🚧 Continuing through 1.4

chapter_1/1_4/additional_exercises.md

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+# 1.4.8 Additional Exercises
+
+1.
+
+Q: For a given predicate $P(x)$, you might believe that the statements
+$\forall x P(x)$ or $\exists x P(x)$ are either true or false. How would you
+decide if you were correct in each case? You have four choices: You could give
+an example of an element $n$ in the domain for which $P(n)$ is true or for which
+$P(n)$ is false, or you could argue that no matter what $n$ is, $P(n)$ is true
+or is false.
+
+(a) What would you need to do to prove $\forall x P(x)$ is true?
+
+(b) What would you need to do to prove $\forall x P(x)$ is false?
+
+\(c\) What would you need to do t prove $\exists x P(x)$ is true?
+
+(d) What would you need to do to prove $\exists x P(x)$ is false?
+
+A:
+
+(a) What would you need to do to prove $\forall x P(x)$ is true?
+
+I would need to prove that for any given $x$, $P(x)$ holds true.
+
+(b) What would you need to do to prove $\forall x P(x)$ is false?
+
+I would need to prove that there is at least one $x$ for which $P(x)$ is false.
+
+\(c\) What would you need to do to prove $\exists x P(x)$ is true?
+
+I would need to prove that there is at least one $x$ for which $P(x)$ holds
+true.
+
+(d) What would you need to do to prove $\exists x P(x)$ is false?
+
+I would need to prove that for any given $x$, $P(x)$ is false.
+
+2.
+
+Q: Consider the statement, "For all integers $a$ and $b$, if $a + b$ is even,
+then $a$ and $b$ are even."
+
+(a) Write the contrapositive of the statement.
+
+(b) Write the converse of the statement.
+
+\(c\) Write the negation of the statement.
+
+(d) Is the original statement true or false? Prove your answer.
+
+(e) Is the contrapositive of the original statement true or false? Prove your
+answer.
+
+(f) Is the converse of the original statement true or false? Prove your answer.
+
+(g) Is the negation of the original statement true or false? Prove your answer.
+
+A:
+
+Let's first establish what the original statement's antecedent and consequent
+is:
+
+"For all integers $a$ and $b$, if $a + b$ is even, then $a$ and $b$ are even."
+
+$P(a, b)$ is "$a + b$ is even."
+
+$Q(a, b)$ is "$a$ and $b$ are even."
+
+$$ P(a, b) \to Q(a, b) $$
+
+(a) Write the contrapositive of the statement.
+
+$$ \neg Q(a, b) \to \neg P(a, b) $$
+
+$\neg Q(a, b) \equiv \neg(E(a) \wedge E(b)) \equiv (\neg E(a) \vee \neg E(b))$
+is "$a$ or $b$ are odd."
+
+$\neg P(a, b)$ is $a + b$ is odd.
+
+"For all integers $a$, and $b$, if $a$ or $b$ is odd, then $a + b$ is odd."
+
+(b) Write the converse of the statement.
+
+The converse is $Q(a, b) \to P(a, b)$.
+
+"For all integers $a$ and $b$, if $a$ and $b$ are even, then $a + b$ is even."
+
+\(c\) Write the negation of the statement.
+
+The negation is $\neg(P(a, b) \to Q(a, b))$
+
+$$ \neg(P \to Q) $$
+
+$$ \neg(\neg P \vee Q) $$
+
+$$ P \wedge \neg Q $$
+
+$$ P(a, b) \wedge \neg Q(a, b) $$
+
+"For all integers, $a$ and $b$, $a + b$ is even and either $a$ or $b$ are odd."
+
+(d) Is the original statement true or false? Prove your answer.
+
+"For all integers $a$ and $b$, if $a + b$ is even, then $a$ and $b$ are even."
+
+Proof by Contradiction:
+
+Let $a$ and $b$ be integers, and assume that $a + b$ is even and either $a$ or
+$b$ are odd.
+
+The sum of an even and an odd integer must be odd.
+
+But then $a + b$ is both even and odd, a contradiction.
+
+The original statement is false.
+
+(e) Is the contrapositive of the original statement true or false? Prove your
+answer.
+
+"For all integers $a$, and $b$, if $a$ or $b$ is odd, then $a + b$ is odd."
+
+Proof by Contradiction:
+
+Let $a$ and $b$ be integers, assume either $a$ or $b$ is odd, and assume that
