From 36f5f17ac826fc421a2551bfabbde8cfa4a4149a Mon Sep 17 00:00:00 2001 From: tomit4 Date: Fri, 12 Jun 2026 10:26:04 -0700 Subject: [PATCH] :construction: Setup for 4.8 --- chapter_4/exercises.md | 192 +++++++++++++++++++++++++++++++++++++ chapter_4/notes.md | 132 +++++++++++++++++++++++++ chapter_4/test_yourself.md | 20 ++++ leftoff.txt | 2 +- 4 files changed, 345 insertions(+), 1 deletion(-) diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 1cb70a5..f730363 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -7859,3 +7859,195 @@ d. 7,917 36. For all odd integers $a$, $b$, and $c$, if $z$ is a solution of $ax^2 + bx + c = 0$ then $z$ is irrational. (In the proof, use the properties of even and odd integers that are listed in Example 4.3.3.) + +--- + +Page 256 + +**Exercise Set 4.8** + +1. A calculator display shows that $\sqrt{2} = 1.414213562$. Because + $1.414213562 = \dfrac{1414213562}{1000000000}$, this suggests that $\sqrt{2}$ + is a rational number, which contradicts Theorem 4.8.1. Explain the + discrepancy. + +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. + +3. Could there be a rational number whose first trillion digits are the same as + the first trillion digits of $\sqrt{2}$? Explain. + +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? + +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. + +**"Proof (by contradiction):** + +Suppose not. Suppose the cube root of every irrational number is rational. But +$2\sqrt{2}$ is irrational because it is a product of a rational and an +irrational number, and the cube root of $2\sqrt{2}$ is $\sqrt{2}$, which is +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." + +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. + +7. $3\sqrt{2} - 7$ is irrational. + +8. $\sqrt{4}$ is irrational. + +9. $\dfrac{\sqrt{2}}{6}$ is irrational. + +10. The sum of any two irrational numbers is irrational. + +11. The difference of any two irrational numbers is irrational. + +12. The positive square root of a positive irrational number is irrational. + +13. If $r$ is any rational number and $s$ is any irrational number, then + $\dfrac{r}{s}$ is irrational. + +14. The sum of any two positive irrational numbers is irrational. + +15. The product of two irrational numbers is irrational. + +16. If an integer greater than $1$ is a perfect square, then its cube root is + irrational. + +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. + +18. + +a. Prove that for every integer $a$, if $a^3$ is even then $a$ is even. + +b. Prove that $\sqrt[3]{2}$ is irrational. + +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 $r_1 \neq r_2$. + +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 $3$. + +c. Prove that $\sqrt{3}$ is irrational. + +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 $d$. + +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 + integers with $d > 0$ and if $q_1$, $r_1$, $q_2$, and $r_2$ are integers + such that + +$$ a = dq_1 + r_1 \quad \text{ where } 0 \leq r_1 < d $$ + +and + +$$ a = dq_2 + r_2 \quad \text{ where } 0 \leq r_2 < d $$ + +then + +$$ q_1 = q_2 \quad \text{ and } r_1 = r_2 $$ + +22. Prove that $\sqrt{5}$ is irrational. + +23. Prove that for any integer $a$, $9 \cancel{\mid} (a^2 - 3)$. + +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. + +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. + +26. Prove that $\sqrt{2} + \sqrt{3}$ is irrational. + +27. Prove that $\log_5(2)$ is irrational. (_Hint:_ Use the unique factorization + of integers theorem.) + +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. + +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? + +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: + +| $p_1$ | $p_2$ | $p_3$ | $p_4$ | $p_5$ | $p_6$ | +| ----- | ----- | ----- | ----- | ----- | ----- | +| $2$ | $3$ | $5$ | $7$ | $11$ | $13$ | + +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 $N_6$. + +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.) + +For exercises 31 and 32, use the fact that for ever integer $n$, + +$$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$ + +31. An alternative proof of the infinitude of the prime numbers begins as + follows: + +**Proof:** Suppose there are only finitely many prime numbers. Then one is the +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. + +32. Prove that for every integer $n$, if $n > 2$ then there is a prime number + $p$ such that $n < p < n!$. + +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 $k$. + +34. + +a. Fermat's last theorem says that for every integer $n > 2$, the equation +$x^n + y^n = z^n$ has no positive integer solution (solution for which $x$, $y$, +and $z$ are positive integers). Prove the following: If for every prime number +$p > 2$, $x^p + y^p = z^p$ 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. + +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. + +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 $A = B$. + +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 $2$. + +36. Prove that there exists a unique prime number of the form $n^2 + 2n - 3$, + where $n$ is a positive integer. + +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_.) + +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_.) diff --git a/chapter_4/notes.md b/chapter_4/notes.md index c8f4450..c7b87a7 100644 --- a/chapter_4/notes.md +++ b/chapter_4/notes.md @@ -978,3 +978,135 @@ odd, $n^2$ is odd. Therefore, $n^2$ is both even and odd. This contradicts Theorem 4.7.2, which states that no integer can be both even and odd. _[This contradiction shows that the supposition is false and, hence, that the proposition is true.]