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341
documents/by-course/math-8/pset-1/2vd.typ
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#import "@preview/ctheorems:1.1.3": *
#import "@preview/showybox:2.0.3": showybox
#let colors = (
rgb("#9E9E9E"),
rgb("#F44336"),
rgb("#E91E63"),
rgb("#9C27B0"),
rgb("#673AB7"),
rgb("#3F51B5"),
rgb("#2196F3"),
rgb("#03A9F4"),
rgb("#00BCD4"),
rgb("#009688"),
rgb("#4CAF50"),
rgb("#8BC34A"),
rgb("#CDDC39"),
rgb("#FFEB3B"),
rgb("#FFC107"),
rgb("#FF9800"),
rgb("#FF5722"),
rgb("#795548"),
rgb("#9E9E9E"),
)
#let dvdtyp(
title: "",
subtitle: "",
author: "",
abstract: none,
bibliography: none,
paper-size: "a4",
date: "today",
body,
) = {
set document(title: title, author: author)
set std.bibliography(style: "springer-mathphys", title: [References])
show: thmrules
set page(
numbering: "1",
number-align: center,
header: locate(loc => {
if loc.page() == 1 {
return
}
box(stroke: (bottom: 0.7pt), inset: 0.4em)[#text(
font: "New Computer Modern",
)[
*#author* --- #datetime.today().display("[day] [month repr:long] [year]")
#h(1fr)
*#title*
]]
}),
paper: paper-size,
// The margins depend on the paper size.
margin: (
left: (86pt / 216mm) * 100%,
right: (86pt / 216mm) * 100%,
),
)
set heading(numbering: "1.")
show heading: it => {
set text(font: "Libertinus Serif")
block[
#if it.numbering != none {
text(rgb("#2196F3"), weight: 500)[#sym.section]
text(rgb("#2196F3"))[#counter(heading).display() ]
}
#it.body
#v(0.5em)
]
}
set text(font: "New Computer Modern", lang: "en")
show math.equation: set text(weight: 400)
// Title row.
align(center)[
#set text(font: "Libertinus Serif")
#block(text(weight: 700, 26pt, title))
#if subtitle != none [#text(12pt, weight: 500)[#(
subtitle
)]]
#if author != none [#text(16pt)[#smallcaps(author)]]
#v(1.2em, weak: true)
#if date == "today" {
datetime.today().display("[day] [month repr:long] [year]")
} else {
date
}
]
if abstract != none [
#v(2.2em)
#set text(font: "Libertinus Serif")
#pad(x: 14%, abstract)
#v(1em)
]
set outline(fill: repeat[~.], indent: 1em)
show outline: set heading(numbering: none)
show outline: set par(first-line-indent: 0em)
show outline.entry.where(level: 1): it => {
text(font: "Libertinus Serif", rgb("#2196F3"))[#strong[#it]]
}
show outline.entry: it => {
h(1em)
text(font: "Libertinus Serif", rgb("#2196F3"))[#it]
}
// Main body.
set par(
justify: true,
spacing: 0.65em,
first-line-indent: 2em,
)
body
// Display the bibliography, if any is given.
if bibliography != none {
show std.bibliography: set text(footnote-size)
show std.bibliography: set block(above: 11pt)
show std.bibliography: pad.with(x: 0.5pt)
bibliography
}
}
#let thmtitle(t, color: rgb("#000000")) = {
return text(
font: "Libertinus Serif",
weight: "semibold",
fill: color,
)[#t]
}
#let thmname(t, color: rgb("#000000")) = {
return text(font: "Libertinus Serif", fill: color)[(#t)]
}
#let thmtext(t, color: rgb("#000000")) = {
let a = t.children
if (a.at(0) == [ ]) {
a.remove(0)
}
t = a.join()
return text(font: "New Computer Modern", fill: color)[#t]
}
#let thmbase(
identifier,
head,
..blockargs,
supplement: auto,
padding: (top: 0.5em, bottom: 0.5em),
namefmt: x => [(#x)],
titlefmt: strong,
bodyfmt: x => x,
separator: [. \ ],
base: "heading",
base_level: none,
) = {
if supplement == auto {
supplement = head
}
let boxfmt(name, number, body, title: auto, ..blockargs_individual) = {
if not name == none {
name = [ #namefmt(name)]
} else {
name = []
}
if title == auto {
title = head
}
if not number == none {
title += " " + number
}
title = titlefmt(title)
body = [#pad(top: 2pt, bodyfmt(body))]
pad(
..padding,
showybox(
width: 100%,
radius: 0.3em,
breakable: true,
padding: (top: 0em, bottom: 0em),
..blockargs.named(),
..blockargs_individual.named(),
[
#title#name#titlefmt(separator)#body
],
),
)
}
let auxthmenv = thmenv(
identifier,
base,
base_level,
boxfmt,
).with(supplement: supplement)
