diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 0ce47d5..9799918 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -1869,36 +1869,123 @@ integers. 1. $-\dfrac{35}{6}$ +$$ -\frac{35}{6} = -\frac{35}{6} $$ + 2. $4.6037$ +$$ 4.6037 = \frac{46037}{10000} $$ + 3. $\dfrac{4}{5} + \dfrac{2}{9}$ +$$ \frac{4}{5} + \frac{2}{9} = \frac{9(4) + 5(2)}{45} = \frac{46}{45} $$ + 4. $0.37373737\dots$ +Let $x = 0.37373737\dots$, then $100x = 37.373737\dots$, so +$100x - x = 99x = 37$. Therefore: + +$$ x = 0.37373737\dots = \frac{37}{99} $$ + 5. $0.56565656\dots$ +Let $x = 0.56565656\dots$, then $100x = 56.565656\dots$, so +$100x - x = 99x = 56$. Therefore: + +$$ x = 0.56565656\dots = \frac{56}{99} $$ + 6. $320.5492492492\dots$ +$$ x = 320.5492492492\dots $$ + +$$ 10000x = 3205492.492492492\dots $$ + +$$ 10x = 3205.492492492\dots $$ + +$$ 10000x - 10x = 9990x = 3202287 $$ + +$$ x = \frac{3202287}{9990} $$ + 7. $52.4672167216721\dots$ +$$ x = 52.4672167216721\dots $$ + +$$ 100000x = 5246721.672167216721\dots $$ + +$$ 10x = 524.672167216721\dots$$ + +$$ 100000x - 10x = 99990x = 5246197 $$ + +$$ x = 52.4672167216721\dots = \frac{5246197}{99990} $$ + 8. The zero product property, says that if a product of two real numbers is $0$, then one of the numbers must be $0$. a. Write this property formally using quantifiers and variables. +Let $P(x)$ be "$x = 0$." + +Let $Q(y)$ be "$y = 0$." + +Let $R(x, y)$ be "$(x)(y) = 0$." + +$$ \forall x \in \mathbb{R} \forall y \in \mathbb{R} (R(x, y) \to (P(x) \vee Q(y))) $$ + b. Write the contrapositive of your answer to part (a). +$$ \forall x \in \mathbb{R} \forall y \in \mathbb{R} ((\neg P(x) \wedge \neg Q(y)) \to \neg R(x, y)) $$ + c. Write an informal version (without quantifier symbols or variables) for your part to part (b). +If any two real numbers do not equal zero, then their product does not equal +zero. + 9. Assume that $a$ and $b$ are both integers and that $a \neq 0$ and $b \neq 0$. Explain why $\dfrac{(b - a)}{(ab^2)}$ must be a rational number. +A rational number is a ratio of integers with a nonzero denominator. The given +fraction + +$$ \frac{(b - q)}{(ab^2)} $$ + +is rational, the numerator is an integer as the difference of integers are +integers, and the denominator is an integer because the product of integers are +integers, also the assumption states that both $a$ and $b$ are not $0$, so the +denominator cannot be $0$ by the zero product property. Hence the given fraction +is a rational number. + 10. Assume that $m$ and $n$ are both integers and that $n \neq 0$. Explain why $\dfrac{(5m - 12n)}{(4n)}$ must be a rational number. +Given that $m$ and $n$ are both integers, in the given fraction + +$$ \frac{(5m -12n)}{4n} $$ + +The numerator $5m - 12n$ is an integer because the difference of integers are +integers. The denominator $4n$ is an integer because the product of integers are +integers. Also, the since $n \neq 0$, $4n \neq 0$ by the zero product property. +Hence the given fraction is a rational number. + 11. Prove that every integer is a rational number. +**Theorem:** + +Suppose $x$ is any integer. + +**Proof:** + +Then: + +$$ x = x \cdot 1 $$ + +$$ \dfrac{x}{1} = x $$ + +Then $x$ is an integer and $1$ is an integer where $1 \neq 0$. Hence $x$ can be +expressed as a quotient of integers with a nonzero denominator and therefore $x$ +is a rational number by definition of a rational number. + +Q.E.D. + 12. Let $S$ be the statement "The square of any rational number is rational." A formal version of $S$ is "For every rational number $r$, $r^2$ is rational." Fill in the blanks in the proof for $S$. @@ -1914,20 +2001,88 @@ Since $a$ and $b$ are both integers, so are the products $a^2$ and __ (d) __. Also $b^2 \neq 0$ by the __ (e) __. Hence $r^2$ is