_ + +--- + +Page 252 + +**Theorem 4.8.1 Irrationality of $\sqrt{2}$** + +$\sqrt{2}$ is irrational. + +**Proof (by contradiction):** + +_[We take the negation and suppose it to be true.]_ Suppose not. That is, +suppose $\sqrt{2}$ is rational. Then there are integers $m$ and $n$ with no +common factors such that + +$$ \sqrt{2} = \frac{m}{n} $$ + +_[by dividing $m$ and $n$ by any common factors if necessary]._ _[We must derive +a contradiction.]_ Squaring both sides of equation (4.8.1) gives + +$$ 2 = \frac{m^2}{n^2} $$ + +Or, equivalently, + +$$ m^2 = 2n^2 $$ + +Note that equation (4.8.2) implies that $m^2$ is even (by definition of even). +It follows that $m$ is even (by Proposition 4.7.4). We file this fact away for +future reference and also deduce (by definition of even) that + +$$ m = 2k \quad \text{ for some integer } k $$ + +Substituting equation (4.8.3) into equation (4.8.2), we see that + +$$ m^2 = (2k)^2 = 4k^2 = 2n^2 $$ + +Dividing both sides of the right-most equation by $2$ gives + +$$ n^2 = 2k^2 $$ + +Consequently, $n^2$ is even, and so $n$ is even (by Proposition 4.7.4). But we +also know that $m$ is even. _[This is the fact we filed away.]_ Hence both $m$ +and $n$ have a common factor of $2$. But this contradicts the supposition that +$m$ and $n$ have no common factors. _[Hence the supposition is false and so the +theorem is true.]_ + +--- + +Page 253 + +**Proposition 4.8.2** + +$1 + 3\sqrt{2}$ is irrational. + +**Proof (by contradiction):** + +Suppose not. Suppose $1 + 3\sqrt{2}$ is rational. _[We must derive a +contradiction.]_ Then by definition of rational, + +$$ 1 + 3\sqrt{2} = \frac{a}{b} \quad \text{ for some integers } a \text{ and } b \text{ with } b \neq 0 $$ + +It follows that + +$$ 3\sqrt{2} = \frac{a}{b} - 1 \quad \text{ by subtracting } 1 \text{ from both sides} $$ + +$$ = \frac{a}{b} - \frac{b}{b} \quad \text{ by substitution} $$ + +$$ = \frac{a - b}{b} $$ + +Hence + +$$ \sqrt{2} = \frac{a - b}{3b} $$ + +But $a - b$ and $3b$ are integers (since $a$ and $b$ are integers and +differences and products of integers are integers), and $3b \neq 0$ by the zero +product property. Hence $\sqrt{2}$ is a quotient of the two integers $a - b$ and +$3b$ with $3b \neq 0$, and so $\sqrt{2}$ is rational (by definition of +rational). This contradicts the fact that $\sqrt{2}$ is irrational. _[The +contradiction shows that the supposition is false.]_ Hence $1 + 3\sqrt{2}$ is +irrational. + +--- + +Page 254 + +**Proposition 4.8.3** + +For any integer $a$ and any prime number $p$, if $p \mid a$ then +$p \cancel{\mid} (a + 1)$. + +**Proof (by contradiction):** + +Suppose not. That is, suppose there exists an integer $a$ and a prime number $p$ +such that $p \mid a$ and $p \mid (a + 1)$. Then, by definition of divisibility, +there exists integers $r$ and $s$ such that $a = pr$ and $a + 1 = ps$. It +follows that + +$$ 1 = (a + 1) - a = ps - pr = p(s - r) $$ + +and so (since $s - r$ is an integer) $p \mid 1$. But, by Theorem 4.4.2, the only +integer divisors of $1$ are $1$ and $-1$, and $p > 1$ because $p$ is prime. Thus +$p \leq 1$ and $p > 1$, which is a contradiction. _[Hence the supposition is +false, and the proposition is true.]_ + +--- + +Page 254 + +**Theorem 4.8.4 Infinitude of the Primes** + +The set of prime numbers is infinite. + +**Proof (by contradiction):** + +Suppose not. That is, suppose the set of prime numbers is finite. _[WE must +deduce a contradiction.]_ Then some prime number $p$ is the largest of all the +prime numbers, and hence we can list the prime numbers in ascending order. + +$$ 2, 3< 5, 7, 11, \dots, p $$ + +Let $N$ be the product of all the prime numbers plus $1$: + +$$ N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p) + 1 $$ + +Then $N > 1$, and so, by Theorem 4.4.4, $N$ is divisible by some prime number +$q$. Because $q$ is prime, $q$ must equal one of the prime numbers +$2, 3, 5< 7, 11, \dots, p$. Thus, by definition of divisibility, $q$ divides +$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p$, and so, by Proposition 4.8.3, $q$ +does not divide $(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p) + 1$, which equals +$N$. Hence $N$ is divisible by $q$ and $N$ is not divisible by $q$, and we have +reached a contradiction. _[Therefore, the supposition is false and the theorem +is true.]_ diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index 8bfd070..d97b7ab 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -223,3 +223,23 @@ $\forall x \in D, \text{ if } \neg Q(x) \text{ then } \neg P(x)$ you suppose that ______ and you show that ______. $Q(x)$ is false; $P(x)$ is false. + +--- + +**Test Yourself** + +Page 256 + +1. The ancient Greeks discovered that in a right triangle where both legs have + 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 ______. + +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 $n$ ______. + +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 ______. diff --git a/leftoff.txt b/leftoff.txt index f06fa6c..cb1a40d 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -241 +250