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chapter_1/1_4/additional_exercises.md

@@ -267,3 +267,727 @@ But $n = 1$ is odd.
 So we have an example where $8n$ is even but $n$ is odd.
 
 Therefore the statement is false.
+
+5.
+
+Q: The game TENZI comes with 40 six-sided dice (each numbered 1 to 6). Suppose
+you roll all 40 dice.
+
+(a) Prove that there will be at least seven dice that land on the same number.
+
+(b) How many dice would you have to roll before you were guaranteed that some
+four of them would all match or all be different? Prove your answer.
+
+A:
+
+(a) Prove that there will be at least seven dice that land on the same number.
+
+Start:
+
+Proof by Contradiction:
+
+Start: Assume that seven of the dice do not land on the same number. This means
+that each number on the die (from 1 to 6) appears at most 6 times among the 40
+dice.
+
+Middle: If each number from 1 to 6 appears at most 6 times, then the total
+number of dice can be calculated as:
+
+$$ 6(\text{appearances per number}) \times 6(\text{numbers}) = 36 $$
+
+We have shown that if each number appears at most 6 times, then the maximum
+number of dice is 36.
+
+Since we are rolling 40 dice, which is more than 36, our initial assumption must
+be false , which is a contradiction.
+
+End: Therefore at least seven dice land on the same number.
+
+(b) How many dice would you have to roll before you were guaranteed that some
+four of them would all match or all be different? Prove your answer.
+
+Proof by Contradiction:
+
+Start: Assume that with 10 dice rolled, there are not four dice that all match
+and there are not four dice that are all different.
+
+Middle: Since there are not four matching dice, each number from 1 to 6 can
+appear at most 3 times.
+
+Also, since there are not four dice that are all different, there can be at most
+3 different numbers appearing among the dice.
+
+Therefore, the maximum number of dice possible is:
+
+$$ 3(\text{copies of each number}) \times 3(\text{different numbers}) = 9 $$
+
+But we rolled 10 dice, which is more than 9. This contradicts our assumption.
+
+End: Therefore, if 10 dice are rolled, then there must be either four dice that
+all match or four dice that are all different.
+
+6.
+
+Q: Prove that for all integers $n$, it is the case that $n$ is even if and only
+if $3n$ is even. That is, prove both implications: If $n$ is even, then $3n$ is
+even, and if $3n$ is even, then $n$ is even.
+
+A:
+
+Let $P(n)$ be "$n$ is even."
+
+Let $Q(n)$ be "$3n$ is even."
+
+We are trying to prove:
+
+$$ \forall n (P(n) \leftrightarrow Q(n)) $$
+
+Let's try to first prove the first implication:
+
+"If $n$ is even, then $3n$ is even."
+
+$$ \forall n (P(n) \to Q(n)) $$
+
+Proof by Direct Proof:
+
+Start: Assume that $n$ is even.
+
+Middle: Let $n = 2k$ where $k$ is any integer, this means:
+
+$$ 3n = 3(2k) = 6k = 2(3k) $$
+
+Here we know that $3k$ can be even or odd, but any number multiplied by $2$ is
+even.
+
+End: Therefore $3n$ is even.
+
+Now let's prove the other implication:
+
+$$ \forall n (Q(n) \leftrightarrow P(n)) $$
+
+"If $3n$ is even, then $n$ is even."
+
+Proof by Contrapositive:
+
+Start: Assume that $n$ is odd.
+
+Middle: Let $n = 2k + 1$ where $k$ is any integer. Then:
+
+$$ 3n = 3(2k + 1) = 6k + 3 = 2(3k + 1) + 1 $$
+
+Since $3k + 1$ is an integer, $3n$ has the form of $2k + 1$, which is odd.
+
+End: Therefore $3n$ is odd.
+
+7.
+
+Q: Prove that $\sqrt{3}$ is irrational.
+
+A:
+
+Proof by Contradiction:
+
+Start: Assume that $\sqrt{3}$ is a rational number, which can be expressed with
+integers $a$ and $b$. Let $a$ and $b$ be integers where $b \neq 0$, and
+$\dfrac{a}{b}$ has $gcd(a, b) = 1$.
