commit 18763aa5420435ec2f695d351ae6097565a893f1 Author: tomit4 Date: Sun May 10 18:53:46 2026 -0700 :tada: Initial commit! diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..a136337 --- /dev/null +++ b/.gitignore @@ -0,0 +1 @@ +*.pdf diff --git a/LICENSE b/LICENSE new file mode 100644 index 0000000..fdddb29 --- /dev/null +++ b/LICENSE @@ -0,0 +1,24 @@ +This is free and unencumbered software released into the public domain. + +Anyone is free to copy, modify, publish, use, compile, sell, or +distribute this software, either in source code form or as a compiled +binary, for any purpose, commercial or non-commercial, and by any +means. + +In jurisdictions that recognize copyright laws, the author or authors +of this software dedicate any and all copyright interest in the +software to the public domain. We make this dedication for the benefit +of the public at large and to the detriment of our heirs and +successors. We intend this dedication to be an overt act of +relinquishment in perpetuity of all present and future rights to this +software under copyright law. + +THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, +EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF +MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. +IN NO EVENT SHALL THE AUTHORS BE LIABLE FOR ANY CLAIM, DAMAGES OR +OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, +ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR +OTHER DEALINGS IN THE SOFTWARE. + +For more information, please refer to diff --git a/README.md b/README.md new file mode 100644 index 0000000..4305ae4 --- /dev/null +++ b/README.md @@ -0,0 +1,4 @@ +# Discrete Mathematics: An Open Introduction + +This repository holds my notes as I do the exercises for +[Discrete Mathematics: An Open Introduction, by Oscar Levin](https://discrete.openmathbooks.org/) diff --git a/chapter_0/0_1/investigate_0.md b/chapter_0/0_1/investigate_0.md new file mode 100644 index 0000000..7e6ec67 --- /dev/null +++ b/chapter_0/0_1/investigate_0.md @@ -0,0 +1,163 @@ +1. + +Q: + +The most popular mathematician in the world is throwing a party for all of his +friednds. To kick things off, they decide that everyone should shake hands. +Assuming all 10 people at the party each shake hands with every other person +(but not themselves, obviously) exactly once, how many handshakes take place? + +1A: + +If we pair off the people into sets, then we get something like this: + +{1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9}, {1, 10} + +{2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9}, {2, 10} + +{3, 4}, {3, 5}, {3, 6}, {3, 7}, {3, 8}, {3, 9}, {3, 10} + +{4, 5}, {4, 6}, {4, 7}, {4, 8}, {4, 9}, {4, 10} + +{5, 6}, {5, 7}, {5, 8}, {5, 9}, {5, 10} + +{6, 7}, {6, 8}, {6, 9}, {6, 10} + +{7, 8}, {7, 9}, {7, 10} + +{8, 9}, {8, 10} + +{9, 10} + +As you can see this just becomes 10+9+8+7+6+5+4+3+2+1=55 handshakes. This likely +is alluding to basic summation mathematical induction. + +2. + +Q: At the warm-up event for Oscar's All-Star Hot Dog Eating Contest, Al ate one +hot dog. Bob then showed him up by eating three hot dogs. Not to be outdone, +Carl ate five. This continued with each contestant eating two more hot dogs than +the previous contestant. How many hot dogs did Zeno (the 26th and final +contestant) eat? How many hot dogs were eaten in total? + +A: + +This is another summation mathematical induction function, just slightly +different: + +(1) + + +(1 + 2) + + +(3 + 2) + + +(5 + 2) + + +(7 + 2) + + +(9 + 2) + + +(11 + 2) + + +(13 + 2) + + +(15 + 2) + + +(17 + 2) + + +(19 + 2) + + +(21 + 2) + + +(23 + 2) + + +(25 + 2) + + +(27 + 2) + + +(29 + 2) + + +(31 + 2) + + +(33 + 2) + + +(35 + 2) + + +(37 + 2) + + +(39 + 2) + + +(41 + 2) + + +(43 + 2) + + +(45 + 2) + + +(47 + 2) + + +(49 + 2) + + +(51 + 2) + + +(53 + 2) = 55 + +(1) + (1 + 2) + (3 + 2) + (5 + 2) + (7 + 2) + (9 + 2) + (11 + 2) + (13 + 2) + +(15 + 2) + (17 + 2) + (19 + 2) + (21 + 2) + (23 + 2) + (25 + 2) + (27 + 2) + +(29 + 2) + (31 + 2) + (33 + 2) + (35 + 2) + (37 + 2) + (39 + 2) + (41 + 2) + +(43 + 2) + (45 + 2) + (47 + 2) + (49 + 2) + (51 + 2) + (53 + 2) = 784 + +55 is how many Zeno ate, and the total hot dogs that were eaten is : 784 + +3. + +Q: After excavating for weeks, you finally arrive at the burial chamber. The +room is empty except for two large chests. On each is carved a message +(strangely English): + +- Exactly one of these chests contains a treasure, while the other is filled + with deadly immortal scorpions. + +- For either chest, if the chest's message is true, then the chest contains + treasure. + +The problem is, you don't know whether the messages are true or false. What do +you do? + +A: Honestly, I'd probably walk away, but since that's likely not the answer, +we'd likely have to consider that these are relating to a form of truth tables. + +The first chest's message claims that one chest contains scorpions, the other +treasure + +So if we say T="contains treasure", and f="contains scorpions", then we can say +that Chest_1 and Chest_2: + +Chest_1 = T Chest_2 = F + +The other chest's message claims that if it's own message or the other chest's +message is true, then the chest contains treasure. + +This is a contradiction, because if the second chest's message is true, then +that chest is the one that contains treasure, and the other chest is the one +that contains scorpions. + +If the second chest's message is false, then the chest doesn't contain treasure, +but then you don't know if the second chest contains treasure or scorpions, +because the chest containing treasure is no longer contingent on whether the +chest's message is true or not. + +This is probably best expressed again, in