diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index f730363..1e7d7e0 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -7807,8 +7807,6 @@ c. State the contrapositive of the result of part (b). The results of exercise Omitted. -RESUME HERE. - **Test for Primality** Given an integer $n > 1$, to test whether $n$ is prime check to see if it is @@ -7871,17 +7869,60 @@ Page 256 is a rational number, which contradicts Theorem 4.8.1. Explain the discrepancy. +The reason for this discrepancy is due to mistaking +$\sqrt{2} \approx 1.41.4213562$ for $\sqrt{2} = 1.414213562$. The calculator +cannot display $\sqrt{2}$ finitely as it is an irrational number and its +non-repeating decimal goes on forever. Therefore, you cannot find equivalencies +for 1.414213562 and express $\sqrt{2}$ as a rational number as it is +"demonstrated" here. + 2. Example 4.3.1(h) illustrates a technique for showing that any repeating decimal number is rational. A calculator display shows the result of a certain calculation as $40.72727272727$. Can you be sure that the result of the calculation is a rational number? Explain. +Yes. The reason you can be sure that the result of the calculation is a rational +number is because repeating decimal places can always be expressed as a rational +number usually by subtracting the repeating decimal places from a larger number +with the same repeating decimal places. For the given example: + +Let $x = 40.72727272727 \dots$, so $100x = 4072.727272727 \dots$. Then: + +$$ 100x - x = 99x = 4072.727272727\dots - 40.72727272727\dots = 4032 $$ + +$$ 99x = 4032 $$ + +$$ x = \frac{4032}{99} $$ + +Which is an expression for 40.72727272727 in rational form. + 3. Could there be a rational number whose first trillion digits are the same as the first trillion digits of $\sqrt{2}$? Explain. +Yes, because the first trillion digits of $\sqrt{2}$ is potentially finite. In +that case it is rational. In another case where the first trillion digits are +then repeated, then we know by 4.3.1(h) that this form of a decimal is also +rational. Similarly, if even smaller portions of those first trillion digits are +then repeated, this same principle applies. + 4. A calculator display shows that the result of a certain calculation is $0.2$. Can you be sure that the result of the calculation is a rational number? +Yes. Since the decimal $0.2$ has a finite amount of decimal places, we can +simply express it as a fraction: + +Let $x = 0.2$ and $10x = 2$. + +Then: + +$$ 10x = 2 $$ + +$$ x = \frac{2}{10} $$ + +$$ x = \frac{1}{5} $$ + +Where $1$ and $5$ are integers and $5 \neq 0$. This is a rational number. + 5. Let $s$ be the statement: The cube root of every irrational number is irrational. This statement is true, but the following "proof" is incorrect. Explain the mistake. @@ -7895,58 +7936,431 @@ irrational. This is a contradiction, and hence it is not true that the cube root of every irrational number is rational. Thus the statement to be proved is true." +This incorrect proof has two problems. One is that in its supposition, the +wording suggests that the cube root of _every_ irrational number is rational, +when the negation of the given statement would be that "Suppose the cube root of +_some_ irrational number is rational." + +The author of this incorrect proof then goes onto use a specific example of +$2\sqrt{2}$ for their proof. While this is fine for a disproof by +counterexample, a proof by contradiction should be more general. It should +instead read as: + +**Proof by contradiction:** + +Suppose not. Suppose the cube root of some irrational number $x$, is rational. + +Since the cube root of $x$ is rational, this means that +$\sqrt[3]{x} = \dfrac{a}{b}$ for some integers $a$ and $b$ where $b \neq 0$. + +Then, by laws of algebra: + +$$ \sqrt[3]{x} = \frac{a}{b} $$ + +$$ x = \left(\frac{a}{b}\right)^3 $$ + +$$ x = \frac{a^3}{b^3} $$ + +Now, $a^3$ and $b^3$ are integers by the product of integers, where $b^3 \neq 0$ +by the zero product property. Thus $x$ is a rational number and an irrational +number. + +This is a contradiction. + +Q.E.D. + Determine which statements in 6-16 are true and which are false. Prove those that are true and disprove those that are false. 6. $6 - 7\sqrt{2}$ is irrational. +**Proof by contradiction:** + +Suppose not. Suppose $6 - 7\sqrt{2}$ is rational. Then by definition of +rational, + +$$ 6 - 7\sqrt{2} = \frac{a}{b} $$ + +For some integers $a$ and $b$ where $b \neq 0$. + +It follows that: + +$$ 6 - 7\sqrt{2} = \frac{a}{b} $$ + +$$ -7\sqrt{2} = \frac{a}{b} - 6 $$ + +$$ \sqrt{2} = \frac{6 - \dfrac{a}{b}}{7} $$ + +$$ \sqrt{2} = \frac{6b - a}{7b} $$ + +Now, since $6b - a$ and $7b$ are integers by the product and difference of +integers and $7b \neq 0$ by the zero product property. This means that +$\sqrt{2}$ is rational. The $\sqrt{2}$, however, is known to not be rational by +Theorem 4.8.1. + +This is a contradiction. + +Q.E.D. + 7. $3\sqrt{2} - 7$ is irrational. +**Proof by contradiction:** + +Suppose not. Suppose $\sqrt{2} - 7$ is rational. + +Since $\sqrt{2} - 7$ is rational, $\sqrt{2} - 7 = \dfrac{a}{b}$ for some +integers $a$ and $b$ where $b \neq 0$. + +Then, by laws of algebra: + +$$ \sqrt{2} - 7 = \frac{a}{b} $$ + +$$ \sqrt{2} = \frac{a}{b} + 7 $$ + +$$ \sqrt{2} = \frac{a + 7b}{b} $$ + +Now, $a + 7b$ is an integer by the product and sum of integers. Thus $\sqrt{2}$ +is rational. We know, however, by Theorem 4.8.1 that $\sqrt{2}$ is irrational. + +This is a contradiction. + +Q.E.D. + 8. $\sqrt{4}$ is irrational. +This is false. $\sqrt{4} = 2 = \dfrac{2}{1}$, which is rational. + 9. $\dfrac{\sqrt{2}}{6}$ is irrational. +**Proof by contradiction:** + +Suppose not. Suppose $\dfrac{\sqrt{2}}{6}$ is rational. Then by the definition +of rational: + +$$ \frac{\sqrt{2}}{6} = \frac{a}{b} $$ + +for some integers $a$ and $b$ where $b \neq 0$. + +Then, by laws of algebra: + +$$ \sqrt{2} = \frac{6a}{b} $$ + +Now, $6a$ is an integer by the product of integers. Thus $\sqrt{2}$ is a +rational number. We know by Theorem 4.8.1 that $\sqrt{2}$ is irrational. + +This is a contradiction. + +Q.E.D. + 10. The sum of any two irrational numbers is irrational. +**Proof by counterexample:** + +Let $a = \sqrt{2}$, and let $b = -\sqrt{2}$. Then, their sum is: + +$$ a + b = \sqrt{2} + (-\sqrt{2}) = 0 = \frac{0}{1} $$ + +Which is rational. This statement is false. + +Q.E.D. + 11. The difference of any two irrational numbers is irrational. +**Proof by counterexample:** + +Let $a = \sqrt{2}$, and let $b = \sqrt{2}$. Then, their sum is: + +$$ a + b = \sqrt{2} - \sqrt{2}) = 0 = \frac{0}{1} $$ + +Which is rational. This statement is false. + +Q.E.D. + 12. The positive square root of a positive irrational