+$a + b$ is even.
+
+In the case that both $a$ and $b$ are odd, then $a + b$ is even (not a
+contradiction).
+
+But, in the case that $a$ is even and $b$ is odd, or in the case that $a$ is odd
+and $b$ is even, then the sum $a + b$ must be odd.
+
+But then $a + b$ is both even and odd, a contradiction.
+
+This statement is false.
+
+(f) Is the converse of the original statement true or false? Prove your answer.
+
+Skipped (too much time)
+
+(g) Is the negation of the original statement true or false? Prove your answer.
+
+Skipped (too much time)
+
+3.
+
+Q: For each of the statements below, say what method of proof you should use to
+prove them. Then say how the proof starts and how it ends. Bonus points for
+filling in the middle.
+
+(a) There are no integers $x$ and $y$ such that $x$ is a prime greater than 5
+and $x = 6y + 3$.
+
+(b) For all integers $n$, if $n$ is a multiple of 3, then $n$ can be written as
+the sum of consecutive integers.
+
+\(c\) For all integers $a$ and $b$, if $a^2 + b^2$ is odd, then $a$ or $b$ is
+odd.
+
+A:
+
+(a) There are no integers $x$ and $y$ such that $x$ is a prime greater than 5
+and $x = 6y + 3$.
+
+Let $P(x)$ be "$x$ is prime", $Q(x)$ be "$x > 5$", $R(x,y)$ be "$x = 6y + 3$".
+
+$$ \neg \exists x \exists y (P(x) \wedge Q(x) \wedge R(x,y)) $$
+
+Method of Proof: Proof by Contradiction.
+
+Start: Suppose there exist integers $x$ and $y$ such that $x$ is prime, $x > 5$,
+and $x = 6y + 3$.
+
+Middle: $x = 6y + 3 = 3(2y+1)$, so $x$ is divisible by 3. Since $x > 5$, we have
+$x \neq 3$, so $x$ is composite, contradicting that $x$ is prime.
+
+End: Therefore, no such integers $x$ and $y$ exist.
+
+(b) For all integers $n$, if $n$ is a multiple of 3, then $n$ can be written as
+the sum of consecutive integers.
+
+$P(n)$ is "$n$ is a multiple of 3"
+
+$Q(n)$ is "$n$ can be written as the sum of consecutive integers."
+
+$$ \forall n (P(n) \to Q(n)) $$
+
+Method of Proof: Direct Proof.
+
+Start: Let $n$ be an integer, assume $n$ is a multiple of 3.
+
+Middle: Say $n = 3k$ where $k$ is some integer. This means that we can sum
+consecutive integers for $n$ as a sum centered around $k$:
+
+$$ 3k = (k - 1) + k + (k + 1) $$
+
+End: Therefore $n$ can be written as the sum of consecutive integers.
+
+\(c\) For all integers $a$ and $b$, if $a^2 + b^2$ is odd, then $a$ or $b$ is
+odd.
+
+$P(a, b)$ is "a^2 + b^2 is odd"
+
+$Q(a, b)$ is "$a$ is odd $\vee b$ is odd"
+
+$$ \forall a \forall b (P(a, b) \to Q(a, b)) $$
+
+Method of Proof: Contrapositive
+
+Start: Let $a$ and $b$ be integers, and assume that both $a$ and $b$ are even.
+
+Middle: An even number squared must be even, and the sum of two even numbers
+also must be even.
+
+End: Therefore $a^2 + b^2$ is even.
+
+4.
+
+Q: Consider the statement, "For all integers $n$, if $n$ is even then $8n$ is
+even."
+
+(a) Prove the statement. What sort of proof are you using?
+
+(b) Is the converse true? Prove or disprove.
+
+A:
+
+(a) Prove the statement. What sort of proof are you using?
+
+Let $P(n)$ be "$n$ is even."
+
+Let $Q(n)$ be "$8n$ is even."
+
+$$ \forall n (P(n) \to Q(n)) $$
+
+Proof by Direct Proof:
+
+Let $n$ be any integer, assume that $n$ is even.
+
+Let $n = 2k$ where $k$ is any integer. We could then write:
+
+$$ 8n = 8(2k) = 16k = 2(8k) $$
+
+Any number that is a multiple of $2$ is even.
+
+Therefore $8n$ is even.
+
+(b) Is the converse true? Prove or disprove.
+
+"For all integers $n$, if $8n$ is even, then $n$ is even."
+
+$P(n)$ is "$8n$ is even."
+
+$Q(n)$ is "$n$ is even."
+
+$$ \forall n (P(n) \to Q(n)) $$
+
+Proof by counterexample:
+
+Let $n = 1$.
+
+Then $8n = 8$, which is even.
+
+But $n = 1$ is odd.
+
+So we have an example where $8n$ is even but $n$ is odd.
+
+Therefore the statement is false.