some form, of truth notation. + +4. + +Q: Back in the days of yore, five small towns decided they wanted to build roads +directly connecting each pair of towns. While the towns had plenty of money to +build roads as long and as winding as they wished, it was very important that +the roads not intersect with each other (as stop signs had not yet been +invented). Also, tunnels and bridges were not allowed, for moral reasons. It is +possible for each of these towns to build a road to each of the four other towns +without creating any intersections? + +A: This seems possible, as you could just create two roads from each town to at +least two other towns (a pair as the question alludes to). But something tells +me this is a trick question. diff --git a/chapter_0/0_1/reading_questions.md b/chapter_0/0_1/reading_questions.md new file mode 100644 index 0000000..ffda080 --- /dev/null +++ b/chapter_0/0_1/reading_questions.md @@ -0,0 +1,16 @@ +1. Right now, how would you describe what **discrete** mathematics is about, if + you were telling your friends about the class you are in? Write one or two + sentences. + +Discrete Mathematics is about performing some mathematical logic where the +inputs and outputs of some function contain only elements that are separate, +_i.e._ they are discrete. The main problems solved by discrete mathematics deal +with cominatorics, sequences, symbolic logic, and graph theory, though there are +others. + +2. What questions do you have after reading this section? Write at least one + question about the content of this section that you are curious about. + +The chest problem interested me as the second chest's message creates a logical +fallacy, such as "This statement is false." I wonder how this problem would be +"solved", if it's even possible, with discrete math. diff --git a/chapter_0/0_2/0_2_8_reading_questions.md b/chapter_0/0_2/0_2_8_reading_questions.md new file mode 100644 index 0000000..9ad3f0f --- /dev/null +++ b/chapter_0/0_2/0_2_8_reading_questions.md @@ -0,0 +1,52 @@ +1. + +Q: Think back to the domino problem, at the beginning of this section. We asked +how many dominoes are in a double-six domino _set_. Is this really a set, in our +mathematical sense? What discrete structure would you use to represent each +domino individually? + +A: No, the domino set is _not_ a set, because the order of the number matters so +you don't repeat any pair. Thusly a sequence, or tuple, would be more +appropriate to represent each domino individually. + +2. + +Q: A double-zero domino set would contain only one domino (both sides showing +0). A double-one set would contain this plus the dominoes (1, 0) and (1, 1). We +can continue in this way, creating a sequence of domino sets. Find the next +three terms of this sequence. + +1, 3, _, _, _ ... + +A: 1, 3, 6, 10, 15 + +The reason for this is we are summing up all the dominos, you can count starting +like we wrote out in our investigation: + +(0, 6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6) + +(0, 5), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5) + +(0, 4), (1, 4), (2, 4), (3, 4), (4, 4) = 5 tuples + +(0, 3), (1, 3), (2, 3), (3, 3) = 4 tuples + +(0, 2), (1, 2), (2, 2) = 3 tuples + +(0, 1), (1, 1) = 2 tuples + +(0, 0) = 1 tuples + +1, (1 + 2), (1 + 2 + 3), (1 + 2 + 3 + 4), (1 + 2 + 3 + 4 + 5) + +3. + +Q: What questions do you have after reading this section? Write at least one +question about the content of this section that you are curious about. + +A: The author makes not of graphs with edges and vertices that are related by +some function. In the example given, there are graphs where edges are made based +off of summing up to 7, or in the other, based off those same sets are connected +by edges based off of if their sum is even. I don't really have a question, but +the logic based off of these relationships are established. In other words, how +the cardinality is established and why. diff --git a/chapter_0/0_2/investigate_0.md b/chapter_0/0_2/investigate_0.md new file mode 100644 index 0000000..36921b0 --- /dev/null +++ b/chapter_0/0_2/investigate_0.md @@ -0,0 +1,36 @@ +2.0: + +Q: A double-six domino set consists of tiles containing paris of numbers, each +from 0 to 6. How many tiles are in a double-six domino set? How many dominoes +are in a double-nine domino set? How many dominoes are in a double-$n$ domino +set? + +This is a combinatorics problem. It is a summation. + +ON a double-six domino set, we can think of each domino being like a bunch of +tuples: + +(0, 6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6) + +(0, 5), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5) + +(0, 4), (1, 4), (2, 4), (3, 4), (4, 4) + +(0, 3), (1, 3), (2, 3), (3, 3) + +(0, 2), (1, 2), (2, 2) + +(0, 1), (1, 1) + +(0, 0) + +A: The author makes not of graphs with edges and vertices that are related by +some function. In the example given, there are graphs where edges are made based +off of summing up to 7, or in the other, based off those same sets are connected +by edges based off of if their sum is even. I don't really have a question, but +the logic based off of these relationships are established. In other words, how +the cardinality is established and why. + +And on a double-nine: + +(0, 9), (1, 9), (2, 9) diff --git a/chapter_1/1_1/1_1_10.md b/chapter_1/1_1/1_1_10.md new file mode 100644 index 0000000..3a3317b --- /dev/null +++ b/chapter_1/1_1/1_1_10.md @@ -0,0 +1,167 @@ +Q: Using the truth conditions for the logical connectives, determine which +statements below are true and which are false. + +1. 17 is prime, and 17 is odd. + +2. 17 is prime, and 18 is prime. + +3. 17 is prime, or 18 is prime. + +4. 