+
+Middle:
+
+This means we can write:
+
+$$ \sqrt{3} = \frac{a}{b}  $$
+
+If we then square both sides of this equation, we get:
+
+$$ 3 = \frac{a^2}{b^2}  $$
+
+$$ a^2 = 3b^2 $$
+
+This implies that if $3$ divides $a$, then $3$ divides $b$.
+
+So we can write $a = 3k$ for some integer $k$. This means that we can substitute
+in $k$:
+
+$$ a = 3k $$
+
+$$ (3k)^2 = 3b^2 $$
+
+$$ 9k^2  = 3b^2 $$
+
+$$ 3k^2  = b^2 $$
+
+So both $a$ and $b$ share a factor of 3. So the $gcd(a, b) \geq 3$.
+
+But we assumed $gcd(a, b) = 1$. This is a contradiction.
+
+End: Therefore, $\sqrt{3}$ is an irrational number.
+
+8.
+
+Q: Consider the statement, "For all integers $a$ and $b$, if $a$ is even and $b$
+is a multiple of $3$, then $ab$ is a multiple of $6$."
+
+(a) Prove the statement. What sort of proof are you using?
+
+(b) State the converse. Is it true? Prove or disprove.
+
+A:
+
+(a) Prove the statement. What sort of proof are you using?
+
+(b) State the converse. Is it true? Prove or disprove.
+
+(a) We'll use a Proof by Direct Proof.
+
+Proof by Direct Proof:
+
+Let $a$ and $b$ be integers, where $a$ is even and $b$ is a multiple of $3$.
+
+Since $a$ is even, we can express $a$ as $a = 2k$, where $k$ is any integer:
+
+$$ a = 2k $$
+
+And since $b$ is a multiple of $3$, we can express $b$ as $b = 3m$ where $m$ is
+any integer.
+
+$$ b = 3m $$
+
+We can then express $ab$ as:
+
+$$ ab = (2k)(3m) = 6km $$
+
+Therefore, $ab$ is a multiple of $6$.
+
+(b) State the converse. Is it true? Prove or disprove.
+
+The converse statement is "For all integers $a$ and $b$, if $ab$ is a multiple
+of $6$, then $a$ is even and $b$ is a multiple of $3$."
+
+This statement is false.
+
+Proof by Counterexample:
+
+Consider $ab = 6$.
+
+Let $a = 3$
+
+$$ ab = 6 $$
+
+$$ 3b = 6 $$
+
+$$ b = 2 $$
+
+So here, $a$ is not even, and $b$ is not a multiple of $3$, but $ab$ is a
+multiple of $6$.
+
+Therefore the statement is false.
+
+9.
+
+Q: Prove the statement, "For all integers $n$, if $5n$ is odd, then $n$ is odd."
+Clearly state the style of proof you are using.
+
+A:
+
+Here instead we will prove the contrapositive statement:
+
+"For all integers $n$, if $n$ is even, then $5n$ is even."
+
+Proof by Contrapositive:
+
+Let $n$ be an integer.
+
+Let $n$ be $2k$ for some integer $k$. We can then write $5n$ as:
+
+$$ 5n = 5(2k) = 10k = 2(5k) $$
+
+So $5n$ is a multiple of $2.
+
+Therefore $5n$ is even.
+
+10.
+
+Q: Prove the statement, "For all integers, $a$, $b$, and $c$, if
+$a^2 + b^2 = c^2$, then $a$ or $b$ is even."
+
+A:
+
+Proof by Contrapositive:
+
+Contrapositive statement:
+
+"For all integers, $a$, $b$, and $c$, if $a$ and $b$ are odd, then
+$a^2 + b^2 \neq c^2$"
+
+Let $a$, $b$, and $c$ be integers, and assume that both $a$ and $b$ are odd.
+
+Let $a = 2k + 1$ for any number $k$.
+
+$$ a^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 $$
+
+So here we have shown that $a^2$ is odd.