chapter_1/1_4/investigate/investigate.md

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+# Investigate!
+
+A **mini sudoku puzzle** is a 4 x 4 grid of squares, divided into four 2 x 2
+boxes. The goal is to fill each square with a digit from 1 to 4, such that no
+digit repeats in any row, any column, or any box.
+
+Here is a simple mini sudoku puzzle you can try to solve.
+
+![image 1_4_1_investigate_1](./1_4_1_investigate_1.png)
+
+You might notice that the solution to the above puzzle has its four outside
+corners all different, and its four middle squares all different.
+
+The goal of this _Investigate!_ question is to prove that this is not a
+coincidence: Suppose a mini sudoku puzzle has all different numbers in its four
+corners (marked with # below). Prove that the center four squares (marked with *
+below) must also contain different numbers.
+
+![image 1_4_1_investigate_2](./1_4_1_investigate_2.png)
+
+## Try it. 1.4.1
+
+Try placing numbers into an empty mini sudoku puzzle. See if you can break the
+statement we were asked to prove in the _Investigate!_ activity. What stops you?
+Briefly explain whether you think the statement is true or false, and why.
+
+A:
+
+I think the statement is true. When I tried to make two of the middle squares
+the same number, the sudoku rules forced a contradiction because the numbers
+would repeat in a row, column, or box. The different corners seem to force the
+middle squares to all be different too.

chapter_1/1_4/investigate/preview_activity.md

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+# Preview Activity
+
+Consider the statement:
+
+If $ab$ is an even number, then $a$ or $b$ is even.
+
+Which of the proofs below appear to be valid proofs of this statement? Note: You
+can assume all the algebra below is correct (because it is).
+
+1.
+
+Suppose $a$ and $b$ are odd. That is, $a = 2k + 1$ and $b = 2m + 1$ for some
+integers $k$ and $m$. Then
+
+$$
+
+ab = (2k + 1)(2m + 1) \\
+
+\quad = 4km + 2k + 2m + 1 \\
+
+\quad = 2(2km + k + m) + 1.
+
+$$
+
+Therefore $ab$ is odd.
+
+2.
+
+Assume that $a$ or $b$ is even -- say it is $a$ (the case where $b$ is even will
+be identical). That is, $a = 2k$ for some integer $k$. Then
+
+$$
+
+ab = (2k)b \\
+
+\quad = 2(kb).
+
+$$
+
+Thus $ab$ is even.
+
+3.
+
+Suppose that $ab$ is even but $a$ and $b$ are both odd. Namely,
+$ab = 2n, a = 2k + 1$ and $b = 2j + 1$ for some integers $n$, $k$, and $j$. Then
+
+$$
+
+2n = (2k + 1)(2j + 1) \\
+
+2n = 4kj + 2k + 2j + 1 \\
+
+n = 2kj + k + j + \frac{1}{2}.
+
+$$
+
+But since $2kj + k + j$ is an integer, this says that the integer $n$ is equal
+to a non-integer, which is impossible.
+
+4.
+
+Let $ab$ be an even number, say $ab = 2n$, and $a$ be an odd number, say
+$a = 2k + 1$.
+
+$$
+
+ab = (2k + 1)b \\
+
+2n = 2kb + b \\
+
+2n - 2kb = b
+
+$$
+
+Therefore $b$ must be even.
+
+A:
+
+1. This one doesn't appear to be a valid proof at first glance to me. Although
+   the logic appears correct, we are not solving for if $ab$ is odd. If we
+   somehow proved that $ab$ was _not_ odd, then maybe this would prove that $ab$
+   is even, but the initial statement "If $ab$ is an even number, then $a$ or
+   $b$ is even.$ is contradicted by the following statement "Suppose $a$ and $b$
+   are odd."
+
+2. This proof appears to be valid for the given statement. Since $a = 2k$ for
+   all natural numbers, this is guaranteed to be even and multiplying it by $b$
+   does not change that this will result in an even number.
+
+3. I don't see the logic in this one. This does not appear to be a valid proof
+   of the statement, but I cannot say as to why.
+
+4. This proof appears to be valid for the given statement, but I am struggling
+   to say as to why.
+
+Note: Not changing my answers here, but I mostly got these all wrong.

