auto-update(nvim): 2025-02-19 04:32:27

documents/by-course/math-8/pset-6/main.typ

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+#import "@youwen/zen:0.1.0": *
+#import "@preview/mitex:0.2.5": *
+
+#show: zen.with(
+  title: "Homework 6",
+  author: "Youwen Wu",
+)
+
+#set heading(numbering: none)
+#show heading.where(level: 2): it => [#it.body.]
+#show heading.where(level: 3): it => [#it.body.]
+
+#set par(first-line-indent: 0pt, spacing: 1em)
+
+
+Problems:
+
+3.1: \#2, 3bd, 5ab, 6b, 7ae, 8ef, 11a
+
+3.2: \#1dfg, 6bcdi, 7
+
+3.3: \#2ace, 3a, 7ac, 9d
+
+#outline()
+
+= 3.1
+
+== 2
+
+=== a
+
+$"Dom"(T) = {3,2,1}$
+
+=== b
+
+$"Rng"(T) = {1,2,3,5,6}$
+
+=== c
+
+$T^(-1) = {(1,3), (3,2), (5,3), (2,2), (6,1), (6,2), (2,1)}$
+
+=== d
+
+$T = {(3,1), (2,3), (3,5), (2,2), (1,6), (2,6), (1,2)}$
+
+== 3
+
+=== b
+
+$
+  "Dom"(W) = RR \
+  "Rng"(W) = RR
+$
+
+=== d
+
+$
+  "Dom"(W) = (-infinity,0)union(0,infinity) \
+  "Rng"(W) = (-infinity,0)union(0,infinity)
+$
+
+== 5
+
+=== a
+
+Consider all $x in RR$. Then there is a $y in RR$ such that $x R y$, namely $y
+= 6x$. Now consider all $y in RR$. Then there is an $x in RR$ such that $x R
+y$, namely $x = y/6$.
+
+=== b
+
+Consider all $x in RR$. Then $x^2$ is either positive or zero. So all $y$ such
+that $x R y$ are in $[0,infinity)$. So the range is $[0,infinity)$. Now
+consider all $y in [0,infinity)$ such that $x R y$. Then we can always find an
+$x in RR$ such that $x = sqrt(y)$ as $y$ is not negative. So the domain is $RR$.
+
+== 6
+
+=== b
+
+$
+  R_2^(-1) = {(x,y) in RR times RR : y = -1 / 5 (x-2)}
+$
+
+== 7
+
+=== a
+
+$
+  R compose S = {(3,5),(5,2)}
+$
+
+=== e
+
+$
+  S compose R = {(1,5),(2,4),(5,4)}
+$
+
+== 8
+
+=== e
+
+$
+  R_2 compose R_4 &= {(x,y) in RR times RR : y = -5(x^2 + 2) + 2} \
+  &= {(x,y) in RR times RR : y = -5x^2 - 8}
+$
+
+=== f
+
+$
+  R_4 compose R_2 &= {(x,y) in RR times RR : (-5x+2)^2 + 2} \
+  &= {(x,y) in RR times RR : 25x^2 - 10x + 6}
+$
+
+== 11
+
+=== a
+
+The domain of $R$ is simply all of the first coordinates and the range of
+$R^(-1)$ is all of the second coordinates of $R^(-1)$. But by the definition of
+$R^(-1)$ all of its second coordinates are the first coordinates of $R$.
+
+More formally, $a in "Dom"(R)$ iff. there exists $a in A$ such that $(a,b) in
+R$ iff. there exists a $(b,a) in R^(-1)$ for every $(a,b) in R$ iff. $a in
+"Rng"(R^(-1))$. Since $a in "Dom"(R) <=> a in "Rng"(R^(-1))$ they are equal.
+
+= 3.2
+
+== 1
+
+=== d
+
+Transitive, but not reflexive or symmetric.
+
+=== f
+
+Symmetric only.
+
+=== g
+
+Transitive and reflexive but not symmetric.
+
+== 6
+
+=== b
+
+For natural numbers $m$ and $m$, both have the same digit in the tens place. So
+$R$ is reflexive. If natural numbers $m$ and $n$ have the same digit in the
+tens place, then $n$ and $m$ have the same digit in the tens place. So $R$ is
+symmetric. Finally, if naturals $n$ and $m$ have the same digit in the tens
