🚧 Setup for 4.8
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@ -7859,3 +7859,195 @@ d. 7,917
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36. For all odd integers $a$, $b$, and $c$, if $z$ is a solution of
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$ax^2 + bx + c = 0$ then $z$ is irrational. (In the proof, use the
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properties of even and odd integers that are listed in Example 4.3.3.)
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---
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Page 256
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**Exercise Set 4.8**
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1. A calculator display shows that $\sqrt{2} = 1.414213562$. Because
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$1.414213562 = \dfrac{1414213562}{1000000000}$, this suggests that $\sqrt{2}$
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is a rational number, which contradicts Theorem 4.8.1. Explain the
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discrepancy.
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2. Example 4.3.1(h) illustrates a technique for showing that any repeating
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decimal number is rational. A calculator display shows the result of a
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certain calculation as $40.72727272727$. Can you be sure that the result of
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the calculation is a rational number? Explain.
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3. Could there be a rational number whose first trillion digits are the same as
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the first trillion digits of $\sqrt{2}$? Explain.
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4. A calculator display shows that the result of a certain calculation is $0.2$.
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Can you be sure that the result of the calculation is a rational number?
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5. Let $s$ be the statement: The cube root of every irrational number is
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irrational. This statement is true, but the following "proof" is incorrect.
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Explain the mistake.
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**"Proof (by contradiction):**
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Suppose not. Suppose the cube root of every irrational number is rational. But
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$2\sqrt{2}$ is irrational because it is a product of a rational and an
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irrational number, and the cube root of $2\sqrt{2}$ is $\sqrt{2}$, which is
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irrational. This is a contradiction, and hence it is not true that the cube root
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of every irrational number is rational. Thus the statement to be proved is
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true."
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Determine which statements in 6-16 are true and which are false. Prove those
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that are true and disprove those that are false.
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6. $6 - 7\sqrt{2}$ is irrational.
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7. $3\sqrt{2} - 7$ is irrational.
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8. $\sqrt{4}$ is irrational.
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9. $\dfrac{\sqrt{2}}{6}$ is irrational.
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10. The sum of any two irrational numbers is irrational.
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11. The difference of any two irrational numbers is irrational.
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12. The positive square root of a positive irrational number is irrational.
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13. If $r$ is any rational number and $s$ is any irrational number, then
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$\dfrac{r}{s}$ is irrational.
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14. The sum of any two positive irrational numbers is irrational.
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15. The product of two irrational numbers is irrational.
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16. If an integer greater than $1$ is a perfect square, then its cube root is
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irrational.
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17. Consider the following sentence: If $x$ is rational then $\sqrt{x}$ is
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irrational. Is this sentence always true, sometimes true and sometimes
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false, or always false? Justify your answer.
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18.
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a. Prove that for every integer $a$, if $a^3$ is even then $a$ is even.
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b. Prove that $\sqrt[3]{2}$ is irrational.
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19.
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a. Use proof by contradiction to show that for any integer $n$, it is impossible
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for $n$ to equal both $3q_1 + r_1$ and $3q_2 + r_2$, where $q_1$, $q_2$, $r_1$,
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and $r_2$ are integers, $0 \leq r_1 < 3$, $0 \leq r_2 < 3$, and $r_1 \neq r_2$.
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b. Use proof by contradiction, the quotient-remainder theorem, division into
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cases, and the result of part (a) to prove that for every integer $n$, if $n^2$
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is divisible by $3$ then $n$ is divisible by $3$.
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c. Prove that $\sqrt{3}$ is irrational.
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20. Give an example to show that if $d$ is not prime and $n^2$ is divisible by
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$d$, then $n$ need not be divisible by $d$.
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21. The quotient-remainder theorem says not only that there exist quotients and
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remainders but also that the quotient and remainder of a division are
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unique. Prove the uniqueness. That is, prove that if $a$ and $d$ are
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integers with $d > 0$ and if $q_1$, $r_1$, $q_2$, and $r_2$ are integers
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such that
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$$ a = dq_1 + r_1 \quad \text{ where } 0 \leq r_1 < d $$
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and
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$$ a = dq_2 + r_2 \quad \text{ where } 0 \leq r_2 < d $$
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then
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$$ q_1 = q_2 \quad \text{ and } r_1 = r_2 $$
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22. Prove that $\sqrt{5}$ is irrational.
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23. Prove that for any integer $a$, $9 \cancel{\mid} (a^2 - 3)$.
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24. An alternative proof of the irrationality of $\sqrt{2}$ counts the number of
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$2$'s on the two sides of the equation $2n^2 = m^2$ and uses the unique
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factorization of integers theorem to deduce a contradiction. Write a proof
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that uses this approach.
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25. Use the proof technique illustrated in exercise 24 to prove that if $n$ is
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any positive integer that is not a perfect square, then $\sqrt{n}$ is
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irrational.
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26. Prove that $\sqrt{2} + \sqrt{3}$ is irrational.