number is irrational. +$$ \forall x \in \mathbb{R} (I(x) \to I(\sqrt{x})) $$ + +Contrapositive: + +$$ \forall x \in \mathbb{R} (\neg I(\sqrt{x}) \to \neg I(x)) $$ + +$$ \forall x \in \mathbb{R} (R(\sqrt{x}) \to R(x)) $$ + +**Proof by contraposition:** + +Suppose $r$ is any positive real number such that $\sqrt{r}$ is rational. + +Since $\sqrt{r}$ is rational, $\sqrt{r} = \dfrac{a}{b}$ where $a$ and $b$ are +integers and $b \neq 0$. + +Then: + +$$ \sqrt{r} = \frac{a}{b} $$ + +$$ r = \left(\frac{a}{b}\right)^2 $$ + +$$ r = \frac{a^2}{b^2} $$ + +Now, $a^2$ and $b^2$ are both integers by the product of integers and +$b^2 \neq 0$ by the zero product property. + +Therefore $r$ is a rational number. + +Q.E.D. + 13. If $r$ is any rational number and $s$ is any irrational number, then $\dfrac{r}{s}$ is irrational. +**Proof by counterexample:** + +Let $r = 0$ and $s = \sqrt{2}$, then +$\dfrac{r}{s} = \dfrac{0}{\sqrt{2}} = 0 = \dfrac{0}{1}$ which is rational. + +Therefore, this statement is false. + +Q.E.D. + 14. The sum of any two positive irrational numbers is irrational. +**Proof by counterexample:** + +Let $x = \sqrt{2}$ and $y = 2 - \sqrt{2}$, then: + +$$ x + y = \sqrt{2} + (2 - \sqrt{2}) = 2 = \dfrac{2}{1} $$ + +Thus, $x + y$ is a rational number. + +Therefore, this statement is false. + +Q.E.D. + 15. The product of two irrational numbers is irrational. +**Proof by counterexample:** + +$$ \sqrt{2} \cdot \sqrt{2} = (\sqrt{2})^2 = 2 = \frac{2}{1} $$ + +Which is a rational number. + +This statement is false. + +Q.E.D. + 16. If an integer greater than $1$ is a perfect square, then its cube root is irrational. +**Proof by counterexample:** + +Consider $64 = 8^2$, then $64 > 1$ and $64$ is a perfect square. Then consider +$\sqrt[3]{64} = 4 = \dfrac{4}{1}$. Thus $\sqrt[3]{64}$ is rational. + +Thus there is at least one integer greater than $1$ that is a perfect square and +its cube root is rational. + +This statement is false. + +Q.E.D. + 17. Consider the following sentence: If $x$ is rational then $\sqrt{x}$ is irrational. Is this sentence always true, sometimes true and sometimes false, or always false? Justify your answer. +This statement is sometimes true and sometimes false. Consider when $x = 2$, +then $\sqrt{2}$ is irrational. This is the case when the statement is true. Then +consider when $x = 9$ then $\sqrt{9} = 3 = \dfrac{3}{1}$ is rational.. This is a +case when the statement is false. Therefore this statement is sometimes true and +sometimes false. + 18. a. Prove that for every integer $a$, if $a^3$ is even then $a$ is even. +**Proof by contrapositive:** + +Suppose $a$ is any integer such that $a$ is odd. + +Since $a$ is odd, $a = 2k + 1$ for some integer $k$. + +Then: + +$$ a^3 = (2k + 1)^3 $$ + +$$ a^3 = 8k^3 + 12k^2 + 6k + 1 $$ + +$$ a^3 = 2(4k^3 + 6k^2 + 3k) + 1 $$ + +Now, $4k^3 + 6k^2 + 3k$ is an integer by the product and sum of integers. +Therefore $a^3$ is odd by definition of odd integers. + +Q.E.D. + b. Prove that $\sqrt[3]{2}$ is irrational. +**Proof by contradiction:** + +Suppose not. That is, suppose $\sqrt[3]{2}$ is rational. Then, by definition of +rational: + +$$ \sqrt[3]{2} = \frac{a}{b} $$ + +for some integers $a$ and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is written in +lowest terms. + +Then: + +$$ \sqrt[3]{2} = \frac{a}{b} $$ + +$$ 2 = \left(\frac{a}{b}\right)^3 $$ + +$$ 2 = \frac{a^3}{b^3} $$ + +$$ 2b^3 = a^3 $$ + +Now, by the definition of even integers, we know that $a^3$ is even. We also +know by part (a), that if $a^3$ is even, then $a$ is even. + +Since $a$ is even, $a = 2k$ for some integer $k$. Then: + +$$ 2b^3 = (2k)^3 $$ + +$$ 2b^3 = 8k^3 $$ + +$$ 2b^3 = 8k^3 $$ + +$$ b^3 = 4k^3 $$ + +$$ b^3 = 2(2k^3) $$ + +Now, $2k^3$ is an integer by the product of integers. Additionally, $b^3$ is +even by the definition of even integers. Additionally, by part (a) $b$ is even +since $b^3$ is even. + +Since both $a$ and $b$ and even, $\sqrt[3]{2}$ is not in lowest terms, so +$\sqrt[3]{2}$ is not rational, which contradicts the supposition. + +Q.E.D. + 19. a. Use proof by contradiction to show that for any integer $n$, it is impossible for $n$ to equal both $3q_1 + r_1$ and $3q_2 + r_2$, where $q_1$, $q_2$, $r_1$, and $r_2$ are integers, $0 \leq r_1 < 3$, $0 \leq r_2 < 3$, and $r_1 \neq r_2$. +**Proof by contradiction:** + +Suppose not. That is, suppose $n$ is some integer such +$n = 3q_1 + r_1 = 3q_2 + r_2$, where $q_1$, $q_2$, $r_1$, and $r_2$ are some +integers such that $0 \leq r_1 < 3$ and $0 \leq r_2 < 3$ and $r_1 \neq r_2$. + +Then: + +$$ 3q_1 + r_1 = 3q_2 + r_2 $$ + +$$ 3q_1 - 3q_2 = r_2 - r_1 $$ + +$$ 3(q_1 - q_2) = r_2 - r_1 $$ + +This means that $3 \mid (r_2 - r_1)$. + +Since $0 \leq r_1,r_2 < 3$, it follows that: + +$$ -2 \leq r_2 - r_1 \leq 2 $$ + +This means that $3 \cancel{\mid} (r_2 - r_1)$, which contradicts the earlier +finding that $3 \mid (r_2 - r_1)$. + +Q.E.D. + b. Use proof by contradiction, the quotient-remainder theorem, division into cases, and the result of part (a) to prove that for every integer $n$, if $n^2$ is divisible by $3$ then $n$ is divisible by $3$. +**Proof by contradiction:** + +Suppose not. That is, suppose $n$ is some integer such that $n^2$ is divisible +by $3$ and $n$ is not divisible by $3$. + +Since $n$ is not divisible by $3$, then $n = 3k + 1$ or $n = 3k + 2$ for some +integer $k$. + +_Case where $n = 3k + 1$:_ + +$$ n^2 = (3k + 1)^2 $$ + +$$ n^2 = 9k^2 + 6k + 1 $$ + +$$ n^2 = 3(3k^2 + 2k) + 1 $$ + +By the quotient-remainder theorem, this means that $3 \cancel{\mid} n^2$. This +contradicts the supposition. + +_Case where $n = 3k + 2$:_ + +$$ n^2 = (3k + 2)^2 $$ + +$$ n^2 = 9k^2 + 12k + 4 $$ + +$$ n^2 = 9k^2 + 12k + 3 + 1 $$ + +$$ n^2 = 3(3k^2 + 4k + 1) + 1 $$ + +By the quotient-remainder theorem, this means that $3 \cancel{\mid} n^2$. This +contradicts the supposition. + +In both cases, the supposition is contradicted. + +Q.E.D. + c. Prove that $\sqrt{3}$ is irrational. +**Proof by contradiction:** + +Suppose not. Suppose that $\sqrt{3}$ is rational. + +Since $\sqrt{3}$ is rational, $\sqrt{3} = \dfrac{a}{b}$ for some integers $a$ +and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is in lowest terms. + +Then: + +$$ \sqrt{3} = \frac{a}{b} $$ + +$$ 3 = \left(\frac{a}{b}\right)^2 $$ + +$$ 3 = \frac{a^2}{b^2} $$ + +$$ 3b^2 = a^2 $$ + +This means that $3 \mid a^2$. By part (b), we then know that $3 \mid a$. This +means that $a = 3k$ for some integer $k$. Then: + +$$ 3b^2 = (3k)^2 $$ + +$$ 3b^2 = 9k^2 $$ + +$$ b^2 = 3k^2 $$ + +This means that $3 \mid b^2$. Then, however, $\dfrac{a}{b}$ is not in lowest +terms since $3 \mid a$ and $3 \mid b$. This is a contradiction. + +Q.E.D. + 20. Give an example to show that if $d$ is not