From d0160fce8cf8d2b27e8afd8ec50722f86f9989ac Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Mon, 13 Jan 2025 02:08:30 -0800 Subject: [PATCH] auto-update(nvim): 2025-01-13 02:08:30 --- documents/by-course/math-8/pset-1/main.typ | 213 ++++++++++++++++++--- 1 file changed, 186 insertions(+), 27 deletions(-) diff --git a/documents/by-course/math-8/pset-1/main.typ b/documents/by-course/math-8/pset-1/main.typ index 3a914a3..978f8af 100644 --- a/documents/by-course/math-8/pset-1/main.typ +++ b/documents/by-course/math-8/pset-1/main.typ @@ -13,6 +13,8 @@ ), ) +#set par(first-line-indent: 0pt, spacing: 1em) + Problems: 1.1: \#1ceij, 2c, 3cdeghjL, 4cdefh, 6, 7cg, 10ce, 11bei, 12a, 13 @@ -27,16 +29,19 @@ Problems: True. ] \ + e. #[ Either $pi$ is rational and $17$ is a prime, or $7 < 13$ and $81$ is a perfect square. True. ] \ + i. #[ It is not the case that $39$ is prime, or that 64 is a power of 2. False. ] \ + j. #[ There are more than three false statements in this book, and this statement is one of them. @@ -160,35 +165,9 @@ Problems: 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. + c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent #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 $ @@ -358,3 +337,183 @@ $ not (not P) or (not Q and not S) $ ] = Exercises + +1. \ + b. If #math.underbrace("the moon is made of cheese", "antedecent"), then #math.underbrace("triangles have four sides", "consequent"). + + d. #math.underbrace([The differentiability of $f$], "antedecent") is sufficient for #math.underbrace[$f$ to be continuous][consequent]. + +2. \ + b. + + Converse: If triangles have four sides, then squares have three sides. + + Contrapositive: If triangles don't have four sides, then squares don't have three sides. + + #linebreak() + + d. + + Converse: The continuity of $f$ is sufficient for $f$ to be differentiable. + + Contrapositive: If $f$ is not continuous, then $f$ is not differentiable. + +3. \ + a. $Q$ must be true. + + #linebreak() + + b. $Q$ must be true. + + #linebreak() + + c. $Q$ is false. + + #linebreak() + + d. $Q$ is false. + + #linebreak() + + e. $Q$ is true. + +5. \ + c. True. + + #linebreak() + + f. True. + + #linebreak() + + g. True. + +6. \ + b. False. + + #linebreak() + + c. False + + #linebreak() + + g. False. + + #linebreak() + +7. \ + b. + + #table( + columns: 6, + align: center, + [$P$], + [$Q$], + [$not P$], + [$not P => Q$], + [$Q <=> P$], + [$(not P => Q) or (Q <=> P)$], + + [T], [T], [F], [T], [T], [T], + [F], [T], [T], [T], [F], [T], + [T], [F], [F], [T], [F], [T], + [F], [F], [T], [F], [T], [T], + ) + + #linebreak() + + e. + #table( + columns: 5, + align: center, + [$P$], [$Q$], [$not Q$], [$Q <=> P$], [$not Q => (Q <=> P)$], + [$T$], [$T$], [F], [T], [T], + [$F$], [$T$], [F], [F], [T], + [$T$], [$F$], [T], [F], [F], + [$F$], [$F$], [T], [T], [T], + ) + +10. \ + b. If $n$ is prime, then $n = 2$ or $n$ is odd. + + $ + n "is prime" => (n=2) or (n mod 2 = 1) + $ + + f. + + $ + 2 < n - 6 => 2n < 4 or n > 4 + $ + + g. + + $ + 6 >= n - 3 <=> n > 4 or n > 10 + $ + +12. \ + b. + + #table( + columns: 9, + align: center, + [$P$], + [$Q$], + [$R$], + [$not R$], + [$not Q$], + [$P and Q$], + [$P and not R$], + [$(P and Q) => R$], + [$(P and not R) => not Q$], + + [T], [T], [T], [F], [F], [T], [F], [T], [T], + [T], [F], [T], [F], [T], [F], [F], [T], [T], + [T], [T], [F], [T], [F], [T], [T], [F], [F], + [T], [F], [F], [T], [T], [F], [T], [T], [T], + [F], [T], [T], [F], [F], [F], [F], [T], [T], + [F], [F], [T], [F], [T], [F], [F], [T], [T], + [F], [T], [F], [T], [F], [F], [F], [T], [T], + [F], [F], [F], [T], [T], [F], [F], [T], [T], + ) + + So they are equivalent. + + #linebreak() + + c. + + #proof[ + $ + P => (Q and R) &<=> not (Q and R) => not P &&("contrapositive") \ + not (Q and R) => not P &<=> (not Q or not R) => not P space &&("DeMorgan's") \ + $ + ] + +13. \ + a. Continuity implies differentiability. + + #linebreak() + + b. Differentiability implies continuity. + + #linebreak() + + c. It is not possible. + + #linebreak() + + d. $P => Q$, where $P$ is true and $Q$ is true. + +16. \ + c. Contradiction. + + #linebreak() + + d. Neither. Statement is equivalent to $P => Q$. + + #linebreak() + + e. Tautology. $P and (Q or not Q)$ is independent of $Q$ so this reduces to + $P <=> P$.