auto-update(nvim): 2025-01-31 01:26:34
This commit is contained in:
parent
68822d7ff0
commit
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GPG key ID:
865658ED1FE61EC3
4 changed files with 317 additions and 343 deletions
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#import "@preview/ctheorems:1.1.3": *
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#import "@preview/showybox:2.0.3": showybox
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#let colors = (
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rgb("#9E9E9E"),
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rgb("#F44336"),
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rgb("#E91E63"),
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rgb("#9C27B0"),
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rgb("#673AB7"),
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rgb("#3F51B5"),
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rgb("#2196F3"),
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rgb("#03A9F4"),
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rgb("#00BCD4"),
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rgb("#009688"),
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rgb("#4CAF50"),
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rgb("#8BC34A"),
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rgb("#CDDC39"),
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rgb("#FFEB3B"),
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rgb("#FFC107"),
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rgb("#FF9800"),
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rgb("#FF5722"),
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rgb("#795548"),
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rgb("#9E9E9E"),
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)
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#let dvdtyp(
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title: "",
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subtitle: "",
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author: "",
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abstract: none,
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bibliography: none,
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paper-size: "a4",
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date: "today",
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body,
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) = {
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set document(title: title, author: author)
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set std.bibliography(style: "springer-mathphys", title: [References])
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show: thmrules
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set page(
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numbering: "1",
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number-align: center,
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header: locate(loc => {
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if loc.page() == 1 {
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return
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}
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box(stroke: (bottom: 0.7pt), inset: 0.4em)[#text(
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font: "New Computer Modern",
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)[
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*#author* --- #datetime.today().display("[day] [month repr:long] [year]")
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#h(1fr)
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*#title*
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]]
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}),
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paper: paper-size,
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// The margins depend on the paper size.
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margin: (
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left: (86pt / 216mm) * 100%,
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right: (86pt / 216mm) * 100%,
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),
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)
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set heading(numbering: "1.")
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show heading: it => {
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set text(font: "Libertinus Serif")
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block[
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#if it.numbering != none {
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text(rgb("#2196F3"), weight: 500)[#sym.section]
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text(rgb("#2196F3"))[#counter(heading).display() ]
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}
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#it.body
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#v(0.5em)
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]
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}
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set text(font: "New Computer Modern", lang: "en")
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show math.equation: set text(weight: 400)
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// Title row.
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align(center)[
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#set text(font: "Libertinus Serif")
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#block(text(weight: 700, 26pt, title))
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#if subtitle != none [#text(12pt, weight: 500)[#(
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subtitle
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)]]
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#if author != none [#text(16pt)[#smallcaps(author)]]
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#v(1.2em, weak: true)
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#if date == "today" {
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datetime.today().display("[day] [month repr:long] [year]")
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} else {
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date
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}
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]
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if abstract != none [
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#v(2.2em)
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#set text(font: "Libertinus Serif")
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#pad(x: 14%, abstract)
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#v(1em)
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]
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set outline(fill: repeat[~.], indent: 1em)
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show outline: set heading(numbering: none)
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show outline: set par(first-line-indent: 0em)
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show outline.entry.where(level: 1): it => {
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text(font: "Libertinus Serif", rgb("#2196F3"))[#strong[#it]]
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}
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show outline.entry: it => {
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h(1em)
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text(font: "Libertinus Serif", rgb("#2196F3"))[#it]
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}
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// Main body.
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set par(
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justify: true,
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spacing: 0.65em,
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first-line-indent: 2em,
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)
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body
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// Display the bibliography, if any is given.
