Add theorem commands

.gitignore

@@ -1,3 +1,4 @@
 *.pdf
 latex.out/
 .direnv
+__pycache__/

CS1527/assessment-2/.envrc

@@ -0,0 +1,2 @@
+export NIX_PATH="nixpkgs=flake:nixpkgs"
+use nix

CS1527/assessment-2/assessment2.py

@@ -0,0 +1,404 @@
+#!/usr/bin/env python
+"""Usage: `python assessment2.py {subcommand} {expressions}`
+
+If expressions are not provided, they are taken from each line of stdin.
+Documentation of subcommands is available from `python assessment2.py -h`.
+
+Examples for marking criteria:
+  1. Eval: `python assessment2.py eval "(((2*(3+2))+5)/2)"`
+  2. Tree: `python assessment2.py render-tree "(((2*(3+2))+5)/2)"`
+  3. Preorder:  `python assessment2.py format-prefix "(((2*(3+2))+5)/2)"`
+     In-order:  `python assessment2.py format-infix "(((2*(3+2))+5)/2)"`
+     Postorder: `python assessment2.py format-postfix "(((2*(3+2))+5)/2)"`
+  4. Errors: `python assessment2.py eval "(4*3*2)" "(4*(2))" "(4*(3+2)*(2+1))" "(2*4)*(3+2)" "((2+3)*(4*5)" "((2+3)*(4*5)" "(2+5)*(4/(2+2)))" "(2+5)*(4/(2+2)))" "(((2+3)*(4*5))+(1(2+3)))"`
+  5. Tests: `python assessment2.py test`
+
+Tests can be run with `pytest assessment2.py` or `python assessment2.py test`.
+
+All of the formatting is implemented with post-order traversals. This is
+necessary to output parenthesis in the pre- and in-order cases. It could also be
+done with a buffer and a combined traversal (with callbacks before, between, and
+after subtree visits), but that requires mutation.
+
+Note: this program requires Python 3.12 due to the generic syntax used to define
+`BTree.traverse`. Pytest is also required. These should be installed already
+on Codio.
+"""
+# stdlib
+from dataclasses import dataclass
+from typing import Literal, Tuple, Callable, cast, override
+from operator import add, sub, mul, truediv as div
+from argparse import ArgumentParser
+import sys
+import re
+from textwrap import dedent
+
+# pytest, from https://pypi.org/project/pytest/
+# authors can be found at https://github.com/pytest-dev/pytest/graphs/contributors,
+import pytest
+
+OPERATORS: dict[Literal["+", "-", "*", "/"], Callable[[float, float], float]] = {
+    "+": add,
+    "-": sub,
+    "*": mul,
+    "/": div,
+}
+
+
+class BTree:
+    value: float
+
+    def traverse[T](
+        self,
+        value: Callable[["Value", int], T],
+        operator: Callable[["Operator", T, T, int], T],
+        depth: int = 0,
+    ) -> T:
+        """Traverse the binary tree with a post-order traversal
+
+        `value` takes the value node and its depth from the root node
+        `operator` takes the operator node, the results of traversing the left
+            and right subtrees, and the depth of the node
+
+        The tree could, for example, be evaluated with `traverse`:
+            ```
+                (...).traverse(
+                    lambda node, _: node.value,
+                    lambda node, left, right, _: OPERATORS[node.operator](left, right)
+                )
+            ```
+        """
+        raise NotImplementedError
+
+    def as_preorder_str(self) -> str:
+        """Format the tree in prefix notation"""
+        return self.traverse(
+            lambda value, _: str(value.value),
+            lambda op, left, right, _: f"({op.operator} {left} {right})",
+        )
+
+    def as_inorder_str(self) -> str:
+        """Format the tree as a typical parenthesized expression"""
+        return self.traverse(
+            lambda value, _: str(value.value),
+            lambda op, left, right, _: f"({left} {op.operator} {right})",
+        )
+
+    __str__ = as_inorder_str
+
+    def as_inorder_lines(self) -> str:
+        """Format the tree, visually as a tree"""
+        return self.traverse(
+            lambda value, depth: " " * depth + str(value.value),
+            lambda op, left, right, depth: f"{left}\n{' ' * depth}{op.operator}\n{right}",
+        )
+
+    def as_postorder_str(self) -> str:
+        """Format the tree in postfix notation (RPN)"""
+        return self.traverse(
+            lambda value, _: str(value.value),
+            lambda op, left, right, _: f"{left} {right} {op.operator}",
+        )
+
+
+@dataclass(frozen=True, slots=True)
+class Value(BTree):
+    value: float
+
+    @override
+    def traverse[T](
+        self,
+        value: Callable[["Value", int], T],
+        operator: Callable[["Operator", T, T, int], T],
+        depth: int = 0,
+    ) -> T:
+        """Traverse the binary tree with a post-order traversal
+
+        `value` takes the value node and its depth from the root node
+        `operator` takes the operator node, the results of traversing the left
+            and right subtrees, and the depth of the node
+
+        The tree could, for example, be evaluated with `traverse`:
+            ```
+                (...).traverse(
+                    lambda node, _: node.value,
+                    lambda node, left, right, _: OPERATORS[node.operator](left, right)
+                )
+            ```
+        """
+
+        return value(self, depth)
+
+
+@dataclass(frozen=True, slots=True)
+class Operator(BTree):
+    operator: Literal["+", "-", "*", "/"]
+    left: BTree
+    right: BTree
+
+    @property
+    def value(self) -> float:  # type: ignore - the field should be read-only
+        return OPERATORS[self.operator](self.left.value, self.right.value)
+
+    @override
+    def traverse[T](
+        self,
+        value: Callable[["Value", int], T],
+        operator: Callable[["Operator", T, T, int], T],
+        depth: int = 0,
+    ) -> T:
+        """Traverse the binary tree with a post-order traversal
+
+        `value` takes the value node and its depth from the root node
+        `operator` takes the operator node, the results of traversing the left
+            and right subtrees, and the depth of the node
+
+        The tree could, for example, be evaluated with `traverse`:
+            ```
+                (...).traverse(
+                    lambda node, _: node.value,
+                    lambda node, left, right, _: OPERATORS[node.operator](left, right)
+                )
+            ```
+        """
+
+        return operator(
+            self,
+            self.left.traverse(value, operator, depth + 1),
+            self.right.traverse(value, operator, depth + 1),
+            depth,
+        )
+
+
+def parse(expr: str) -> BTree:
+    """Parse a parenthesized expression as in the spec.
+
+    Requires a single parenthesized string, potentially with whitespace
+    Raises ValueError on parse errors
+    """
+
+    def number(expr: str) -> Tuple[BTree, str] | None:
+        if not expr[0].isdigit():
