From 5c8a213de8aa6b59f2d36e27f9264202fc9592b6 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Sat, 16 May 2026 22:36:42 -0700 Subject: [PATCH] :construction: Continuing through 1.4 --- chapter_1/1_4/additional_exercises.md | 269 +++++++++++++++++ .../discrete_math_starts_and_ends_proofs.png | Bin 0 -> 43868 bytes .../1_4/investigate/1_4_1_investigate_1.png | Bin 0 -> 2619 bytes .../1_4/investigate/1_4_1_investigate_2.png | Bin 0 -> 3179 bytes chapter_1/1_4/investigate/investigate.md | 32 ++ chapter_1/1_4/investigate/preview_activity.md | 96 ++++++ chapter_1/1_4/practice_problems.md | 285 ++++++++++++++++++ chapter_1/1_4/reading_questions.md | 95 ++++++ chapter_1/1_4/start_and_end_proofs.md | 21 ++ leftoff.txt | 2 +- 10 files changed, 799 insertions(+), 1 deletion(-) create mode 100644 chapter_1/1_4/additional_exercises.md create mode 100644 chapter_1/1_4/discrete_math_starts_and_ends_proofs.png create mode 100644 chapter_1/1_4/investigate/1_4_1_investigate_1.png create mode 100644 chapter_1/1_4/investigate/1_4_1_investigate_2.png create mode 100644 chapter_1/1_4/investigate/investigate.md create mode 100644 chapter_1/1_4/investigate/preview_activity.md create mode 100644 chapter_1/1_4/practice_problems.md create mode 100644 chapter_1/1_4/reading_questions.md create mode 100644 chapter_1/1_4/start_and_end_proofs.md diff --git a/chapter_1/1_4/additional_exercises.md b/chapter_1/1_4/additional_exercises.md new file mode 100644 index 0000000..5a5d94a --- /dev/null +++ b/chapter_1/1_4/additional_exercises.md @@ -0,0 +1,269 @@ +# 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? 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zzqW+|tUY+j77$3rVs8Sk0_AY`m=1x(<#G?@J~*XPGC0Cu6sD$0>*07!v`O-{8Cw^Z zvG?!g2Ht#@N@}Bzj1W>Ak&B7Khx?62C}?%iiR8);8KM8q%y{KwJQMp4l1{q-Oz>z<6q!;rsgFi&ro0;vMbH3-o&9Q*K0fz9< zP^Yf+xUtZG6P|x(=M>bYI2V}5hOs<-rfuc&#ulU8)wRm|E>idi7;L+@na7<2_=kh& z7$~$VEGUTKt@|`eeg?!ZApr`x&yMJ}wN%WIoJ$<+@X|na?V*^sRdC8k06bFr{YesV vm&#_Ee%g$J=x;CGzX|HUOX0sF;_VeSjpiS+Jxu`q9{6CUm!M@PZt?#D=1?Gx literal 0 HcmV?d00001 diff --git a/chapter_1/1_4/investigate/investigate.md b/chapter_1/1_4/investigate/investigate.md new file mode 100644 index 0000000..186e83f --- /dev/null +++ b/chapter_1/1_4/investigate/investigate.md @@ -0,0 +1,32 @@ +# 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. diff --git a/chapter_1/1_4/investigate/preview_activity.md b/chapter_1/1_4/investigate/preview_activity.md new file mode 100644 index 0000000..928e4ef --- /dev/null +++ b/chapter_1/1_4/investigate/preview_activity.md @@ -0,0 +1,96 @@ +# 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. diff --git a/chapter_1/1_4/practice_problems.md b/chapter_1/1_4/practice_problems.md new file mode 100644 index 0000000..55608c5 --- /dev/null +++ b/chapter_1/1_4/practice_problems.md @@ -0,0 +1,285 @@ +# 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: + +$\text{Assume} |A| \neq |B| \quad \text{ Proof by contrapositive}$ diff --git a/chapter_1/1_4/reading_questions.md b/chapter_1/1_4/reading_questions.md new file mode 100644 index 0000000..966397f --- /dev/null +++ b/chapter_1/1_4/reading_questions.md @@ -0,0 +1,95 @@ +# 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. diff --git a/chapter_1/1_4/start_and_end_proofs.md b/chapter_1/1_4/start_and_end_proofs.md new file mode 100644 index 0000000..a769c89 --- /dev/null +++ b/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. diff --git a/leftoff.txt b/leftoff.txt index dde92dd..29d6383 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -82 +100