+place, and naturals $m$ and $p$ have the same digit in the tens place, than $n$
+and $p$ have the same digit in the tens place. So the relation $R$ is
+reflexive, symmetric, and transitive, so it's an equivalence relation on $NN$.
+
+An element of $overline(106)$ less than 50 is 05. Between 150 and 300 is 250.
+Greater than 1000 is 2000. Three such elements in the equivalence class
+$overline(635)$ are 35, 235, and 1035.
+
+=== c
+
+For $x in RR$, x = x. So $V$ is reflexive. For $x,y in RR$, if $x = y$, then $y
+= x$, or if $x y = 1$, then $y x = 1$ by the commutativity of multiplication.
+So $V$ is symmetric. For $x,y,z in RR$, if $x = y$ and $y = z$, then $x = z$.
+
+If $x y = 1$ and $y z = 1$, then either
+
+1. $y = 1$ which implies $x = 1$ and $z = 1$ so $x z = 1$ and $x V z$.
+2. $y != 1$ and $x = 1/y$ and $z = 1/y$ and we have $x =
+  z$ so $x V z$.
+
+So the relation is reflexive, symmetric, and transitive, and it's an
+equivalence relation.
+
+- the equivalence class of 3, $overline(3)$ is ${3, 1/3}$
+- the equivalence class of $-2/3$, $overline(-2/3)$ is ${-2/3, -3/2}$
+- the equivalence class of $0$, $overline(0)$, is just ${0}$
+
+=== d
+
+For any $a$, $a$ has a unique prime factorization so $a R a$. So $R$ is
+reflexive. If $a R b$, then $b R a$ since $a$ and $b$ have unique prime
+factorizations and have the same amount of twos in their unique prime factors.
+So $R$ is symmetric. If $a R b$, and $b R c$, then the unique prime
+factorization of $a$ has the same amount of 2s as the prime factorization of
+$b$ which has the same amount of 2s as the prime factorization of $c$. So $a$
+and $c$ have the same amount of 2s in their prime factorization and $a R c$. So $R$ is transitive. Therefore $R$ is an equivalence relation.
+
+- In $overline(7)$, we have no 2s in the prime factorization, ${3,5,9} subset overline(7)$
+- In $overline(10)$, there is one 2, so ${4,6,14} subset overline(10)$
+- In $overline(72)$, there are three 2s, so ${8, 24, 40} subset overline(72)$
+
+=== i
+
+For any $x in RR$, $x T x$ iff. $sin(x) = sin(x)$ which is true, so $T$ is
+reflexive. For $x, y in RR$, $x T y$ iff. $sin(x) = sin(y)$ iff. $sin(y) =
+sin(x)$ iff. $y T x$. So $T$ is symmetric. For $x,y,z in RR$, $x T y$ and $y T
+z$ iff. $sin(x) = sin(y)$ and $sin(y) = sin(z)$ iff. $sin(x) = sin(z)$ iff. $x
+T z$ so $T$ is transitive. Therefore $T$ is an equivalence relation on $RR$.
+
+- $overline(0)$ is given by all $y in RR$ where $sin(y) = 0$, so ${pi n : n in ZZ}$
+- $overline(pi/2)$ is given by all $y in RR$ where $sin(y) = sin(pi/2) = 1$, so ${pi/2 + 2pi n : n in ZZ}$
+- $overline(pi/4)$ is given by all $y in RR$ where $sin(y) = sin(pi/4)$, so ${pi/4 + 2pi n, (3pi)/4 + 2pi n : n in ZZ}$

documents/by-course/math-8/pset-6/package.nix

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+{
+  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;
+  }
+)