+
+Let $b = 2m + 1$ for any number $m$.
+
+$$ b^2 = (2m + 1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1 $$
+
+So here we have shown that $b^2$ is odd.
+
+And we know that any two odd numbers when added, their sum is an even number:
+
+$$ \text{odd number} + \text{ odd number } = \text{ even number} $$
+
+So, we can say that:
+
+$$ a^2 + b^2 = \text{ even number} $$
+
+And since $c^2$ must be an even number, then $c$ must also be an even number.
+
+This shows that if $a$ and $b$ are both odd, then $c$ is even, and therefore
+$a^2 + b^2 = c^2$ cannot contradict the claim that $a$ or $b$ is even. In other
+words, either $a$ or $b$ must be even
+
+Thus, the statement is true.
+
+11.
+
+Q: Suppose that you would like to prove the following implication:
+
+    For all numbers, $n$, if $n$ is prime, then $n$ is solitary.
+
+Write out the beginning and end of the argument if you were to prove the
+statement,
+
+(a) Directly
+
+(b) By contrapositive
+
+\(c\) By contradiction
+
+You do not need to provide the middle parts of the proofs (since you do not know
+what solitary means). However, make sure that you give the first few and last
+lines of the proofs so that we can see the logical structure you would follow.
+
+A:
+
+(a) Directly
+
+Beginning: Let $n$ be any integer where $n$ is prime.
+
+End: Therefore $n$ is solitary.
+
+(b) By contrapositive
+
+Beginning: Let $n$ be any integer where $n$ is not solitary.
+
+End: Therefore $n$ is not prime.
+
+\(c\) By contradiction
+
+Beginning: Let $n$ be any integer where $n$ is prime, assume that $n$ is not
+solitary.
+
+End: This is a contradiction.
+
+12.
+
+Q: Suppose you have a collection of rare 5-cent stamps and 8-cent stamps. You
+desperately need to mail a letter and, having no other stamps available, decide
+to dip into your collection. The question is, what amounts of postage can you
+make?
+
+(a) Prove that if you only use an even number of both types of stamps, the
+amount of postage you make must be even.
+
+(b) Suppose you made an even amount of postage. Prove that you used an even
+number of at least one of the types of stamps.
+
+\(c\) Suppose you made exactly 72 cents of postage. Prove that you used at least
+6 of at least one type of stamp.
+
+A:
+
+(a) Prove that if you only use an even number of both types of stamps, the
+amount of postage you make must be even.
+
+Proof by Direct Proof:
+
+Let $n$ be an even integer that represents the number of 5-cent stamps used, and
+$m$ be an even integer that represents the number of 8-cent stamps used.
+
+The sum for the amount of postage, $S$, then is:
+
+$$ S = 5n + 8m $$
+
+Let $n$ be $2k$ for any integer $k$, and $m$ be $2p$ for any integer $p$.
+
+The sum for the amount of postage, $S$ then can be seen as:
+
+$$ S = 5(2k) + 8(2p) = 10k + 16p = 2(5k + 8p) $$
+
+Since $S$ is $2$ times an integer, it is even.
+
+Therefore the amount of postage made must be even.
+
+(b) Suppose you made an even amount of postage. Prove that you used an even
+number of at least one of the types of stamps.
+
+This is the converse of (a).
+
+Proof by Contrapositive:
+
+Let $n$ and $m$ both be odd integers
+
+Let $S = 5n + 8m$ be the total postage.
+
+Let $n$ be $2k + 1$ for any $k$ and $m$ be $2p + 1$ for any $p$.
+
+Our sum becomes:
+
+$$ S = 5(2k + 1) + 8(2p + 1) = 10k + 5 + 16p + 8 = 10k + 16p + 13 = 2(5k + 8p) + 13 $$
+
+So $S$ is an odd number.
+
+Therefore the amount of postage is not even.
+
+\(c\) Suppose you made exactly 72 cents of postage. Prove that you used at least
+6 of at least one type of stamp.
