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Example 8.1 Let’s investigate elements of \( A = \{ x : x \in \mathbb{N} \) and \( 7 \mid x\} \) .
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This set has form \( A = \{ x : P\left( x\right) \} \) where \( P\left( x\right) \) is the open sentence \( \left( {x \in \mathbb{N}}\right) \land \left( {7 \mid x}\right) \) . Thus \( {21} \in A \) because \( P\left( {21}\right) \) is true. Similarly, \( 7,{14},{28},{35} \), etc., are all elements of \( A \) . But \( 8 \notin A \) (for example) because \( P\left( 8\right) \) is false. Likewise \( - {14} \notin A \) because \( P\left( {-{14}}\right) \) is false.
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Yes
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Consider the set \( B = \{ \left( {x, y}\right) \in \mathbb{Z} \times \mathbb{Z} : x \equiv y\left( {\;\operatorname{mod}\;5}\right) \} \). Now suppose \( n \in \mathbb{Z} \) and consider the ordered pair \( \left( {{4n} + 3,{9n} - 2}\right) \). Does this ordered pair belong to \( B \)?
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To answer this, we first observe that \( \left( {{4n} + 3,{9n} - 2}\right) \in \mathbb{Z} \times \mathbb{Z} \). Next, we observe that \( \left( {{4n} + 3}\right) - \left( {{9n} - 2}\right) = - {5n} + 5 = 5\left( {1 - n}\right) \), so \( 5 \mid \left( {\left( {{4n} + 3}\right) - \left( {{9n} - 2}\right) }\right) \), which means \( \left( {{4n} + 3}\right) \equiv \left( {{9n} - 2}\right) \left( {\;\operatorname{mod}\;5}\right) \). Therefore we have established that \( \left( {{4n} + 3,{9n} - 2}\right) \) meets the requirements for belonging to \( B \), so \( \left( {{4n} + 3,{9n} - 2}\right) \in B \) for every \( n \in \mathbb{Z} \).
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Yes
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Example 8.5 Prove that \( \{ x \in \mathbb{Z} : {18}\;\left| {\;x\} \subseteq \{ x \in \mathbb{Z} : 6\;}\right| \;x\} . \)
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Proof. Suppose \( a \in \{ x \in \mathbb{Z} : {18} \mid x\} \) . This means that \( a \in \mathbb{Z} \) and \( {18} \mid a \) . By definition of divisibility, there is an integer \( c \) for which \( a = {18c} \) . Consequently \( a = 6\left( {3c}\right) \), and from this we deduce that \( 6 \mid a \) . Therefore \( a \) is one of the integers that 6 divides, so \( a \in \{ x \in \mathbb{Z} : 6 \mid x\} \) . We’ve shown \( a \in \{ x \in \mathbb{Z} : {18}|x\} \) implies \( a \in \{ x \in \mathbb{Z} : 6|x\} \), so it follows that \( \{ x \in \mathbb{Z} : {18} \mid x\} \subseteq \{ x \in \mathbb{Z} : 6 \mid x\} . \)
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Yes
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Prove that if \( A \) and \( B \) are sets, then \( \mathcal{P}\left( A\right) \cup \mathcal{P}\left( B\right) \subseteq \mathcal{P}\left( {A \cup B}\right) \) .
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Proof. Suppose \( X \in \mathcal{P}\left( A\right) \cup \mathcal{P}\left( B\right) \) . \n\nBy definition of union, this means \( X \in \mathcal{P}\left( A\right) \) or \( X \in \mathcal{P}\left( B\right) \) . \n\nTherefore \( X \subseteq A \) or \( X \subseteq B \) (by definition of power sets). We consider cases. \n\nCase 1. Suppose \( X \subseteq A \) . Then \( X \subseteq A \cup B \), and this means \( X \in \mathcal{P}\left( {A \cup B}\right) \) . \n\nCase 2. Suppose \( X \subseteq B \) . Then \( X \subseteq A \cup B \), and this means \( X \in \mathcal{P}\left( {A \cup B}\right) \) . \n\n(We do not need to consider the case where \( X \subseteq A \) and \( X \subseteq B \) because that is taken care of by either of cases 1 or 2.) The above cases show that \( X \in \mathcal{P}\left( {A \cup B}\right) \) . \n\nThus we’ve shown that \( X \in \mathcal{P}\left( A\right) \cup \mathcal{P}\left( B\right) \) implies \( X \in \mathcal{P}\left( {A \cup B}\right) \), and this completes the proof that \( \mathcal{P}\left( A\right) \cup \mathcal{P}\left( B\right) \subseteq \mathcal{P}\left( {A \cup B}\right) \) .
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Yes
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Example 8.9 Suppose \( A \) and \( B \) are sets. If \( \mathcal{P}\left( A\right) \subseteq \mathcal{P}\left( B\right) \), then \( A \subseteq B \) .
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Proof. We use direct proof. Assume \( \mathcal{P}\left( A\right) \subseteq \mathcal{P}\left( B\right) \). Based on this assumption, we must now show that \( A \subseteq B \). To show \( A \subseteq B \), suppose that \( a \in A \). Then the one-element set \( \{ a\} \) is a subset of \( A \), so \( \{ a\} \in \mathcal{P}\left( A\right) \). But then, since \( \mathcal{P}\left( A\right) \subseteq \mathcal{P}\left( B\right) \), it follows that \( \{ a\} \in \mathcal{P}\left( B\right) \). This means that \( \{ a\} \subseteq B \), hence \( a \in B \). We’ve shown that \( a \in A \) implies \( a \in B \), so therefore \( A \subseteq B \).
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Yes
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Example 8.11 Suppose \( A, B \), and \( C \) are sets, and \( C \neq \varnothing \) . Prove that if \( A \times C = B \times C \), then \( A = B \) .