prime and $n^2$ is divisible by $d$, then $n$ need not be divisible by $d$. +We need to find a non prime number $d$ and a real number $n$ such that: + +$$ d \mid n^2 $$ + +$$ d \cancel{\mid} n $$ + +Let $d = 4$ and $n = 2$. + +Then: + +$$ 4 \mid 4 $$ + +$$ 4 \cancel{\mid} 2 $$ + 21. The quotient-remainder theorem says not only that there exist quotients and remainders but also that the quotient and remainder of a division are unique. Prove the uniqueness. That is, prove that if $a$ and $d$ are @@ -7963,30 +8377,326 @@ then $$ q_1 = q_2 \quad \text{ and } r_1 = r_2 $$ +**Proof:** + +Suppose $a$, $d$ are any integers and $q_1$, $q_2$, $r_1$, and $r_2$ are some +integers such that $a = dq_1 + r_1$ where $0 \leq r_1 < d$ and $a = dq_2 + r_2$ +where $0 \leq r_2 < d$. + +Then: + +$$ dq_1 + r_1 = dq_2 + r_2 $$ + +$$ dq_1 - dq_2 = r_2 - r_1 $$ + +$$ d(q_1 - q_2) = r_2 - r_1 $$ + +Since $0 \leq r_1,r_2 < d$, this means that: + +$$ -(d - 1) \leq r_2 - r_1 \leq d - 1 $$ + +And this means: + +$$ |r_2 - r_1| < d $$ + +Since $d(q_1 - q_2) = r_2 - r_1$ means $d \mid (r_2 - r_1)$, but we also know +that $|r_2 - r_1| < d$, then it follows that $r_2 - r_1 = 0$. Thus, it then +follows that $r_1 = r_2$ + +Then by substitution: + +$$ d(q_1 - q_2) = r_2 - r_1 $$ + +$$ q_1 - q_2 = 0 $$ + +Thus it follows that $q_1 = q_2$. + +Therefore, it has been shown that $q_1 = q_2$ and $r_1 = r_2$. + +Q.E.D. + 22. Prove that $\sqrt{5}$ is irrational. +**Lemma 22:** + +If $5$ divides $a^2$, then $5$ divides $a$. + +**Proof by contradiction:** + +Suppose not, that is suppose $a$ is some integer such that $5 \mid a^2$ and +$5 \cancel{\mid} a$. + +Since $5 \cancel{\mid} a$, this means that $a = 5q + r$ for some unique integers +$q$ and $r$, such that $1 \leq r < 5$. + +Then, by substitution: + +$$ a^2 = (5q + r)^2 $$ + +$$ a^2 = 25q^2 + 10qr + r^2 $$ + +$$ a^2 = 5(5q^2 + 2qr) + r^2 $$ + +So: + +$$ a^2 \equiv r^2 (\mod 5) $$ + +Then by cases: + +$$ r = 1 \to r^2 (\mod 5) = 1 $$ + +$$ r = 2 \to r^2 (\mod 5) = 4 $$ + +$$ r = 3 \to r^2 (\mod 5) = 4 $$ + +$$ r = 4 \to r^2 (\mod 5) = 1 $$ + +So in all cases $r^2 \cancel{\equiv} 0 (\mod 5)$ + +Therefore $5 \cancel{\mid} a^2$, which contradicts the supposition.$ + +Q.E.D. + +**Proof by contradiction:** + +Suppose not. Suppose that $\sqrt{5}$ is rational. + +Since $\sqrt{5}$ is rational, $\sqrt{5} = \dfrac{a}{b}$ for some integers $a$ +and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is in lowest terms. + +Then: + +$$ \sqrt{5} = \frac{a}{b} $$ + +$$ 5 = \left(\frac{a}{b}\right)^2 $$ + +$$ 5 = \frac{a^2}{b^2} $$ + +$$ 5b^2 = a^2 $$ + +Thus we know that $5 \mid a^2$, and by Lemma 22, we then know that $5 \mid a$. + +So, since $5 \mid a$, $a = 5k$ for some integer $k$. + +Then by substitution: + +$$ 5b^2 = (5k)^2 $$ + +$$ 5b^2 = 25k^2 $$ + +$$ b^2 = 5k^2 $$ + +So then we know that $5 \mid b^2$, and by Lemma 22, we then know that +$5 \mid b$. Then, however, $\dfrac{a}{b}$ is not in lowest terms. Therefore +$\sqrt{5}$ is irrational, which contradicts the supposition. + +Q.E.D. + 23. Prove that for any integer $a$, $9 \cancel{\mid} (a^2 - 3)$. +_Hint:_ This statement is true. If $a^2 - 3 = 9b$, then +$a^2 = 9b + 3 = 3(3b + 1)$, and so $a^2$ is divisible by $3$. Hence, by exercise +19(b), $a$ is divisible by $3$. Thus $a^2 =(3c)^2$ for some integer $c$. + +**Proof by contradiction:** + +Suppose