+
+Proof by Contradiction:
+
+Let $n$ be an integer that represents the number of 5-cent stamps used, and $m$
+be an integer that represents the number of 8-cent stamps used. Assume that
+$5n + 8m = 72$. Assume that $n < 6$ and $m < 6$.
+
+If $n < 6$ and $m < 6$, this means that:
+
+$$ n <= 5 $$
+
+And
+
+$$ m <= 5 $$
+
+By setting $n$ and $m$ to their maximum values, we can evaluate $5n + 8n = 72$:
+
+$$ 5(5) + 8(5) = 25 + 40 = 65 \neq 72  $$
+
+But we assumed that $5n + 8m = 72$. This is a contradiction since $72 > 65$.
+
+Therefore at least 6 of at least one type of stamp was used.
+
+13.
+
+Q: Prove: $x = y$ if and only if $xy = \dfrac{(x + y)^2}{4}$. Note, you will
+need to prove two "directions" here, the "if" and the "only if" part.
+
+A:
+
+Let's start with the first "if" part:
+
+Prove: $x = y$ if $xy = \dfrac{(x + y)^2}{4}$.
+
+If $x = y$, then $xy = \dfrac{(x + y)^2}{4}$.
+
+Proof by Direct Proof:
+
+Let $x$ and $y$ be integers where $x = y$.
+
+Since $x = y$, the following equation:
+
+$$ xy = \frac{(x + y)^2}{4} $$
+
+Can be written as:
+
+$$ x^2 = \frac{(x + x)^2}{4} $$
+
+Evaluation of the right-hand side of the equation shows:
+
+$$ x^2 = \frac{(2x)^2}{4} $$
+
+$$ x^2 = \frac{4x^2}{4} $$
+
+$$ x^2 = x^2 $$
+
+Therefore $xy = \dfrac{(x + y)^2}{4}$ when $x = y$.
+
+And now let's prove the "only if" part:
+
+Prove: $xy = \dfrac{(x + y)^2}{4}$ if $x = y$.
+
+If $xy = \dfrac{(x + y)^2}{4}$, then $x = y$.
+
+Proof by Direct Proof:
+
+Let $x$ and $y$ be integers where $xy = \dfrac{(x + y)^2}{4}$.
+
+Evaluating the equation:
+
+$$ xy = \frac{(x + y)^2}{4} $$
+
+$$ xy = \frac{(x + y)(x + y)}{4} $$
+
+$$ xy = \frac{x^2 + 2xy + y^2}{4} $$
+
+$$ 4xy = x^2 + 2xy + y^2 $$
+
+$$ x^2 + 2xy + y^2 - 4xy = 0 $$
+
+$$ x^2 - 2xy + y^2 = 0 $$
+
+$$ (x - y)^2 = 0 $$
+
+$$ x - y = 0 $$
+
+$$ x = y $$
+
+Therefore $x = y$ when $xy = \dfrac{(x + y)^2}{4}$.
+
+14.
+
+Q: Prove that $\log(7)$ is irrational.
+
+A:
+
+Proof by Contradiction:
+
+Let $a$ and $b$ be integers where $b \neq 0$. Assume that
+$\log(7) = \dfrac{a}{b}$.
+
+Evaluating this equation:
+
+$$ \log(7) = \frac{a}{b} $$
+
+$$ 7 = 10^{\frac{a}{b}} $$
+
+$$ 7 = \sqrt[b]{10^a}$$
+
+$$ 7^b = 10^a $$
+
+$7^b$ only has a prime factor of $7$, while $10^a$ has prime factors of $2$ and
+$5$. But two integers with different prime factorizations cannot be equal.
+
+Therefore $\log(7)$ is irrational.
+
+15.
+
+Q: Prove that there are no integer solutions to the equation $x^2 = 4y + 3$.
+
+A:
+
+Proof by Contradiction:
+
+Let $x$ and $y$ be integers, and assume that $x^2 = 4y + 3$.
