auto-update(nvim): 2025-01-22 03:26:59

documents/by-course/math-8/pset-2/main.typ

@@ -0,0 +1,444 @@
+#import "@youwen/zen:0.1.0": *
+
+#show: zen.with(
+  title: "Homework 2",
+  author: "Youwen Wu",
+)
+
+#set heading(
+  numbering: (
+    num => {
+      return "1." + str(num)
+    }
+  ),
+)
+
+#set par(first-line-indent: 0pt, spacing: 1em)
+
+Problems:
+
+1.3: \#1fgjkLmp, 2fgjkLmp, 6, 8, 9, 10, 13
+
+1.4: \#5cdghi, 6bd, 7defghijk, 8, 9a (use any proof method, not necessarily proof by working backward), 11bcd
+
+*1.3*
+
+1f.
+
+Let $H(p)$ be true if a person is honest and false otherwise.
+
+$ ((forall x)H(x)) or ((forall x)(not H(x))) $
+
+1g.
+
+Again let $H(p)$ be true if a person is honest and false otherwise.
+
+$ (exists x)(exists y)(H(x) and not H(y)) $
+
+1j.
+
+$ (forall x)(exists y)(x > y) $
+
+1k.
+
+$ (exists.not x)(forall y)(x > y) $
+
+1L.
+
+$ (x in ZZ)(y in ZZ)(y > x)(exists z in RR)(x < z < y) $
+
+1m.
+
+$ (exists x in ZZ^+)(exists.not y in ZZ^+)(y < x) $
+
+1p.
+
+$ (forall x)(x > 0)(exists y)(2^y = x) $
+
+2f.
+
+Let $H(p)$ be true if a person is honest and false otherwise.
+
+$ (exists x)(exists y)(H(x) and not H(y)) $
+
+In English: Some people are honest and some people are not honest.
+
+2g.
+
+Let $H(p)$ be true if a person is honest and false otherwise.
+
+$ ((forall x)H(x)) or ((forall x)(not H(x))) $
+
+In English: all people are honest or no one is honest.
+
+2j.
+
+
+$ (exists x)(forall y) not (x > y) $
+
+2k.
+
+$ (forall x)(exists y)(y > x) $
+
+2L.
+
+$ (exists x in ZZ)(exists y in ZZ)(forall z)((z > x) and (z > y)) $
+
+2m.
+
+$ (forall x in ZZ^+)(exists y in ZZ^+)(y < x) $
+
+2p.
+
+$ (exists x)(x > 0)(forall y) not (2^y = x) $
+
+6a.
+
+$T$, $U$, $V$.
+
+6b.
+
+$T$ only.
+
+6c.
+
+$T$, $U$, $V$.
+
+6d.
+
+$T$ only.
+
+8a. False. Consider $x = -10$.
+
+8b. True.
+
+8c. True.
+
+8d. True.
+
+8e. False.
+
+8f. True.
+
+8g. True.
+
+8h. True.
+
+8i. True ($x=0)$.
+
+8j. False.
+
+8k. False.
+
+8L. True.
+
+9a. All natural numbers are at least one.
+
+9b. There is exactly one and only one real number that is both greater than or
+equal to 0 and less than or equal to 0.
+
+9c. All natural numbers are prime, and if they are not two then they are odd.
+
+9d. There is one and only one real number that is $e$ raised to the power of
+one.
+
+9e. There is no real number such that its square is negative.
+
+9f. There is a unique real number that is the square root of 0.
+
+9g. Any odd natural number is also odd when squared.
+
+10a. True.
+
+10b. False.
+
+10c. False.
+
+10d. False.
+
+10e. True ($x=0$).
+
+10f. False.
+
+10g. True.
+
+10h. False.
+
+10i. False.
+
+10j. True.
+
+10k. False.
+
+\13. (b), (c)
+
+*1.4*
+
+5c. If $x$ and $y$ are even, then $x y$ is divisible by 4.
+
+#proof[
+  $
+    exists j,k in ZZ, x = 2j, y = 2k \
+    x y = 4 j k
+  $
+
+  Clearly $x y$ has $4$ in its factors and so $x y | 4$.