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Proof. Suppose \( A \times C = B \times C \) . We must now show \( A = B \) .\n\nFirst we will show \( A \subseteq B \) . Suppose \( a \in A \) . Since \( C \neq \varnothing \), there exists an element \( c \in C \) . Thus, since \( a \in A \) and \( c \in C \), we have \( \left( {a, c}\right) \in A \times C \), by definition of the Cartesian product. But then, since \( A \times C = B \times C \), it follows that \( \left( {a, c}\right) \in B \times C \) . But \( \left( {a, c}\right) \in B \times C \) means \( a \in B \), by definition of the Cartesian product. We have shown \( a \in A \) implies \( a \in B \), so \( A \subseteq B \) .\n\nNext we show \( B \subseteq A \) . We use the same argument as above, with the roles of \( A \) and \( B \) reversed. Suppose \( a \in B \) . Since \( C \neq \varnothing \), there exists an element \( c \in C \) . Thus, since \( a \in B \) and \( c \in C \), we have \( \left( {a, c}\right) \in B \times C \) . But then, since \( B \times C = A \times C \), we have \( \left( {a, c}\right) \in A \times C \) . It follows that \( a \in A \) . We have shown \( a \in B \) implies \( a \in A \), so \( B \subseteq A \) .\n\nThe previous two paragraphs have shown \( A \subseteq B \) and \( B \subseteq A \), so \( A = B \) . In summary, we have shown that if \( A \times C = B \times C \), then \( A = B \) . This completes the proof.
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Yes
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Given sets \( A, B \) and \( C \), prove \( A \times \left( {B \cap C}\right) = \left( {A \times B}\right) \cap \left( {A \times C}\right) .
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Proof. First we will show that \( A \times \left( {B \cap C}\right) \subseteq \left( {A \times B}\right) \cap \left( {A \times C}\right) \). Suppose \( \left( {a, b}\right) \in A \times \left( {B \cap C}\right) \). By definition of the Cartesian product, this means \( a \in A \) and \( b \in B \cap C \). By definition of intersection, it follows that \( b \in B \) and \( b \in C \). Thus, since \( a \in A \) and \( b \in B \), it follows that \( \left( {a, b}\right) \in A \times B \) (by definition of \( \times \) ). Also, since \( a \in A \) and \( b \in C \), it follows that \( \left( {a, b}\right) \in A \times C \) (by definition of \( \times \) ). Now we have \( \left( {a, b}\right) \in A \times B \) and \( \left( {a, b}\right) \in A \times C \), so \( \left( {a, b}\right) \in \left( {A \times B}\right) \cap \left( {A \times C}\right) . We’ve shown that \( \left( {a, b}\right) \in A \times \left( {B \cap C}\right) \) implies \( \left( {a, b}\right) \in \left( {A \times B}\right) \cap \left( {A \times C}\right) \) so we have \( A \times \left( {B \cap C}\right) \subseteq \left( {A \times B}\right) \cap \left( {A \times C}\right) . Next we will show that \( \left( {A \times B}\right) \cap \left( {A \times C}\right) \subseteq A \times \left( {B \cap C}\right) \). Suppose \( \left( {a, b}\right) \in \left( {A \times B}\right) \cap \left( {A \times C}\right) \). By definition of intersection, this means \( \left( {a, b}\right) \in A \times B \) and \( \left( {a, b}\right) \in A \times C \). By definition of the Cartesian product, \( \left( {a, b}\right) \in A \times B \) means \( a \in A \) and \( b \in B \). By definition of the Cartesian product, \( \left( {a, b}\right) \in A \times C \) means \( a \in A \) and \( b \in C \). We now have \( b \in B \) and \( b \in C \), so \( b \in B \cap C \), by definition of intersection. Thus we’ve deduced that \( a \in A \) and \( b \in B \cap C \), so \( \left( {a, b}\right) \in A \times \left( {B \cap C}\right) \). In summary, we’ve shown that \( \left( {a, b}\right) \in \left( {A \times B}\right) \cap \left( {A \times C}\right) \) implies \( \left( {a, b}\right) \in A \times \left( {B \cap C}\right) \) so we have \( \left( {A \times B}\right) \cap \left( {A \times C}\right) \subseteq A \times \left( {B \cap C}\right) \). The previous two paragraphs show that \( A \times \left( {B \cap C}\right) \subseteq \left( {A \times B}\right) \cap \left( {A \times C}\right) \) and \( \left( {A \times B}\right) \cap \left( {A \times C}\right) \subseteq A \times \left( {B \cap C}\right) \), so it follows that \( \left( {A \times B}\right) \cap \left( {A \times C}\right) = A \times \left( {B \cap C}\right) .
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Yes
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Given sets \( A, B \), and \( C \), prove \( A \times \left( {B \cap C}\right) = \left( {A \times B}\right) \cap \left( {A \times C}\right) \) .
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Just observe the following sequence of equalities.\n\n\[ A \times \left( {B \cap C}\right) = \{ \left( {x, y}\right) : \left( {x \in A}\right) \land \left( {y \in B \cap C}\right) \}\n\]\n\n(def. of \( \times \) )\n\n\[ = \;\left\{ {\left( {x, y}\right) : \left( {x \in A}\right) \land \left( {y \in B}\right) \land \left( {y \in C}\right) }\right\}\]\n\n(def. of \( \cap \) )\n\n\[ = \left\{ {\left( {x, y}\right) : \left( {x \in A}\right) \land \left( {x \in A}\right) \land \left( {y \in B}\right) \land \left( {y \in C}\right) }\right\}\]\n\n\( \left( {P = P \land P}\right) \)\n\n\[ = \left\{ {\left( {x, y}\right) : \left( {\left( {x \in A}\right) \land \left( {y \in B}\right) }\right) \land \left( {\left( {x \in A}\right) \land \left( {y \in C}\right) }\right) }\right\}\]\n\n(rearrange)\n\n\[ = \left\{ {\left( {x, y}\right) : \left( {x \in A}\right) \land \left( {y \in B}\right) }\right\} \cap \left\{ {\left( {x, y}\right) : \left( {x \in A}\right) \land \left( {y \in C}\right) }\right\}\]\n\n(def. of \( \cap \) )\n\n\[ = \left( {A \times B}\right) \cap \left( {A \times C}\right)\]\n\n(def. of \( \times \) )\n\nThis completes the proof.
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Yes
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Theorem 8.1 If \( A = \\left\\{ {{2}^{n - 1}\\left( {{2}^{n} - 1}\\right) : n \\in \\mathbb{N}}\\right. \\), and \( \\left. {{2}^{n} - 1\\text{is prime}}\\right\\} \) and \( P = \) \( \\{ p \\in \\mathbb{N} : p \) is perfect \( \\} \), then \( A \\subseteq P \) .