a ratio of two integers with a non-zero denominator,n and so __ (f) __ by definition of rational. +a. a rational number + +b. integers $a$ and $b$ + +c. $\left(\frac{a}{b}\right)^2$ + +d. $b^2$ + +e. zero product property + +f. $r^2$ is a rational number + 13. Consider the following statement: The negative of any rational number is rational. a. Write the statement formally using a quantifier and a variable. +$$ \forall q \in \mathbb{Q} (-q \in \mathbb{Q}) $$ + +Alternatively: + +$$ \forall q ((q \in \mathbb{Q}) \in (-q \in \mathbb{Q})) $$ + b. Determine whether the statement is true or false and justify your answer. +**Theorem:** + +Suppose $q$ is any rational number. + +**Proof:** + +Since $q$ is a rational number, $q$ can be expressed as $\dfrac{a}{b}$ where $a$ +and $b$ are integers and $b \neq 0$. + +Then: + +$$ -q = -\left(\frac{a}{b}\right) \quad \text{ by substitution} $$ + +$$ -q = \frac{-a}{b} $$ + +Then the numerator $-a$ is an integer because the product of integers are +integers. The denominator $b$ is an integer and $b \neq 0$ by assumption of $q$ +as a rational number. Hence $-q$ can be expressed as the ratio of two integers +with a nonzero denominator, and therefore $-q$ is a rational number by +definition of rational numbers. + +Q.E.D. + 14. Consider the statement: The cube of any rational number is a rational number. a. Write the statement formally using a quantifier and a variable. +$$ \forall q ((q \in \mathbb{Q}) \to (q^3 \in \mathbb{Q})) $$ + b. Determine whether the statement is true or false and justify your answer. +**Theorem:** + +Suppose $q$ is any rational number. + +**Proof:** + +Since $q$ is a rational number, $q = \dfrac{a}{b}$ where $a$ and $b$ are +integers and $b \neq 0$. + +Then: + +$$ q^3 = \left(\frac{a}{b}\right)^3 \quad \text{ by substitution} $$ + +$$ q^3 = \frac{a^3}{b^3} \quad \text{ by power of a quotient} $$ + +Then the numerator $a^3$ is an integer because the products of integers are +integers. Also the denominator $b^3$ is an integer because the product of +integers are integers and $b^3 \neq 0$ by the zero product property. + +Thus $q^3$ can be expressed as a ratio of two integers with a nonzero +denominator and therefore $q^3$ is a rational number by definition of a rational +number. + +Q.E.D. + Determine which of the statements in 15-19 are true and which are false. Prove each true statement directly from the definitions, and give a counterexample for each false statement. For a statement that is false, determine whether a small @@ -1936,44 +2091,383 @@ Follow the directions for writing proofs on page 173. 15. The product of any two rational numbers is a rational number. +**Theorem:** + +Suppose $q$ and $r$ are rational numbers. + +**Proof:** + +Since $q$ and $r$ are rational numbers, then $q = \dfrac{a}{b}$ and +$r = \dfrac{c}{d}$ where $a$, $b$, $c$, and $d$ are some integers and $b \neq 0$ +and $d \neq 0$. + +Then: + +$$ qr = \left(\frac{a}{b}\right)\left(\frac{c}{d}\right) \quad \text{ by substitution} $$ + +$$ qr = \frac{ac}{bd} $$ + +Then the numerator $ac$ is an integer because the product of integers are +integers. The denominator $bd$ is an integer because the product of integers are +integers, and $bd \neq 0$ by the of the zero product property. + +Thus $qr$ can be expressed as a ratio of two integers with a nonzero denominator +and therefore $qr$ is a rational number by the definition of a rational number. + +Q.E.D. + 16. The quotient of any two rational numbers is a rational number. +This is false. + +**Claim:** + +There exists some rational number $q$ and some rational number $r$ such that +$\dfrac{q}{r}$ is not a rational number. + +**Proof:** + +Let $q = \dfrac{1}{2}$ and $r = \dfrac{0}{1}$. + +Then: + +$$ \frac{q}{r} = \frac{\dfrac{1}{2}}{\dfrac{0}{1}} \quad \text{ by substitution} $$ + +$$ \quad = \frac{1}{2 \cdot 0} $$ + +$$ \quad = \text{ undefined} $$ + +So the numerator of the given $\dfrac{q}{r}$ is $1$ which is an integer, but the +denominator is $0$, which means $\dfrac{q}{r}$ is not any number, and therefore +not a rational number. + +Thus there exists two rational numbers whose quotients are not a rational +number, therefore the statement is false. + +Q.E.D. + +A small change that would make this true were if the statement were reworded as: + +For any two rational numbers, the quotient of those two numbers is a rational +number as long as the rational number in the divisor doesn't equal $0$. + 17. The difference of any two rational numbers is a rational number. +**Theorem:** + +Suppose that $q$ and $r$ are any rational numbers. + +**Proof:** + +Since $q$ and $r$ are rational numbers, $q = \dfrac{a}{b}$ and +$r = \dfrac{c}{d}$ where $a$, $b$, $c$, and $d$ are some integers and $b \neq 0$ +and $d \neq 0$. + +Then: + +$$ q - r = \frac{a}{b} - \frac{c}{d} \text{ by substitution} $$ + +$$ \quad = \frac{ad - cb}{bd} $$ + +Then the numerator $ad - cb$ is an integer because the difference and products +of integers are integers. The denominator $bd$ is an nonzero integer because the +products of integers are integers and because of the zero product property. + +Thus $q - r$ can be expressed as a ratio of two integers with a nonzero +denominator, and therefore $q - r$ is a rational number by the definition of a +rational number. + +Q.E.D. + 18. If $r$ and $s$ are any two rational numbers, then $\dfrac{r + s}{2}$ is rational. +**Theorem:** + +Suppose $r$ and $s$ are any two rational numbers. + +**Proof:** + +Since $r$ and $s$ are rational numbers, then $r = \dfrac{a}{b}$ and +$s = \dfrac{c}{d}$ where $a$, $b$, $c$, and $d$ are integers and $b \neq 0$ and +$d \neq 0$. + +Then: + +By substitution: + +$$ \frac{r + s}{2} = \frac{\dfrac{a}{b} + \dfrac{c}{d}}{2} $$ + +$$ \quad = \frac{1}{2}\left(\frac{a}{b} + \frac{c}{d}\right) $$ + +$$ \quad = \frac{a}{2b} + \frac{c}{2d} $$ + +$$ \quad = \frac{ad + bc}{2bd} $$ + +Then the numerator $ad + bc$ is an integer because the products and sums of +integers are integers. The denominator $2bd$ is a nonzero integer because the +products of integers are integers and because of the zero product property. + +Thus $\dfrac{r + s}{2}$ can be expressed as the ratio of two integers with a +nonzero denominator, and therefore $\dfrac{r + s}{2}$ is a rational number by +the definition of a rational number. + +Q.E.D. + 19. For all real numbers $a$ and $b$, if $a < b$ then $a < \dfrac{a + b}{2} < b$. (You may use the properties of inequalities in T17-T27 of Appendix A.) +**Theorem:** + +Suppose $a$ and $b$ are any real numbers and that $a < b$. + +**Proof:** + +Then: + +By T19: + +$$ a + a < a + b $$ + +$$ 2a < a + b $$ + +By T20: + +$$ a < \frac{a + b}{2} $$ + +And: + +By T19: + +$$ a + b < b + b $$ + +$$ a + b < 2b $$ + +By T20: + +$$ \frac{a + b}{2} < b $$ + +Therefore $a < \dfrac{a + b}{2} < b$. + +Q.E.D. + 20. Use the results of exercises 18 and 19 to prove that given any two rational numbers $r$ and $s$ with $r < s$, there is another rational number between $r$ and $s$. An important consequence is that there are infinitely many rational numbers in between any two distinct rational numbers. See Section 7.4. +**Theorem:** + +Suppose $r$ is any rational number and $s$ is any rational number where $r < s$. + +**Proof:** + +By 18, we know that $\dfrac{r + s}{2}$ is a rational number. + +By 19, we know that if $r < s$, then $r < \dfrac{r + s}{2} < s$. + +Therefore there exists some rational number $\dfrac{r + s}{2}$ that is between +$r$ and $s$, _[as was to be shown]_. + +Q.E.D. + Use the properties of even and odd integers that are listed in Example 4.3.3 to do exercises 21-23. Indicate which properties you use to justify your reasoning. 