return auxthmenv.with(numbering: "1.1")
}
#let styled-thmbase = thmbase.with(
titlefmt: thmtitle,
namefmt: thmname,
bodyfmt: thmtext,
)
#let builder-thmbox(color: rgb("#000000"), ..builderargs) = styled-thmbase.with(
titlefmt: thmtitle.with(color: color.darken(30%)),
bodyfmt: thmtext.with(color: color.darken(70%)),
namefmt: thmname.with(color: color.darken(30%)),
frame: (
body-color: color.lighten(92%),
border-color: color.darken(10%),
thickness: 1.5pt,
inset: 1.2em,
radius: 0.3em,
),
..builderargs,
)
#let builder-thmline(
color: rgb("#000000"),
..builderargs,
) = styled-thmbase.with(
titlefmt: thmtitle.with(color: color.darken(30%)),
bodyfmt: thmtext.with(color: color.darken(70%)),
namefmt: thmname.with(color: color.darken(30%)),
frame: (
body-color: color.lighten(92%),
border-color: color.darken(10%),
thickness: (left: 2pt),
inset: 1.2em,
radius: 0em,
),
..builderargs,
)
#let problem-style = builder-thmbox(
color: colors.at(11),
shadow: (offset: (x: 2pt, y: 2pt), color: luma(70%)),
)
#let exercise = problem-style("item", "Exercise")
#let problem = exercise
#let theorem-style = builder-thmbox(
color: colors.at(6),
shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
)
#let example-style = builder-thmbox(
color: colors.at(16),
shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
)
#let theorem = theorem-style("item", "Theorem")
#let lemma = theorem-style("item", "Lemma")
#let corollary = theorem-style("item", "Corollary")
#let definition-style = builder-thmline(color: colors.at(8))
// #let definition = definition-style("definition", "Definition")
#let proposition = definition-style("item", "Proposition")
#let remark = definition-style("item", "Remark")
#let observation = definition-style("item", "Observation")
// #let example-style = builder-thmline(color: colors.at(16))
#let example = example-style("item", "Example")
#let proof(body, name: none) = {
v(0.5em)
[_Proof_]
if name != none {
[ #thmname[#name]]
}
[.]
body
h(1fr)
// Add a word-joiner so that the proof square and the last word before the
// 1fr spacing are kept together.
sym.wj
// Add a non-breaking space to ensure a minimum amount of space between the
// text and the proof square.
sym.space.nobreak
$square.stroked$
v(0.5em)
}
#let fact = thmplain(
"item",
"Fact",
titlefmt: content => [*#content.*],
namefmt: content => [_(#content)._],
separator: [],
inset: 0pt,
padding: (bottom: 0.5em, top: 0.5em),
)
#let abuse = thmplain(
"item",
"Abuse of Notation",
titlefmt: content => [*#content.*],
namefmt: content => [_(#content)._],
separator: [],
inset: 0pt,
padding: (bottom: 0.5em, top: 0.5em),
)
#let definition = thmplain(
"item",
"Definition",
titlefmt: content => [*#content.*],
namefmt: content => [_(#content)._],
separator: [],
inset: 0pt,
padding: (bottom: 0.5em, top: 0.5em),
)

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documents/by-course/math-8/pset-1/main.typ
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#import "./dvd.typ": *
#import "@youwen/zen:0.1.0": *
#show: dvdtyp.with(
#show: zen.with(
title: "Homework 1",
author: "Youwen Wu",
)

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#import "@youwen/zen:0.1.0": *
#import "@preview/mitex:0.2.5": *
#show: zen.with(
title: "Homework 3",
author: "Youwen Wu",
)
#set heading(numbering: none)
#set par(first-line-indent: 0pt, spacing: 1em)
#let nonzero = $ZZ_(!=0)$
Problems:
1.5: $hash$ 3cdefgh, 4de, 6de, 7ab, 9, 10, 11, 12a
1.6: $hash$ 1bd, 4abcd, 6abefik
2.1: $hash$ 4, 5, 6abcd, 8, 11ab, 14ab, 15abcd
= 1.5
*3c.*
If $x^2$ is not divisible by 4, then $x$ is odd.
#proof[
Suppose $x$ is not odd. We seek to show that $x^2$ is divisible by $4$.
Then $x$ is even and $exists k in ZZ_(!=0), x = 2k$. Thus
$x^2 = 4k^2$, which implies $4 | x^2$. So $x$ is not odd implies $x^2$ is
divisible by 4, and therefore the contrapositive (which is our original
statement) is also true.
