auto-update(nvim): 2025-02-27 15:38:55

documents/by-course/math-8/course-notes/main.typ

@@ -955,9 +955,10 @@ times B$, which has 6 elements. $A$ and $B$ admit $2^6 = 64$ distinct
 ]
 
 #definition[
-  A function $f : A -> B$ is called a *map/mapping*. $A$ is the *domain*, $B$ is
-  the *codomain*. The *range* of $f$ $"Rng"(f) = {y in B : (exists x in A)(f(x) =
-  y)}$.
+  A function $f : A -> B$ is called a *map/mapping*. $A$ is the *domain*, $B$
+  is the *codomain*.
+
+  If $B = A$, then we say $f$ is a function on $A$.
 ]
 
 More terminology. Let $f : A -> B$, $(a,b) in f$, $f(x) = y$. Then
@@ -965,6 +966,8 @@ More terminology. Let $f : A -> B$, $(a,b) in f$, $f(x) = y$. Then
 - $y$ is the *value* of $f$ at $x$
 - $y$ is the *image* of $x$ under $f$
 - $x$ is the *preimage* of $y$ under $f$
+- When $(x,y) in f$, we write $y = f(x)$. The *range* of $f$ $"Rng"(f) = {y in
+  B : (exists x in A)(f(x) = y)}$.
 
 #fact[
   $f : A -> B = g : C -> D$ if and only if $f subset.eq g$ and $g subset.eq f$.
@@ -1008,6 +1011,84 @@ iota$ which is an inclusion function $iota' : A -> C$.
   overline(2)$, and so on. So $f$ is not _one-to-one_.
 ]
 
+Two functions $f$ and $g$ are equal if and only if
+1. $"Dom"(f) = "Dom"(g)$
+2. $forall x in "Dom"(f)$, $f(x) = g(x)$
+
+#proof[
+  Suppose $x in "Dom"(f)$. Then $(x,y) in f$ for some $y$, and because $f = g$, we have $(x,y) in g$. Therefore $x in "Dom"(g)$. This shows $"Dom"(f) subset.eq "Dom"(g)$. Similarly, we show that $"Dom"(g) subset.eq "Dom"(f)$. Therefore $"Dom"(g) = "Dom"(f)$.
+
+  Suppose that $x in "Dom"(f)$. Then for some $y$, we have $(x,y) in f$. Because $f = g$, $(x,y) in g$. Therefore $f(x) = y = g(x)$.
+]
+
+#definition[
+  A function $x$ with domain $NN$ is called an infinite sequence, or simply a
+  sequence. The image of $n$ is written $x_n$ and called the $n$th term of the
+  sequence.
+]
+
+= Modular arithmetic
+
+We discuss the arithmetic of congruence classes.
+
+For a fixed natural number $m$, we may define the relation of congruence modulo $m$ on $ZZ$ as
+$
+  a = b (mod m) "if" m | b - a
+$
+#fact[
+  The only possible remainders when an integer is divided by $m$ are
+  $0,1,2...,m-1$, so there are $m$ congruence classes $mod m$. We call $ZZ_m$
+  the set of integers modulo $m$.
+]
+
+For each natural number $m$, it is possible to do arithmetic with numbers in $ZZ_m$ using addition and multiplication that depends on the modulus $m$.
+
+#example[
+  Think of 5 and 3 as the representatives of the equivalence classes $overline(5) + overline(3)$, then the sum of those classes is the class containing $5 + 3$. That is $overline(2)$ in $ZZ_6$.
+]
+
+#definition[
+  The sum of $overline(x)$ and $overline(y)$ in $ZZ_m$ is
+  $
+    overline(x) + overline(y) = overline(x+y)
+  $
+]
+
+#definition[
+  The product of $overline(x)$ and $overline(y)$ in $ZZ_m$ is
+  $
+    overline(x) dot overline(y) = overline(x dot y)
+  $
+]
+
+#theorem[
+  Let $m$ be a positive integer and $a$, $b$, $c$, and $d$ be integers. If $a = c ("mod" m)$ and $b = d ("mod" m)$, then $a + b = c + d ("mod" m)$.
+]
+
+#theorem[
+  Let $m$ be a positive integer and $a$, $b$, $c$, and $d$ be integers. If $a = c ("mod" m)$ and $b = d ("mod" m)$, then $a dot b = c dot d ("mod" m)$.
+]
+
+#theorem[
+  Let $m$ be a positive composite integer. Then there exist nonzero equivalence classes $overline(x)$ and $overline(y)$ in $ZZ_m$ such that $overline(x) overline(y) = 0$.
+]
+
+#proof[
+  $m$ is composite so $m = x y$ for integers $x$ and $y$, let $x$ and $y$ be positive, nonzero and less than $m$, then neither $overline(x)$ nor $overline(y)$ are zero but $x y = m$, so $x y = 0$.
+]
+
+#theorem[
+  Let $p$ be a prime. Whenever $overline(x) overline(y) = overline(0)$ in $ZZ_p$, then either $overline(x) = overline(0)$ or $overline(y) = overline(0)$.
+]
+
+#proof[
+  Let $overline(x)$ and $overline(y)$ be elements of $ZZ_p$, and suppose $overline(x) overline(y) = overline(0)$. Then $x y = 0 (mod p)$. Therefore $p | x y$. By Euclid's Lemma, either $p | x$ or $p | y$. Thus either $x = 0 (mod p)$ or $y = 0 (mod p)$. Therefore the theorem is true.
+]
+
+#theorem[Cancellation law for $ZZ_p$][
+  Let $p$ be prime. If $x y = x z$ in $ZZ_p$ and $x !=0$, then $y = z$.
+]
+
 == Logistics: exam 2
 