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27. Prove that $\log_5(2)$ is irrational. (_Hint:_ Use the unique factorization
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of integers theorem.)
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28. Let $N = 2 \cdot 3 \cdot 5 \cdot 7 + 1$. What remainder is obtained when $N$
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is divided by $2$?$3$?$5$?$7$? Justify your answer.
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29. Suppose $a$ is an integer and $p$ is a prime number such that $p \mid a$ and
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$p \mid (a + 3)$. What can you deduce about $p$? Why?
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30. Let $p_1, p_2, p_3, \dots$ be a list of all prime numbers in ascending
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order. Here is a table of the first six:
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| $p_1$ | $p_2$ | $p_3$ | $p_4$ | $p_5$ | $p_6$ |
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| ----- | ----- | ----- | ----- | ----- | ----- |
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| $2$ | $3$ | $5$ | $7$ | $11$ | $13$ |
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a. Let
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$N_1 = p_1, N_2 = p_1 \cdot p_2, N_3 = p_1 \cdot p_2 \cdot p_3, \dots N_6 = p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6$.
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Calculate $N_1$, $N_2$, $N_3$, $N_4$, $N_5$, and $N_6$.
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b. For each $i = 1, 2, 3, 4, 5, 6$, find whether $N_i$ is itself prime or just
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has a prime factor less than itself. (_Hint:_ Use the test for primality from
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exercise 31 in Section 4.7 to determine your answers.)
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For exercises 31 and 32, use the fact that for ever integer $n$,
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$$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$
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31. An alternative proof of the infinitude of the prime numbers begins as
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follows:
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**Proof:** Suppose there are only finitely many prime numbers. Then one is the
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largest. Call it $p$. Let $M = p! + 1$. We will show that there is a prime
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number $q$ such that $q > p$. Complete this proof.
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32. Prove that for every integer $n$, if $n > 2$ then there is a prime number
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$p$ such that $n < p < n!$.
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33. Prove that if $p_1, p_2, \dots$, and $p_n$ are distinct prime numbers with
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$p_1 = 2$ and $n > 1$, then $p_1, p_2, \dots, p_n + 1$ can be written in the
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form $4k + 3$ for some integer $k$.
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34.
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a. Fermat's last theorem says that for every integer $n > 2$, the equation
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$x^n + y^n = z^n$ has no positive integer solution (solution for which $x$, $y$,
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and $z$ are positive integers). Prove the following: If for every prime number
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$p > 2$, $x^p + y^p = z^p$ has no positive integer solution, then for any
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integer $n > 2$ that is not a power of $2$, $x^n + y^n = z^n$ has no positive
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integer solution.
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b. Fermat proved that there are no integers $x$, $y$, and $z$ such that
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$x^4 + y^4 = z^4$. Use this result to remove the restriction in part (a) that
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$n$ not be a power of $2$. That is, prove that if $n$ is a power of $2$ and
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$n > 4$, then $x^n + y^n = z^n$ has no positive integer solution.
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For exercises 35-38 note that to show there is a unique object with a certain
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property, show that (1) there is an object with the property and (2) if objects
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$A$ and $B$ have the property, then $A = B$.
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35. Prove that there exists a unique prime number of the form $n^2 - 1$, where
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$n$ is an integer that is greater than or equal to $2$.
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36. Prove that there exists a unique prime number of the form $n^2 + 2n - 3$,
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where $n$ is a positive integer.
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37. Prove that there is at most one real number $a$ with the property that
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$a + r = r$ for every real number $r$. (Such a number is called an _additive
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identity_.)
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38. Prove that there is at most one real number $b$ with the property that
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$br = r$ for every real number $r$. (Such a number is called a
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_multiplicative identity_.)
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@ -978,3 +978,135 @@ odd, $n^2$ is odd. Therefore, $n^2$ is both even and odd. This contradicts
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Theorem 4.7.2, which states that no integer can be both even and odd. _[This
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contradiction shows that the supposition is false and, hence, that the
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proposition is true.]_
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---
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Page 252
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**Theorem 4.8.1 Irrationality of $\sqrt{2}$**
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$\sqrt{2}$ is irrational.
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**Proof (by contradiction):**
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_[We take the negation and suppose it to be true.]_ Suppose not. That is,
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suppose $\sqrt{2}$ is rational. Then there are integers $m$ and $n$ with no
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common factors such that
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$$ \sqrt{2} = \frac{m}{n} $$
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_[by dividing $m$ and $n$ by any common factors if necessary]._ _[We must derive
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a contradiction.]_ Squaring both sides of equation (4.8.1) gives
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$$ 2 = \frac{m^2}{n^2} $$
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Or, equivalently,
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$$ m^2 = 2n^2 $$
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Note that equation (4.8.2) implies that $m^2$ is even (by definition of even).