17 is prime, or 19 is prime. + +5. If 17 is prime, then 19 is prime. + +6. If 18 is prime, then my favorite number is 17. + +7. 17 is prime if and only if 19 is prime. + +8. 17 is not prime if and only if 19 is not prime. + +A: + +1. 17 is prime, and 17 is odd. + +This equates to a $P \wedge Q$ statement. + +$P = \text{ 17 is prime}$ + +$Q = \text{ 17 is odd}$ + +Recall the truth table for $P \wedge Q$, which equates to the expected output +that 17 is prime, which is true: + +$P = \text{ 17 is prime} = \text{ T}$ + +and also the statement 17 is odd, which is also true: + +$Q = \text{ 17 is odd} = \text{ T}$. + +This equates to the truth table row: + +| $P$ | $Q$ | $P\wedge Q$ | +| --- | --- | ----------- | +| T | T | T | + +There for this statement is true. + +2. 17 is prime, and 18 is prime. + +This can be followed similar to one, we already know the first statement is +true, but the second statement, $Q$, is not true (_i.e._ 18 is not prime, it can +be divided by a whole number). + +Therefore: + +$P = \text{ 17 is prime} = \text{ T}$ + +$Q = \text{ 18 is prime} = \text{ F}$. + +From our truth table we know that: + +| $P$ | $Q$ | $P\wedge Q$ | +| --- | --- | ----------- | +| T | F | F | + +The statement is false. + +3. 17 is prime, or 18 is prime. + +This correlates to $P \vee Q$, and in this case only $P$ or $Q$ must be true, +and since these are the same atomic statements as number 2, we know that $P$ is +true, and $Q$ is false, and so therefore: + +$P = \text{ 17 is prime} = \text{ T}$ + +$Q = \text{ 18 is prime} = \text{ F}$. + +From our truth table we know that: + +| $P$ | $Q$ | $P\vee Q$ | +| --- | --- | --------- | +| T | F | T | + +The statement is true + +4. 17 is prime, or 19 is prime. + +Similar to 3, except here we also know that 19 is prime, so our statements look +like this: + +$P = \text{ 17 is prime} = \text{ T}$ + +$Q = \text{ 19 is prime} = \text{ T}$. + +From our truth table we know that: + +| $P$ | $Q$ | $P\vee Q$ | +| --- | --- | --------- | +| T | T | T | + +The statement is true. + +5. If 17 is prime, then 19 is prime. + +This is referencing $P\to Q$, which does confuse us a bit, so be careful here. + +Here we know both statements are true, so this corresponds to our truth table +as: + +$P = \text{ 17 is prime} = \text{ T}$ + +$Q = \text{ 19 is prime} = \text{ T}$. + +From our truth table we know that: + +| $P$ | $Q$ | $P\to Q$ | +| --- | --- | -------- | +| T | T | T | + +The statement is true. + +6. If 18 is prime, then my favorite number is 17. + +Interestingly here, we have no way of validating whether 17 is our favorite +number or not, so we cannot really validate if $Q$ is true or false. Or can we? + +Consulting the table for $P\to Q$, we find: + +| $P$ | $Q$ | $P\to Q$ | +| --- | --- | -------- | +| T | T | T | +| T | F | F | +| F | T | T | +| F | F | T | + +And we already know that $P$ is false, because 18 is not a prime number! +Regardless, if $P$ is false, the conclusion of $P\to Q$ is always true, as you +can see in the above table. + +The statement is true. + +7. 17 is prime if and only if 19 is prime. + +This is a $P \leftrightarrow Q$ statement. + +$P = \text{ 17 is prime} = \text{ T}$ + +$Q = \text{ 19 is prime} = \text{ T}$. + +From our truth table we know that: + +| $P$ | $Q$ | $P\leftrightarrow Q$ | +| --- | --- | -------------------- | +| T | T | T | + +The statement is true. + +8. 17 is not prime if and only if 19 is not prime. + +This technically is still a $P \leftrightarrow Q$ statement. Let's see what the +solution has to say here, because honestly I'm a little confused. + +Okay, so this is true as well. Consider number 7 now and realize that we're +saying that $P$ is false, since 17 is a prime number, but our assertion states +that it is not (a false statement). As mentioned in 7, if $P$ in $P\to Q$ is +false, then $Q$ is always true. + +The statement is true. diff --git a/chapter_1/1_1/1_1_11.md b/chapter_1/1_1/1_1_11.md new file mode 100644 index 0000000..86755d0 --- /dev/null +++ b/chapter_1/1_1/1_1_11.md @@ -0,0 +1,47 @@ +Identify the logical structure of each of the following statements. + +1. 4 and 5 are both prime. + +2. Only one of 4 or 5 is prime. + +3. You must attend every day and do the homework to pass this class. + +4. Every number is even or odd. + +A: + +1. 4 and 5 are both prime. + +This follows the $P \wedge Q$ statement (in this case it is false). It is +asserting that 4 is prime and that 5 is prime. 4 is not prime, and therefore the +statement is false. + +2. Only one of 4 or 5 is prime. + +We can't just put an "or" on one side of the statement here. we have to take +each proposition as it's own. + +Where $P = \text{ 4 is prime}$ and $Q = \text{ 5 is prime}$ + +$$ (P \vee Q) \wedge \neg (P \wedge Q) $$ + +This equates to: + +$$ \text{T} \wedge \neg \text{F} = \text{T} $$ + +3. You must attend every day and do the homework to pass this class. + +This one actually the assertion is at the end. + +If you passed the class ($P$), then we conclude that you must have attended +every day ($Q$) and also done the homework ($R$): + +$$ P \to (Q \wedge R) $$ + +4. Every number is even or odd. + +We don't yet have the terminology to express whether a number is even or odd +(this likely means we express it as being in some set or the output of some +function), but we can ascertain that this is a standard or statement. + +$$ P \vee Q $$ diff --git a/chapter_1/1_1/1_1_13.md b/chapter_1/1_1/1_1_13.md new file mode 100644 index 0000000..8be09fa --- /dev/null +++ b/chapter_1/1_1/1_1_13.md @@ -0,0 +1,18 @@ +Translate the statement "Every number is even or odd" into symbols. + +If we define: + +$$ E(x) = x \text{ is even} $$ + +$$ O(x) = x \text{ is odd} $$ + +Then