From bceed18c0463048001674e9cc13d50e835a02b75 Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Wed, 19 Feb 2025 04:32:27 -0800 Subject: [PATCH] auto-update(nvim): 2025-02-19 04:32:27 --- documents/by-course/math-8/pset-6/main.typ | 202 ++++++++++++++++++ documents/by-course/math-8/pset-6/package.nix | 37 ++++ 2 files changed, 239 insertions(+) create mode 100644 documents/by-course/math-8/pset-6/main.typ create mode 100644 documents/by-course/math-8/pset-6/package.nix diff --git a/documents/by-course/math-8/pset-6/main.typ b/documents/by-course/math-8/pset-6/main.typ new file mode 100644 index 0000000..b453d2b --- /dev/null +++ b/documents/by-course/math-8/pset-6/main.typ @@ -0,0 +1,202 @@ +#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 ${pi/4 + 2pi n, (3pi)/4 + 2pi n : n in ZZ}$ diff --git a/documents/by-course/math-8/pset-6/package.nix b/documents/by-course/math-8/pset-6/package.nix new file mode 100644 index 0000000..7879e7e --- /dev/null +++ b/documents/by-course/math-8/pset-6/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; + } +)