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Proof. Assume \( A \) and \( P \) are as stated. To show \( A \\subseteq P \), we must show that \( p \\in A \) implies \( p \\in P \). Thus suppose \( p \\in A \). By definition of \( A \), this means\n\n\[ p = {2}^{n - 1}\\left( {{2}^{n} - 1}\\right) \]\n\n(8.2)\n\nfor some \( n \\in \\mathbb{N} \) for which \( {2}^{n} - 1 \) is prime. We want to show that \( p \\in P \), that is, we want to show \( p \) is perfect. Thus, we need to show that the sum of the positive divisors of \( p \) that are less than \( p \) add up to \( p \). Notice that since \( {2}^{n} - 1 \) is prime, any divisor of \( p = {2}^{n - 1}\\left( {{2}^{n} - 1}\\right) \) must have the form \( {2}^{k} \) or \( {2}^{k}\\left( {{2}^{n} - 1}\\right) \) for \( 0 \\leq k \\leq n - 1 \). Thus the positive divisors of \( p \) are as follows:\n\n\[ {2}^{0},\\;{2}^{1},\\;{2}^{2},\\;\\ldots \\;{2}^{n - 2},\\;{2}^{n - 1}, \]\n\n\[ {2}^{0}\\left( {{2}^{n} - 1}\\right) ,\\;{2}^{1}\\left( {{2}^{n} - 1}\\right) ,\\;{2}^{2}\\left( {{2}^{n} - 1}\\right) ,\\;\\ldots \\;{2}^{n - 2}\\left( {{2}^{n} - 1}\\right) ,\\;{2}^{n - 1}\\left( {{2}^{n} - 1}\\right) . \]\n\nNotice that this list starts with \( {2}^{0} = 1 \) and ends with \( {2}^{n - 1}\\left( {{2}^{n} - 1}\\right) = p \).\n\nIf we add up all these divisors except for the last one (which equals \( p \)) we get the following:\n\n\[ \\mathop{\\sum }\\limits_{{k = 0}}^{{n - 1}}{2}^{k} + \\mathop{\\sum }\\limits_{{k = 0}}^{{n - 2}}{2}^{k}\\left( {{2}^{n} - 1}\\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{{n - 1}}{2}^{k} + \\left( {{2}^{n} - 1}\\right) \\mathop{\\sum }\\limits_{{k = 0}}^{{n - 2}}{2}^{k} \]\n\n\[ = \\left( {{2}^{n} - 1}\\right) + \\left( {{2}^{n} - 1}\\right) \\left( {{2}^{n - 1} - 1}\\right) \]\n\n(by Equation (8.1))\n\n\[ = \\left\\lbrack {1 + \\left( {{2}^{n - 1} - 1}\\right) }\\right\\rbrack \\left( {{2}^{n} - 1}\\right) \]\n\n\[ = {2}^{n - 1}\\left( {{2}^{n} - 1}\\right) \]\n\n\[ = p \]\n\n(by Equation (8.2)).\n\nThis shows that the positive divisors of \( p \) that are less than \( p \) add up to \( p \). Therefore \( p \) is perfect, by definition of a perfect number. Thus \( p \\in P \), by definition of \( P \).\n\nWe have shown that \( p \\in A \) implies \( p \\in P \), which means \( A \\subseteq P \).
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Yes
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Conjecture For every \( n \in \mathbb{Z} \), the integer \( f\left( n\right) = {n}^{2} - n + {11} \) is prime.
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Disproof. The statement \
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No
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Conjecture If \( A, B \) and \( C \) are sets, then \( A - \left( {B \cap C}\right) = \left( {A - B}\right) \cap \left( {A - C}\right) \) .
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Disproof. This conjecture is false because of the following counterexample. Let \( A = \{ 1,2,3\} ,\;B = \{ 1,2\} \) and \( C = \{ 2,3\} . \) Notice that \( A - \left( {B \cap C}\right) = \{ 1,3\} \) and \( \left( {A - B}\right) \cap \left( {A - C}\right) = \varnothing \), so \( A - \left( {B \cap C}\right) \neq \left( {A - B}\right) \cap \left( {A - C}\right) \) .
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Yes
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Conjecture There is a real number \( x \) for which \( {x}^{4} < x < {x}^{2} \) .
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Let's see if we can disprove it. According to our strategy for disproof, to disprove it we must prove its negation. Symbolically, the statement is \( \exists x \in \mathbb{R},{x}^{4} < x < {x}^{2} \), so its negation is\n\n\[ \sim \left( {\exists x \in \mathbb{R},{x}^{4} < x < {x}^{2}}\right) = \forall x \in \mathbb{R}, \sim \left( {{x}^{4} < x < {x}^{2}}\right) .\n\]\n\nThus, in words the negation is\n\nFor every real number \( x \), it is not the case that \( {x}^{4} < x < {x}^{2} \) .\n\nThis can be proved with contradiction, as follows. Suppose for the sake of contradiction that there is an \( x \) for which \( {x}^{4} < x < {x}^{2} \) . Then \( x \) must be positive since it’s greater than the non-negative number \( {x}^{4} \) . Dividing all parts of \( {x}^{4} < x < {x}^{2} \) by the positive number \( x \) produces \( {x}^{3} < 1 < x \) . Now subtract 1 from all parts of \( {x}^{3} < 1 < x \) to obtain \( {x}^{3} - 1 < 0 < x - 1 \) and reason as follows:\n\n\[ {x}^{3} - 1 < 0 < x - 1 \]\n\n\[ \left( {x - 1}\right) \left( {{x}^{2} + x + 1}\right) < 0 < \left( {x - 1}\right) \]\n\n\[ {x}^{2} + x + 1\; < \;0\; < \;1 \]\n\n(Division by \( x - 1 \) did not reverse the inequality \( < \) because the second line above shows \( 0 < x - 1 \), that is, \( x - 1 \) is positive.) Now we have \( {x}^{2} + x + 1 < 0 \) , which is a contradiction because \( x \) being positive forces \( {x}^{2} + x + 1 > 0 \) .\n\nWe summarize our work as follows.\n\nThe statement \
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Yes
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Conjecture There exist three integers \( x, y, z \), all greater than 1 and no two equal, for which \( {x}^{y} = {y}^{z} \) .
