diff --git a/documents/by-course/math-8/pset-1/dvd.typ b/documents/by-course/math-8/pset-1/dvd.typ index cb4d89b..1b6ae29 100644 --- a/documents/by-course/math-8/pset-1/dvd.typ +++ b/documents/by-course/math-8/pset-1/dvd.typ @@ -286,14 +286,23 @@ #let example = example-style("item", "Example") #let proof(body, name: none) = { - thmtitle[Proof] + [_Proof_] if name != none { [ #thmname[#name]] } - thmtitle[.] + [.] body h(1fr) - $square$ + + // 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 + + $square.stroked$ } #let fact = thmplain( diff --git a/documents/by-course/math-8/pset-1/main.typ b/documents/by-course/math-8/pset-1/main.typ index 40f9cdb..3a914a3 100644 --- a/documents/by-course/math-8/pset-1/main.typ +++ b/documents/by-course/math-8/pset-1/main.typ @@ -103,7 +103,258 @@ Problems: [F], [F], [T], [F], [T], [T], [F], [F], [F], [F], [F], [F], ) + ] \ + h. $not P and not Q$ + #table( + align: center, + columns: 5, + [$P$], [$Q$], [$not P$], [$not Q$], [$not P and not Q$], + [T], [T], [F], [F], [F], + [T], [F], [F], [T], [F], + [F], [T], [T], [F], [F], + [F], [F], [T], [T], [T], + ) \ + j. $(P and Q) or (P and R)$ + #table( + align: center, + columns: 6, + [$P$], [$Q$], [$R$], [$P and Q$], [$P and R$], [$(P and Q) or (P and R)$], + [T], [T], [T], [T], [T], [T], + [T], [F], [T], [F], [T], [T], + [T], [T], [F], [T], [F], [T], + [T], [F], [F], [F], [F], [F], + [F], [T], [T], [F], [F], [F], + [F], [F], [T], [F], [F], [F], + [F], [T], [F], [F], [F], [F], + [F], [F], [F], [F], [F], [F], + ) \ + L. $(P and Q) or (P and not S)$ + #table( + align: center, + columns: 7, + [$P$], + [$Q$], + [$S$], + [$not S$], + [$P and Q$], + [$P and not S$], + [$(P and Q) or (P and not S)$], + + [T], [T], [T], [F], [T], [F], [T], + [T], [F], [T], [F], [F], [F], [F], + [T], [T], [F], [T], [T], [T], [T], + [T], [F], [F], [T], [F], [T], [T], + [F], [T], [T], [F], [F], [F], [F], + [F], [F], [T], [F], [F], [F], [F], + [F], [T], [F], [T], [F], [F], [F], + [F], [F], [F], [T], [F], [F], [F], + ) + +4. \ + c. $(P or Q) and (R or S)$ is true. \ + d. $(not P or not Q) or (not R or not S)$ is true. \ + e. $not P or not Q$ is false. \ + f. $(not Q or S) and (Q or S)$ is false. \ + h. $K and not (S or Q)$ is false. + +6. \ + a. $not P and not Q$, $not (P and not Q)$ are not equivalent because we can choose $P$ and $Q$ to both be true which gives false for proposition 1 and true for proposition 2. \ + b. $(not P) or (not Q), not (P or not Q)$ are not equivalent because we can choose $P$ to be true and $Q$ to be false, and proposition 1 is true while proposition 2 is false. \ + c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent, by a truth table. + + #proof[ + Direct computation, by a truth table. Hopefully this suffices to show the absurdity of computing nontrivial boolean results by truth table. + #table( + align: center, + columns: (1fr, 1fr, 1fr, 1fr, 1fr, 2fr, 2fr), + [$P$], + [$Q$], + [$R$], + [$P and Q$], + [$Q or R$], + [$(P and Q) or R$], + [$P and (Q or R)$], + + [T], [T], [T], [T], [T], [T], [T], + [T], [F], [T], [F], [T], [T], [T], + [T], [T], [F], [F], [T], [T], [T], + [T], [F], [F], [F], [F], [F], [F], + [F], [T], [T], [F], [T], [T], [F], + [F], [F], [T], [], [], [], [], + [F], [T], [F], [], [], [], [], + [F], [F], [F], [], [], [], [], + ) + + We need not complete the table as we have already identified a contradiction for when $P$ and $Q$ are false, $R$ true. ] + #remark(numbering: none)[ + We can also show this result directly by + + $ + (P and Q) or R &= (P or R) and (Q or R) \ + P and (Q or R) &= (P or Q) and (P or R) + $ + + Clearly the choices of $P$ and $Q$ are false, $R$ is true again yields a contradictory result where proposition 1 is true while proposition 2 is false. + ] + + d. $not (P and Q), not P and not Q$ \ + + The statements are not equivalent. + + #proof[ + By DeMorgan's, + $ not (P and Q) = not P or not Q $ + Therefore, choosing $P,Q$ with $P != Q$ gives a contradictory result between the two propositions. + ] + + e. $(P and Q) or R, P or (Q and R)$ + + They are not equivalent. + + #proof[ + Symbolic manipulation gives + $ + (P and Q) or R &= (P or R) and (Q or R) \ + P or (Q and R) &= (P or Q) and (P or R) + $ + Then select the conjunctive statement in parentheses that occurs + exclusively in one of the statements and set both of its propositions to + false. In this case we take $Q$ and $R$ false while $P$ is true, which + makes proposition 1 false while proposition 2 is true. + ] + + f. $(P and Q) or P, P$ + + They are equivalent, trivially. Clearly $Q$ is independent of the outcome's truth value which only depends on $P$. + +7. \ + + c. Julius Caesar was born in 1492 or 