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if bibliography != none {
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show std.bibliography: set text(footnote-size)
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show std.bibliography: set block(above: 11pt)
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show std.bibliography: pad.with(x: 0.5pt)
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bibliography
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}
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}
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#let thmtitle(t, color: rgb("#000000")) = {
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return text(
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font: "Libertinus Serif",
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weight: "semibold",
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fill: color,
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)[#t]
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}
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#let thmname(t, color: rgb("#000000")) = {
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return text(font: "Libertinus Serif", fill: color)[(#t)]
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}
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#let thmtext(t, color: rgb("#000000")) = {
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let a = t.children
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if (a.at(0) == [ ]) {
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a.remove(0)
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}
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t = a.join()
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return text(font: "New Computer Modern", fill: color)[#t]
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}
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#let thmbase(
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identifier,
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head,
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..blockargs,
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supplement: auto,
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padding: (top: 0.5em, bottom: 0.5em),
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namefmt: x => [(#x)],
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titlefmt: strong,
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bodyfmt: x => x,
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separator: [. \ ],
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base: "heading",
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base_level: none,
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) = {
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if supplement == auto {
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supplement = head
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}
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let boxfmt(name, number, body, title: auto, ..blockargs_individual) = {
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if not name == none {
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name = [ #namefmt(name)]
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} else {
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name = []
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}
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if title == auto {
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title = head
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}
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if not number == none {
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title += " " + number
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}
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title = titlefmt(title)
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body = [#pad(top: 2pt, bodyfmt(body))]
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pad(
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..padding,
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showybox(
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width: 100%,
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radius: 0.3em,
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breakable: true,
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padding: (top: 0em, bottom: 0em),
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..blockargs.named(),
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..blockargs_individual.named(),
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[
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#title#name#titlefmt(separator)#body
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],
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),
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)
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}
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let auxthmenv = thmenv(
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identifier,
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base,
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base_level,
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boxfmt,
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).with(supplement: supplement)
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return auxthmenv.with(numbering: "1.1")
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}
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#let styled-thmbase = thmbase.with(
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titlefmt: thmtitle,
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namefmt: thmname,
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bodyfmt: thmtext,
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)
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#let builder-thmbox(color: rgb("#000000"), ..builderargs) = styled-thmbase.with(
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titlefmt: thmtitle.with(color: color.darken(30%)),
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bodyfmt: thmtext.with(color: color.darken(70%)),
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namefmt: thmname.with(color: color.darken(30%)),
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frame: (
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body-color: color.lighten(92%),
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border-color: color.darken(10%),
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thickness: 1.5pt,
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inset: 1.2em,
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radius: 0.3em,
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),
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..builderargs,
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)
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#let builder-thmline(
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color: rgb("#000000"),
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..builderargs,
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) = styled-thmbase.with(
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titlefmt: thmtitle.with(color: color.darken(30%)),
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bodyfmt: thmtext.with(color: color.darken(70%)),
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namefmt: thmname.with(color: color.darken(30%)),
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frame: (
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body-color: color.lighten(92%),
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border-color: color.darken(10%),
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thickness: (left: 2pt),
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inset: 1.2em,
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radius: 0em,
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),
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..builderargs,
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)
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#let problem-style = builder-thmbox(
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color: colors.at(11),
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shadow: (offset: (x: 2pt, y: 2pt), color: luma(70%)),
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)
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#let exercise = problem-style("item", "Exercise")
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#let problem = exercise
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#let theorem-style = builder-thmbox(
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color: colors.at(6),
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shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
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)
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#let example-style = builder-thmbox(
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color: colors.at(16),
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shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
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)
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#let theorem = theorem-style("item", "Theorem")
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#let lemma = theorem-style("item", "Lemma")
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#let corollary = theorem-style("item", "Corollary")
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#let definition-style = builder-thmline(color: colors.at(8))
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// #let definition = definition-style("definition", "Definition")
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#let proposition = definition-style("item", "Proposition")
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#let remark = definition-style("item", "Remark")
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#let observation = definition-style("item", "Observation")
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// #let example-style = builder-thmline(color: colors.at(16))
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#let example = example-style("item", "Example")
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#let proof(body, name: none) = {
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v(0.5em)
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[_Proof_]
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if name != none {
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[ #thmname[#name]]
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}
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[.]
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body
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h(1fr)
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// Add a word-joiner so that the proof square and the last word before the
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// 1fr spacing are kept together.