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Proof. Note that if \( x = 2, y = {16} \) and \( z = 4 \), then \( {x}^{y} = {2}^{16} = {\left( {2}^{4}\right) }^{4} = {16}^{4} = {y}^{z} \) .
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Yes
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Conjecture There is a real number \( x \) for which \( {x}^{4} < x < {x}^{2} \) .
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Disproof. Suppose for the sake of contradiction that this conjecture is true. Let \( x \) be a real number for which \( {x}^{4} < x < {x}^{2} \) . Then \( x \) is positive, since it is greater than the non-negative number \( {x}^{4} \) . Dividing all parts of \( {x}^{4} < x < {x}^{2} \) by the positive number \( x \) produces \( {x}^{3} < 1 < x \) . Now subtract 1 from all parts of \( {x}^{3} < 1 < x \) to obtain \( {x}^{3} - 1 < 0 < x - 1 \) and reason as follows:\n\n\[ \n{x}^{3} - 1 < 0 < x - 1 \n\]\n\n\[ \n\left( {x - 1}\right) \left( {{x}^{2} + x + 1}\right) < 0 < \left( {x - 1}\right) \n\]\n\n\[ \n{x}^{2} + x + 1 < 0 < 1 \n\]\n\nNow we have \( {x}^{2} + x + 1 < 0 \), which is a contradiction because \( x \) is positive. Thus the conjecture must be false.
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Yes
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Proposition 10.1 Suppose \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) are \( n \) integers, where \( n \geq 2 \) . If \( p \) is prime and \( p \mid \left( {{a}_{1} \cdot {a}_{2} \cdot {a}_{3}\cdots {a}_{n}}\right) \), then \( p \mid {a}_{i} \) for at least one of the \( {a}_{i} \) .
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Proof. The proof is induction on \( n \) .\n\n(1) The basis step involves \( n = 2 \) . Let \( p \) be prime and suppose \( p \mid \left( {{a}_{1}{a}_{2}}\right) \) . We need to show that \( p \mid {a}_{1} \) or \( p \mid {a}_{2} \), or equivalently, if \( p \nmid {a}_{1} \), then \( p \mid {a}_{2} \) . Thus suppose \( p \nmid {a}_{1} \) . Since \( p \) is prime, it follows that \( \gcd \left( {p,{a}_{1}}\right) = 1 \) . By Proposition 7.1 (on page 152), there are integers \( k \) and \( \ell \) for which \( 1 = {pk} + {a}_{1}\ell \) . Multiplying this by \( {a}_{2} \) gives\n\n\[ \n{a}_{2} = {pk}{a}_{2} + {a}_{1}{a}_{2}\ell .\n\]\n\nAs we are assuming that \( p \) divides \( {a}_{1}{a}_{2} \), it is clear that \( p \) divides the expression \( {pk}{a}_{2} + {a}_{1}{a}_{2}\ell \) on the right; hence \( p \mid {a}_{2} \) . We’ve now proved that if \( p \mid \left( {{a}_{1}{a}_{2}}\right) \), then \( p \mid {a}_{1} \) or \( p \mid {a}_{2} \) . This completes the basis step.\n\n(2) Suppose that \( k \geq 2 \), and \( p \mid \left( {{a}_{1} \cdot {a}_{2}\cdots {a}_{k}}\right) \) implies then \( p \mid {a}_{i} \) for some \( {a}_{i} \) . Now let \( p \mid \left( {{a}_{1} \cdot {a}_{2}\cdots {a}_{k} \cdot {a}_{k + 1}}\right) \) . Then \( p \mid \left( {\left( {{a}_{1} \cdot {a}_{2}\cdots {a}_{k}}\right) \cdot {a}_{k + 1}}\right) \) . By what we proved in the basis step, it follows that \( p \mid \left( {{a}_{1} \cdot {a}_{2}\cdots {a}_{k}}\right) \) or \( p \mid {a}_{k + 1} \) . This and the inductive hypothesis imply that \( p \) divides one of the \( {a}_{i} \).
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Yes
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Example 11.1 Let \( A = \{ 1,2,3,4\} \), and consider the following set:\n\n\[ R = \{ \left( {1,1}\right) ,\left( {2,1}\right) ,\left( {2,2}\right) ,\left( {3,3}\right) ,\left( {3,2}\right) ,\left( {3,1}\right) ,\left( {4,4}\right) ,\left( {4,3}\right) ,\left( {4,2}\right) ,\left( {4,1}\right) \} \subseteq A \times A. \]
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The set \( R \) is a relation on \( A \), by Definition 11.1. Since \( \left( {1,1}\right) \in R \), we have 1R1. Similarly \( {2R1} \) and \( {2R2} \), and so on. However, notice that (for example) \( \left( {3,4}\right) \notin R \), so \( {3R4} \) . Observe that \( R \) is the familiar relation \( \geq \) for the set \( A \) .
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Yes
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Example 11.2 Let \( A = \{ 1,2,3,4\} \), and consider the following set:\n\n\[ S = \{ \left( {1,1}\right) ,\left( {1,3}\right) ,\left( {3,1}\right) ,\left( {3,3}\right) ,\left( {2,2}\right) ,\left( {2,4}\right) ,\left( {4,2}\right) ,\left( {4,4}\right) \} \subseteq A \times A. \]
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Here we have \( {1S1},{1S3},{4S2} \), etc., but \( {3S4} \) and \( {2S1} \) . What does \( S \) mean? Think of it as meaning \
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No
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Example 11.4 Let \( B = \{ 0,1,2,3,4,5\} \), and consider the following set:\n\n\[ \nU = \{ \left( {1,3}\right) ,\left( {3,3}\right) ,\left( {5,2}\right) ,\left( {2,5}\right) ,\left( {4,2}\right) \} \subseteq B \times B.\n\]\n\nThen \( U \) is a relation on \( B \) because \( U \subseteq B \times B \).