]
*3d.*
If $x y$ is even, then either $x$ or $y$ is even.
#proof[
Suppose that $not (x "or" y" is even")$. In other words, $x$ and $y$ are both
odd. We seek to show that $x y$ is odd.
Then $exists j,k in ZZ_(n!=0), x = 2j + 1, y = 2k + 1$. So $x y = 4 j k + 2j +
2k + 1 = 2 (2 j k + j + k) + 1$ where $2 j k + j + k$ is an integer so $x y$
is odd. This is the contrapositive, so the original statement is also true.
]
*3e.*
If $x + y$ is even, then $x$ and $y$ have the same parity.
#proof[
Suppose $x$ and $y$ had different parities. We seek to show that $x + y$ is
odd.
Without loss of generality, let us inspect the case where $x$ is odd. Then
$y$ must be even and
$
exists j,k in ZZ_(!=0), x = 2j + 1, y = 2k \
x + y = 2j + 1 + 2k = 2(j + k) + 1 = 2n + 1, n in ZZ_(n!=0)
$
Repeat the same reasoning for when $x$ is even and $y$ is odd. We've shown
that $x$ and $y$ having different parities implies that $x + y$ is odd.
Therefore the contrapositive is also true.
]
*3f.*
If $x y$ is odd, then both $x$ and $y$ are odd.
#proof[
Suppose that both $x$ and $y$ are odd was not true, in other words $x$ or $y$
are even (here or is the logical $or$). We seek to show that this implies $x y
$ is even. Then we have two cases: either $x$ or
$y$, but not both, are even, and the case where $x$ and $y$ are both even.
We look at the first case, without loss of generality, assume $x$ is even and
$y$ is odd. Then
$
exists j,k in ZZ_(!= 0), x = 2j, y = 2k + 1 \
x y = 2j(2k + 1) = 4 j k + 2j = 2(2 j k + j) = 2n, n in ZZ_(!=0)
$
So $x$ and $y$ having different parities implies $x y$ is even. The argument holds identically when instead $y$ is even and $x$ is odd. Now we turn our attention to the case when $x$ and $y$ are both even. Then
$
exists j,k in ZZ_(!=0), x = 2j, y = 2k \
x y = 2j dot 2k = 4 j k = 2 (2 j k) = 2n, n in ZZ_(!=0)
$
So this also implies that $x y$ is even. Therefore $x$ or $y$ being even indeed
implies $x y$ is also even, and the contrapositive is also true.
]
*3g.*
If 8 does not divide $x^2 - 1$, then $x$ is even.
#proof[
Suppose that $x$ is odd. We seek to show that 8 does in fact divide $x^2 - 1$.
$
exists k in nonzero, x = 2k + 1 \
x^2 - 1 = 4k^2 + 4k = 4(k^2 + k)
$
Now we need to see if it's divisible by 8. First consider the unique case $k = 0$, where $x^2 - 1 = 0$. Clearly $8 | 0$ so 8 does divide $x^2-1$.
Now we consider all other values. Notice that for all other possible values
of $k$, $k^2 + k$ is greater than 1. We see this by noting that $k^2 + k$ is
a quadratic with its absolute minima at $k = 1/2$, therefore we can check the
two non-zero integers closest to this value. For $k = 1$, $k^2 + k = 2$.
Since we already checked $k=-1$, let's check $k=-2$, which gives $k^2 + k =
2$. For all values of $k$ greater than 1 or less than $-2$, $k^2 + k$ must be
greater than 2 (because it's a quadratic).
Therefore $4(k^2 + k)$ can be written as $8n$ for some integer $n$, so 8
indeed divides $x^2 - 1$ for all possible $k$, and the contrapositive is also
true.
]
*3h.*
If $x$ does not divide $y z$, then $x$ does not divide $z$.
#proof[
Assume $x$ does divide $z$. We seek to show that $x$ does divide $y z$. If $x
| z$ then $exists k in nonzero, k x = z$. So $y z$ can be written as $y k x$.
But this shows $exists j in nonzero, j x = y z$, namely $j = y k$, which
means $x | y z$, and the contrapositive is also true.
]
*4d.*
If $(x+1)(x-1) < 0$, then $x < 1$.
#proof[
Suppose that $x >= 1$. We seek to show that $(x+1)(x-1) >= 0$. Expanding out
factors,
$
(x+1)(x-1) = x^2 - 1>= 0
$
This quadratic is zero at exactly $x = 1$ and positive for all $x > 1$. So
it's true for all possible values of $x$. Therefore our original statement is
also true.