21. True or false? If $m$ is any even integer and $n$ is any odd integer, then $m^2 + 3n$ is odd. Explain. +**Theorem:** + +Suppose $m$ is any even integer and $n$ is any odd integer. + +**Proof:** + +By 3, the product of any two odd integers is odd, then: + +$$ 3n \text{ is odd} $$ + +Since $m$ is even, $m = 2k$ for some integer $k$. + +Then: + +By substitution: + +$$ m^2 = (2k)^2 $$ + +$$ \quad = 4k^2 $$ + +$$ \quad = 2(2k^2) $$ + +Then $m^2$ is even by the definition of an even integer. + +$$ m^2 \text{ is even} $$ + +By 5, the sum of any odd integer and any even integer is odd. + +Thus $m^2 + 3n$ is odd, therefore the statement is true. + +Q.E.D. + 22. True or false? If $a$ is any odd integer, then $a^2 + a$ is even. Explain. +**Theorem:** + +Suppose $a$ is any odd integer. + +**Proof:** + +Then: + +$$ a^2 = a \cdot a $$ + +By 3, the product of any two odd integers is odd, then: + +$$ a^2 \text{ is odd} $$ + +By 2, the sum and difference of any two odd integers are even, then: + +$$ a^2 + a \text{ is even} $$ + +Therefore the statement is true. + +Q.E.D. + 23. True or false? If $k$ is any even integer and $m$ is any odd integer, then $(k + 2)^2 - (m - 1)^2$ is even. Explain. +**Theorem:** + +Suppose $k$ is any even integer and $m$ is any odd integer. + +**Proof:** + +By 1, the sum of any two even integers is even, then: + +$$ k + 2 \text{ is even} $$ + +By 1, the product of any two even integers is even, then: + +$$ (k + 2)^2 = (k + 2)(k + 2) $$ + +$$ (k + 2)^2 \text{ is even} $$ + +By 2 the difference of any two odd integers is even, then: + +$$ m - 1 \text{ is even} $$ + +By 1 the product of any two even integers is even, then: + +$$ (m - 1)^2 = (m - 1)(m - 1) $$ + +$$ (m - 1)^2 \text{ is even} $$ + +By 1 the difference of any two even integers is even, then: + +$$ (k + 2)^2 - (m - 1)^2 \text{ is even} $$ + +Therefore the statement is true. + +Q.E.D. + Derive the statements in 24-26 as corollaries of Theorems 4.3.1, 4.3.2, and the results of exercises 12, 13, 14, 15, and 17. 24. For any rational numbers $r$ and $s$, $2r + 3s$ is rational. +**Theorem:** + +By 15, the product of any two rational numbers is a rational number, then: + +$$ 2r \text{ is rational} $$ + +and + +$$ 3s \text{ is rational} $$ + +By Theorem 4.3.2, the sum of any two rational numbers is rational, then: + +$$ 2r + 3s \text{ is rational} $$ + +Therefore the statement is true. + +Q.E.D. + +**Proof:** + 25. If $r$ is any rational number, then $3r^2 - 2r + 4$ is rational. +By 15, the product of any two rational numbers is a rational number, then: + +$$ r^2 = r \cdot r $$ + +$$ r^2 \text{ is rational} $$ + +$$ 3r^2 = 3 \cdot r^2 $$ + +$$ 3r^2 \text{ is rational} $$ + +and + +$$ 2r = 2 \cdot r $$ + +$$ 2r \text{ is rational} $$ + +By 17, the difference of any two rational numbers is a rational number, then: + +$$ 3r^2 - 2r \text{ is rational} $$ + +By Theorem 4.3.2, the sum of any two rational numbers is rational, then: + +$$ (3r^2 - 2r) + 4 \text{ is rational} $$ + +Therefore the statement is true. + +Q.E.D. + 26. For any rational number $s$, $5s^3 + 8s^2 - 7$ is rational. +**Theorem:** + +Suppose $s$ is any rational number. + +**Proof:** + +By 15, the product of any two rational numbers is a rational number, then: + +$$ s^2 = s \cdot s $$ + +$$ s^2 \text{ is rational} $$ + +$$ s^3 = s^2 \cdot s $$ + +$$ s^3 \text{ is rational} $$ + +$$ 5s^3 = 5 \cdot s^3 $$ + +$$ 5s^3 \text{ is rational} $$ + +and + +$$ 8s^2 = 8 \cdot s^2 $$ + +$$ 8s^2 \text{ is rational} $$ + +By 17, the difference of any two rational numbers is a rational number, then: + +$$ 8s^2 - 7 \text{ is