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You may be hard-pressed to invent any \
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No
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Example 11.7 Here \( A = \{ b, c, d, e\} \), and \( R \) is the following relation on \( A \) : \( R = \{ \left( {b, b}\right) ,\left( {b, c}\right) ,\left( {c, b}\right) ,\left( {c, c}\right) ,\left( {d, d}\right) ,\left( {b, d}\right) ,\left( {d, b}\right) ,\left( {c, d}\right) ,\left( {d, c}\right) \} .
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This relation is not reflexive, for although \( {bRb},{cRc} \) and \( {dRd} \), it is not true that \( {eRe} \) . For a relation to be reflexive, \( {xRx} \) must be true for all \( x \in A \) .\n\nThe relation \( R \) is symmetric, because whenever we have \( {xRy} \), it follows that \( {yRx} \) too. Observe that \( {bRc} \) and \( {cRb};{bRd} \) and \( {dRb};{dRc} \) and \( {cRd} \) . Take away the ordered pair \( \left( {c, b}\right) \) from \( R \), and \( R \) is no longer symmetric.\n\nThe relation \( R \) is transitive, but it takes some work to check it. We must check that the statement \( \left( {{xRy} \land {yRz}}\right) \Rightarrow {xRz} \) is true for all \( x, y, z \in A \) . For example, taking \( x = b, y = c \) and \( z = d \), we have \( \left( {{bRc} \land {cRd}}\right) \Rightarrow {bRd} \) , which is the true statement \( \left( {T \land T}\right) \Rightarrow T \) . Likewise, \( \left( {{bRd} \land {dRc}}\right) \Rightarrow {bRc} \) is the true statement \( \left( {T \land T}\right) \Rightarrow T \) . Take note that if \( x = b, y = e \) and \( z = c \) , then \( \left( {{bRe} \land {eRc}}\right) \Rightarrow {bRc} \) becomes \( \left( {F \land F}\right) \Rightarrow T \), which is still true. It’s not much fun, but going through all the combinations, you can verify that \( \left( {{xRy} \land {yRz}}\right) \Rightarrow {xRz} \) is true for all choices \( x, y, z \in A \) . (Try at least a few of them.)
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Yes
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Proposition Let \( n \in \mathbb{N} \) . The relation \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) on the set \( \mathbb{Z} \) is reflexive, symmetric and transitive.
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Proof. First we will show that \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is reflexive. Take any integer \( x \in \mathbb{Z} \) , and observe that \( n \mid 0 \), so \( n \mid \left( {x - x}\right) \) . By definition of congruence modulo \( n \), we have \( x \equiv x\left( {\;\operatorname{mod}\;n}\right) \) . This shows \( x \equiv x\left( {\;\operatorname{mod}\;n}\right) \) for every \( x \in \mathbb{Z} \), so \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is reflexive.\n\nNext, we will show that \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is symmetric. For this, we must show that for all \( x, y \in \mathbb{Z} \), the condition \( x \equiv y\left( {\;\operatorname{mod}\;n}\right) \) implies that \( y \equiv x\left( {\;\operatorname{mod}\;n}\right) \) . We use direct proof. Suppose \( x \equiv y\left( {\;\operatorname{mod}\;n}\right) \) . Thus \( n \mid \left( {x - y}\right) \) by definition of congruence modulo \( n \) . Then \( x - y = {na} \) for some \( a \in \mathbb{Z} \) by definition of divisibility. Multiplying both sides by -1 gives \( y - x = n\left( {-a}\right) \) . Therefore \( n \mid \left( {y - x}\right) \), and this means \( y \equiv x\left( {\;\operatorname{mod}\;n}\right) \) . We’ve shown that \( x \equiv y\left( {\;\operatorname{mod}\;n}\right) \) implies that \( y \equiv x\left( {\;\operatorname{mod}\;n}\right) \), and this means \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is symmetric.\n\nFinally we will show that \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is transitive. For this we must show that if \( x \equiv y\left( {\;\operatorname{mod}\;n}\right) \) and \( y \equiv z\left( {\;\operatorname{mod}\;n}\right) \), then \( x \equiv z\left( {\;\operatorname{mod}\;n}\right) \) . Again we use direct proof. Suppose \( x \equiv y\left( {\;\operatorname{mod}\;n}\right) \) and \( y \equiv z\left( {\;\operatorname{mod}\;n}\right) \) . This means \( n \mid \left( {x - y}\right) \) and \( n \mid \left( {y - z}\right) \) . Therefore there are integers \( a \) and \( b \) for which \( x - y = {na} \) and \( y - z = {nb} \) . Adding these two equations, we obtain \( x - z = {na} + {nb} \) . Consequently, \( x - z = n\left( {a + b}\right) \), so \( n \mid \left( {x - z}\right) \), hence \( x \equiv z\left( {\;\operatorname{mod}\;n}\right) \) . This completes the proof that \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is transitive.\n\nThe past three paragraphs have shown that the relation \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is reflexive, symmetric and transitive, so the proof is complete.
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Yes
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Consider the relation \( {R}_{1} \) in Figure 11.2. The equivalence class containing 2 is the set \( \left\lbrack 2\right\rbrack = \left\{ {x \in A : x{R}_{1}2}\right\} \) .
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Because in this relation the only element that relates to 2 is 2 itself, we have \( \left\lbrack 2\right\rbrack = \{ 2\} \) . Other equivalence classes for \( {R}_{1} \) are \( \left\lbrack {-1}\right\rbrack = \{ - 1\} ,\left\lbrack 1\right\rbrack = \{ 1\} ,\left\lbrack 3\right\rbrack = \{ 3\} \) and \( \left\lbrack 4\right\rbrack = \{ 4\} \) . Thus this relation has five separate equivalence classes.
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Yes
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Consider the relation \( {R}_{2} \) in Figure 11.2. The equivalence class containing 2 is the set [2] \( = \left\{ {x \in A : x{R}_{2}2}\right\} \) .
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Because only 2 and 4 relate to 2, we have [2] = \{2,4\}. Observe that we also have [4] = \{ \( x \in A : x{R}_{2}4 \) \} = \{2,4\}, so [2] = [4]. Another equivalence class for \( {R}_{2} \) is [1] = \( \{ x \in A : x{R}_{2}1\} = \{ - 1,1,3\} \) . In addition, note that \( \left\lbrack 1\right\rbrack = \left\lbrack {-1}\right\rbrack = \left\lbrack 3\right\rbrack = \{ - 1,1,3\} . \) Thus this relation has just two equivalence classes, namely \( \{ 2,4\} \) and \( \{ - 1,1,3\} \) .