we can say or statement is closer to: + +$$ E(x) \vee O(x) $$ + +A: + +But we need to express that this is "every number", which we can do with +$\forall$, _i.e._ "for all": + +$$ \forall x (E(x) \vee O(x)) $$ diff --git a/chapter_1/1_1/1_1_14.md b/chapter_1/1_1/1_1_14.md new file mode 100644 index 0000000..1969ad7 --- /dev/null +++ b/chapter_1/1_1/1_1_14.md @@ -0,0 +1,5 @@ +## Definition 1.1.14 + +Given a sentence with free variables, the **universal generalization** of that +sentence is the statement obtained bya dding enough universal quantifiers to the +beginning of the sentence so that all free variables become bound. diff --git a/chapter_1/1_1/discrete_math_truth_tables.png b/chapter_1/1_1/discrete_math_truth_tables.png new file mode 100644 index 0000000..88c302e Binary files /dev/null and b/chapter_1/1_1/discrete_math_truth_tables.png differ diff --git a/chapter_1/1_1/investigate_1_1_1.md b/chapter_1/1_1/investigate_1_1_1.md new file mode 100644 index 0000000..6e301fa --- /dev/null +++ b/chapter_1/1_1/investigate_1_1_1.md @@ -0,0 +1,42 @@ +Q: While walking through a fictional forest, you encounter three trolls guarding +a bridge. Each is either a _knight_, who always tells the truth, or a _knave_, +who always lies. The trolls will not let you pass until you correctly identify +each as either a knight or a knave. Each troll makes a single statement. + +Troll 1: If I am a knave, then there are exactly two knights here. + +Troll 2. Troll 1 is lying. + +Troll 3: Either we are all knaves, or at least one of us is a knight. + +Which troll is which? + +A: + +Let's think this through step by step by assuming the first Troll is lying. + +If "If I am a knave, then there are exactly two knights here." = False: + +Then: Troll 1 = knave, BUT that doesn't meant that there are exactly two +knights. In other words, the other two could be either knaves or knights. But we +know that this troll is a knave. + +Moving on, we have: + +Troll 2: "Troll 1 is lying" + +If Troll 2 is lying, then Troll 1 is not lying (i.e. a knight), and this +invalidates our previous assumption that Troll 1 is lying, and therefore we +still don't know if the other two trolls are knights. + +Moving on to troll 3: + +Troll 3: "Either we are all knaves, or at least one of us is a knight." + +If we assume troll 3 is also lying, then all three trolls are knaves, and we +have solved their riddle and pass, because he is no knight. + +These are just my initial thoughts on this based off the prompt that we could +start just thinking about this if Troll 1 were lying. + +$$ \therefore $$ diff --git a/chapter_1/1_1/latex_terms.md b/chapter_1/1_1/latex_terms.md new file mode 100644 index 0000000..abe1d42 --- /dev/null +++ b/chapter_1/1_1/latex_terms.md @@ -0,0 +1,71 @@ +## Definition 1.1.7 Logical Connectives. + +We define the following **logical connectives.** + +- $P \wedge Q$ is read "P and Q", and is called a **conjunction**. + +- $P \vee Q$ is read "P or Q", and is called a **disjunction**. + +- $P \to Q$ is read "if P then Q", and is called an **implication** or + **conditional**. + +- $P \leftrightarrow Q$ is read "P if and only if Q" and is called a + **biconditional**. + +- $$ \neg P $$ + is read "not P", and is called a **negation**. + +## Definitions 1.1.8 Truth Conditions for Connectives. + +The **truth conditions** for the logical connectives are defined as follows. + +- $P \wedge Q$ is true when both $P$ and $Q$ are true. + +- $P \vee Q$ is trueh when $P$ or $Q$ or both are true. + +- $P \to Q$ is true when $P$ is false or $Q$ is true (or both). + +- $P \leftrightarrow Q$ is true when $P$ and $Q$ are both true, or both false. + +- $\neg P$ is true when $P$ is false. + +**Personal Notes** + +The statement $P \to Q$ was confusing for me. ChatGPT expalains it as: + +The rule given in your book is: + +$P \to Q$ is true when $P$ is false, or $Q$ is true (or both). + +That sounds weird, so let’s rephrase: + +The only time “If P, then Q” is false is if P happens but Q doesn’t happen. In +every other situation, the statement is considered true. + +Think of it as a promise: + +“If I do X, then Y will happen.” + +The promise is broken only if I do X and Y fails to happen. Otherwise, the +promise hasn’t been broken. + +Step 3: + +Truth table example + +| P (It rains) | Q (Ground wet) | P → Q (If it rains, then ground gets wet) | +| ------------ | -------------- | ----------------------------------------------- | +| True | True | True (rain happened, ground got wet ✅) | +| True | False | False (rain happened, ground stayed dry ❌) | +| False | True | True (didn’t rain, but ground is wet anyway ✅) | +| False | False | True (didn’t rain, ground isn’t wet ✅) | + +Notice that whenever it didn’t rain ($P$ is false), we consider the statement +“If it rains, then the ground gets wet” as true, because the promise about rain +hasn’t been broken. + +## Definition 1.1.12 Quantifiers + +The **universal quantifier** is written as $\forall$ and is read, "for all." The +**existential quantifier** is written $\exists$ and is read, "there exists" or +"for some." diff --git a/chapter_1/1_1/practice_problems.md b/chapter_1/1_1/practice_problems.md new file mode 100644 index 0000000..6e5a24b --- /dev/null +++ b/chapter_1/1_1/practice_problems.md @@ -0,0 +1,426 @@ +1. + +Q: For each sentence below, decide whether it is an atomic statement, a +molecular statement, or not a statement at all. + +(a) Some say the end is near, and some say we'll see Armageddon soon. + +(b) Mom's coming 'round to put it back the way it ought to be. + +(c) Learn to swim. + +A: + +(a) This is moleculr statement, two predicates are asserted, one is that "Some +say the end is near" and the