+            return None
+        rest = expr.lstrip("0123456789")
+        return Value(int(expr[: len(expr) - len(rest)])), rest.lstrip()
+
+    def parenthesized(expr: str) -> Tuple[BTree, str] | None:
+        if not expr.startswith("("):
+            return None
+        expr = expr[1:].lstrip()
+
+        left, expr = operand(expr)
+
+        operator, expr = expr[0], expr[1:]
+        if operator not in OPERATORS:
+            raise ValueError(f"Unknown operator {operator}")
+        operator = cast(Literal["+", "-", "*", "/"], operator)
+
+        right, expr = operand(expr.lstrip())
+
+        if not expr.startswith(")"):
+            if expr and expr[0] in OPERATORS:
+                raise ValueError("Too many operands in expression")
+            raise ValueError("Expected closing parenthesis")
+        return Operator(operator, left, right), expr[1:].lstrip()
+
+    def operand(expr: str) -> Tuple[BTree, str]:
+        v = number(expr) or parenthesized(expr)
+        if not v:
+            raise ValueError(
+                "Expected a number or the beginning of a parenthesized expression"
+            )
+        return v
+
+    result = parenthesized(expr.lstrip())
+    if not result or result[1]:
+        # If rest is non-empty, then something was after the parenthesis starting the expression
+        # i.e. the full expression was not parenthesized.
+        raise ValueError("Expected parenthesized expression at the top level")
+
+    return result[0]
+
+
+@pytest.mark.parametrize(
+    ("expr", "ast"),
+    [
+        ("(1 + 1)", Operator("+", Value(1), Value(1))),
+        (
+            "(2*(3+ 2))",
+            Operator("*", Value(2), Operator("+", Value(3), Value(2))),
+        ),
+        (" (1 +1) ", Operator("+", Value(1), Value(1))),
+    ],
+)
+def test_parse(expr: str, ast: BTree):
+    """Test parsing of expressions"""
+    assert parse(expr) == ast
+
+
+@pytest.mark.parametrize(
+    ("expr", "value"),
+    [
+        ("(1 + 2)", 3),
+        ("(12 + (((4 / 2) * 3) + (3 * 2)))", 24),
+        ("(((2 * (3 + 2)) + 5) / 2)", 7.5),
+    ],
+)
+def test_eval(expr: str, value: float):
+    """Test evaluation"""
+    assert parse(expr).value == value
+
+
+@pytest.mark.parametrize(
+    ("expr", "tree"),
+    [
+        (
+            "(1 + 2)",
+            """
+                 1
+                +
+                 2
+            """,
+        ),
+        (
+            "(1 + (((4 / 2) * 3) + (3 * 2)))",
+            """
+                 1
+                +
+                    4
+                   /
+                    2
+                  *
+                   3
+                 +
+                   3
+                  *
+                   2
+            """,
+        ),
+        (
+            "(((2*(3+2))+5)/2)",
+            """
+                   2
+                  *
+                    3
+                   +
+                    2
+                 +
+                  5
+                /
+                 2
+            """
+        )
+    ],
+)
+def test_inorder_lines(expr: str, tree: str):
+    """Test tree rendering"""
+    assert parse(expr).as_inorder_lines() == dedent(tree).strip("\n")
+
+
+@pytest.mark.parametrize(
+    ("expr", "prefix"),
+    [
+        ("(1 + 2)", "(+ 1 2)"),
+        ("(1 + (((4 / 2) * 3) + (3 * 2)))", "(+ 1 (+ (* (/ 4 2) 3) (* 3 2)))"),
+    ],
+)
+def test_prefix(expr: str, prefix: str):
+    """Test prefix formatting"""
+    assert parse(expr).as_preorder_str() == prefix
+
+
+@pytest.mark.parametrize(
+    ("expr"),
+    [
+        "(1 + 2)",
+        "(1 + (((4 / 2) * 3) + (3 * 2)))",
+        "(((3 + 2) * 2) + (((7 / 3) * 5) - 3))",
+    ],
+)
+def test_infix_roundtrip(expr: str):
+    """Test infix formatting roundtrips"""
+    assert str(parse(expr)) == expr
+
+
+@pytest.mark.parametrize(
+    ("expr", "postfix"),
+    [
+        ("(1 + 2)", "1 2 +"),
+        ("(1 + (((4 / 2) * 3) + (3 * 2)))", "1 4 2 / 3 * 3 2 * + +"),
+    ],
+)
+def test_postfix(expr: str, postfix: str):
+    """Test postfix formatting"""
+    assert parse(expr).as_postorder_str() == postfix
+
+
+@pytest.mark.parametrize(
+    ("expr", "err"),
+    [
+        ("(4 * 3 * 2)", "Too many operands in expression"),
+        ("(4 * (2))", "Unknown operator )"),
+        ("(4 * (3 + 2) * (2 + 1))", "Too many operands in expression"),
+        ("(2 *4)*(3+2)", "Expected parenthesized expression at the top level"),
+        ("(2+5)*(4/(2+2))", "Expected parenthesized expression at the top level"),
+        ("(((2+3)*(4*5))+(1(2+3)))", "Unknown operator ("),
+    ],
+)
+def test_error(expr: str, err: str):
+    """Test error messages"""
+    with pytest.raises(ValueError, match=re.escape(err)):
+        parse(expr)
+
+
+if __name__ == "__main__":
+    parser = ArgumentParser(usage=__doc__)
+    subcommands = parser.add_subparsers(title="Subcommands", required=True)
+
+    def with_expressions(parser):
+        parser.add_argument(
+            "expressions",
+            nargs="*",
+            help="The expressions to operate on. If none are provided, operate on lines of stdin.",
+        )
+        return parser
+
+    with_expressions(
+        subcommands.add_parser("eval", help="Evaluate the expression")
+    ).set_defaults(func=lambda expr: expr.value)
+    with_expressions(
+        subcommands.add_parser("render-tree", help="Render the expression as a tree")
+    ).set_defaults(func=BTree.as_inorder_lines)
+    with_expressions(
+        subcommands.add_parser(
+            "format-prefix", help="Format the expression in prefix notation"
+        )
+    ).set_defaults(func=BTree.as_preorder_str)
+    with_expressions(
+        subcommands.add_parser(
+            "format-infix",
+            help="Format the expression in typical parenthesized infix notation",
+        )
+    ).set_defaults(func=BTree.as_inorder_str)
+    with_expressions(
+        subcommands.add_parser(
+            "format-postfix",
+            help="Format the expression in postfix notation, i.e. as RPN",
+        )
+    ).set_defaults(func=BTree.as_postorder_str)
+    subcommands.add_parser("test", help="Run tests").set_defaults(
+        func=lambda: pytest.main([__file__])
+    )
+    args = parser.parse_args()
+    if "expressions" in args:
+        for expression in args.expressions or sys.stdin:
+            if len(args.expressions) > 1:
+                print(expression + ":")
+            try:
+                parsed = parse(expression)
+            except ValueError as e:
+                print(f"Error: {e}")
+                continue
+            print(args.func(parsed))
+    else:
+        args.func()