chapter_1/1_4/practice_problems.md

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+# Practice Problems
+
+1.
+
+Q: Arrange some of the statements below to form a correct proof of the following
+statement: "For any integer $n$, if $n$ is even, then $7n$ is even."
+
+- Let $n$ be an arbitrary integer, and assume $7n$ is even.
+
+- Let $n$ be an arbitrary integer, and assume $7n$ is odd.
+
+- Since $7$ is odd and the product of an odd number and an odd number is odd,
+
+- Since an even number divided by $7$ must be odd,
+
+- $n$ must be odd.
+
+- $7n$ must be odd.
+
+- Let $n$ be an arbitrary integer, and assume $n$ is even.
+
+- Since the product of any number with an even number is even,
+
+- $7n$ must be even.
+
+A:
+
+Let's establish the antecedent and consequent here.
+
+$P(n)$ is "$n$ is even"
+
+$Q(n)$ is "$7n$ is even"
+
+$$ P(n) \to Q(n) $$
+
+In a direct proof, we start with "assume the antecedent" and we end with
+"therefore the consequent".
+
+Proof by direct proof:
+
+- Let $n$ be an arbitrary integer, and assume $n$ is even.
+
+- Since the product of any number with an even number is even,
+
+- $7n$ must be even.
+
+2.
+
+Q: Arrange some of the statements below to form a correct proof of the following
+statement: "For any integer $n$, if $7n$ is even, then $n$ is even."
+
+- Let $n$ be an arbitrary integer, and assume $n$ is even.
+
+- Since the $7$ is odd and the product of an odd number with an even number is
+  even,
+
+- $7n$ must be even.
+
+- Let $n$ be an arbitrary integer, and assume $7n$ is even.
+
+- Since an even number divided by $7$ must be even,
+
+- $n$ must be even.
+
+- Let $n$ be an arbitrary integer, and assume $n$ is odd.
+
+- Since $7$ is odd and the product of an odd number and an odd number is odd,
+
+- $7n$ must be odd.
+
+A:
+
+The contrapositive proof starts with Assume the conclusion is false, and
+conclude therefore that the assumption is false.
+
+$$ \neg Q \to \neg P $$
+
+So our proof by contrapositive would look like:
+
+- Let $n$ be an arbitrary integer, and assume $n$ is odd.
+
+- Since $7$ is odd and the product of an odd number and an odd number is odd,
+
+- $7n$ must be odd.
+
+3.
+
+Q: Consider the statement, "For any numbers $a$ and $b$, if $a + b$ is odd, then
+either $a$ or $b$ is odd."
+
+Give a valid proof of the statement using a _proof by contrapositive_. Arrange
+some statements below to complete the proof.
+
+- Let $a$ and $b$ be integers, and assume that $a + b$ is odd.
+
+- Let $a$ and $b$ be integers, and assume that if $a + b$ is odd, then either
+  $a$ or $b$ is odd.
+
+- Let $a$ and $b$ be integers, and assume both are even.
+
+- The sum of two even integers must also be even.
+
+- Therefore $a + b$ is even.
+
+- Let $a$ and $b$ be integers and assume that $a + b$ is odd but $a$ and $b$ are
+  both even.
+
+- The sum of two odd integers must be even.
+
+- But then $a + b$ is both even and odd, a contradiction.
+
+A:
+
+A proof by contrapositive starts with assuming that the original statement's
+consequent is false, and concludes that the original statement's antecedent is
+false.
+
+$$ \neg Q \to \neg P $$
+
+$\neg Q$ would be something like "For any numbers $a$ and $b$, it is false that
+either are odd, so both $a$ and $b$ are even."
+
+$\neg P$ would be something like "$a + b$ is even."
+
+So the order of our proof would be:
+
+- Let $a$ and $b$ be integers, and assume both are even.
+
+- The sum of two even integers must also be even.
+
+- Therefore $a + b$ is even.
+
+4.
+
+Q: Consider the same statement, "For any numbers $a$ and $b$, if $a + b$ is odd,
+then either $a$ or $b$ is odd."
+
+Give a valid proof of the statement, this time using a _proof by contradiction_
+using some of the statements below.
+
+- Let $a$ and $b$ be integers, and assume that $a + b$ is odd.
+
+- Let $a$ and $b$ be integers, and assume that if $a + b$ is odd, then either
+  $a$ or $b$ is odd.
+
+- Let $a$ and $b$ be integers, and assume both are even.
+
+- The sum of two even integers must also be even.