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Yes
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In Example 11.8 we proved that for a given \( n \in \mathbb{N} \) the relation \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is reflexive, symmetric and transitive. Thus, in our new parlance, \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) is an equivalence relation on \( \mathbb{Z} \) . Consider the case \( n = 3 \) . Let’s find the equivalence classes of the equivalence relation \( \equiv \left( {\;\operatorname{mod}\;3}\right) \) .
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The equivalence class containing 0 seems like a reasonable place to start. Observe that\n\n\[ \left\lbrack 0\right\rbrack = \left\{ {x \in \mathbb{Z} : x \equiv 0\left( {\mathrm{{mod}}\;3}\right) }\right\} = \]\n\n\[ \left\{ {x \in \mathbb{Z} : 3 \mid \left( {x - 0}\right) }\right\} = \left\{ {x \in \mathbb{Z} : 3 \mid x}\right\} = \{ \ldots , - 3,0,3,6,9,\ldots \} . \]\n\nThus the class [0] consists of all the multiples of 3. (Or, said differently, [0] consists of all integers that have a remainder of 0 when divided by 3 .) Note that \( \left\lbrack 0\right\rbrack = \left\lbrack 3\right\rbrack = \left\lbrack 6\right\rbrack = \left\lbrack 9\right\rbrack \), etc. The number 1 does not show up in the set \( \left\lbrack 0\right\rbrack \) so let's next look at the equivalence class [1]:\n\n\[ \left\lbrack 1\right\rbrack = \left\{ {x \in \mathbb{Z} : x \equiv 1\left( {\text{mod }3}\right) }\right\} = \left\{ {x \in \mathbb{Z} : 3 \mid \left( {x - 1}\right) }\right\} = \left\{ {\ldots , - 5, - 2,1,4,7,{10},\ldots }\right\} . \]\n\nThe equivalence class [1] consists of all integers that give a remainder of 1 when divided by 3 . The number 2 is in neither of the sets [0] or [1], so we next look at the equivalence class [2]:\n\n\[ \left\lbrack 2\right\rbrack = \left\{ {x \in \mathbb{Z} : x \equiv 2\left( {\text{mod }3}\right) }\right\} = \left\{ {x \in \mathbb{Z} : 3 \mid \left( {x - 2}\right) }\right\} = \left\{ {\ldots , - 4, - 1,2,5,8,{11},\ldots }\right\} . \]\n\nThe equivalence class [2] consists of all integers that give a remainder of 2 when divided by 3. Observe that any integer is in one of the sets [0], [1] or [2], so we have listed all of the equivalence classes. Thus \( \equiv \left( {\;\operatorname{mod}\;3}\right) \) has exactly three equivalence classes, as described above.
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Yes
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Theorem 11.1 Suppose \( R \) is an equivalence relation on a set \( A \) . Suppose also that \( a, b \in A \) . Then \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) if and only if \( {aRb} \) .
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Proof. Suppose \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . Note that \( {aRa} \) by the reflexive property of \( R \), so \( a \in \{ x \in A : {xRa}\} = \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack = \{ x \in A : {xRb}\} . \) But \( a \) belonging to \( \{ x \in A : {xRb}\} \) means \( {aRb} \) . This completes the first part of the if-and-only-if proof.\n\nConversely, suppose \( {aRb} \) . We need to show \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . We will do this by showing \( \left\lbrack a\right\rbrack \subseteq \left\lbrack b\right\rbrack \) and \( \left\lbrack b\right\rbrack \subseteq \left\lbrack a\right\rbrack \) .\n\nFirst we show \( \left\lbrack a\right\rbrack \subseteq \left\lbrack b\right\rbrack \) . Suppose \( c \in \left\lbrack a\right\rbrack \) . As \( c \in \left\lbrack a\right\rbrack = \{ x \in A : {xRa}\} \), we get \( {cRa} \) . Now we have \( {cRa} \) and \( {aRb} \), so \( {cRb} \) because \( R \) is transitive. But \( {cRb} \) implies \( c \in \{ x \in A : {xRb}\} = \left\lbrack b\right\rbrack \) . This demonstrates that \( c \in \left\lbrack a\right\rbrack \) implies \( c \in \left\lbrack b\right\rbrack \) , so \( \left\lbrack a\right\rbrack \subseteq \left\lbrack b\right\rbrack \) .\n\nNext we show \( \left\lbrack b\right\rbrack \subseteq \left\lbrack a\right\rbrack \) . Suppose \( c \in \left\lbrack b\right\rbrack \) . As \( c \in \left\lbrack b\right\rbrack = \{ x \in A : {xRb}\} \), we get \( {cRb} \) . Remember that we are assuming \( {aRb} \) , so \( {bRa} \) because \( R \) is symmetric. Now we have \( {cRb} \) and \( {bRa} \), so \( {cRa} \) because \( R \) is transitive. But \( {cRa} \) implies \( c \in \{ x \in A : {xRa}\} = \left\lbrack a\right\rbrack \) . This demonstrates that \( c \in \left\lbrack b\right\rbrack \) implies \( c \in \left\lbrack a\right\rbrack \) ; hence \( \left\lbrack b\right\rbrack \subseteq \left\lbrack a\right\rbrack \) .\n\nThe previous two paragraphs imply that \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) .
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Yes
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Our examples and experience suggest that the equivalence classes of an equivalence relation on a set form a partition of that set. This is indeed the case, and we now prove it.
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Theorem 11.2 Suppose
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No
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Theorem 11.2 Suppose \( R \) is an equivalence relation on a set \( A \) . Then the set \( \{ \left\lbrack a\right\rbrack : a \in A\} \) of equivalence classes of \( R \) forms a partition of \( A \) .