rational} $$ + +By Theorem 4.3.2, the sum of any two rational numbers is rational, then: + +$$ 5s^3 + (8s^2 - 7) \text{ is rational} $$ + +Therefore the statement is true. + +Q.E.D. + 27. It is a fact that if $n$ is any nonnegative integer, then $$ 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^n} = \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{1 - \left(\dfrac{1}{2}\right)} $$ @@ -1982,13 +2476,80 @@ $$ 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^n} = \fr right-hand side of this equation rational? If so, express it as a ratio of two integers. +**Theorem:** + +Suppose $n$ is any nonnegative integer. + +**Proof:** + +Consider: + +$$ \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{1 - \left(\dfrac{1}{2}\right)} $$ + +The denominator can be simpilifed as: + +$$ 1 - \frac{1}{2} = \frac{1}{2} $$ + +Then: + +$$ \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{\frac{1}{2}} $$ + +$$ \quad = 2\left(1 - \frac{1}{2^{n + 1}}\right)$$ + +$$ \quad = 2 - \frac{2}{2^{n + 1}} $$ + +$$ \quad = 2 - \frac{1}{2^n} $$ + +$$ \quad = \frac{2 \cdot 2^n - 1}{2^n}$$ + +$$ \quad = \frac{2^{n + 1} - 1}{2^n}$$ + +Since $n$ is a nonnegative integer, the numerator $2^{n + 1} - 1$ is an integer +because the products and differences of integers are integers. And the +denominator $2^n$ is a positive integer because of the products of integers and +because $n$ is a nonnegative integer. + +Therefore right-hand side of the given equation is a rational number by the +definition of rational numbers. + +Q.E.D. + 28. Suppose $a$, $b$, $c$, and $d$ are integers and $a \neq c$. Suppose also that $x$ is a real number that satisfies the equation -$$ \frac{ax + b}{cs + d} = 1 $$ +$$ \frac{ax + b}{cx + d} = 1 $$ Must $x$ be rational? If so, express $x$ as a ratio of two integers. +**Theorem:** + +Suppose $a$, $b$, $c$, and $d$ are any integers and suppose $x$ is a real number +that satisfies the equation: + +$$ \frac{ax + b}{cx + d} = 1 $$ + +_Proof:_* + +Consider: + +$$ \frac{ax + b}{cx + d} = 1 $$ + +$$ ax + b = cx + d $$ + +$$ ax - cx = d - b $$ + +$$ x(a - c) = d - b $$ + +$$ x = \frac{d - b}{a - c} $$ + +Then the numerator $d - b$ is an integer because the difference of integers are +integers. The denominator $a - c$ must be a nonzero integer because the +difference of integers are integers and because $a \neq c$. + +Therefore $x$ must be a rational number by the definition of rational numbers. + +Q.E.D. + 29. Suppose $a$, $b$, and $c$ are integers and $x$, $y$, and $z$ are nonzero real numbers that satisfy the following equations: @@ -1996,11 +2557,68 @@ $$ \frac{xy}{x + y} = a \quad \text{ and } \quad \frac{xz}{x + z} = b \quad \tex Is $x$ rational? If so, express it as ratio of two integers. +Omitted. + 30. Prove that if one solution for a quadratic equation of the form $x^2 + bx + c = 0$ is rational (where $b$ and $c$ are rational), then the other solution is also rational. (Use the fact that if the solutions of the equation are $r$ and $s$, then $x^2 + bx + c = (x - r)(x - s)$.) +**Theorem:** + +Suppose there is any rational number $r$ that is a solution to a quadratic +equation of the form: + +$$ x^2 + bx + c = 0 $$ + +Where $b$ and $c$ are rational. + +And suppose $s$ is the other solution to the given equation. + +**Proof:** + +Given that both $r$ and $s$ are solutions to the given equation, then: + +$$ x^2 + bx + c = (x - r)(x - s) = x^2 - rx - sx + rs $$ + +This means that: + +$$ bx = (-r - s)x $$ + +$$ b = -1(r + s) $$ + +And: + +$$ c = rs $$ + +Let's analyze $b$ and isolate $s$. + +$$ b = -1(r + s) $$ + +$$ -b = r + s $$ + +$$ -b - r = s $$ + +Since both $b$ and $r$ are rational numbers, then $b = \dfrac{g}{h}$ and +$r = \dfrac{i}{j}$ where $g$, $h$, $i$, and $j$ are some integers and $h \neq 0$ +and $j \neq 0$. + +Then: + +$$ s = -b - r = -\left(\frac{g}{h}\right) - \left(\frac{i}{j}\right) \text{ by substitution} $$ + +$$ \quad = \frac{-1(gj + hi)}{hj} $$ + +The numerator $-1(gj + hi)$ is an integer because the sum and products of +integers are integers. The denominator is a nonzero integer because the products +of integers are integers and because of the zero product property. + +Therefore $s$ can be expressed as a ratio of two integers where the denominator +is nonzero. Thus $s$ is a rational number by the definition of rational numbers, +therefore the other solution is rational. + +Q.E.D. + 31. Prove that if a real number $c$ satisfies a polynomial equation of the form $$ r_3x^3 + r_2x^2 + r_1x + r_0 = 0 $$ @@ -2015,10 +2633,77 @@ where $n_0$, $n_1$, $n_2$, and $n_3$ are integers. **Definition:** A number $c$ is called a **root** of a polynomial $p(x)$ if, and only if, $p(c) = 0$. +**Theorem:** + +Suppose $c$ is any real number that satisfies a polynomial equation of the form + +$$ r_3x^3 + r_2x^2 + r_1x + r_0 = 0 $$ + +where $r_0$, $r_1$, $r_2$, and $r_3$ are rational numbers. + +**Proof:** + +Since $c$ is a real number that satisfies the given equation, then: + +$$ r_3c^3 + r_2c^2 + r_1c + r_0 = 0 $$ + +Since $r_3$, $r_2$, $r_1$, and $r_0$ are rational numbers, then +$r_3 = \dfrac{a_3}{b_3}$, $r_2 = \dfrac{a_2}{b_2}$, $r_1 = \dfrac{a_1}{b_1}$, +and $r_0 = \dfrac{a_0}{b_0}$ where $a_3$, $a_2$, $a_1$, $a_0$ are some integers +and $b_3$, $b_2$, $b_1$, $b_0$ are some nonzero integers. + +Then, by substitution: + +$$ r_3c^3 + r_2c^2 + r_1c + r_0 = \left(\frac{a_3}{b_3}\right)c^3 + \left(\frac{a_2}{b_2}\right)c^2 + \left(\frac{a_1}{b_1}\right)c + \frac{a_0}{b_0} = 0 $$ + +$$ \quad = \frac{a_3b_2b_1b_0c^3 + a_2b_3b_1b_0c^2 + a_1b_3b_2b_0c + a_0b_3b_2b_1}{b_3b_2b_1b_0} = 0 $$ + +$$ \quad = a_3b_2b_1b_0c^3 + a_2b_3b_1b_0c^2 + a_1b_3b_2b_0c + a_0b_3b_2b_1 = 0 $$ + +Let $n_3 = a_3b_2b_1b_0$, and $n_2 = a_2b_3b_1b_0$, and $n_1 = a_1b_3b_2b_0$, +and $n_0 = a_0b_3b_2b_1$. + +Then $n_3$, $n_2$, $n_1$, and $n_0$ are integers because of the product of +integers. + +Thus we can write the given equation as: + +$$ n_3c^3 + n_2c^2 + n_1c + n_0 = 0 $$ + +Where $c$ is a real number that satisfies the equation: + +$$ n_3x^3 + n_2x^2 + n_1x + n_0 = 0 $$ + +Q.E.D. + 32. Prove that for every real number $c$, if $c$ is a root of a polynomial with rational coefficients, then $c$ is a root of a polynomial with integer coefficients. +**Theorem:** + +Suppose $c$ is a root of a polynomial with rational coefficients. + +**Proof:** + +Then + +$$ r_nx^n + r_{n - 1}x^{n - 1} + \dots + r_1x + r_0 = 0 $$ + +where each $r_i$ is rational. + +Then each $r_i$ can be written as a ratio of integers with nonzero denominators. +Let $D$ be a common multiple of all denominators of the $r_i$. Multiplying the +equation by $D$ gives + +$$ s_nx^n + s_{n - 1}x^{n - 1} + \dots + s_1x + s_0 = 0 $$ + +where each $s_i$ is an integer. + +Thus $c$ is a root of a polynomial with integer coefficients. + +Q.E.D. + Use the properties of even and odd integers that are listed in Example 4.3.3 to do exercises 33 and 34. @@ -2031,9 +2716,86 @@ multiplying out $(x - r)(x - s)$ when both $r$ and $s$ are odd integers? When both $r$ and $s$ are even integers? When one of $r$ and $s$ is even and the other odd? +_Case when both $r$ and $s$ are odd integers:_ + +**Theorem:** + +Suppose $r$ and $s$ are odd integers. + +**Conclusion:** + +Let $x$ be some real number. + +Then: + +$$ (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs $$ + +We know the coefficient of $x^2$ is $1$. + +By 2, we know the sum of any two odd integers are even, then: + +We know the coefficient of $-1(r + s)$ is even. + +By 3, we know the product of any two odd