From dfbe35333bfb6380982b8ea9757322348f9f3b3c Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Thu, 6 Feb 2025 00:26:56 -0800 Subject: [PATCH] auto-update(nvim): 2025-02-06 00:26:56 --- content/Math/pstat.md | 7 + documents/by-course/math-8/pset-4/main.typ | 273 ++++++++++++++++++ documents/by-course/math-8/pset-4/package.nix | 37 +++ 3 files changed, 317 insertions(+) create mode 100644 documents/by-course/math-8/pset-4/main.typ create mode 100644 documents/by-course/math-8/pset-4/package.nix diff --git a/content/Math/pstat.md b/content/Math/pstat.md index e69de29..0964549 100644 --- a/content/Math/pstat.md +++ b/content/Math/pstat.md @@ -0,0 +1,7 @@ +--- +id: pstat +aliases: [] +tags: [] +--- + + diff --git a/documents/by-course/math-8/pset-4/main.typ b/documents/by-course/math-8/pset-4/main.typ new file mode 100644 index 0000000..54a4b2e --- /dev/null +++ b/documents/by-course/math-8/pset-4/main.typ @@ -0,0 +1,273 @@ +#import "@youwen/zen:0.1.0": * + +#show: zen.with( + title: "Homework 4", + author: "Youwen Wu", +) + +#set heading(numbering: none) + +#set par(first-line-indent: 0pt, spacing: 1em) + +#let nonzero = $ZZ_(!=0)$ + +Problems: + +2.2: \#1dfh, 2bdf, 5, 6d, 7q, 8h, 9b, 10e, 11b, 13a, 15b, 16 + +2.3: \#1bhLmn, 2bhLmn, 3, 12, 16d, 18bd + += 2.2 + +*1d.* + +$A - (B - C) = {1,3,5,7,9}$ + +*1f.* + +$A union (C sect D) = {1,2,3,5,7,8,9}$ + +*1h.* + +${1,5,7}$ + +*2b.* + +$[2,8)$ + +*2d.* + +$[3,6]$ + +*2f.* + +$(6,8)$ + +*5.* + +Only $C$ and $D$ are disjoint. + +*6d.* + +$A = {1, 2}, B = {3, 4}, C = {2, 3}$ + +*7q.* + +Claim: if $A subset.eq B$, then $A union C subset.eq B union C$. + +#proof[ + Suppose $A subset.eq B$. Therefore $forall a in A, a in B$. This implies that + $forall a in A, a in A union C$, and $forall a in A, a in B union C$. + Note that $forall c in C$, both $c in A union C$ and $c in B union C$. Since + $forall x in A union C$, either $x in A$ or $x in C$, and either way $x in B$ + and $x in C$, then $A union B subset.eq A union C$. +] + +*8h.* + +Claim: $(A union B)^c = A^c sect B^c$. + +#proof[ + For all $x$ in $(A union B)^c$, $x$ is not in $A union B$. Therefore $x$ is + not in $A$ and $x$ is not in $B$. In other words $(A union B)^c$ is comprised of all $x$ not in $A$ and not in $B$, which are all the elements in both $A^c$ and $B^c$, which is $A^c sect B^c$. +] + +*9b.* + +Claim: If $A subset.eq B union C$ and $A sect B = emptyset$, then $A subset.eq C$. + +#proof[ + Suppose $A subset.eq B union C$. Then $forall a in A$, either $a in B$ or $a + in C$. However $a$ is never in $B$, as $A$ and $B$ are disjoint and $a in B$ + would be a contradiction of this fact. So in fact the only case is $a in C$. + Therefore we have $forall a in A, a in C$, which is the definition of $A + subset.eq C$. +] + +*10e.* + +Claim: if $A union B subset.eq C union D$, $A sect B = emptyset$, and $C subset.eq A$, then $B subset.eq D$. + +#proof[ + Suppose $C subset.eq A$. Then $forall c in C, c in A$. Since $A$ and $B$ are + disjoint, $forall b in B, b in.not C$. Because we assume $A union B subset.eq + C union D$, $forall x in A union B, x in C union D$. Then this implies + $forall a in A$ and $forall b in B$, either $a in C$, or $b in C$, or $a in + D$, or $b in D$. However we established earlier that $forall b in B, b in.not + C$. This means that $forall b in B, b in D$. + Therefore $B subset.eq D$. +] + +*11b.* + +Claim: if $A sect C subset.eq B sect C$, then $A subset.eq B$. + +Counterexample: $A = {1,2,3}, B = {2,3,4}, C = {2,3,4}$. + +*13a.