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Proof. To show that \( \{ \left\lbrack a\right\rbrack : a \in A\} \) is a partition of \( A \) we need to show two things: We need to show that the union of all the sets \( \left\lbrack a\right\rbrack \) equals \( A \), and we need to show that if \( \left\lbrack a\right\rbrack \neq \left\lbrack b\right\rbrack \), then \( \left\lbrack a\right\rbrack \cap \left\lbrack b\right\rbrack = \varnothing \) . Notationally, the union of all the sets \( \left\lbrack a\right\rbrack \) is \( \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \), so we need to prove \( \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack = A \) . Suppose \( x \in \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \) . This means \( x \in \left\lbrack a\right\rbrack \) for some \( a \in A \) . Since \( \left\lbrack a\right\rbrack \subseteq A \), it then follows that \( x \in A \) . Thus \( \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \subseteq A \) . On the other hand, suppose \( x \in A \) . As \( x \in \left\lbrack x\right\rbrack \), we know \( x \in \left\lbrack a\right\rbrack \) for some \( a \in A \) (namely \( a = x \) ). Therefore \( x \in \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \), and this shows \( A \subseteq \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \) . Since \( \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \subseteq A \) and \( A \subseteq \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack \), it follows that \( \mathop{\bigcup }\limits_{{a \in A}}\left\lbrack a\right\rbrack = A \) . Next we need to show that if \( \left\lbrack a\right\rbrack \neq \left\lbrack b\right\rbrack \) then \( \left\lbrack a\right\rbrack \cap \left\lbrack b\right\rbrack = \varnothing . \) Let’s use contrapositive proof. Suppose it’s not the case that \( \left\lbrack a\right\rbrack \cap \left\lbrack b\right\rbrack = \varnothing \), so there is some element \( c \) with \( c \in \left\lbrack a\right\rbrack \cap \left\lbrack b\right\rbrack \) . Thus \( c \in \left\lbrack a\right\rbrack \) and \( c \in \left\lbrack b\right\rbrack \) . Now, \( c \in \left\lbrack a\right\rbrack \) means \( {cRa} \), and then \( {aRc} \) since \( R \) is symmetric. Also \( c \in \left\lbrack b\right\rbrack \) means \( {cRb} \) . Now we have \( {aRc} \) and \( {cRb} \), so \( {aRb} \) (because \( R \) is transitive). By Theorem 11.1, \( {aRb} \) implies \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . Thus \( \left\lbrack a\right\rbrack \neq \left\lbrack b\right\rbrack \) is not true. We’ve now shown that the union of all the equivalence classes is \( A \), and the intersection of two different equivalence classes is \( \varnothing \) . Therefore the set of equivalence classes is a partition of \( A \) .
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Yes
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Consider the function \( f : \mathbb{Z} \rightarrow \mathbb{N} \), where \( f\left( n\right) = \left| n\right| + 2 \), from Example 12.1. The domain is \( \mathbb{Z} \) and the codomain is \( \mathbb{N} \). The range of this function is the set \( \{ f\left( a\right) : a \in \mathbb{Z}\} = \{ \left| a\right| + 2 : a \in \mathbb{Z}\} = \{ 2,3,4,5,\ldots \} \).
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The range of this function is the set \( \{ f\left( a\right) : a \in \mathbb{Z}\} = \{ \left| a\right| + 2 : a \in \mathbb{Z}\} = \{ 2,3,4,5,\ldots \} \).
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Yes
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Example 12.2 Let \( A = \{ p, q, r, s\} \) and \( B = \{ 0,1,2\} \), and \[ f = \{ \left( {p,0}\right) ,\left( {q,1}\right) ,\left( {r,2}\right) ,\left( {s,2}\right) \} \subseteq A \times B. \] This is a function \( f : A \rightarrow B \) because each element of \( A \) occurs exactly once as a first coordinate of an ordered pair in \( f \) .
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Observe that we have \( f\left( p\right) = 0 \) , \( f\left( q\right) = 1, f\left( r\right) = 2 \) and \( f\left( s\right) = 2 \) . The domain of this function is \( A = \{ p, q, r, s\} \) . The codomain and range are both \( B = \{ 0,1,2\} \) .
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Yes
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Example 12.3 Say a function \( \varphi : {\mathbb{Z}}^{2} \rightarrow \mathbb{Z} \) is defined as \( \varphi \left( {m, n}\right) = {6m} - {9n} \) . Note that as a set, this function is \( \varphi = \left\{ {\left( {\left( {m, n}\right) ,{6m} - {9n}}\right) : \left( {m, n}\right) \in {\mathbb{Z}}^{2}}\right\} \subseteq {\mathbb{Z}}^{2} \times \mathbb{Z}. \) What is the range of \( \varphi \) ?
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To answer this, first observe that for any \( \left( {m, n}\right) \in {\mathbb{Z}}^{2} \), the value \( \varphi \left( {m, n}\right) = \) \( {6m} - {9n} = 3\left( {{2m} - {3n}}\right) \) is a multiple of 3 . Thus every number in the range is a multiple of 3, so the range is a subset of the set of all multiples of 3 . On the other hand if \( b = {3k} \) is a multiple of 3 we have \( \varphi \left( {-k, - k}\right) = 6\left( {-k}\right) - 9\left( {-k}\right) = \) \( {3k} = b \), which means any multiple of 3 is in the range of \( \varphi \) . Therefore the range of \( \varphi \) is the set \( \{ {3k} : k \in \mathbb{Z}\} \) of all multiples of 3 .
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Yes
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Show that the function \( f : \mathbb{R} - \{ 0\} \rightarrow \mathbb{R} \) defined as \( f\left( x\right) = \frac{1}{x} + 1 \) is injective but not surjective.
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We will use the contrapositive approach to show that \( f \) is injective. Suppose \( a,{a}^{\prime } \in \mathbb{R} - \{ 0\} \) and \( f\left( a\right) = f\left( {a}^{\prime }\right) \) . This means \( \frac{1}{a} + 1 = \frac{1}{{a}^{\prime }} + 1 \) . Subtracting 1 from both sides and inverting produces \( a = {a}^{\prime } \) . Therefore \( f \) is injective.\n\nThe function \( f \) is not surjective because there exists an element \( b = 1 \in \mathbb{R} \) for which \( f\left( x\right) = \frac{1}{x} + 1 \neq 1 \) for every \( x \in \mathbb{R} - \{ 0\} \) .