 - One non-proof free response on set operations and families (2.2-2.3, $union$,
@@ -1020,7 +1101,10 @@ iota$ which is an inclusion function $iota' : A -> C$.
 - T/F questions
 - Extra credit question
 
+
 = Solutions to selected exercises and problems
+in fact the real numbers are uncountable
+[9:42 PM]
 
 Solutions to selected problems and exercises.
 

documents/by-course/math-8/pset-7/main.typ

@@ -0,0 +1,273 @@
+#import "@youwen/zen:0.1.0": *
+#import "@preview/mitex:0.2.5": *
+
+#show: zen.with(
+  title: "Homework 7",
+  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:
+
+3.2: \#9cd, 10ac, 16a
+
+3.4: \#1, 7ad, 11ac
+
+4.1: \#1cef, 2, 3b, 4d, 7, 9, 12b, 13, 16, 17ab, 18ab
+
+#outline()
+
+= 3.2
+
+== 9
+
+=== c
+
+$
+  overline(0) = {...,0,1,2,3...} = ZZ
+$
+
+=== d
+
+$
+  &overline(0) = {...,-7,0,7,14,...} \
+  &overline(1) = {...,-8,-1,1,8,15,...} \
+  &overline(2) = {...,-9,-2,2,9,16,...} \
+  &overline(3) = {...,-10,-3,3,10,17,...} \
+  &overline(4) = {...,-11,-4,4,11,18,...} \
+  &overline(5) = {...,-12,-5,5,12,19,...} \
+  &overline(6) = {...,-13,-6,6,13,20,...} \
+$
+
+== 10
+
+=== a
+
+5 and -5.
+
+=== c
+
+14 and -10
+
+== 16
+
+=== a
+
+$ 6 equiv 1 space (mod 5)$, but $6 equiv 6 space (mod 10)$.
+
+= 3.4
+
+== 1
+
+=== a
+
+$6 + 6 equiv 5$
+
+=== b
+
+$5 dot 4 equiv 6$
+
+=== c
+
+$3 dot 3 + 5 dot 2 equiv 5$
+
+=== d
+
+$2 dot 4 + 3 dot 5 equiv 2$
+
+=== e
+
+$5 dot 1 + 3 dot 2 equiv 4$
+
+=== f
+
+$0 dot 3 + 2 dot 4 equiv 1$
+
+== 7
+
+=== a
+
+$(238 + 496 - 44) mod 9 = 6$
+
+=== d
+
+$317 dot 403 mod 9 = 5$
+
+== 11
+
+=== a
+
+No
+
+=== c
+
+Yes
+
+= 4.1
+
+== 1
+
+=== c
+
+It is a function. Domain: ${1,2}$, codomain: ${1,2}$.
+
+=== e
+
+No.
+
+=== f
+
+No.
+
+== 2
+
+It fails the "vertical line test" when graphed.
+
+Consider the input 1 so $f(1) = plus.minus sqrt(1)$. Then $f(1) = 1$ and $f(1) = -1$, so it relates 1 to multiple distinct values. Therefore it is not a function.
+
+== 3
+
+=== b
+
+- Domain: $RR$
+- Range: $y >= 5$, $y in RR$
+- Another codomain: $RR^+$ (positive real numbers)
+
+== 4
+
+=== d
+
+Domain: ${-2}$
+
+== 7
+
+#proof[