not. That is, suppose that for some integer $a$, $9 \mid (a^2 - 3)$. + +Since $9 \mid (a^2 - 3)$, then $a^2 - 3 = 9b$ for some integer $b$. This then +becomes: + +$$ a^2 - 3 = 9b $$ + +$$ a^2 = 9b + 3 $$ + +$$ a^2 = 3(3b + 1) $$ + +So $3 \mid a^2$, by 9(b), we then know that $3 \mid a$. Since $3 \mid a$, +$a = 3c$ for some integer $c$. By substitution: + +$$ (3c)^2 = 3(3b + 1) $$ + +$$ 9c^2 = 3(3b + 1) $$ + +$$ 3c^2 = 3b + 1 $$ + +$$ 3c^2 - 1 = 3b $$ + +But since $ 3 \cancel{\mid} 3c^2 - 1$ and $3 \mid 3b$, this statement can never +hold for $b$ and $c$. This is a contradiction. + +Q.E.D. + 24. An alternative proof of the irrationality of $\sqrt{2}$ counts the number of $2$'s on the two sides of the equation $2n^2 = m^2$ and uses the unique factorization of integers theorem to deduce a contradiction. Write a proof that uses this approach. +Statement: $\sqrt{2}$ is irrational. + +**Proof by contradiction:** + +Suppose not. That is, suppose $\sqrt{2}$ is rational. Then there are some +integers $m$ and $n$ where $n \neq 0$ such that: + +$$ \sqrt{2} = \frac{m}{n} $$ + +where $\dfrac{m}{n}$ have no common factors. Then: + +$$ \sqrt{2} = \frac{m}{n} $$ + +$$ 2 = \frac{m^2}{n^2} $$ + +$$ 2n^2 = m^2 $$ + +By the unique factorization theorem, we can express $n$ as the product of prime +numbers as: + +$$ n = 2^a \cdot p_1^{e_1} \dots \cdot p_k^{e_k} $$ + +Then, we can express $n^2$ as: + +$$ n^2 = 2^{2a} $$ + +Since the exponent is even, we know the number of $2s$ in $n^2$ is even. It then +follows that $2n^2$ adds one more $2$: + +$$ 2n^2 = 2^{2a + 1} $$ + +So $2n^2$ has an odd number of $2$s. + +We can then apply the same logic to $m^2$. + +$$ m = 2^b \cdot p_1^{e_1} \dots \cdot p_q^{e_q} $$ + +$$ m^2 = 2^{2b} $$ + +So $m^2$ has an even number of $2$s. Thus $2n^2$ has an odd number of $2$s and +$m^2$ has an even number of $2$s, which contradicts the statement $2n^2 = m^2$. + +Q.E.D. + 25. Use the proof technique illustrated in exercise 24 to prove that if $n$ is any positive integer that is not a perfect square, then $\sqrt{n}$ is irrational. +**Proof by contradiction:** + +Suppose not. That is, suppose $n$ is some positive integer such that $n$ is not +a perfect square and $\sqrt{n}$ is rational. + +Since $\sqrt{n}$ is rational, $\sqrt{n} = \dfrac{a}{b}$ for some integers $a$ +and $b$, where $b \neq 0$ and where $\dfrac{a}{b}$ is in lowest terms. + +Then: + +$$ \sqrt{n} = \frac{a}{b} $$ + +$$ n = \left(\frac{a}{b}\right)^2 $$ + +$$ n = \frac{a^2}{b^2} $$ + +$$ nb^2 = a^2 $$ + +By the unique factorization theorem, $a$ can be written as a product of prime +numbers: + +$$ a = p_1^{e_1}p_2^{e_2} \cdot \dots \cdot p_k^{e_k} $$ + +And: + +$$ a^2 = p_1^{2e_1}p_2^{2e_2} \cdot \dots \cdot p_k^{2e_k} $$ + +This means that all prime exponents in $a^2$ are even. + +Similarly: + +$$ b = q_1^{f_1}q_2^{f_2} \cdot \dots \cdot q_m^{f_m} $$ + +And: + +$$ b^2 = q_1^{2e_1}q_2^{2e_2} \cdot \dots \cdot q_k^{2e_m} $$ + +This means that all prime exponents in $b^2$ are even. + +Since $nb^2 = a^2$, all prime exponents in $nb^2$ are even since all the prime +exponents in $a^2$ are even. Therefore $n$ is a perfect square + +This contradicts the supposition that $n$ is not a perfect square. + +Q.E.D. + 26. Prove that $\sqrt{2} + \sqrt{3}$ is irrational. +**Proof by contradiction:** + +Suppose not. That is suppose that $\sqrt{2} + \sqrt{3}$ is rational. + +Since $\sqrt{2} + \sqrt{3}$ is rational, then +$\sqrt{2} + \sqrt{3} = \dfrac{a}{b}$ where $a$ and $b$ are some integers and +$b \neq 0$ and $\dfrac{a}{b}$ has no common factors. + +Then, by substitution: + +$$ \sqrt{2} + \sqrt{3} = \frac{a}{b} $$ + +$$ \left(\sqrt{2} + \sqrt{3}\right)^2 = \left(\frac{a}{b}\right)^2 $$ + +$$ \left(\sqrt{2} + \sqrt{3}\right)\left(\sqrt{2} + \sqrt{3}\right) = \frac{a^2}{b^2} $$ + +$$ 2 + 2\sqrt{2}\sqrt{3} + 3 = \frac{a^2}{b^2} $$ + +$$ 2\sqrt{6} + 5 = \frac{a^2}{b^2} $$ + +$$ 2\sqrt{6} = \frac{a^2}{b^2} - 5 $$ + +$$ \sqrt{6} = \frac{\dfrac{a^2}{b^2} - 5}{2} $$ + +$$ \sqrt{6} = \frac{a^2 - 5b^2}{2b^2} $$ + +By the proof given in exercise 25, we know that $\sqrt{6}$ is irrational. + +Now, $a^2 - 5b^2$ is an integer by the product and difference of integers. Also +$2b^2$ is an integer by the product of integers and $2b^2 \neq 0$ by the zero +product property. + +Therefore $\sqrt{6}$ is a rational number which contradicts what we derived from +the proof given in exercise 25 that $\sqrt{6}$ is irrational. + +Q.E.D. + 27. Prove that $\log_5(2)$ is irrational. (_Hint:_ Use the unique factorization of integers theorem.) +Omitted. + 28. Let $N = 2 \cdot 3 \cdot 5 \cdot 7 + 1$. What remainder is obtained when $N$ is divided by $2$?$3$?$5$?$7$? Justify your answer. +$$ N = 2(3 \cdot 5 \cdot 7) + 1 $$ + +$$ N \mod 2 = 1 $$ + +$$ N = 3(2 \cdot 5 \cdot 7) + 1 $$ + +$$ N \mod 3 = 1 $$ + +$$ N = 5(2 \cdot 3 \cdot 7) + 1 $$ + +$$ N \mod 5 = 1 $$ + +$$ N = 7(2 \cdot 3 \cdot 5) + 1 $$ + +$$ N \mod 7 = 1 $$ + 29. Suppose $a$ is an integer and $p$ is a prime number such that $p \mid a$ and $p \mid (a + 3)$. What can you deduce about $p$? Why? +Since $p \mid a$ and $p \mid (a + 3)$, then $p \mid ((a + 3) - a)$. It follows +then that $p \mid 3$, and the only prime number that divides $3$ is $3$ itself. + +$$ p = 3 $$ + 30. Let $p_1, p_2, p_3, \dots$ be a list of all prime numbers in ascending order. Here is a table of the first six: @@ -7998,10 +8708,32 @@ a. Let $N_1 = p_1, N_2 = p_1 \cdot p_2, N_3 = p_1 \cdot p_2 \cdot p_3, \dots N_6 = p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6$. Calculate $N_1$, $N_2$, $N_3$, $N_4$, $N_5$, and $N_6$. +| $N_1$ | $N_2$ | $N_3$ | $N_4$ | $N_5$ | $N_6$ | +| ----- | ----- | ----- | ----- | ------ | ------- | +| $2$ | $6$ | $30$ | $210$ | $2310$ | $30030$ | + b. For each $i = 1, 2, 3, 4, 5, 6$, find whether $N_i$ is itself prime or just has a prime factor less than itself. (_Hint:_ Use the test for primality from exercise 31 in Section 4.7 to determine your answers.) +**Test for Primality** + +Given an integer $n > 1$, to test whether $n$ is prime check to see if it is +divisible by a prime number less than or equal to its square root. If it is not +divisible by any of these numbers, then it is prime. + +$N_1$ is prime. + +$N_2$ is not prime. + +$N_3$ is not prime. + +$N_4$ is not prime. + +$N_5$ is not prime. + +$N_6$ is not prime. + For exercises 31 and 32, use the fact that for ever integer $n$, $$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$ @@ -8013,13 +8745,41 @@ $$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$ largest. Call it $p$. Let $M = p! + 1$. We will show that there is a prime number $q$ such that $q > p$. Complete this proof. +**Proof by contradiction:** + +Suppose not. That is suppose there are only finitely many prime numbers. Then +one is the largest. Call it $p$. Let $M = p! + 1$. + +Let $q$ be a prime number such that $q \leq p$. + +Since $q \leq p$, $q \mid p!