other is "some say we'll see armageddon". + +If we say $P(x)$ is "some say the end is near", and $Q(x)$ is "some say we'll +see Armageddon soon", we can write this as: + +$$ \exists x \left(P(x) \wedge Q(x)\right) $$ + +(b) This can be reversed as "if there's a way it ought to be, then Mom is coming +'round to put it back." + +If we say that $x$ is "it", and $P(x)$ is "the way it ought to be", and $Q(x)$ +is "Mom is coming round to put it back," then we can write this as: + +$$ P(x) \to Q(x) $$ + +This is an atomic statement, as both $P(x)$ and $Q(x)$ cannot be divided into +smaller elements. + +(c) Learn to swim. + +This is not a statement, there is no assertion made. + +--- + +2. + +Q: Classify each of the sentences below as an atomic statement, a molecular +statement, or not a statement at all. If the statement is molecular, say what +kind it is (conjunction, disjunction, conditional, biconditional, negation). + +(a) Everybody can be fooled sometimes. + +This is molecular due to the "sometimes" statement. + +If we say that $x$ is somebody, and $y$ is some point in time. And if we say +that $P(x)$ is some situation somebody is in, and $Q(y)$ is "being fooled at +some point in time", then we can express this as: + +$$ \forall x \exists y\left(P(x) \to Q(y)\right) $$ + +And this is a conjunction. + +W: atomic + +(b) Every natural number greater than 1 is either prime or composite. + +This is also molecular. Let's break this down. If we say that $x$ is some +natural number, and $P(x)$ is "$x$ is prime", and $Q(x)$ is "$x$ is composite", +then we can say that: + +$$ \forall (x > 1) \in \mathbb{N} \left(P(x) \vee Q(x)\right) $$ + +This is an disjunction. + +(c) Go to your room! + +This is not a statement as no assertion is made. + +(d) The Broncos will win the Super Bowl, or I'll eat my hat. + +This is an atomic statement. If we say that $P$ is "The Broncos will win the +Super Bowl", and we say that $Q$ is "I'll eat my hat". Then we can express this +as: + +$$ P \vee Q $$ + +W: this is molecular. + +(e) This shirt is not black. + +Here an assertion is made, but no conclusion is drawn, so this is not a +statement. If we say that $P$ is "This shirt is black", then we are only left +with: + +$$ \neg P $$ + +And there is no conclusion $Q$ to be drawn. + +This is not a statement. + +W: this is a statement. $\neg P$ is a statement. + +--- + +3. + +Q: Determine whether each molecular statement below is true or false, or whether +it is impossible to determine. Assume you do not know what my favorite number is +(but you do know which numbers are prime). + +(a) If 4 is my favorite number, then 4 + 1 is my favorite number. + +(b) 8 is my favorite number, and 3 is not prime. + +(c) 4 is my favorite number, or 4 is prime. + +(d) If 4 is prime then 2 $\cdot$ 4 is prime. + +(e) If 3 is prime, then 3 is my favorite number. + +(f) 8 is my favorite number, and 4 is not prime. + +A: + +(a) If 4 is my favorite number, then 4 + 1 is my favorite number. + +We can not determine this as we do not know if $4$ is my favorite number. + +$$ P(4) \to P(5) $$ + +(b) 8 is my favorite number, and 3 is not prime. + +False, although we are saying that $P(8)$ is true, our original statement says +we cannot know that, and we say that Q(3) is "3 is prime", then we can say + +$$ P(8) \wedge \neg Q(3) $$ + +Which is a false statement as, again $P(8)$ is unknown and $\neg Q(3)$ is false. + +(c) 4 is my favorite number, or 4 is prime. + +This is a false statement, we are saying that $P(4)$ is true, but we cannot know +that (like in problem 3). And we're saying that $Q(4)$ as being "4 is prime". +While $Q(4)$ is not true, $P(4)$ is unknown, and this implies the following: + +$$ P(4) \vee Q(4) $$ + +Which is a false statement since $P(4)$ is unknown and $Q(4)$ is false. + +(d) If 4 is prime then 2 $\cdot$ 4 is prime. + +This is a true statement. If we say that $P(x)$ is true if 4 is prime, then we +can say that $Q(x)$ is true if 2 $\cdot$ 4 is prime. + +$$ P(x) \to Q(x) $$ + +Since $P(x)$ and $Q(x)$ is false, this means that $P(x) \to Q(x)$ is true (see +truth tables). + +(e) If 3 is prime, then 3 is my favorite number. + +This is a not possible to determine. We say that $P$ is "3 is prime", which is +true, but then we say that $Q$ is "3 is my favorite number", it follows that: + +$$ P \to Q $$ + +Since $P$ is true, but $Q$ is unknown, we cannot determine the validity of this +statement + +(f) 8 is my favorite number, and 4 is not prime. + +We can say that this is not possible to determine. We can declare $P(8)$ is "8 +is my favorite number" as true, but we cannot know that. Also "4 is prime" is +expressed as $Q(4)$, but is false so therefore: + +$$ P(8) \wedge \neg Q(4) $$ + +Is unkonwn because $P(8)$ is unknown and $\neq Q(4)$ is true, but both must be +true for this statement to be true, and so therefore this statement is not +possible to determine. + +4. + +Q: Let $P(x, y)$ be the predicate, "person $x$ can be fooled at time $y$." + +Match each statement with its representation in symbols. + +| | | +| ------------------------------------------------ | ----------------------------- | +| It is always true that some people can be fooled | $\exists x \forall y P(x, y)$ | +| Sometimes everyone can be fooled. | $\forall x \exists y P(x, y)$ | +| Everyone can be fooled sometimes. | $\forall y \exists x P(x, y)$ | +| Some people can be fooled all of the time. | $\exists y\forall x P(x, y)$ | + +A: + +| | | +| ------------------------------------------------ | ----------------------------- | +| It is always true that some people can be fooled | $\forall y \exists x P(x, y)$ | +| Sometimes everyone can be fooled. | $\exists y \forall x P(x, y)$ | +| Everyone can be fooled sometimes. | $\forall x \exists y P(x, y)$ | +| Some people can be fooled all of the time. | $\exists x\forall y P(x, y)$ | + +5. + +Q: Your friend believes that you cannot fool everyone at the same time. What is +another way of saying this, and how would you write that in symbols (using +$P(x, y)$ to say you can fool $x$ at time $y$). + +A. Someone is never fooled. $\exists x \forall y \neg P(x, y)$ + +B. Everyone is never fooled. $\forall x \forall y \neg P(x, y)$ + +C. Someone is not fooled sometimes. $\exists x \exists y \neg P (x, y)$ + +D. Everyone is not fooled sometimes. $\forall x \exists y \neg P (x, y)$ + +A: + +I'd say D is correct. + +6. + +Q: Regardless of your beliefs of how many people can be fooled at various times, +what could you conclude if we reinterpret $P(x, y)$ to mean $x < y$ and only +quantify over the natural numbers (so $\forall x$ means "For all natural +numbers," and $\exists x$ means "There exists a natural number")? Select all of +the following that apply. + +A. $\forall x \exists y P(x, y)$ is true. + +B. $\exists x \forall y P(x, y)$ is true. + +C. $\forall y \exists x P(x, y)$ is true. + +D. $\exists y \forall x P(x, y)$ is true. + +E. No matter what $P(x, y)$ means, we can conclude that +$\forall x \exists y P(x, y)$ and $\exists y \forall x$ are NOT _logically +equivalent_ + +A: + +A. This is true. We interpret this as "For all numbers for $x$, there exists +some number, $y$ where $x < y$". If $x \in \mathbb{N}$, then we know that this +is the same meaning as $x \in [0, \infty)$. We're essentially saying that every +possible value for $x$, there exists some value for $y$ that must be greater +than $x$. That's true. + +B. This is false. "There exists some number, $x$, for all possible numbers for +$y$ where $x < y$". This means that some value of $x$ will always be less than +every possible $y$, that cannot be true. + +C. This is true. "For all possible values for $y$, there exists some value $x$ +where $x < y$." This is saying that for any given value for $y$, we can always +find at least one value for $x$ where $x < y$. + +D. This is false. "There exists some value for $y$ such that for all possible +values of $x$, $x < y$." This would mean there is a natural number $y$ that is +greater than every natural number $x$, but this is impossible because for any +$y$, we can choose $x = y$, which makes $x < y$ false. + +E. This is true. The statements $\forall x \exists y\, P(x,y)$ and +$\exists y \forall x\, P(x,y)$ are not logically equivalent. + +The first means: for every value of $x$, there exists some value of $y$ such +that $x < y$. This is true for natural numbers since we can always choose a +larger number. + +The second means: there exists some value of $y$ such that for all values of +$x$, $x < y$. This is false, because for any choice of $y$, we can take $x = y$, +which makes $x < y$ false. + +The two statements, in essence, logically contradict each other. + +7. + +Q: Let $P(x)$ be the predicate, "$17x + 1$ is even." + +(a) Is $P(15)$ true or false? + +(b) What, if anything, can yhou conclude about $\exists x P(x)$ from the truth +value of $P(15)$? + +(c) What, if anything, can you conclude about $\forall x P(x)$ from the truth +value of $P(15)$? + +A: + +(a) Yes, simple arithmetic shows that if the predicate states: + +$$ P(x) = (17x + 1) \mod 2 = 0 $$ + +Then: + +$$ P(15) = (17(15) + 1) = 256 \mod 2 = 0 $$ + +Therefore the statement $P(15)$ is true. + +(b) + +We can conclude that this statement, $\exists x P(x)$ is true. This is saying +"There exists some value for $x$ where $17x + 1$ is even." This is definitely +true, as there only has to exist 1 value for $x$ for this statement to be true, +and in part a we showed just that. + +(c) This is false. The statement $\forall x P(x)$ is saying "For every possible +value in $x$, it is true that $17x + 1$ is even". That cannot be true. Just take +$P(2) = 35$ as a simple example. Basically when you start off with a $\forall$ +predicate, every single value passed for $x$ (in this case) has to hold for the +assertion. + +8. + +Q: Let $P(x)$ be the predicate, "$18x + 1$ is even." + +(a) Is $P(15)$ true or false? + +(b) What, if anything, can you conclude about $\exists x P(x)$ from the truth +value of $P(15)$? + +(c) What, if anything, can you conclude about $\forall x P(x)$ from the truth +value of $P(15)$? + +A: + +(a) Let's evaluate $P(15)$: + +$$ P(15) = 18(15) + 1 = 271 $$ + +But obviously 271 is odd. This is a false statement. + +(b) Because $18$ is an even number, and adding $1$ onto an even number will +always result in an odd number. We can conclude that the statement +$\exists x P(x)$, which means "There exists at least one value for $x$ where +$18x + 1$ is even." is false. There is no value for $x$ where $P(x)$ is true. + +(c) The statement $\forall x P(x)$, which states "For every value of $x$, +$18x + 1$ is even" is also false. In fact, the exact opposite is true. As we +explained in part b, $P(x)$ can never be true for _any_ value of $x$. Thusly, to +state that it must be true for every possible value of $x$ as $\forall x P(x)$ +asserts, simply cannot be the case. + +9. + +Q: Consider the sentence, $\exists x P(x, y) \to \forall x P(x, y)$. What can we +say about this sentence? Select all that apply. + +A. The sentence is a statement because it contains quantifiers. + +B. The sentence is not a statement because $x$ and $z$ are free variables. + +C. The sentence is not a statement because $y$ is a free variable. + +D. The universal generalization of the sentence is a statement. + +A: + +A. This is false, a sentence becomes a statement when a quantifier makes the +resulting