+]
+
+5d.
+
+#proof[
+  $
+    exists j,k in ZZ, x = 2j, y = 2k \
+    3x - 5y = 6j - 10k = 2(3j - 5k) = 2n, n in ZZ
+  $
+]
+
+5e.
+
+#proof[
+  $
+    exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
+    x + y = 2j + 1 + 2k + 1 = 2(j + k + 1) = 2n, n in ZZ
+  $
+]
+
+5f.
+
+#proof[
+  $
+    exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
+    3x - 5y = 6j + 3 - 10k - 5 = 2(3j-5k-1) = 2n, n in ZZ
+  $
+]
+
+5g.
+
+#proof[
+  $
+    exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
+    x y = (2j + 1)(2k + 1) &= 4 j k + 2j + 2k + 1 \
+    = 2(2 j k + j + k) + 1 &= 2n + 1, n in ZZ
+  $
+]
+
+5i.
+
+#proof[
+  If one of $x$, $y$, and $z$ are odd, then let there be $i,j,k in ZZ$, such
+  that for the two even terms are represented as $2i$ and $2j$, and the single
+  odd term represented as $2k + 1$. Then clearly
+  $
+    x + y + z = 2i + 2j + 2k + 1 = 2(i + j + k) + 1 = 2n + 1, n in ZZ
+  $
+]
+
+6b.
+
+#proof[
+  Start by rewriting
+  $
+    |b - a| = |-(a - b)|
+  $
+
+  But by the definition of the absolute value,
+  $ |-(a - b)| = |a-b| $
+  So indeed
+  $ |a - b| = |b - a| $
+]
+
+6d.
+
+#proof[
+  In the case where $a$ and $b$ are positive or zero the two sides are equal.
+  $
+    |a + b| <= |a| + |b|
+  $
+  Otherwise, consider if $a$ is positive, $b$ is negative. We replace all
+  occurrences of $b$ with $-|b|$.
+
+  $
+    |a - |b|| <= |a| + |-|b|| \
+    |a - |b|| <= |a| + |b|
+  $
+
+  Clearly for any $b$ the left side is strictly lower than the right. Repeat
+  this exact for $a$ is negative.
+]
+
+7d.
+
+#proof[
+  Consider two cases, $a$ is even and $a$ is odd. First assume $a$ is even:
+  $
+    exists k in ZZ, a = 2k \
+    2k(2k+1) = 4k^2 + 2k = 2(k^2 + 1) = 2n, n in ZZ
+  $
+  Then assume $a$ is odd:
+  $
+    exists k in ZZ, a = 2k + 1 \
+    2k(2k+1+1) = 4k^2 + 4k = 2(k^2 + 2) = 2n, n in ZZ
+  $
+]
+
+7e.
+
+#proof[
+  If $1 | a$, then we can find some $k in ZZ$ such that $1k = a$. Such a $k$ is
+  $a$, because $1 dot a = a$. Therefore $1 | a$.
+]
+
+7f.
+
+#proof[
+  If $a | a$, we can find some $k in ZZ$ such that $a k = a$. Such a $k$ is $1$,
+  because $a dot 1 = a$. Therefore $a | a$.
+]
+
+7g.
+
+#proof[
+  If $a | b$, then there is a $k in ZZ$ such that $a k = b$. So $a = b/k$. If
+  $k = 1$, then $b/k = b$. Otherwise for $k > 0$ or $k < 0$ $b/k$ is less than
+  $b$. So $b/k = a <= b$.
+]
+
+7h.
+
+#proof[
+  Since $a | b$, we know $exists k in ZZ, k a = b$. We also know
+  that
+  $ (exists j in ZZ) (j a = b c) => (a | b c) $
+
+  We can find $j$:
+  $
+    j a = b c \
+    j cancel(a) = k cancel(a) c \
+    j = k c
+  $
+
+  Therefore $a | b c$.
+]
+
+7i.
+
+#proof[
+  Suppose that $a = b = 1$ was not the case. Then either $a$ or $b$ must be
+  greater than 1, or both $a$ and $b$ are greater than 1.
+
+  If $a > 1$ and $b = 1$, then $a b = a$. But is not 1, and $a b = 1$, so there