+
+Since both $x$ and $y$ are integers, we can take $\mod 4$ of both sides and
+evaluate:
+
+$$ 4y \equiv 0 (\mod 4) $$
+
+$$ 3 \equiv 3 (\mod 4) $$
+
+$$ x^2 \equiv 3 $$
+
+But we know that $x^2 \equiv (0 \vee 1) \mod 4$, so $x^2 \equiv 3$ is a
+contradiction.
+
+Therefore there are no integer solutions to the equation $x^2 = 4y + 3$.
+
+16.
+
+Q: Prove that every prime number greater than 3 is either one more or one less
+than a multiple of 6.
+
+A:
+
+Proof by Direct Proof:
+
+Let $p$ be some prime number greater where $p > 3$.
+
+Consider $p = 6k + r$ where $k$ is any integer and $r$ represents some remainder
+within the set $\{0, 1, 2, 3, 4, 5\}$.
+
+$$ r \in \{0, 1, 2, 3, 4, 5\} $$
+
+To be clear, $r$ represents all possible remaining values when an integer is
+divided by 6.
+
+Since $p$ is a prime number where $p > 3$, $p$ cannot be divisible by $2$ or
+$3$.
+
+Consider the following implications:
+
+- $r = 0 \to p = 6k \to \text{ not prime}$
+
+- $r = 2 \to p = 6k + 2  = 2(3k + 1) \to \text{ even } \to \text{ not prime}$
+
+- $r = 3 \to p = 6k + 3 = 3(2k + 1) \to \text{ divisible by } 3 \to \text{ not prime}$
+
+- $r = 4 \to p = 6k + 4  = 2(3k + 2) \to \text{ even } \to \text{ not prime}$
+
+From the cases above, the only cases that do not make $p$ composite are $r = 1$
+or $r = 5$.
+
+This means that $p = 6k + 1$ is one more than a multiple of 6, and
+$p = 6k + 5 = 6k - 1 + 6 = 6(k + 1) - 1$ is one less than a multiple of $6$.
+
+Therefore every prime number greater than 3 is either one more or one less than
+a multiple of 6.
+
+17.
+
+Q: Your "friend" has shown you a "proof" he wrote to show that $1 = 3$. Here is
+the proof:
+
+Proof: I claim that $1 = 3$. Of course we can do anything to one side of an
+equation as long as we also do it to the other side. So subtract $2$ from both
+sides. This gives $-1 = 1$. Now square both sides, to get $1 = 1$. And we all
+agree this is true.
+
+What is going on here? Is your friend's argument valid? Is the argument a proof
+of the claim $1 = 3$? Carefully explain using what we know about logic.
+
+A:
+
+Our friend's argument is invalid. What is happening here is that our friend is
+that our friend is making a mistake in their logic specifically when they square
+both sides. They assume that if $a = b$, then $a^2 = b^2$, but the converse
+statement of if $b^2 = a^2$, then $b = a$ does not hold (consider $(-1)^2$ as an
+example).
+
+18.
+
+Q: A standard deck of 52 cards consists of 4 suits (hearts, diamonds, spades,
+and clubs) each containing 13 different values (Ace, 2, 3, ..., 10, J, Q, K). If
+you draw some number of cards at random, you might or might not have a pair (two
+cards with the same value) or three cards all of the same suit. However, if you
+draw enough cards, you will be guaranteed to have these. For each of the
+following, find the smallest number of cards you would need to draw to be
+guaranteed having the specified cards. Prove your answers.
+
+(a) Three of a kind (for example, three 7's).
+
+(b) A flush of five cards (for example, five hearts).
+
+\(c\) Three cards that are either all the same suit or all different suits.
+
+A:
+
+(a) Three of a kind (for example, three 7's).
+
+Proof by Direct Proof:
+
+Let 13 be the number of different values in a standard deck of 52 cards
+consisting of 4 suits.
+
+Consider we try and avoid drawing three of a kind for as long as possible. This
+means we would draw at most 2 cards into each of the 13 values. This gives a
+maximum of:
+
+$$ 2 \times 13 = 26 $$
+
+Here we would have 26 cards without having 3 cards of the same values.