+
+- Therefore $a + b$ is even.
+
+- Let $a$ and $b$ be integers and assume that $a + b$ is odd but $a$ and $b$ are
+  both even.
+
+- The sum of two odd integers must be even.
+
+- But then $a + b$ is both even and odd, a contradiction.
+
+A:
+
+A proof by contradiction starts with assuming the negation of our original
+statement, and then concluding with a statement that is a contradiction.
+
+$$ \neg(P \to Q) $$
+
+It might be helpful to use negation is disjunction and De Morgan's laws here:
+
+$$ \neg(\neg P \vee Q) $$
+
+$$ P \wedge \neg Q $$
+
+Which reads something like "For any numbers $a$ and $b$, $a + b$ is odd and
+$a + b$ is even." That's a contradiction. Let's see how we can order our given
+choices to create a proof statement.
+
+Proof by contradiction:
+
+- Let $a$ and $b$ be integers and assume that $a + b$ is odd but $a$ and $b$ are
+  both even.
+
+- The sum of two even integers must also be even.
+
+- But then $a + b$ is both even and odd, a contradiction.
+
+5.
+
+Q: Below are three statements that together with a possible first line of a
+proof of that statement. In each case, say whether the first line is the start
+of a direct proof, a proof by contrapositive, or a proof by contradiction.
+
+(a)
+
+**_Statement:_** For every integer $n$, the number $7n - 1$ is divisible by $6$.
+
+**_First line:_** Suppose there were some integer $n$ for which $7n - 1$ was not
+divisible by $6$.
+
+(b)
+
+**_Statement:_** For any integer $n$, if $n$ is prime, then $n$ is solitary.
+
+**_First line:_** Let $n$ be an integer, and assume $n$ is not solitary.
+
+\(c\)
+
+**_Statement:_** If a shape is a pentagon, then its interior angles add up to
+480 degrees.
+
+**_First line:_** Consider an arbitrary shape, and assume it is a pentagon.
+
+A:
+
+(a)
+
+**_Statement:_** For every integer $n$, the number $7n - 1$ is divisible by $6$.
+
+**_First line:_** Suppose there were some integer $n$ for which $7n - 1$ was not
+divisible by $6$.
+
+We can approach this by first defining what the antecedent and the consequent
+is:
+
+$P(n)$ is "Suppose there were some integer $n$"
+
+$Q(n)$ is "There exists a number $7n - 1$ that is divisible by $6$."
+
+The first line states a negation of the original statement, this is the start to
+a Proof by Contradiction.
+
+(b)
+
+**_Statement:_** For any integer $n$, if $n$ is prime, then $n$ is solitary.
+
+**_First line:_** Let $n$ be an integer, and assume $n$ is not solitary.
+
+$P(n)$ is "$n$ is prime."
+
+$Q(n)$ is "$n$ is solitary."
+
+The first line starts with a negation of the original consequent, this is the
+start to a Proof by Contrapositive.
+
+\(c\)
+
+**_Statement:_** If a shape is a pentagon, then its interior angles add up to
+480 degrees.
+
+**_First line:_** Consider an arbitrary shape, and assume it is a pentagon.
+
+$P$ is "the shape is a pentagon."
+
+$Q$ is "its interior angles add up to 480 degrees."
+
+The first line assumes the assumption of the original statement, this is a
+Direct Proof.
+
+6.
+
+Q: What would the first line be for a proof in each style, of the following
+statement:
+
+"If a function $f : A \to B$ is a bijection, then |A| = |B|."
+
+$\text{Assume } f : A \to B \text{ is a bijection} \quad \text{ Direct proof}$
+
+$\text{Assume } f: A \to B \text{ is a bijection and } |A| \neq |B| \quad \text{ Proof by contrapositive}$
+
+$\text{Assume} |A| \neq |B| \quad \text{ Proof by contradiction}$
+
+A:
+
+Yes, this first one needs no adjustment, this is the first line of a Direct
+proof:
+
+$\text{Assume } f : A \to B \text{ is a bijection} \quad \text{ Direct proof}$
+
+The next one negates the entire original statement, that is a first line of a
+Proof by Contradiction:
+
+$\text{Assume } f: A \to B \text{ is a bijection and } |A| \neq |B| \quad \text{ Proof by contradiction}$
+
+The next one negates the original consequent, and so is the first line of a
+Proof by Contrapositive:
+
+$\text{Assume} |A| \neq |B| \quad \text{ Proof by contrapositive}$