1493 and died in 1776. \ + + This proposition is of the form $(P or Q) and R$, where + + $ + P &= #[Caesar was born in 1492] \ + Q &= #[Caesar was born in 1493] \ + R &= #[Caesar died in 1776] \ + $ + + It is false, clearly, as Caesar certainly could not have died in 1776. + + g. It is not the case that $-5$ and $13$ are elements of $NN$, but $4$ is in the set of rational numbers. + + This proposition takes the form $not (P and Q) and R$, where + + $ + P &= -5 in NN \ + Q &= 13 in NN \ + R &= 4 in QQ + $ + + It is true, since $P$ is false, $Q$ is true, $R$ is true, which implies $not + (P and Q)$ is true, thus making the proposition as a whole true. + +10. \ + c. $(P and Q) or (not P or not Q)$ \ + + This is a tautology that is true $forall P,Q$. + + #proof[ + #table( + align: center, + columns: 5, + [$P$], + [$Q$], + [$P and Q$], + [$not P or not Q$], + [$(P and Q) or (not P or not Q)$], + + [T], [T], [T], [F], [T], + [T], [F], [F], [T], [T], + [F], [T], [F], [T], [T], + [F], [F], [T], [T], [T], + ) + ] + + e. $(Q and not P) and not (P and R)$ \ + + This is neither a tautology nor a contradiction. + + #proof[ + #table( + align: center, + columns: 5, + [$P$], + [$Q$], + [$Q and not P$], + [$not (P and R)$], + [$(Q and not P) and not (P and R)$], + + [T], [T], [F], [F], [F], + [F], [T], [F], [T], [F], + [T], [F], [T], [T], [T], + [F], [F], [F], [T], [F], + ) + ] + +11. \ + + b. Cleveland will win the first game or the second game. + + Cleveland loses both the first and second games. + + e. Roses are red and violets are blue. + + Roses are green. + + i. The function $g$ has a relative maximum at $x = 2$ or $x = 4$ and a relative minimum at $x = 3$. + + $ g(x) "is given by" integral^x_0 (t - 5)(t-6) dif t $ + +12a. + +Restore parentheses to $not not P or not Q and not S$. + +$ not (not P) or (not Q and not S) $ + +13. \ + + a. Make a truth table for the exclusive or ($xor$) + + #table( + align: center, + columns: 3, + [$P$], [$Q$], [$P xor Q$], + [T], [T], [F], + [F], [T], [T], + [T], [F], [T], + [F], [F], [F], + ) + + b. Show that $A xor B = (A or B) and not (A and B)$. + + Intuitively this fact makes sense. Either $A$ or $B$ are true, but $A$ and $B$ can't _both_ be true. + + #proof[ + By direct computation, + + #table( + align: center, + columns: 6, + [$P$], + [$Q$], + [$P xor Q$], + [$A or B$], + [$not (A and B)$], + [$(A or B) and not (A and B)$], + + [T], [T], [F], [T], [F], [F], + [F], [T], [T], [T], [T], [T], + [T], [F], [T], [T], [T], [T], + [F], [F], [F], [F], [T], [F], + ) + ] = Exercises diff --git a/documents/by-course/pstat-120a/course-notes/main.typ b/documents/by-course/pstat-120a/course-notes/main.typ index 7d35b13..8211fba 100644 --- a/documents/by-course/pstat-120a/course-notes/main.typ +++ b/documents/by-course/pstat-120a/course-notes/main.typ @@ -4,7 +4,7 @@ #show: dvdtyp.with( title: "PSTAT120A Course Notes", author: "Youwen Wu", - date: "Winter 2024", + date: "Winter 2025", subtitle: "Taught by Brian Wainwright", ) @@ -27,11 +27,11 @@ upon them. #remark[ Keep in mind that without $cal(Z F C)$ or another model of set theory that resolves fundamental issues, our set theory is subject to paradoxes like - Russell's. + Russell's. Whoops, the universe doesn't exist. ] #definition[ - A Set is a collection of elements. + A *Set* is a collection of elements. ] #example[Examples of sets][ @@ -101,9 +101,9 @@ as the notation for $n$ dimensional spaces in $RR$? + $(A sect B)' = A' union B'$ ] -#remark[Generalized DeMorgan's][ - + $(union_i A_i)' = sect_i A_i'$ - + $(sect_i A_i)' = union_i A_i'$ +#fact[Generalized DeMorgan's][ + + $(union_i A_i)' = sect_i A_i '$ + + $(sect_i A_i)' = union_i A_i '$ ] == Sizes of infinity @@ -125,14 +125,26 @@ When a set is uncountably infinite, its cardinality is greater than $aleph_0$. + The natural numbers $NN$. + The rationals $QQ$. + The natural numbers $ZZ$. + + The set of all logical tautologies. ] #example("Uncountable sets")[ + The real numbers $RR$. + The real numbers in the interval $[0,1]$. + + The _power set_ of $ZZ$, which is the set of all subsets of $ZZ$. +] + +#remark[ + All the uncountable sets above have cardinality $2^(aleph_0)$ or $aleph_1$ or + $frak(c)$ or $beth_1$. This is the _cardinality of the continuum_, also + called "aleph 1" or "beth 1". + + However, in general uncountably infinite sets do not have the same + cardinality. ] #fact[ If a set is countably infinite, then it has a bijection with $ZZ$. This means - every set with cardinality $aleph_0$ has a bijection to $ZZ$. + every set with cardinality $aleph_0$ has a bijection to $ZZ$. More generally, + any sets with the same cardinality have a bijection between them. ]