integers is odd, then: + +We know the coefficient of $rs$ is odd. + +_Case when both $r$ and $s$ are even integers:_ + +**Theorem:** + +Suppose $r$ and $s$ are even integers. + +**Conclusion:** + +Let $x$ be some real number. + +Then: + +$$ (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs $$ + +We know the coefficient of $x^2$ is $1$. + +By 1 we know the sum of any two even integers is even, then: + +We know the coefficient of $(-1)(r + s)x$ is even. + +By 1 we know the product of any two even integers is even, then: + +We know the coefficient of $(-1)rs$ is even. + +_Case where $r$ and $s$ is even and the other odd:_ + +**Theorem:** + +Suppose $r$ and $s$ are any integers where one is even and the other is odd. + +**Conclusion:** + +Let $x$ be some real number. + +Then: + +$$ (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs $$ + +We know the coefficient of $x^2$ is $1$. + +By 5, we know the sum of any odd integer and any even integer is odd, then: + +We know the coefficient of $(-1)(r +s)x$ is odd. + +By 4, we know the product of any even integer and any odd integer is even, then: + +We know the coefficient of $(-1)rs$ is even. + b. It follows from part (a) that $x^2 - 1253x + 255$ cannot be written as a product of two polynomials with integer coefficients. Explain why this is so. +Because in all cases from part (a), the middle coefficient and the third +coefficient were always either even and odd or odd and even. Since both 1253 and +255 are odd, this expression cannot be expressed as the product of two +polynomials with integer coefficients. + 34. Observe that $$ (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst $$ @@ -2041,9 +2803,151 @@ $$ (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst $$ a. Derive a result for cubic polynomials similar to the result in part (a) of exercise 33 for quadratic polynomials. +_Case where $r$ and $s$ and $t$ are all even_: + +The coefficient of $x^3$ is $1$. + +By 1, the sum of any two even integers is even, then: + +$$ r + s \text{ is even} $$ + +$$ (r + s) + t \text{ is even} $$ + +The coefficient of $(r + s + t)x^2$ is even. + +By 1, the sum and product of any two even integers is even, then: + +$$ rs \text{ is even} $$ + +$$ st \text{ is even} $$ + +$$ (rs + rs + rt) \text{ is even} $$ + +The coefficient of $(rs + rs + st)x$ is even. + +By 1, the product of any two even integers is even, then: + +$$ rs \text{ is even} $$ + +$$ rst \text{ is even} $$ + +The coefficient of $rst$ is even. + +_Case where $r$ is odd and $s$ and $t$ are even_: + +The coefficient of $x^3$ is $1$. + +By 1 the sum of any two even integers is even, then: + +$$ s + t \text{ is even} $$ + +By 5 the sum of any odd integer and any even integer is odd. + +$$ r + (s + t) \text{ is odd} $$ + +The coefficient of $(r + s + t)x^2$ is odd. + +By 4, the product of any even integer and any odd integer is even, then: + +$$ rs \text{ is even} $$ + +By 1, the sum and product of any two even integers is even. + +$$ st \text{ is even} $$ + +$$ rs + st \text{ is even} $$ + +$$ rs + (rs + st) \text{ is even} $$ + +The coefficient of $(rs + rs + st)x$ is even. + +$$ (rs)t \text{ is even} $$ + +The coefficient of $rst$ is even. + +_Case where $r$ and $s$ are odd and $t$ is even_: + +The coefficient of $x^3$ is $1$. + +By 2, the sum of any two odd integers is even. + +$$ r + s \text{ is even} $$ + +By 1, the sum of any two even integers is even. + +$$ (r + s) + t \text{ is even} $$ + +The coefficient of $(r + s + t)x^2$ is even. + +By 3, the product of any two odd integers is odd. + +$$ rs \text{ is odd} $$ + +By 4, the product of any even integer and any odd integer is even. + +$$ st \text{ is even} $$ + +By 2, the sum of any two odd integers is even. + +$$ rs + rs \text{ is even} $$ + +By 1, the sum of any two even integers is even. + +$$ (rs + rs) + st \text{ is even} $$ + +The coefficient of $(rs + rs + st)x$ is even + +By 4, the product of any even integer and any odd integer is even. + +$$ (rs)t \text{ is even} $$ + +The coefficient of $rst$ is even. + +_Case where $r$ and $s$ and $t$ are all odd_: + +The coefficient of $x^3$ is $1$. + +By 2, the sum of any two odd integers is even. + +$$ r + s \text{ even} $$ + +By 5, the sum of any odd integer and any even integer is odd. + +$$ (r + s) + t \text{ is odd} $$ + +The coefficient of $(r + s + t)x^2$ is odd. + +By 3, The product of any two odd integers is odd. + +$$ rs \text{ is odd} $$ + +$$ st \text{ is odd} $$ + +By 2, the sum of any two odd integers is even. + +$$ rs + rs \text{ is even} $$ + +By 5, the sum of any odd integer and any even integer is odd. + +$$ (rs + rs) + st \text{ is odd} $$ + +The coefficient of $(rs + rs + st)x$ is odd. + +$$ (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst $$ + +By 3, The product of any two odd integers is odd. + +$$ (rs)t \text{ is odd.} $$ + +The coefficient of $rst$ is odd. + b. Can $x^3 + 7x^2 - 8x - 27$ be written as a product of three polynomials with integer coefficients? Explain. +In all cases, the order of the second through fourth terms are never: "odd, +even, odd". Therefore the given polynomial $x^3 + 7x^2 - 8x - 27$ can be written +as a product of three polynomials with integer coefficients. + In 35-39 find the mistakes in the "proofs" that the sum of any two rational numbers is a rational number. @@ -2053,12 +2957,16 @@ numbers is a rational number. together. So if $r$ and $s$ are particular but arbitrarily chosen rational numbers, then $r + s$ is rational." +This proof assumes what is to be proved. + 36. **"Proof:** Let rational numbers $r = \dfrac{1}{4}$ and $s = \dfrac{1}{2}$ be given. Then $r + s = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}$, which is a rational number. This is what was to be shown." +This proof is arguing from examples. + 37. **"Proof:** Suppose $r$ and $s$ are rational numbers. By definition of rational, @@ -2072,6 +2980,8 @@ $r + s = \dfrac{p}{b}$, where $p$ and $b$ are integers and $b \neq 0$. Thus $r + s$ is a rational number by definition of rational. This is what was to be shown." +This incorrect proof uses the same letter to mean two different things. + 38. **"Proof:** Suppose $r$ and $s$ are rational numbers. Then $r = \dfrac{a}{b}$ @@ -2083,6 +2993,10 @@ $$ r + s = \frac{a}{b} + \frac{c}{d} $$ But this is a sum of two fractions, which is a fraction. So $r - s$ is a rational number since a rational number is a fraction." +This incorrect proof exhibits confusion between what is known and what is still +to be shown. Additionally, they simply abandon what is to be shown since what is +to be shown is $r + s$ is rational, not $r - s$ is rational. + 39. **"Proof:** Suppose $r$ and $s$ are rational numbers. If $r + s$ is rational, @@ -2095,3 +3009,5 @@ $$ r + s = \frac{i}{j} + \frac{m}{n} = \frac{a}{b} $$ which is a quotient of two integers with a nonzero denominator. Hence it is a rational number. This is what is to be shown. + +This incorrect prove is assuming what is to be proved. diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index 29759e9..db6f218 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -92,6 +92,13 @@ Page 210 1. To show that a real number is rational, we must show that we can write it as ______. +The ratio of integers, where the denominator is not 0. + 2. An irrational number is a ______ that is ______. +real number; not rational + 3. Zero is a rational number because ______. + +zero is an integer that is a ratio of integers where the denominator is not +zero, $0 = \dfrac{0}{1}$.