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It follows that $m$ is even (by Proposition 4.7.4). We file this fact away for
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future reference and also deduce (by definition of even) that
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$$ m = 2k \quad \text{ for some integer } k $$
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Substituting equation (4.8.3) into equation (4.8.2), we see that
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$$ m^2 = (2k)^2 = 4k^2 = 2n^2 $$
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Dividing both sides of the right-most equation by $2$ gives
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$$ n^2 = 2k^2 $$
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Consequently, $n^2$ is even, and so $n$ is even (by Proposition 4.7.4). But we
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also know that $m$ is even. _[This is the fact we filed away.]_ Hence both $m$
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and $n$ have a common factor of $2$. But this contradicts the supposition that
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$m$ and $n$ have no common factors. _[Hence the supposition is false and so the
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theorem is true.]_
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---
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Page 253
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**Proposition 4.8.2**
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$1 + 3\sqrt{2}$ is irrational.
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**Proof (by contradiction):**
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Suppose not. Suppose $1 + 3\sqrt{2}$ is rational. _[We must derive a
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contradiction.]_ Then by definition of rational,
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$$ 1 + 3\sqrt{2} = \frac{a}{b} \quad \text{ for some integers } a \text{ and } b \text{ with } b \neq 0 $$
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It follows that
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$$ 3\sqrt{2} = \frac{a}{b} - 1 \quad \text{ by subtracting } 1 \text{ from both sides} $$
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$$ = \frac{a}{b} - \frac{b}{b} \quad \text{ by substitution} $$
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$$ = \frac{a - b}{b} $$
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Hence
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$$ \sqrt{2} = \frac{a - b}{3b} $$
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But $a - b$ and $3b$ are integers (since $a$ and $b$ are integers and
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differences and products of integers are integers), and $3b \neq 0$ by the zero
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product property. Hence $\sqrt{2}$ is a quotient of the two integers $a - b$ and
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$3b$ with $3b \neq 0$, and so $\sqrt{2}$ is rational (by definition of
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rational). This contradicts the fact that $\sqrt{2}$ is irrational. _[The
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contradiction shows that the supposition is false.]_ Hence $1 + 3\sqrt{2}$ is
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irrational.
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---
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Page 254
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**Proposition 4.8.3**
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For any integer $a$ and any prime number $p$, if $p \mid a$ then
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$p \cancel{\mid} (a + 1)$.
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**Proof (by contradiction):**
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Suppose not. That is, suppose there exists an integer $a$ and a prime number $p$
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such that $p \mid a$ and $p \mid (a + 1)$. Then, by definition of divisibility,
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there exists integers $r$ and $s$ such that $a = pr$ and $a + 1 = ps$. It
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follows that
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$$ 1 = (a + 1) - a = ps - pr = p(s - r) $$
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and so (since $s - r$ is an integer) $p \mid 1$. But, by Theorem 4.4.2, the only
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integer divisors of $1$ are $1$ and $-1$, and $p > 1$ because $p$ is prime. Thus
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$p \leq 1$ and $p > 1$, which is a contradiction. _[Hence the supposition is
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false, and the proposition is true.]_
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---
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Page 254
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**Theorem 4.8.4 Infinitude of the Primes**
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The set of prime numbers is infinite.
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**Proof (by contradiction):**
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Suppose not. That is, suppose the set of prime numbers is finite. _[WE must
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deduce a contradiction.]_ Then some prime number $p$ is the largest of all the
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prime numbers, and hence we can list the prime numbers in ascending order.
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$$ 2, 3< 5, 7, 11, \dots, p $$
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Let $N$ be the product of all the prime numbers plus $1$:
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$$ N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p) + 1 $$
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Then $N > 1$, and so, by Theorem 4.4.4, $N$ is divisible by some prime number
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$q$. Because $q$ is prime, $q$ must equal one of the prime numbers
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$2, 3, 5< 7, 11, \dots, p$. Thus, by definition of divisibility, $q$ divides
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$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p$, and so, by Proposition 4.8.3, $q$
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does not divide $(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p) + 1$, which equals
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$N$. Hence $N$ is divisible by $q$ and $N$ is not divisible by $q$, and we have
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reached a contradiction. _[Therefore, the supposition is false and the theorem
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is true.]_
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@ -223,3 +223,23 @@ $\forall x \in D, \text{ if } \neg Q(x) \text{ then } \neg P(x)$
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you suppose that ______ and you show that ______.
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$Q(x)$ is false; $P(x)$ is false.
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---
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**Test Yourself**
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Page 256
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1. The ancient Greeks discovered that in a right triangle where both legs have
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length $1$, the ratio of the length of the hypotenuse to the length of one of
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the legs is not equal to a ratio of ______.
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2. One way to prove that $\sqrt{2}$ is an irrational number is to assume that
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$\sqrt{2} = \dfrac{m}{n}$ for some integers $m$ and $n$ that have no common
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factor greater than $1$, use the lemma that says that if the square of an
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integer is even then ______, and eventually show that $m$ and $n$ ______.
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3. One way to prove that there are infinitely many prime numbers is to assume
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that there is a largest prime number $p$, construct the number ______, and
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then show that this number has to be divisible by a prime number that is
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greater than ______.
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@ -1 +1 @@
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241
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250
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