$. This means that: + +$$ p! \equiv 0 (\mod q) $$ + +It then follows that: + +$$ M = p! + 1 \equiv 1 (\mod q) $$ + +So, $q \cancel{\mid} M$. + +Since $q$ is a prime number that cannot divide $M$, either $M$ is prime itself +or $M$ has a prime factor greater than $p$, which contradicts the supposition. + +Q.E.D. + +We will show that there is a prime number $q$ such that $q > p$. + 32. Prove that for every integer $n$, if $n > 2$ then there is a prime number $p$ such that $n < p < n!$. +Omitted. + 33. Prove that if $p_1, p_2, \dots$, and $p_n$ are distinct prime numbers with $p_1 = 2$ and $n > 1$, then $p_1, p_2, \dots, p_n + 1$ can be written in the form $4k + 3$ for some integer $k$. +Omitted. + 34. a. Fermat's last theorem says that for every integer $n > 2$, the equation @@ -8029,11 +8789,15 @@ $p > 2$, $x^p + y^p = z^p$ has no positive integer solution, then for any integer $n > 2$ that is not a power of $2$, $x^n + y^n = z^n$ has no positive integer solution. +Omitted. + b. Fermat proved that there are no integers $x$, $y$, and $z$ such that $x^4 + y^4 = z^4$. Use this result to remove the restriction in part (a) that $n$ not be a power of $2$. That is, prove that if $n$ is a power of $2$ and $n > 4$, then $x^n + y^n = z^n$ has no positive integer solution. +Omitted. + For exercises 35-38 note that to show there is a unique object with a certain property, show that (1) there is an object with the property and (2) if objects $A$ and $B$ have the property, then $A = B$. @@ -8041,13 +8805,21 @@ $A$ and $B$ have the property, then $A = B$. 35. Prove that there exists a unique prime number of the form $n^2 - 1$, where $n$ is an integer that is greater than or equal to $2$. +Omitted. + 36. Prove that there exists a unique prime number of the form $n^2 + 2n - 3$, where $n$ is a positive integer. +Omitted. + 37. Prove that there is at most one real number $a$ with the property that $a + r = r$ for every real number $r$. (Such a number is called an _additive identity_.) +Omitted. + 38. Prove that there is at most one real number $b$ with the property that $br = r$ for every real number $r$. (Such a number is called a _multiplicative identity_.) + +Omitted. diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index d97b7ab..e7c1897 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -234,12 +234,18 @@ Page 256 length $1$, the ratio of the length of the hypotenuse to the length of one of the legs is not equal to a ratio of ______. +two integers. + 2. One way to prove that $\sqrt{2}$ is an irrational number is to assume that $\sqrt{2} = \dfrac{m}{n}$ for some integers $m$ and $n$ that have no common factor greater than $1$, use the lemma that says that if the square of an integer is even then ______, and eventually show that $m$ and $n$ ______. +that integer is even; have a common factor greater than 1. + 3. One way to prove that there are infinitely many prime numbers is to assume that there is a largest prime number $p$, construct the number ______, and then show that this number has to be divisible by a prime number that is greater than ______. + +$N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot \dots \cdot p) + 1$; $p$