CS1527/assessment-2/shell.nix

@@ -0,0 +1,3 @@
+{ pkgs ? import <nixpkgs> {} }: pkgs.mkShell {
+  packages = [(pkgs.python312.withPackages (p: with p; [ python-lsp-server pytest pytest-watch black ])) pkgs.nodePackages.pyright];
+}

CS1527/notes-2024-02-27.md

@@ -0,0 +1 @@
+

CS1534/notes-2024-02-15.md

@@ -0,0 +1 @@
+

MA1006/decls.tex

@@ -9,6 +9,7 @@
 \author{}
 
 \newcommand{\paren}[1]{\left(#1\right)}
+\newcommand{\powerset}[1]{\mathcal{P}\paren{#1}}
 \renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
 \renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
 \newcommand{\C}{\mathbb{C}}
@@ -22,9 +23,16 @@
 \newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
 \newcommand{\adj}[1]{\operatorname{adj}#1}
 \newcommand{\card}[1]{\left|#1\right|}
+\newcommand{\littletaller}{\mathchoice{\vphantom{\big|}}{}{}{}}
+\newcommand{\restr}[2]{{% we make the whole thing an ordinary symbol
+  \left.\kern-\nulldelimiterspace % automatically resize the bar with \right
+  #1 % the function
+  \littletaller % pretend it's a little taller at normal size
+  \right|_{#2} % this is the delimiter
+  }}
 
 \makeatletter
-\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
+\renewcommand*{\env@matrix}[1][*\c@MaxMatrixCols c]{%
   \hskip -\arraycolsep
   \let\@ifnextchar\new@ifnextchar
   \array{#1}}

MA1511/decls.tex

@@ -9,6 +9,7 @@
 \author{}
 
 \newcommand{\paren}[1]{\left(#1\right)}
+\newcommand{\powerset}[1]{\mathcal{P}\paren{#1}}
 \renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
 \renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
 \newcommand{\C}{\mathbb{C}}
@@ -22,9 +23,16 @@
 \newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
 \newcommand{\adj}[1]{\operatorname{adj}#1}
 \newcommand{\card}[1]{\left|#1\right|}
+\newcommand{\littletaller}{\mathchoice{\vphantom{\big|}}{}{}{}}
+\newcommand{\restr}[2]{{% we make the whole thing an ordinary symbol
+  \left.\kern-\nulldelimiterspace % automatically resize the bar with \right
+  #1 % the function
+  \littletaller % pretend it's a little taller at normal size
+  \right|_{#2} % this is the delimiter
+  }}
 
 \makeatletter
-\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
+\renewcommand*{\env@matrix}[1][*\c@MaxMatrixCols c]{%
   \hskip -\arraycolsep
   \let\@ifnextchar\new@ifnextchar
   \array{#1}}

