$])[
+  $FF^n$ denotes the set of all lists of length $n$ with entries in $FF$
+  (either $RR$ or $CC$).
+]
+
+// ─────────────────────────────────────────────────────────────────────────────
+// Questions
+// ─────────────────────────────────────────────────────────────────────────────
+
+#qs(title: [Let $V$ be a vector space over $FF$. Prove each of the following.])[
+  #pt(title: [The additive identity $0 in V$ is unique.])[
+    #prf[
+      Suppose $0$ and $0'$ are both additive identities in $V$. Then
+      $ 0 = 0 + 0' = 0', $
+      where the first equality uses the fact that $0'$ is an identity and
+      the second uses the fact that $0$ is an identity. Hence $0 = 0'$.
+    ]
+  ]
+
+  #pt(title: [Every element of $V$ has a unique additive inverse.])[
+    #prf[
+      Let $v in V$ and suppose $w, w'$ are both additive inverses of $v$. Then
+      $ w = w + 0 = w + (v + w') = (w + v) + w' = 0 + w' = w'. $
+      Hence the additive inverse is unique.
+    ]
+  ]
+
+  #pt(title: [$0 v = 0$ for every $v in V$, where the left $0$ is the scalar and the right $0$ is the zero vector.])[
+    #prf[
+      For any $v in V$,
+      $ 0 v = (0 + 0) v = 0 v + 0 v. $
+      Adding $-(0 v)$ to both sides gives $0 = 0 v$.
+    ]
+  ]
+]
+#v(10em)
+#qs(title: [Let $U = {(x_1, x_2, x_3, x_4) in FF^4 : x_1 + 2 x_2 = 0 "and" x_3 = 5 x_4}$.])[
+  #pt(title: [Show that $U$ is a subspace of $FF^4$.])[
+    #ans[
+      We verify the three subspace conditions.
+
+      #pt(title: [Additive identity])[
+        $(0,0,0,0)$ satisfies $0 + 2(0) = 0$ and $0 = 5(0)$, so $0 in U$.
+      ]
+
+      #pt(title: [Closed under addition])[
+        Let $(x_1,x_2,x_3,x_4),(y_1,y_2,y_3,y_4) in U$. Then
+        $ (x_1+y_1) + 2(x_2+y_2) = (x_1+2x_2) + (y_1+2y_2) = 0+0 = 0, $
+        and $x_3+y_3 = 5x_4+5y_4 = 5(x_4+y_4)$. So the sum is in $U$.
+      ]
+
+      #pt(title: [Closed under scalar multiplication])[
+        Let $lambda in FF$ and $(x_1,x_2,x_3,x_4) in U$. Then
+        $ lambda x_1 + 2(lambda x_2) = lambda(x_1 + 2x_2) = 0, $
+        and $lambda x_3 = lambda(5 x_4) = 5(lambda x_4)$. So $lambda (x_1,dots,x_4) in U$.
+      ]
+
+      Hence $U$ is a subspace of $FF^4$.
+    ]
+  ]
+
+  #pt(title: [Find a basis for $U$ and state $dim U$.])[
+    #ans[
+      The constraints $x_1 = -2x_2$ and $x_3 = 5x_4$ leave $x_2$ and $x_4$
+      as free variables. Setting $(x_2, x_4) = (1,0)$ and $(0,1)$ gives
+      $
+        e_1 = (-2, 1, 0, 0), quad e_2 = (0, 0, 5, 1).
+      $
+      These two vectors span $U$ and are linearly independent, so
+      ${e_1, e_2}$ is a basis and $dim U = 2$.
+    ]
+  ]
+]
+#v(17em)
+#qs(title: [Suppose $T : V -> W$ is a linear map. Using the theorem below, answer the following.])[
+  #thm(title: [fundamental theorem of linear maps])[
+    $ dim V = dim "null" T + dim "range" T. $
+  ]
+
+  #pt(title: [If $dim V = 7$ and $dim W = 3$, what are the possible values of $dim "null" T$?])[
+    #ans[
+      Since $dim "range" T <= dim W = 3$, we have $dim "range" T in {0,1,2,3}$.
+      By the fundamental theorem,
+      $ dim "null" T = 7 - dim "range" T in {4, 5, 6, 7}. $
+    ]
+  ]
+
+  #pt(title: [Can $T$ be injective if $dim V > dim W$?])[
+    #ans[
+      No. If $T$ is injective then $dim "null" T = 0$, so $dim "range" T = dim V > dim W$,
+      contradicting $dim "range" T <= dim W$.
+    ]
+  ]
+]
+#qs(title: [Let $v_1, dots, v_m in V$ and define the linear map $T : FF^m -> V$ by $T(c_1, dots, c_m) = c_1 v_1 + dots.c + c_m v_m$.])[
+  #eg(title: [span as range])[
+    The range of $T$ is exactly $"span"(v_1, dots, v_m)$. For instance, if
+    $v_1 = (1,0)$ and $v_2 = (0,1)$ in $FF^2$, then $T : FF^2 -> FF^2$ is
+    the identity and $"range" T = FF^2$.
+  ]
+
+  #pt(title: [Show that $v_1, dots, v_m$ spans $V$ if and only if $T$ is surjective.])[
+    #ans[
+      $T$ is surjective $<==>$ $"range" T = V$ $<==>$ every $v in V$ is a linear
+      combination of $v_1, dots, v_m$ $<==>$ $"span"(v_1, dots, v_m) = V$.
+    ]
+  ]
+
+  #pt(title: [Show that $v_1, dots, v_m$ is linearly independent if and only if $T$ is injective.])[
+    #ans[
+      $T$ is injective $<==>$ $"null" T = {0}$ $<==>$ the only solution to
+      $c_1 v_1 + dots.c + c_m v_m = 0$ is $c_1 = dots.c = c_m = 0$
+      $<==>$ $v_1, dots, v_m$ is linearly independent.
+    ]
+  ]
+]
+
+#qs(title: [Let $vc(u) = (1, 2, -1)$ and $vc(v) = (3, 0, 2)$ in $RR^3$. Compute $vc(u) + 2 vc(v)$ and verify it lies in $"span"{vc(u), vc(v)}$.])[
+  #note[
+    The notation $vc(w)$ denotes a vector $w$ with an arrow, used here to
+    distinguish vectors from scalars.
+  ]
+
+  #ans[
+    $
+      vc(u) + 2 vc(v) = (1,2,-1) + (6,0,4) = (7, 2, 3).
+    $
+    Since $(7,2,3) = 1 dot vc(u) + 2 dot vc(v)$, it is a linear combination
+    of $vc(u)$ and $vc(v)$, so it lies in $"span"{vc(u), vc(v)}$.
+  ]
+]