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sym.wj
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// Add a non-breaking space to ensure a minimum amount of space between the
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// text and the proof square.
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sym.space.nobreak
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$square.stroked$
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v(0.5em)
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}
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#let fact = thmplain(
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"item",
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"Fact",
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titlefmt: content => [*#content.*],
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namefmt: content => [_(#content)._],
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separator: [],
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inset: 0pt,
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padding: (bottom: 0.5em, top: 0.5em),
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)
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#let abuse = thmplain(
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"item",
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"Abuse of Notation",
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titlefmt: content => [*#content.*],
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namefmt: content => [_(#content)._],
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separator: [],
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inset: 0pt,
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padding: (bottom: 0.5em, top: 0.5em),
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)
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#let definition = thmplain(
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"item",
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"Definition",
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titlefmt: content => [*#content.*],
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namefmt: content => [_(#content)._],
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separator: [],
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inset: 0pt,
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padding: (bottom: 0.5em, top: 0.5em),
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)
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@ -1,6 +1,6 @@
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#import "./dvd.typ": *
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#import "@youwen/zen:0.1.0": *
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#show: dvdtyp.with(
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#show: zen.with(
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title: "Homework 1",
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title: "Homework 1",
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author: "Youwen Wu",
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author: "Youwen Wu",
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)
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)
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278
documents/by-course/math-8/pset-3/main.typ
Normal file
278
documents/by-course/math-8/pset-3/main.typ
Normal file
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@ -0,0 +1,278 @@
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#import "@youwen/zen:0.1.0": *
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#import "@preview/mitex:0.2.5": *
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#show: zen.with(
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title: "Homework 3",
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author: "Youwen Wu",
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)
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#set heading(numbering: none)
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#set par(first-line-indent: 0pt, spacing: 1em)
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#let nonzero = $ZZ_(!=0)$
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Problems:
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1.5: $hash$ 3cdefgh, 4de, 6de, 7ab, 9, 10, 11, 12a
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1.6: $hash$ 1bd, 4abcd, 6abefik
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2.1: $hash$ 4, 5, 6abcd, 8, 11ab, 14ab, 15abcd
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= 1.5
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*3c.*
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If $x^2$ is not divisible by 4, then $x$ is odd.
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#proof[
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Suppose $x$ is not odd. We seek to show that $x^2$ is divisible by $4$.
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||||||
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Then $x$ is even and $exists k in ZZ_(!=0), x = 2k$. Thus
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||||||
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$x^2 = 4k^2$, which implies $4 | x^2$. So $x$ is not odd implies $x^2$ is
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divisible by 4, and therefore the contrapositive (which is our original
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statement) is also true.
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]
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*3d.*
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If $x y$ is even, then either $x$ or $y$ is even.
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||||||
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#proof[
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||||||
|
Suppose that $not (x "or" y" is even")$. In other words, $x$ and $y$ are both
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odd. We seek to show that $x y$ is odd.
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||||||
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||||||
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Then $exists j,k in ZZ_(n!=0), x = 2j + 1, y = 2k + 1$. So $x y = 4 j k + 2j +
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||||||
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2k + 1 = 2 (2 j k + j + k) + 1$ where $2 j k + j + k$ is an integer so $x y$
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||||||
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is odd. This is the contrapositive, so the original statement is also true.
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||||||
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]
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*3e.*
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||||||
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||||||
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If $x + y$ is even, then $x$ and $y$ have the same parity.
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||||||
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||||||
|
#proof[
|
||||||
|
Suppose $x$ and $y$ had different parities. We seek to show that $x + y$ is
|
||||||
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odd.
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||||||
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||||||
|
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.*
|
||||||
37
documents/by-course/math-8/pset-3/package.nix
Normal file
37
documents/by-course/math-8/pset-3/package.nix
Normal file
|
|
@ -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;
|
||||||
|
}
|
||||||
|
)
|
||||||
Loading…
Reference in a new issue