MA1511/sets.tex

@@ -29,18 +29,18 @@ A family of sets indexed by a set $I$ (the indexing set): $A_i ~~\forall~i\in I
 $A_i$ is a set for every element $i \in I$ \\
 A family of sets indexed by $\N$ is called a sequence of sets. Also written $(B_i)^{\inf}_{i=0}$ or $(B_i)_{i \geq 0}$
 \begin{align*}
-\bigcup_{i \in I}~A_i \equiv \{x | \exists i \in I. x \in A_i \} \\
-\bigcap_{i \in I}~A_i \equiv \{x | \forall i \in I. x \in A_i \} & \text{ Exists iff } \exists~i\in I
+	\bigcup_{i \in I}~A_i \equiv \{x | \exists i \in I. x \in A_i \}                                      \\
+	\bigcap_{i \in I}~A_i \equiv \{x | \forall i \in I. x \in A_i \} & \text{ Exists iff } \exists~i\in I
 \end{align*}
 \begin{align*}
-	& \forall i \in I. A_i \subseteq \cup_{j \in I}A_j \\
-	& \forall i \in I. A_i \subseteq B \implies \cup_{j \in I}A_j \subseteq B \\
-	& \forall i \in I.\cap_{j \in I}A_j \subseteq A_i  \\
-	& \forall i \in I. B \subseteq A_i \implies B \subseteq \cap_{j \in I}A_j \\ \\
-	& B \cup \cap_{i \in I}A_i = \cup_{i \in I}(B \cap A_i) \\
-	& B \cap \cup_{i \in I}A_i = \cap_{i \in I}(B \cup A_i) \\
-	& B \setminus \cup_{i \in I}A_i = \cap_{i\in I}(B\setminus A_i) \\
-	& B \setminus \cap_{i \in I}A_i = \cup_{i\in I}(B \setminus A_i)
+	 & \forall i \in I. A_i \subseteq \cup_{j \in I}A_j                        \\
+	 & \forall i \in I. A_i \subseteq B \implies \cup_{j \in I}A_j \subseteq B \\
+	 & \forall i \in I.\cap_{j \in I}A_j \subseteq A_i                         \\
+	 & \forall i \in I. B \subseteq A_i \implies B \subseteq \cap_{j \in I}A_j \\ \\
+	 & B \cup \cap_{i \in I}A_i = \cup_{i \in I}(B \cap A_i)                   \\
+	 & B \cap \cup_{i \in I}A_i = \cap_{i \in I}(B \cup A_i)                   \\
+	 & B \setminus \cup_{i \in I}A_i = \cap_{i\in I}(B\setminus A_i)           \\
+	 & B \setminus \cap_{i \in I}A_i = \cup_{i\in I}(B \setminus A_i)
 \end{align*}
 \section*{Cartesian Products}
 Ordered pairs can be represented as $(x, y) \equiv \{x, \{x, y\}\}$ \\
@@ -51,9 +51,312 @@ $\card{X\times Y} = \card{X} \times \card{Y}$ (for finite $X$, $Y$)
 For sets $X$, $Y$:
 \begin{description}
 	\item[A function $F: X \to Y$] $\subseteq X\times Y$ \text { where } \\
-		$\forall x \in X.~\exists \text{ a unique } y \in Y.~ (x, y) \in F$ \\
-		$F(x)$ denotes the unique element $y \in Y$ for which $(x, F(x)) \in F$
+		$\forall x \in X.~\exists \text{ a unique } F(x) \in Y.~ (x, F(x)) \in F$ \\
+		There exist $\card{Y}^{\card{X}}$ functions $F: X \to Y$
 	\item[$\operatorname{dom}(F)$] The domain of $F$, i.e. $X$
-	\item[$\operatorname{incl}^{X}_{A} : A \to X$] $= a \quad \forall A X. \text{ where } A \subseteq X$
+	\item[$\operatorname{incl}^X_A : A \to X$] $= a \quad \forall A,X. \text{ where } A \subseteq X$
+	\item[Constant function] $\exists y_0 \in Y.~\forall x \in X.~ f(x) = y_0$
+	\item[Characteristic Function of a set $A \subseteq X$: $\chi_A: X \to \{0, 1\}$]
+		\[
+			\chi_A: X \to \{0, 1\} = \left\{\begin{array}{lr}
+				0 & \text{ if } x \not\in A \\
+				1 & \text{ if } x \in A
+			\end{array} \right.
+		\]
+	\item[Restriction of a function $f: X \to Y$] $\restr{f}{A}$ is $f$ specialized contravariantly to $A \subseteq X$
+	\item[$f(A)$: Image of $A$ under $f$] $f$ mapped over $A$ \quad for function $f: X \to Y$, $A \subseteq X$
+	\item[$\operatorname{ran}(f)$ / image of $f$ / range of $f$] $\{ f(x) | x \in X \}$, $f(X)$, i.e. all possible values of $f(x)$
+	\item[Preimage of $B$ under $f$] $\{ x \in X ~|~ f(x) \in B \}$ \\
+		written $f^{-1}(B)$, but is \emph{not} the inverse of f
 \end{description}
+For $f: X \to Y$, $A \subseteq A' \subseteq X$, $B \subseteq B' \subseteq Y$:
+\begin{align*}
+	f(A) \subseteq~          & f(A')      \\
+	f^{-1}(B) \subseteq ~    & f^{-1}(B') \\
+	f^{-1}(f(A)) \supseteq ~ & A          \\
+	f(f^{-1}(B)) \subseteq ~ & B
+\end{align*}
+For set families $(A_i \subseteq X)_{i \in I}, (B_j \subseteq Y)_{j \in J}$:
+\begin{align*}
+	f(\cup_{i \in I} A_i) = ~         & \cup_{i \in I}f(A_i)       \\
+	f(\cap_{i \in I} A_i) \subseteq ~ & \cap_{i \in I}f(A_i)       \\
+	f^{-1}(\cup_{j \in J} B_j) = ~    & \cup_{j \in J} f^{-1}(B_j) \\
+	f^{-1}(\cap_{j \in J} B_j) = ~    & \cap_{j \in J} f^{-1}(B_j)
+\end{align*}
+\section*{Function Composition}
+For functions $f: X \to Y$, $g: Y \to Z$:
+\begin{align*}
+	 & (g \circ f): X \to Z                      \\
+	 & (g\circ f)(x) = g(f(x)) & \forall x \in X \\ \\
+\end{align*}
+\section*{Surjection and Injection}
+For $f: X \to Y$, $f$ is
+\begin{description}
+	\item[surjective] iff $f(X) = Y$, i.e. $\forall y \in Y.~ \exists x \in X.~ f(x) = y$ \\
+		Range is codomain, 'onto'
+	\item[injective] iff $\forall x, x' \in X. ~ f(x) = f(x') \implies x = x'$
+	\item[bijective] iff $f$ is injective and $f$ is surjective \\
+		one-to-one
+\end{description}
+Given $f: X \to Y$, $g: Y \to Z$: \\
+\quad If f and g are injective, so is $g\circ f$ \\
+\quad If f and g are surjective, so is $g \circ f$ \\
+\quad If $g \circ f$ is injective, so is $f$ \\
+\quad If $g \circ f$ is surjective, so is $g$
+\section*{Inverse Functions}
+The inverse function $f^{-1}$ of $f: X \to Y$ exists iff $f$ is bijective, and is defined by \[f^{-1}(y) = x \text{ where } \exists! x \in X. f(x) = y \qquad \forall y \in Y\]
+\begin{align*}
+	(f \circ f^{-1}) = \operatorname{Id} \\
+	(f^{-1} \circ f) = \operatorname{Id}