statement true or false. But free variables still matter. The above +sentence essentially says "If there exists some value for $x$ where $P(x, y)$ is +true, then for every value for $x$, $P(x, y)$ must be true." But a sentence +cannot be a statement until, in this case, both $x$, and $y$ are replaced by +constants. + +B. $z$ is never mentioned, so this statement is not true. + +C. This is true. A statement cannot contain a free variable. Only when you +replace a free variable with a constant of some sort does a sentence become a +statement proper. + +D. No, this statement is false. Although $x$ is bound by the universal +quantifier $\forall$, $y$ is not bound, and so remains a free variable. The +universal generalization of a sentence can only be a statement if the universal +quantifiers at the beginning of the sentence bind (restrict the domain) of all +free variables. $x$ is bound in this way after the predicate, but $y$ is not. + +10. + +Q: Suppose $P(x, y)$ is some binary predicate defined on a very small domain of +discourse: just the integers 1, 2, 3, 4. For each of the 16 pairs of these +numbers, $P(x, y)$ is either true or false, according to the following table +($x$ values are rows, $y$ values are columns). + +| | 1 | 2 | 3 | 4 | +| - | - | - | - | - | +| 1 | T | F | F | F | +| 2 | F | T | T | F | +| 3 | T | T | T | T | +| 4 | F | F | F | F | + +For example, $P(1, 3)$ is false, as indicated by the $F$ in the first row, third +column. Use the table to decide whether the following statements are true or +false. + +(a) $\forall y \exists x P(x, y)$. + +(b) $\exists x \forall y P(x, y)$. + +(c) $\forall x \exists y P(x, y)$. + +(d) $\exists y \forall x P(x, y)$. + +A: + +(a) This is true. "For all columns, y, there exists some row, $x$, where +$P(x, y)$ returns true." This means that there must be some T somewhere in the +column. + +$$ y = 1: T exists (1, 3) $$ + +$$ y = 4: T exists (4, 1) $$ + +(b) "There exists at least one row, $x$, where all of the column values, $y$, +return T." + +This is true. + +$$ x = 3: T exists (3, 1), (3, 2), (3, 3), (3, 4) $$ + +(c) "For all rows, $x$, there exists at least one column $y$, that returns T." + +This is false, this is essentially saying every row has at least one T value, +but row 4 does not. Consider the following statement: + +$$ x = 4: T \text{ exists in at least one of: } (4, 1), (4, 2), (4, 3), (4, 4) $$ + +This isn't true, row 4 returns all F. + +(d) "There exists at least one column, $y$, where all rows $x$ return T" + +This is false. There is no column $y$ where all rows are filled with Ts. diff --git a/chapter_1/1_1/preview_activity.md b/chapter_1/1_1/preview_activity.md new file mode 100644 index 0000000..b8a7326 --- /dev/null +++ b/chapter_1/1_1/preview_activity.md @@ -0,0 +1,95 @@ +1. + +Q: Which of the following sentences should count as statements? That is, for +which of the sentences below could you _potentially_ claim the sentence was +eithr true or false? Select all that apply. + + A. The sum of the first 100 positive integers. + + B. What is the sum of the first 100 positive integers? + + C. The sum of the first 100 positive integers is 5050. + + D. Is the sum of the first 100 positive integers 5050? + + E. The sum of the first 100 positive integers is 17. + +A: C, E + +These are potentially true/false, as each makes a definitive claim. The rest are +either statements of being/existence (A), or questions with a specific answer +that is not true/false (B), or a question that could be answered by a true/false +statement, but doesn't definitively declare a true/false assertion (D). + +2. + +Q: You and your roommate are arguing, and they make the audacious claim that +pineapple is good both on pizza and in smoothies. Which of the following are +reasonable responses to this claim, from a logical point of view? + + A. The statement is false because even though pineapple is good in smoothies, it is NOT good on pizza. + + B. The statement is false because while pineapple is good on pizza and pineapple is good in smoothies, a pizza smoothie is never good. + + C. The statement is half true because regardless of what you think about pineapple on pizza, we can all agree at least that pineapple is good in smoothies. + + D. The statement is false because everyone who likes pineapple on pizza does NOT like pineapple in smoothies. + +A: From a logical point of view, statements A, C, and D are all reasonable +responses. Only B is not reasonable because it is a conclusion based on an +assertion that was never made (i.e. the roommate never claimed that pizza +smoothie is good). + +3. + +Q: Your roommate now makes an even more outrageous claim: If a superhero movie +is part of the Marvel Cinematic Universe, then it is good. Which of the +following are reasonable responses to this claim, from a logical point of view? + + A. This is false because there are good superhero movies, like Wonder Woman and Dark Knight, that are based on DC Comics, and so not part of the Marvel Cinematic Universe. + + B. The statement is false because there is at least one superhero movie that is part of the Marvel Cinematic Universe that is also not good. + + C. The statement is false because, for example, Green Lantern is neither Marvel nor good. + + D. The statement is true because more than half of the Marvel movies are good. + +A: Only B is a valid logical point of view. This is because the roommate claims +that if a movie in the Marvel Universe, it is good, but the counterargument is +that there is at least one superhero movie in the Marvel Universe that is not +good, which invalidates his claim. + +The first option, A, is not a valid logical point of view because just