+  Now we show the converse. Suppose the conditions hold, that is,
+  1. $"Dom"(f) = "Dom"(g)$
+  2. $forall x in "Dom"(f)$, $f(x) = g(x)$
+
+  For any $(x,y) in f$, $y = f(x)$. Then by (1) $x$ is in the domain of $g$.
+  By (2), $g(x) = f(x) = y$, so $(x,y) in g$. So $f subset.eq g$. Repeat an
+  identical reasoning showing $g subset.eq f$. Therefore $f = g$.
+]
+
+== 9
+
+=== a
+
+$chi_A (1) = 1$
+
+=== b
+
+$chi_A (3) = 0$
+
+=== c
+
+$chi_A (pi) = 0$
+
+=== d
+
+$chi_A (2) - chi_A (0.2) = 1$
+
+== 12
+
+=== b
+
+- 1st term: 0
+- 5th term: 4
+- 10th term: 11
+
+== 13
+
+=== a
+
+$f(3) = 3$
+
+=== b
+
+0
+
+=== c
+
+3
+
+=== d
+
+${x : x - 1 | 6 = 0}$
+
+== 16
+
+=== a
+
+$"Rng"(pi_1) = {a in A : exists b in B "such that" (a,b) in S}$
+
+=== b
+
+
+$"Rng"(pi_1) = {b in B : exists a in A "such that" (a,b) in S}$
+
+== 17
+
+=== a
+
+There are $n^m$ functions.
+
+=== b
+
+
+There are $m n$ functions.
+
+== 18
+
+=== a
+
+#mitext(`
+Let $f: A \to B$ be a function and define the relation $T$ on $A$ by
+\[
+x \,T\, y \quad \Longleftrightarrow \quad f(x) = f(y).
+\]
+We must show that $T$ is reflexive, symmetric, and transitive.
+
+For any $x \in A$, we have $f(x) = f(x)$. Hence,
+\[
+x \,T\, x,
+\]
+so $T$ is reflexive.
+
+Suppose $x \,T\, y$. By definition, this means $f(x) = f(y)$. Since equality in $B$ is symmetric, we also have $f(y) = f(x)$. Therefore,
+\[
+y \,T\, x,
+\]
+so $T$ is symmetric.
+
+Suppose $x \,T\, y$ and $y \,T\, z$. Then $f(x) = f(y)$ and $f(y) = f(z)$. By transitivity of equality in $B$, $f(x) = f(z)$. Hence,
+\[
+x \,T\, z,
+\]
+so $T$ is transitive.
+
+Since $T$ is reflexive, symmetric, and transitive, it is an equivalence relation on $A$.
+`)
+
+=== b
+
+#mitext(`
+We want to describe the equivalence classes of $0$, $2$, and $4$ under the relation $x \,T\, y$ if and only if $x^2 = y^2$.
+
+\[
+[0] = \{y \in \mathbb{R}: y^2 = 0^2\} = \{0\}.
+\]
+
+\[
+[2] = \{y \in \mathbb{R}: y^2 = 2^2\} = \{2, -2\}.
+\]
+
+\[
+[4] = \{y \in \mathbb{R}: y^2 = 4^2\} = \{4, -4\}.
+\]
+`)

documents/by-course/math-8/pset-7/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;
+  }
+)