+\end{align*}
+\section*{Power Sets}
+Powerset of S: $\powerset{S}$ has $2^{\card{S}}$ elements is the set of all subsets of $S$ \\
+$Fun(X, \{0, 1\})$ is the set of functions $X \to \{0, 1\}$ \\
+$\Phi: Fun(X, \{0, 1\}) \to \powerset{X}$ \\
+$\Phi(f) = \{ x \in X | f(x) = 1 \}$ \\
+$\Phi$ is bijective.
+\section*{Binary Operators}
+A binary operator is $X^2 \to X$ \\
+Union is $\powerset{X}^2 \to \powerset{X}$ \\
+$\square$ is the unknown or indeterminate binop sigil \\
+Unital: $\forall x \in X. \exists u \in X. u \square x = x \square u = x$
+\section*{Construction of the Natural Numbers}
+\begin{align*}
+	x^{+} & = x \cup \{x\} & \text{successor of $x$}
+\end{align*}
+A set $X$ is \emph{inductive} if $\varnothing \in X \land (a \in X \implies a^+ \in X)$ \\
+Axiom: There exists an inductive set. \\
+Definition: A natural number is a set that is an element of all inductive sets. \\
+Theorem: There exists a set whose elements are the natural numbers. \\
+\begin{align*}
+	 & \text{Given an inductive set $A$}                                                       \\
+	 & \omega = \{ x \in A | x \text{ is a natural number}\}                                   \\
+	 & \text{any natural number is in $A$ (since $A$ is inductive), and therefore in $\omega$} \\
+	 & \omega \text{ is the natural numbers}
+\end{align*}
+\\
+Define:
+\begin{align*}
+	0 & = \varnothing \\
+	1 & = 0^+         \\
+	2 & = 1^+         \\
+	\ldots
+\end{align*}
+$\in$ is an ordering over $\omega$, as is $\subseteq$ \\
+$\omega$ is an inductive set. Proof: \\
+$\varnothing$ is a natural number.
+$x \in \omega$ implies $x^+ \in \omega$, as $x$ is natural and is therefore in
+every inductive set, and so $x^+$ is in every inductive set and is therefore a
+natural number.
+\subsection*{Principle of Induction}
+If $A \subseteq \omega$ and $A$ is inductive, $A = \omega$.
+Since $A$ is inductive, it contains every natural number, so $\omega \subseteq A$. \\
+Since $A \subseteq \omega \land \omega \subseteq A$, $A = \omega$
+\section*{Recursion on $\omega$}
+\subsection*{Principle of Recursion}
+For a set $X$, $x_0 \in X$, $h: X \to X$, there exists a unique function $f: \omega \to X$ where $f(0) = x_0$, $f(n^+) = h(f(n))$.
+\section*{Relations}
+Total order: $(X, \prec)$, requires $\forall x, y, z \in X$: \\
+\begin{align*}
+	 & x \prec y \land y \prec z \implies x \prec z                                         \\
+	 & x = y \oplus x \prec y \oplus y \prec x      & \text{ (where $\oplus$ denotes XOR) }
+\end{align*}
+Lexicographic order $x <_L y$ on $\N\times\N = x_0 < y_0 \lor (x_0 = y_0 \land x_1 < y_1)$ \\
+For a non-empty subset $A$ of $X$ given a total order $(X, \prec)$, a minimum/least element $a_0 \in A$
+exists where $\forall a \in A.~a_0 \preceq a$.
+\section*{Ordering on $\omega$}
+$(\omega, \in)$ and $(\omega, \subset)$ are both total orderings on $\omega$, such that $\varnothing$ is the minimum and $\forall x \in \omega.~x < x^+$
+\section*{Strong Induction}
+Every non-empty subset of $\N$ has a minimum. \\ \\
+Let $\phi(x)$ be a predicate over $\N$ where $\forall n\in \N.~ (\forall m \in \N.~ m < n \implies \phi(m)) \implies \phi(n)$. \\
+Then $\phi(0)$ holds, as $\neg\exists n \in \N.~n<0$, $\phi(1)$ holds as $\phi(0)$ holds, etc.
+\begin{align*}
+	 & \neg\exists x \in \N.~\neg\phi(x)                                                                                                   \\
+	 & \text{Let } A \text{ be a subset of }N\text{ such that }\phi(n)\text{ is false }\forall n \in A                                     \\
+	 & \text{If not, $\exists a_0 \in A$}                                                                                                  \\
+	 & \forall n \in N.~n < a_0 \implies n \not\in A \implies \phi(n)                                                                      \\
+	 & \text{Then $\phi(n)$ holds $\forall n < a_0$, so $\phi(a_0)$, then $a_0 \not\in A$, which is a contradiction, so $A = \varnothing$}
+\end{align*}
+\section*{Fibonacci}
+Fibonacci Sequence: $F_n$ \\
+Roots of $x^2 - x - 1$ are $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$ \\
+$F_n = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}}$ \\
+For $n=0$:
+
+$F_0 = 1 = \frac{\frac{1 + \sqrt{5}}{2} - \frac{1-\sqrt{5}}{2}} = \frac{2\sqrt{5}}{2} = 1$ \\
+
+For $n=1$: \\
+
+$F_1 = 1 = \frac{\paren{\frac{1 + \sqrt{5}}{2}}^2 - \paren{\frac{1 - \sqrt{5}}{2}}^2}{\sqrt{5}} = \frac{\phi + 1 - \psi - 1}{\sqrt{5}}$ \\
+
+For $n\geq 2$:
+\begin{align*}
+	 & F_n = F_{n-1} + F_{n-2} = \frac{\phi^n - \psi^n}{\sqrt{5}} + \frac{\phi^{n-1} - \psi^{n-1}}{\sqrt{5}}                      \\
+	 & = \frac{\phi^{n-1}\paren{\phi + 1} - \psi^{n-1}\paren{\psi + 1}}{\sqrt{5}}                                                 \\
+	 & = \frac{\phi^{n-1}\phi^2 - \psi^{n-1}\psi^2}{\sqrt{5}}                                                                     \\
+	 & = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}}                                                            & \text{as required}
+\end{align*}
+
+\subsection*{Zeckendorff's Theorem}
+Every natural number can be written as a sum of non-adjacent Fibonacci numbers in a unique way (excluding $F_0$). \\
+A finite subset $I$ of $\N^+$ is \emph{Zeckendorff} if it contains no adjacent elements ($\forall x \in I.~x^+ \not\in I$) \\