because +there are DC movies that are good has no bearing on whether Marvel movies are +good as there is no relation between them. + +C is also not a valid option because it points out a DC movie is bad, which +again, has no bearing on the assertion's statement that all Marvel movies are +good. + +D is not a valid logical conclusion because it only claims that more than half +of the Marvel movies are good, which does not answer whether or not the rest of +them are, which is the only way that the assertion could be concluded as being +true. + +4. + +Q: Your roommate just won't let up with their outrageous claims. Now they claim +that either every troll is a knave, or there is at least one troll that is a +knight. What can you say to this? + + A. Yes, this is true because every troll is either a knight or a knave. If it is not the case that _all_ trolls are knaves, then there must be _some_ troll that is a knight. + + B. This is false because some trolls are knights and some other trolls are knaves. + + C. The statement is false because there is no way to verify which of the two options is the case. + + D. The statement is false because no troll could say that all trolls are knaves, since knaves always lie + +A: + +C is the only verifiable logical argument. Since we only have each troll's word +on whether they are a knave(lying) or a knight (telling the truth), there is +always the potential that they are lying. Without another external way of +validating whether they are a knave or a knight, we cannot validate the +roommate's assertion. diff --git a/chapter_1/1_1/reading_questions.md b/chapter_1/1_1/reading_questions.md new file mode 100644 index 0000000..43c91b0 --- /dev/null +++ b/chapter_1/1_1/reading_questions.md @@ -0,0 +1,97 @@ +1. + +Q: Match each statement in symbols with it stype of statement. + +| | | +| ------------ | ------------------------------- | +| $P \to Q$ | $P$ and $Q$ (conjunction) | +| $P \vee Q$ | If $P$, then $Q$, (implication) | +| $P \wedge Q$ | $P$ or $Q$ (disjunction) | +| $\neg P$ | Not $P$ (negation) | + +A: + +| | | +| ------------ | ------------------------------- | +| $P \to Q$ | If $P$, then $Q$, (implication) | +| $P \wedge Q$ | $P$ and $Q$ (conjunction) | +| $P \vee Q$ | $P$ or $Q$ (disjunction) | +| $\neg P$ | Not $P$ (negation) | + +2. + +Q: Consider the sentence, "If $x > 3$, then $x$ is even." + +Which of the following statements are true about the sentence? Select all that +apply. + + A. The sentence is a false statement since it has a free variable. + + B. The universal generalization of the sentence is a statement. + + C. If you substitute 10 for $x$, the resulting statement is true. + + D. The sentence becomes a true statement no matter what natural number you substitute for $x$. + +A: + +B and C are true about the sentence. + +A. No, this sentence is not a statement because it has a free variable, but that +doesn't automatically make the statement true or false. + +B. Given a sentence with free variables, the **universal generalization** of a +sentence is the statement obtained by adding enough universal quantifiers to the +beginning of the sentence so that all free variables become bound. + +The beginning of the sentence states "If $x > 3$", this is a universal +quantifier that bounds our free variable, $x$, and therefore by the definition +of the **universal generalization**, this sentence is therefore a statement. + +C. This is also true, though it doesn't follow in "regular" day to day logic, +but recall our truth table for $P \to Q$: + +| $P$ | $Q$ | $P \to $Q$ | +| --- | --- | ---------- | +| T | T | T | + +Again our sentence is: + +"If $x > 3$, then $x$ is even." + +We can say that $P(x)$ is "If $x > 3$", and that $Q(x)$ is "$x$ is even." + +$$ P(x) \to Q(x) $$ + +Since $x = 10$ according to C, this means that: + +$$ P(10) = \text{ If }10 > 3 = \text{ True} $$ + +$$ Q(10) = 10 \text{ is even} = \text{ True} $$ + +So therefore it follows that: + +$$ P(10) \to Q(10) $$ + +Is a true statement. + +3. What questions do you have after reading this section? Write at least one + question about the content of this section that you are curious about. + +This section is quite interesting, I suppose I'm curious about how negation +adjusts the truth tables of other qualifiers. In one of the examples, this +statement is made: + +17 is not prime if and only if 19 is not prime. + +To which the answer is: + +True. Now both parts are false (since both are the negation of a true +statement), so the entire statement is true. + +The $P \leftrightarrow Q$ statement does confuse me, and so therefore we have: + +$$ \neg P \leftrightarrow \neg Q $$ + +and I suppose I just need to sit down and have some clarification on how +negation affects this statement. diff --git a/chapter_1/1_1/rules_of_thumb.md b/chapter_1/1_1/rules_of_thumb.md new file mode 100644 index 0000000..5cb1e08 --- /dev/null +++ b/chapter_1/1_1/rules_of_thumb.md @@ -0,0 +1,17 @@ +## Every blank is blank. + +Any statement of the form, "Every $P$-thing is a $Q$-thing" can be written as: + +$$ \forall x \left(P(x) \to Q(x)\right) $$ + +Example: all mammals have hair, becomes $\forall x \left(M(x) \to H(x)\right)$, +where $M(x)$ means $x$ is a mammal, and $H(x)$ means $x$ has hair. + +## Some blanks are blank + +Any statement of the form, "Some $P$-things are $Q$-things", can be written as: + +$$ \exists x\left(P(x) \wedge Q(x)\right) $$ + +Example: Some cats can swim, becomes $\exists x\left(C(x) \vee S(x)\right)$, +where $C(x)$ means $x$ is a cat, and $S(x)$ means $x$ can swim. diff --git a/leftoff.txt b/leftoff.txt new file mode 100644 index 0000000..21e72e8 --- /dev/null +++ b/leftoff.txt @@ -0,0 +1 @@ +48