+  is a contradiction.
+
+  Repeat the same argument for $b >1$, $a = 1$.
+
+  In the case that both $a > 1$ and $b > 1$, $a b > 1$. Again, $a b = 1$ so
+  this cannot be true.
+
+  There are no cases where $a$ and $b$ can be anything other than 1,
+  so $a = b = 1$.
+]
+
+7j.
+
+#proof[
+  If $a | b$ then $exists k in ZZ, k a = b$. If $b | a$ then $exists j in ZZ, j
+  b = a$.
+
+  $
+    k j cancel(b) = cancel(b) \
+    k j = 1
+  $
+
+  By the fact proven above in Problem 7i, this implies $k = j = 1$.
+
+  So $1 a = b => a = b$.
+]
+
+7k.
+
+#proof[
+  $
+    a | b => exists k in ZZ, k a = b \
+    c | d => exists j in ZZ, j c = d \
+  $
+
+  Multiply the equations together:
+
+  $ k j a c = b d $
+
+  If $exists n in ZZ, n a c = b d$, then $a c | b d$. We see that $n = k j$.
+  Therefore $a c$ indeed divides $b d$.
+]
+
+8a.
+
+#proof[
+  Consider the case where $n$ is even. Then $exists k in ZZ, n = 2k$.
+
+  $
+    n^2 + n + 3 &= (2k)^2 + 2k + 3 \
+    = 4k^2 + 2k + 3 &= 2(2k^2 + k + 1) + 1 = 2m + 1, m in ZZ
+  $
+
+  Consider the case where $n$ is odd. Then $exists k in ZZ, n = 2k + 1$.
+
+  $
+    n^2 + n + 3 &= (2k+1)^2 + 2k+1 + 3 \
+    = 4k^2 + 6k + 5 &= 2(2k^2 + 3k + 2) + 1 = 2m + 1, m in ZZ
+  $
+
+  So in both cases $n^2 + n + 3$ is odd.
+]
+
+8b.
+
+#proof[
+  First we rewrite $n^2 + n + 3$.
+
+  $
+    n^2 + n + 3 = n(n + 1) + 3
+  $
+
+  By 7(d), we know that $n(n+1)$ is even $forall n$. Then by 5(h) we can take
+  $x := n(n+1)$ and $y := 3$. Since $x$ is even, $y$ is odd, $x + y$ is odd. So
+  $n^2 + n + 3$ is odd.
+]
+
+9a.
+
+#proof[
+  In the case of $x=0$ and $y=0$ we trivially have $0 >= 0$. Otherwise,
+
+  $
+    (x+y) / 2 &>= sqrt(x y) \
+    x+y &>= 2sqrt(x y) \
+    (x+y)^2 &>= 4x y \
+    x^2 + 2 x y + y^2 &>= 4 x y \
+    x^2 - 2 x y + y^2 &>= 0
+  $
+
+  THIS IS WRONG FIX IT!!!
+
+  You can think of $x^2 - 2 x y + y^2$ as quadratic with $y$ as a constant. For
+  all possible values of $y$ the equation is nonnegative (since the absolute
+  minimum occurs at the vertex). Therefore the inequality holds true.
+]
+
+11b.
+
+Grade: C.
+
+The assertions that $exists q, b = a q$ and $exists q, c = a q$ are essentially
+correct but these $q$ are not the same. This can be corrected fairly
+straightforwardly by replacing one of the $q$ with another variable serving the
+same purpose, then proceeding.
+
+11c.
+
+Grade: A.
+
+11d.
+
+Grade: F.
+
+This only shows $m "is odd" => m^2 "is odd"$, when the claim is the other way around. The converse is not automatically true.

documents/by-course/math-8/pset-2/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;
+  }
+)

documents/by-course/pstat-120a/course-notes/main.typ

@@ -771,3 +771,7 @@ us generalize to more than two colors.
 
   Both approaches given the same answer.
 ]
+
+= Discussion section #datetime(day: 22, month: 1, year: 2025).display()
+
+