+Define $\mathcal{Z}$ as the set of all Zeckendorff sets, and $\sigma: \mathcal{Z} \to \N^+$ by $\sigma(I) = \sum_{i \in I}F_i$.
+The theorem claims $\sigma$ is bijective. \\ \\
+For nonempty $I \in \mathcal{Z}$ with largest element $k$:
+\begin{align*}
+	 & \text{Let } J = I \setminus \{k\}                                                  \\
+	 & \sigma(I) = F_k + \sigma(J) \geq F_k                                               \\
+	 & \text{If } J = \varnothing\text{,  } \sigma(I) = F_k \leq F_{k + 1}                \\
+	 & \text{Otherwise, we must show } \sigma(I) < F_{k + 1}                              \\
+	 & \equiv F_k + \sigma(J) < F_{k + 1}                                                 \\
+	 & \equiv \sigma(J) < F_{k + 1} - F_k = F_{k-1}                                       \\
+	 & \text{But if $k' = \operatorname{max}(J)$, }                                       \\
+	 & \sigma(J) < F_{k' + 1} \land k' < k - 2 \text{  (since $I$ is \emph{Zeckendorff})} \\
+	 & \sigma(J) < F_{k - 1} \text{  as required.}
+\end{align*}
+
+Proof of theorem: $\forall n \in N.~ \sigma$ is bijective.
+\begin{align*}
+	n = 0: \quad & I = \varnothing                                  \\
+	n > 0: \quad & \text{Let } F_k \leq n < F_{k + 1},~ m = n - F_k \\
+	             & m = \sigma(J) \text{ for some } J                \\
+	             & \text{If } J = \varnothing, I = \{k\}.           \\
+	             & \text{Otherwise, } k' = \operatorname{max}(J).   \\
+	             & \text{If } k' \leq k - 2, I = J \cup \{k\}       \\
+\end{align*}
+\section*{Equivalence Relations}
+Reflexive, Symmetric, Transitive \\
+Equivalence relations are usually called $\sim$ \\
+In a set $X$,
+\[ [x] = \{ y \in X | x \sim y \} \]
+\[ [x] = [y] \equiv x \sim y \]
+Any two equivalence classes of $X$ are either disjoint or equal. \\
+\subsection*{Quotient Sets}
+$X/\sim~ = \{[x] | x \in X \} \subset \powerset{X}$ \\
+A complete set of representatives is a subset $A$ of $X$ where $\forall x \in X. \exists! a \in A. a \in [x].$ \\
+I.E. a complete set of representatives contains exactly one element from each element of $X/\sim$ \\
+$f: A \to X/\sim$ defined by $f(a) = [a]$
+\\\\\\
+A function $f: X \to Y$ is \emph{compatible} iff $x \sim y \implies f(x) = f(y)~\forall x, y \in X$ \\
+For a \emph{compatible} function, $\Bar{f}: X/\sim~\to Y$ exists and is defined by $\Bar{f}([x]) = f(x)$
+\subsection*{Integers modulo $k$}
+Fix some $k \in \N^+$ \\
+Define $\sim$ on $\Z$ by $n \sim m \iff n - m$ is a multiple of $k$ \\
+$\sim$ is an equivalence relation \\
+$[0, k)\cap\N$ is a complete set of representatives for $\sim$ \\
+$Z/k$ is the set of integers modulo $k$, $n \equiv m~(\operatorname{mod} k) \iff n \sim m$ \\
+\([m] = [m]_k\) \\
+$+$ and $\times$ on $\Z/k$:
+\begin{align*}
+	[n] + [m] = [n + m] \\
+	[n][m] = [nm]
+\end{align*}
+\section*{Countable Sets}
+A set $X$ is finite if $\exists n \geq 0. $ a bijection $\{1, ...n\} \to X$ \\
+Pigeonhole Principle: for finite $X$, any injective $f: X \to X$ is also surjective. \\
+$\N$ is infinite. Proof: $f: \N \to \N $ defined by $ f(x) = x + 1$ is trivially injective, and $\neg\exists x.~f(x) = 0$, and so not surjective. \\
+By the inverse of the Pigeonhole Principle, $\N$ is infinite.\\
+A set $X$ is \emph{countably infinite} iff there exists a bijection $\N \to X$. \\
+A set is \emph{countable} iff it is finite or countably infinite. \\
+Any subset of $\N$ is countable. Proof: Let $X \subseteq \N$.\\
+If $X$ is finite, it's trivially countable.
+Otherwise, $X$ is infinite and it must be shown that $X$ is countably infinite. \\
+For $k \in \N$, $X_{>k} = \{ n \in X | n > k \}$. Then $X_{>k} \not= \varnothing$, as $X$ would be a subset of $\{1..k\}$ and would be finite. \\
+Then $min(X_{>k})$ exists, and $h: X \to X$ can be defined by $h(x) = min(X_{>x})$, and $f: \N \to X$ by recursion on $h$ with $f(0) = min(X)$. \\
+\\
+If an injection $f: A \to X$ exists, $A$ is countable if $X$ is. Proof: \\
+If $A$ is finite, it's countable. Otherwise, \\
+Since $X$ is countable, there exists a bijection $g: X \to \N$, and $g \circ f: A \to \N$
+exists and is a composite of 2 injective functions, and therefore is itsself injective. \\
+\\
+Any subset of a countable set is countable, by above with the inclusion function.
+
+$N^2$ is countable. Proof: \\
+Take $f: \N^2 \to \N$ defined by $f(a, b) = 2^a3^b$. \\
+$f$ is injective. Proof is simple -- prime factor decompositions are unique. \\
+
+If a function $f: X \to Y$ exists where $X$ is countable, $f(X)$ is countable. Proof: \\
+For $y \in f(X)$, \emph{choose} an $x_y \in f^{-1}(\{y\})$ and define $g: f(X) \to X$ by $g(y) = x_y$. \\
+$g$ is injective, so $f(X)$ injects into the countable set $X$ and is itself countable. \\
+In particular, for any surjection $f: X \to Y$, $Y$ is countable if $X$ is \\
+\\
+The union over a countable set of countable sets is countable. Proof: For family $X_{i \in I}$, $X_i$ countable, $I$ countable \\
+There is an injection $h: I \to \N$, and $f_i: X_i \to \N \forall i \in I$, as these are countable sets. \\
+Define $Y = \bigcup_{i \in I}X_i$, and $g: Y \to \N$ by $g(y) = (h(i), f_i(y))$ where $i \in I$ is chosen so that $y \in X_i$.
+Then $g$ is injective because equality distributes over pairs, and $h$ and $f_i$ are injective.\\
+
+If $X$ and $Y$ are countable, so is $X\times Y$. Proof: \\