* + +(I wrote a Vim macro to compute these because it was so annoying). +$ + A times B = \ + {(1,a), (1, e), (1, k), (1, n), (1, r), (3,a), (3, e), (3, k), (3, n), (3, r), (5,a), (5, e), (5, k), (5, n), (5, r)} \ + B times A = \ + {(a,1), (e, 1), (k, 1), (n, 1), (r, 1), (a,3), (e, 3), (k, 3), (n, 3), (r, 3), (a,5), (e, 5), (k, 5), (n, 5), (r, 5)} \ +$ + +*15b.* + +Claim: $A times emptyset = emptyset$. + +#proof[ + Let there be a set $X$ such that $A times emptyset = X$. Then any element in + $X$ is an ordered pair $(a,b)$ such that $a in A$ and $b in emptyset$. + But $b in emptyset$ is a contradiction no ordered pair can exist. Therefore + $X$ contains no elements and $X = emptyset$, so $A times emptyset = + emptyset$. +] + +*16a.* + +$ + A = {1}, B = {2}, C = {3}, D = {4} +$ + +*16b.* + +$ + A = {1}, B = {2}, C = {2} +$ + +*16c.* + +$ + A = {1}, B = {2}, C = {3} +$ + += 2.3 + +*1b.* + +$ + union.big_(A in cal(A)) A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} \ + sect.big_(A in cal(A)) A = emptyset +$ + +*1h.* + +$ + union.big_(A in cal(A)) A = [-pi, infinity) \ + sect.big_(A in cal(A)) A = [-pi, 0] +$ + +*1L.* + +$ + union.big_(L in cal(l)) L = (-infinity, infinity) \ + sect.big_(L in cal(l)) L = emptyset +$ + +*1m.* + +$ + union.big_(A in cal(A)) A = (-infinity, infinity) - ZZ \ + sect.big_(A in cal(A)) A = emptyset +$ + +*1n.* + +$ + union.big_(D in cal(D)) D = (-infinity, 1) \ + sect.big_(D in cal(D)) D = (-1, 0] +$ + +*2b.* + +No. + +*2h.* + +No. + +*2L.* + +Yes. + +*2m.* + +Yes. + +*2n.* + +No. + +*3a.* + +Claim: let $cal(A)$ be a family of sets, for every set B in the family +$cal(A)$, $B subset.eq union.big _(A in cal(A)) A$ + +#proof[ + By the definition of the union of sets, $forall b in B$, $b in union.big _(A + in cal(A)) A$. Therefore $B subset.eq union.big _(A in cal(A))$. +] + +*3b.* + +Claim: let $cal(A)$ be a nonempty family of sets and $B$ be a set. Then if $A +subset.eq B$ for all $A in cal(A)$, then $union.big _(A in cal(A)) A subset.eq +B$. + +#proof[ + Suppose $A subset.eq B$ for all $A in cal(A)$. Then $forall a in A, a in B$ + for all $A in cal(A)$. Then note that $forall in union.big _(A in cal(A)) + A$, there exists some $A$ with $x in A$. This implies that any element of + $union.big _(A in cal(A)) A$ is an element of some $A in cal(A)$. By our + earlier assumption this implies that any element in $union.big _(A in cal(A)) + A$ is an element of $B$. This is the definition of $union.big _(A in cal(A)) + A subset.eq B$. +] + +*12a.* + +$ + cal(A) = {{1, x} : x in {2, 3, dots.c, 20}} +$ + +*12b.* + +$ + cal(B) = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}} +$ + +*12c.* + +$ + cal(l) = {{x} : x in {1, 2, 3, dots.c, 20}} +$ + +*16d.* + +Claim: $sect.big _(i = 1) ^infinity A_i subset.eq sect.big _(i=k) ^m A_i$. + +#proof[ + Note that + + $ + sect.big_(i = 1)^infinity A_i = (sect.big_(i = 1)^(k-1) A_i) sect (sect.big_(i = k)^m A_i) sect (sect.big_(i = m + 1)^infinity A_i) + $ + + This implies that the set $sect.big _(i=1) ^infinity A_i$ contains only + elements belonging to (by the definition of the set intersection) $sect.big + _(i=k) ^m A_i$. Therefore $sect.big _(i=1) ^infinity A_i$ is indeed a subset + of $sect.big _(i=k) ^m A_i$. +] + +*18b.* + +$ + {(-infinity ,infinity), (-infinity, 1]} +$ + +*18d.* + +${{1, 2, 3}, {1, 2}, {1}, emptyset}$ diff --git a/documents/by-course/math-8/pset-4/package.nix b/documents/by-course/math-8/pset-4/package.nix new file mode 100644 index 0000000..7879e7e --- /dev/null +++ b/documents/by-course/math-8/pset-4/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; + } +)