]
*4e.*
If $x(x-4) > -3$, then $x < 1$ or $x > 3$.
#proof[
Suppose that $x >= 1$ and $x <= 3$. We seek to show $x(x-4) <= -3$. Expanding
out factors,
$
x(x-4) = x^2 - 4x > -3
$
This quadratic has its stationary point at $x = 2$. Let's check its value at $x
= 1$ and $x = 3$.
At $x = 1$, $x(x-4) = -3$. So for all values greater than 1 until $x = 2$, $x^2 - 4x$ is less than $-3$. Our inequality is satisfied.
At $x = 3$, $x(x-4) = -3$ again. So for all values less than 3 until $x = 2$,
$x^2 - 4x$ is less than $-3$. Our inequality is satisfied for both $x >=1$
and $x <=3$, so $x^2 -4x > -3$ is true for all possible $x$ and the
contrapositive is also true.
]
*6d.*
If $a - b$ is odd, then $a + b$ is odd.
#proof[
Suppose, seeking a contradiction, that if $a - b$ is even, then $a + b$ is odd.
Then $exists k in nonzero, a - b = 2k$. Which means we can write $a + b = 2k
+ 2b$. But we can factor this as $2(k + b)$ and so $a + b = 2n, n in
nonzero$, implying it is even. However we assumed that $a + b$ should be odd,
a contradiction. Therefore $a - b$ must be odd.
]
*6e.*
If $a < b$ and $a b < 3$, then $a = 1$.
#proof[
Suppose, seeking a contradiction, that $a >= b$ and $a b >= 3$ implies $a =
1$. Consider the specific cases $a = b$, $a b = 3$. Then $a = 3/a$, and $a =
plus.minus sqrt(3)$. However we assumed $a = 1$ is implied, a contradiction.
Therefore we must have $a < b$ and $a b < 3$.
]
*7a.*
$a c$ divides $b$ and $b$ divides $b + 3$ if and only if $a = 2$ and $b = 3$.
We first show the result left to right, namely, $a c | b c => a | b$.
$
exists k in nonzero, k a c = b c \
k a = b
$
which is the definition of $a | b$.
Now we show the right to left direction, namely, $a | b => a c | b c$.
$
exists k in nonzero, k a = b \
k a dot c = b dot c
$
So $a c | b c$ by the definition of divisibility. Therefore the biconditional
is true, as we have shown both directions.
*7b.*
The right to left direction is very easy in this case. By directly plugging in $a = 2$, $b = 3$, we see that $2 + 1 | 3$ and $3 | 3 + 3$. To show the left to right case,
*9.*
#proof[
Suppose that instead $n/(n+1) <= n/(n+2)$. We can be assured the following
operations do not flip the inequality as $n$ cannot be negative.
$
n(n+2) <= n(n+1) \
n + 2 <= n + 1 \
2 <= 1
$
So $n/(n+1) > n/(n+2)$.
]
*10.*
#proof[
Suppose that $sqrt(5)$ was rational. That is, $sqrt(5) = p/q$ for nonzero
integers $p$ and $q$. Additionally, assume that $p/q$ is in its most reduced
form, that is, $p$ and $q$ share no common factors besides 1. Then
$
p^2 = 5q \
$
implies that
$ 5 | p^2 $
We need to show that $5 | p^2 => 5 | p$.
By the fundamental theorem of arithmetic, $p^2$ has 5 as one of its unique
prime factors. If 5 was not a factor of $p$, then $p^2 = p dot p$ would not
have 5 in its factors either. So 5 is a factor of $p$ and thus $5 | p$. Note
that 5 appears at least twice amongst the prime factors of $p^2$. Then $5q$
should also have at least two 5s in its prime factorization. Then $q$ has at
least one 5 in its prime factorization. However we assumed that $p$ and $q$
share no common factors besides 1, so this is a contradiction. Therefore
$sqrt(5)$ is not rational.
]
*11.*
#proof[
We say that two numbers $x$ and $y$ are within $1/2$ unit from one another if
$|x - y| < 1/2$. Consider the distance between $z$, and $y$, if it is within
$1/2$, then we are done. Otherwise $z - y >= 1/2$.
We know that
$ (1 - z) + x + (y-x) + (z-y) = 1 $
because this is the length of all the line segments partitioned by $x,y,z$,
which is the interval 1. If $(z - y) >= 1/2$, then everything else must be
less than $1/2$. So the maximum value of $y-x$, the distance between $x$ and
$y$, is less than $1/2$. Therefore either $y$ and $z$ are within $1/2$ unit
of each other or $y$ and $x$ are.
]
*12a.*

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documents/by-course/math-8/pset-3/package.nix Normal file
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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;
}
)