+Define for $x \in X$ a subset $Y_x = \{(x, y) | y \in Y\}$ of $X \times Y$, then $\{Y_x\}_{x \in X}$ is a countable family of countable sets.
+\\
+$\Z$ is countable, as a union of $\N$ and $(\times {-1})(\N^+)$, or $\N \times \{0, 1\}$ \\
+
+$\Q$ is countable. Proof:
+\begin{align*}
+	 & \text{Define } f: \Z\times (\Z \setminus \{ 0 \}) \text{ by } f(a, b) = \frac{a}{b}      \\
+	 & f \text{ is surjective, by definition of } \Q.                                           \\
+	 & \Z \text{ is countable, as above }                                                       \\
+	 & \Z \setminus \{ 0 \} \text{ is countable, as a subset of a countable set } \Z            \\
+	 & \Z\times (\Z \setminus \{ 0 \}) \text{ is countable, as a product of countable sets }    \\
+	 & \text{Since } f \text{ is surjective with a countable domain, } \Q \text{ is countable }
+\end{align*}
+
+For set family $X_{n \in \N^+}$, $\times_{n \in \N^+} X_n$ is countable if $\forall n \in \N.~X_n$ is countable. Proof: \\
+Base Case: $n = 1$: $X_1$ is countable, since $X_1$ is countable. \\
+Induction: Assume $\times_{n \in \N^+, \leq k} X_n$ is countable. \\
+Then the result for $k + 1$ is $\times_{n \in \N^+, \leq k + 1} X_n$, which is $(\times_{n \in \N^+, \leq k} X_n) \times X_{k + 1}$, which is the product of countable sets and is therefore countable. \\
+By induction, the result holds for $n \in N^+$ \\
+This generalizes to all countable indexing sets $I$, by constructing an injection $f: I \to \N$.
+\\ \\
+Let $X$ be countable. Define $\mathbb{P}_{<\infty}\paren{X}$ as the set of all finite subsets of $X$. $\mathbb{P}_{<\infty}\paren{X}$ is countable. Proof: \\
+For $n \in \N$, let $\mathbb{P}_{\leq n}(X)$ denote the set of all nonempty subsets of $X$ with $n$ elements or less. \\
+The function $p_n: X^n \to \mathbb{P}_{\leq n}(X)$ defined by $p_n(x_1, x_2, ... x_n) = \{ x_1, x_2, ... x_n\}$ is surjective. \\
+Then, since $X$ is countable, so is $\mathbb{P}_{\leq n}(X)$ (as $X^n$ is countable and there exists a surjection $X^n \to \mathbb{P}_{\leq n}(X)$) \\
+\[\mathbb{P}_{\leq\infty}(X) = \bigcup_{n \in \N^+}\mathbb{P}_{\leq n}(X) \cup \{\varnothing\} \]
+This is a countably-sized union of countable sets and so is itself countable, as required.
+
+\subsection*{Cantor's Theorem}
+Let $X$ be a set. Then there exists no surjective function $f: X \to \powerset{X}$. Proof: \\
+Let $f: X \to \powerset{X}$ be a function. We will prove that $f$ is not surjective. \\
+Define $D \subseteq X$ by $\{ x \in X | x \not\in f(x) \}$ \\
+Then $D \in \powerset{X}$, and we will show that there is no element of $X$ for which $f(x) = D$. \\
+Suppose there was such an $x$. Either: \\
+
+$x \in D$. Then $x \in f(x)$, but by definition of $D$, $x \not\in f(x)$, which is a contradiction.
+
+$x \not\in D$. Then $x \not\in f(x)$, so $x \in D$ by definition of $D$, which is a contradiction. \\\\
+Since both cases give a contradiction, there exists no such $x$, and $f$ is not surjective. \\
+This is the less famous diagonal argument. \\
+\\
+$\R$ is uncountable. Proof:
+
+Define $f: \powerset{\N} \to \R$ by $f(A) = \sum_{n \in A} 2(3^{-n}) \quad \forall A \subseteq \N$.
+Then $f$ is injective: \\
+
+Take $\alpha, \beta \in \powerset{\N}$, where $\alpha \not = \beta$, and we will show $f(\alpha) \not = f(\beta)$. \\
+Then, take $k \in \N$ as the smallest natural number in exactly one of $\alpha$, $\beta$, and assume it's in $\beta$ WLOG. \\
+Then
+\begin{align*}
+	f(\alpha) & = \sum_{n \in \alpha}\frac{2}{3^n}                                                              \\
+	          & = \sum_{n \in \alpha, < k} \frac{2}{3^n} + \sum_{n \in \alpha, > k} \frac{2}{3^n}               \\
+	          & \leq \sum_{n \in \alpha, < k} \frac{2}{3^n} + \sum_{n > k} \frac{2}{3^n}                        \\
+	          & = \sum_{n \in \alpha, < k} \frac{2}{3^n} + \frac{1}{3^k}                                        \\
+	          & = \sum_{n \in \beta, < k} \frac{2}{3^n} + \frac{2}{3^k}                                         \\
+	          & < \sum_{n \in \beta, < k} \frac{2}{3^n} + \frac{2}{3^k} + \sum_{n \in \beta, > k} \frac{2}{3^n} \\
+	          & = f(\beta) \\ \\
+	f(\alpha) < f(\beta) & \implies f(\alpha) \neq f(\beta)
+\end{align*}
+Then there is no surjection $\N \to \powerset{\N}$ and so $\powerset{\N}$ is uncountable.
+Since there's an injection $\powerset{\N} \to \R$, $\R$ is uncountable, as if $\R$ was countable, $\powerset{\N}$ would be countable
+\\ \\
+The set of all polynomials with rational coefficients is countable. Proof:
+
+	Let $P$ be the set of all polynomials with rational coefficients, $P_n$ be the set of all polynomials with rational coefficients and degree $\leq$ n.
+
+	Then $P = \bigcup_{n \in \N} P_n$, $\N$ is countable, and $P_n$ is countable as there exists a surjection $\Q^n \to P_n$ by assigning each element of the tuple to a coefficient. \\
+
+\subsection*{Algebraic Numbers}
+The algebraic numbers ($\mathcal{A}$) are the real numbers which are roots of polynomials with rational coefficients.
+
+$\mathcal{A}$ is countable, as \[ \mathcal{A} = \bigcup_{p \in P} \{ x \in \R |~p(x) = 0 \} \] is a countable union of finite sets. \\
+Then $\R\setminus\mathcal{A}$ is uncountable, as if it were, $\R = (\R \setminus \mathcal{A}) \cup \mathcal{A}$, a union of countable sets.
 \end{document}