+
+Consider what would happen if we drew one more card.
+
+$$ 26 + 1 = 27 $$
+
+By the Pigeonhole Principle, when we distribute 27 cards among 13 ranks, at
+least one rank must contain at least 3 cards:
+
+$$ \left\lceil\frac{27}{13}\right\rceil \geq 3 $$
+
+Therefore, drawing 27 cards would be needed to guarantee having three of a kind.
+
+(b) A flush of five cards (for example, five hearts).
+
+Proof by Direct Proof:
+
+Let 4 be the number of different suits in a standard deck of 52 cards.
+
+Consider we try and avoid drawing 5 cards of the same suit for as long as
+possible. This means we would draw at most 4 cards into each of the 4 suits.
+This gives a maximum of:
+
+$$ 4 \times 4 = 16 $$
+
+Here we would have 16 cards without having 5 cards of the same suit.
+
+Now consider what would happen if we drew one more card:
+
+$$ 16 + 1 = 17 $$
+
+By the Pigeonhole Principle, when we distribute 17 cards among 4 suits, at least
+one suit must contain at least 5 cards:
+
+$$ \left\lceil\frac{17}{4}\right\rceil \geq 5 $$
+
+Therefore, drawing 17 cards would be needed to guarantee having a flush of five
+cards.
+
+\(c\) Three cards that are either all the same suit or all different suits.
+
+Proof by Direct Proof:
+
+First, 4 cards are not sufficient: choose 2 hearts and 2 spades. Then no suit
+has 3 cards, and only two suits appear, so it is impossible to have 3 cards all
+of different suits. Hence, 4 does not guarantee the condition.
+
+Now consider any set of 5 cards. If any suit appears at least 3 times, we are
+done. Otherwise, each suit appears at most twice. Since $5 > 2 \cdot 2 = 4$, at
+least three different suits must appear among the five cards, so we can choose
+one card from each of three suits to obtain three cards all of different suits.
+
+Thus, every set of 5 cards satisfies the condition, and 5 is minimal.
+
+19.
+
+Q: Suppose you are at a party with 19 of your closest friends (so, including
+you, there are 20 people there). Explain why there must be at least two people
+at the party who are friends with the same number of people at the party. Assume
+friendship is always reciprocated.
+
+A:
+
+Proof by Direct Proof:
+
+Let there be 20 people at the party. For each person, let $d$ be the number of
+friends they have at the party. Then $d \in \{0, 1, 2, \dots, 19\}$.
+
+Consider that it is impossible for both a person with 19 friends and a person
+with 0 friends to exist simultaneously, since friendship is reciprocal.
+Therefore, at most 19 distinct values of $d$ can occur among the 20 people.
+
+By the Pigeonhole Principle, two people must have the same value of $d$.
+
+Therefore, at least two people have the same number of friends at the party.
+
+20.
+
+Q: Your friend has given you his list of 115 best Doctor Who episodes (in order
+of greatness). It turns out that you have seen 60 of them. Prove that there are
+at least two episodes you have seen that are exactly four episodes apart on your
+friend's list.
+
+A:
+
+Proof by Direct Proof:
+
+Label the episodes $1, 2, \dots, 115$, Let $S$ be the set of indices of the 60
+episodes you have seen.
+
+Partition the 115 indices into 4 residue classes modulo 4:
+
+$$ \{1, 5, 9, \dots\}, \quad \{2, 6, 10, \dots\}, \quad \{3, 7, 11, \dots\}, \quad \{4, 8, 12, \dots\} $$
+
+These are 4 disjoint classes.
+
+Since there are 60 indices in $s$ and only 4 classes, by the Pigeonhole
+Principle, one of these classes contains at least two elements of $S$.
+
+So there exists two seen episodes whose indices differ by a multiple of 4.
+
+Now, within a single residue class, the elements occur in increasing order
+spaced exactly by 4, so the closest such pair must differ by exactly 4.
+
+Therefore, there exists two episodes you have seen that are exactly four
+episodes apart on the list.