chapter_1/1_4/reading_questions.md

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+# 1.4.6 Reading Questions
+
+1.
+
+Q: Which of the following would be the best first line of a _direct proof_ if
+you wanted to prove the statement, "For all sets $A$ of single-digit numbers, if
+$|A| = 6$, then $A$ contains an even number."
+
+A. Suppose there exists a set $A$ of single-digit numbers with $|A| = 6$ but
+that contains only odd numbers.
+
+B. Fix an arbitrary set $A$ of single-digit numbers, and assume $|A| = 6$.
+
+C. Suppose $A$ is a set of single-digit numbers with $|A| \neq 6$.
+
+D. Let $A$ be a set of single-digit numbers that contains an even number.
+
+E. Let $A$ be a set of single-digit numbers, and assume that $A$ does not
+contain any even numbers.
+
+A:
+
+To me B would be the best first line of a _direct proof_. In a direct proof, we
+assume the antecedent, $P$ and prove the consequent, $Q$.
+
+Here the antecedent, $P(A)$, is "|A| = 6" And $Q(A)$ is "A contains an even
+number."
+
+$$ P(A) \to Q(A) $$
+
+2.
+
+Q: Which of the following would be the best first line of a _proof by
+contrapositive_ if you wanted to prove the statement, "For all sets $A$ of a
+single-digit numbers, if $|A| = 6$, then $A$ contains an even number."
+
+A. Suppose there exists a set $A$ of single-digit numbers with $|A| = 6$ but
+that contains only odd numbers.
+
+B. Fix an arbitrary set $A$ of single-digit numbers, and assume $|A| = 6$.
+
+C. Suppose $A$ is a set of single-digit numbers with $|A| \neq 6$.
+
+D. Let $A$ be a set of single-digit numbers that contains an even number.
+
+E. Let $A$ be a set of single-digit numbers, and assume that $A$ does not
+contain any even numbers.
+
+A:
+
+In a proof by contrapositive, we assume the conclusion is false and conclude
+that the antecedent is false:
+
+$$ \neg Q \to \neg P $$
+
+The conclusion is $Q(A)$ is "$A$ contains an even number", and the antecedent is
+"$|A| = 6$".
+
+Therefore our choice should read something along the lines of "If $A$ does not
+contain an even number, then $|A| \neq 6$."
+
+We only want to know about the best first line, so we want to choose the one
+that starts with "$A$ does not contain an even number."
+
+This is reflected most by choice E.
+
+3.
+
+Q: Which of the following would be the best first line of a _proof by
+contradiction_ if you wanted to prove the statement, "For all sets $A$ of a
+single-digit numbers, if $|A| = 6$, then $A$ contains an even number."
+
+A. Suppose there exists a set $A$ of single-digit numbers with $|A| = 6$ but
+that contains only odd numbers.
+
+B. Fix an arbitrary set $A$ of single-digit numbers, and assume $|A| = 6$.
+
+C. Suppose $A$ is a set of single-digit numbers with $|A| \neq 6$.
+
+D. Let $A$ be a set of single-digit numbers that contains an even number.
+
+E. Let $A$ be a set of single-digit numbers, and assume that $A$ does not
+contain any even numbers.
+
+A:
+
+In a proof by contradiction, we assume $\neg(P \to Q)$ and conclude with a
+contradiction.
+
+Since $P \to Q$ reads along the lines of "If $|A| = 6$, then $A$ contains an
+even number", $\neg(P \to Q)$ would read along the lines of "It is false that if
+$|A| = 6$, then $A$ contains an even number." Or, one could say "It is true that
+if $|A| = 6$, then $A$ contains an odd number."
+
+This antecedent looks most similar to A.

chapter_1/1_4/start_and_end_proofs.md

@@ -0,0 +1,21 @@
+# Starts and Ends Proofs.
+
+To prove an implication $P \to Q$:
+
+**Direct**
+
+Start: Assume $P$.
+
+End: Therefore $Q$.
+
+**Contrapositive**
+
+Start: Assume $\neg Q$.
+
+End: Therefore $\neg P$.
+
+**Contradiction**
+
+Start: Assume $\neg (P \to Q)$.
+
+End: ...which is a contradiction.

leftoff.txt

@@ -1 +1 @@
-82
+100