MA2008/2024-09-24

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+

MA2008/decls.tex

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+\documentclass[fleqn]{article}
+\usepackage{amsmath,amssymb,amsthm}
+\usepackage[margin=0.25in]{geometry}
+\usepackage{enumitem}
+\usepackage{systeme}
+\usepackage{mathtools}
+
+\date{}
+\author{}
+
+\newcommand{\paren}[1]{\left(#1\right)}
+\newcommand{\powerset}[1]{\mathcal{P}\paren{#1}}
+\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
+\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\conj}[1]{\overline{#1}}
+\renewcommand{\mod}[1]{\left|#1\right|}
+\newcommand{\abs}[1]{\left|#1\right|}
+\newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
+\newcommand{\adj}[1]{\operatorname{adj}#1}
+\newcommand{\card}[1]{\left|#1\right|}
+\newcommand{\littletaller}{\mathchoice{\vphantom{\big|}}{}{}{}}
+\newcommand{\restr}[2]{{% we make the whole thing an ordinary symbol
+  \left.\kern-\nulldelimiterspace % automatically resize the bar with \right
+  #1 % the function
+  \littletaller % pretend it's a little taller at normal size
+  \right|_{#2} % this is the delimiter
+  }}
+
+\makeatletter
+\renewcommand*{\env@matrix}[1][*\c@MaxMatrixCols c]{%
+  \hskip -\arraycolsep
+  \let\@ifnextchar\new@ifnextchar
+  \array{#1}}
+\makeatother
+
+
+% https://gitlab.com/jim.hefferon/linear-algebra/-/blob/master/src/sty/linalgjh.sty
+\newlength{\grsteplength}
+\setlength{\grsteplength}{1.5ex plus .1ex minus .1ex}
+
+\newcommand{\grstep}[2][\relax]{%
+   \ensuremath{\mathrel{
+       \hspace{\grsteplength}\mathop{\longrightarrow}\limits^{#2\mathstrut}_{
+                                     \begin{subarray}{l} #1 \end{subarray}}\hspace{\grsteplength}}}}
+\newcommand{\repeatedgrstep}[2][\relax]{\hspace{-\grsteplength}\grstep[#1]{#2}}
+
+\newcommand{\swap}{\leftrightarrow}
+
+% https://tex.stackexchange.com/a/198806
+\makeatletter
+\newcommand{\subalign}[1]{%
+  \vcenter{%
+    \Let@ \restore@math@cr \default@tag
+    \baselineskip\fontdimen10 \scriptfont\tw@
+    \advance\baselineskip\fontdimen12 \scriptfont\tw@
+    \lineskip\thr@@\fontdimen8 \scriptfont\thr@@
+    \lineskiplimit\lineskip
+    \ialign{\hfil$\m@th\scriptstyle##$&$\m@th\scriptstyle{}##$\hfil\crcr
+      #1\crcr
+    }%
+  }%
+}
+\makeatother
+
+\theoremstyle{definition}
+\newtheorem*{theorem}{Theorem}
+\newtheorem*{lemma}{Lemma}
+\newtheorem*{corollary}{Corollary}
+
+\theoremstyle{remark}
+\newtheorem*{note}{Note}

MA2008/linear-transforms.tex

@@ -0,0 +1,27 @@
+\input{decls.tex}
+\title{Vector Spaces and Linear Transformations}
+\begin{document}
+\maketitle
+
+\begin{description}
+\item[Linear Transformation] A function $\phi: V \to W$ between vector spaces $V$ and $W$ (over some field $K$), such that
+	\begin{align*}
+		\phi(v + w) & \equiv \phi(v) + \phi(w) \\
+		\phi(x \cdot v) & \equiv x \cdot \phi(v) \tag{For $x \in K$}
+	\end{align*}
+\end{description}
+
+Differentiation is a linear transformation.
+Solutions to $f'' + f = 0$ for function $f$ are a vector space.
+
+\begin{theorem}
+For any scalars $\lambda, \mu \in \R$, there is a unique solution such that $f(0) = \mu$ and $f'(0) = \lambda$
+\end{theorem}
+The vector space is then two-dimensional, with basis $sin(x), cos(x)$
+
+
+\subsection*{}
+Vector spaces are used over finite fields in \emph{Algebraic Coding Theory}. The field is $\mathbb{F}_2 = \{0, 1\}$ - the integers mod 2.
+Binary strings of length $n$ are then a vector space over $\mathbb{F}_2^n$.
+ECC can be based on vector subspaces of $F_2^n$. (Vector subspaces are closed subsets of a vector space).
+\end{document}

MA2008/tmpl.tex

@@ -0,0 +1,6 @@
+\input{decls.tex}
+\title{}
+\begin{document}
+\maketitle
+
+\end{document}