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Yes
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Show that the function \( f : \mathbb{R} - \{ 0\} \rightarrow \mathbb{R} - \{ 1\} \) where \( f\left( x\right) = \frac{1}{x} + 1 \) is injective and surjective (hence bijective).
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This is just like the previous example, except that the codomain has been changed. The previous example shows \( f \) is injective. To show that it is surjective, take an arbitrary \( b \in \mathbb{R} - \{ 1\} \) . We seek an \( a \in \mathbb{R} - \{ 0\} \) for which \( f\left( a\right) = b \), that is, for which \( \frac{1}{a} + 1 = b \) . Solving for \( a \) gives \( a = \frac{1}{b - 1} \), which is defined because \( b \neq 1 \) . In summary, for any \( b \in \mathbb{R} - \{ 1\} \), we have \( f\left( \frac{1}{b - 1}\right) = b \) , so \( f \) is surjective.
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Yes
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Show that the function \( g : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z} \) defined by the formula \( g\left( {m, n}\right) = \left( {m + n, m + {2n}}\right) \), is both injective and surjective.
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We will use the contrapositive approach to show that \( g \) is injective. Thus we need to show that \( g\left( {m, n}\right) = g\left( {k,\ell }\right) \) implies \( \left( {m, n}\right) = \left( {k,\ell }\right) \) . Suppose \( \left( {m, n}\right) ,\left( {k,\ell }\right) \in \mathbb{Z} \times \mathbb{Z} \) and \( g\left( {m, n}\right) = g\left( {k,\ell }\right) \) . Then \( \left( {m + n, m + {2n}}\right) = \left( {k + \ell, k + 2\ell }\right) \) . It follows that \( m + n = k + \ell \) and \( m + {2n} = k + 2\ell \) . Subtracting the first equation from the second gives \( n = \ell \) . Next, subtract \( n = \ell \) from \( m + n = k + \ell \) to get \( m = k \) . Since \( m = k \) and \( n = \ell \), it follows that \( \left( {m, n}\right) = \left( {k,\ell }\right) \) . Thus \( g \) is injective.\n\nTo see that \( g \) is surjective, consider an arbitrary element \( \left( {b, c}\right) \in \mathbb{Z} \times \mathbb{Z} \) . We need to show that there is some \( \left( {x, y}\right) \in \mathbb{Z} \times \mathbb{Z} \) for which \( g\left( {x, y}\right) = \left( {b, c}\right) \) . To find \( \left( {x, y}\right) \), note that \( g\left( {x, y}\right) = \left( {b, c}\right) \) means \( \left( {x + y, x + {2y}}\right) = \left( {b, c}\right) \) . This leads to the following system of equations:\n\n\[ x + y = b \]\n\n\[ x + {2y} = c.\]\n\nSolving gives \( x = {2b} - c \) and \( y = c - b \) . Then \( \left( {x, y}\right) = \left( {{2b} - c, c - b}\right) \) . We now have \( g\left( {{2b} - c, c - b}\right) = \left( {b, c}\right) \), and it follows that \( g \) is surjective.
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Yes
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Example 12.7 Consider function \( h : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Q} \) defined as \( h\left( {m, n}\right) = \frac{m}{\left| n\right| + 1} \) . Determine whether this is injective and whether it is surjective.
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This function is not injective because of the unequal elements \( \left( {1,2}\right) \) and \( \left( {1, - 2}\right) \) in \( \mathbb{Z} \times \mathbb{Z} \) for which \( h\left( {1,2}\right) = h\left( {1, - 2}\right) = \frac{1}{3} \) . However, \( h \) is surjective: Take any element \( b \in \mathbb{Q} \) . Then \( b = \frac{c}{d} \) for some \( c, d \in \mathbb{Z} \) . Notice we may assume \( d \) is positive by making \( c \) negative, if necessary. Then \( h\left( {c, d - 1}\right) = \frac{c}{\left| {d - 1}\right| + 1} = \frac{c}{d} = b \) .
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Yes
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Suppose \( A = \{ a, b, c\}, B = \{ 0,1\}, C = \{ 1,2,3\} \) . Let \( f : A \rightarrow B \) be the function \( f = \{ \left( {a,0}\right) ,\left( {b,1}\right) ,\left( {c,0}\right) \} \), and let \( g : B \rightarrow C \) be \( g = \{ \left( {0,3}\right) ,\left( {1,1}\right) \} \) . Then \( g \circ f = \{ \left( {a,3}\right) ,\left( {b,1}\right) ,\left( {c,3}\right) \} \).
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Then \( g \circ f = \{ \left( {a,3}\right) ,\left( {b,1}\right) ,\left( {c,3}\right) \} \).
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Yes
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Example 12.9 Say \( A = \{ a, b, c\}, B = \{ 0,1\}, C = \{ 1,2,3\} \) . Let \( f : A \rightarrow B \) be the function \( f = \{ \left( {a,0}\right) ,\left( {b,1}\right) ,\left( {c,0}\right) \} \), and let \( g : C \rightarrow B \) be the function \( g = \{ \left( {1,0}\right) ,\left( {2,1}\right) ,\left( {3,1}\right) \} \) . In this situation the composition \( g \circ f \) is not defined because the codomain \( B \) of \( f \) is not the same set as the domain \( C \) of \( g \) .
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Remember: In order for \( g \circ f \) to make sense, the codomain of \( f \) must equal the domain of \( g \) . (Or at least be a subset of it.)
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No
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Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be defined as \( f\left( x\right) = {x}^{2} + x \), and \( g : \mathbb{R} \rightarrow \mathbb{R} \) be defined as \( g\left( x\right) = x + 1 \) . Then \( g \circ f : \mathbb{R} \rightarrow \mathbb{R} \) is the function defined by the formula \( g \circ f\left( x\right) = g\left( {f\left( x\right) }\right) = g\left( {{x}^{2} + x}\right) = {x}^{2} + x + 1.
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Since the domains and codomains of \( g \) and \( f \) are the same, we can in this case do a composition in the other order. Note that \( f \circ g : \mathbb{R} \rightarrow \mathbb{R} \) is the function defined as \( f \circ g\left( x\right) = f\left( {g\left( x\right) }\right) = f\left( {x + 1}\right) = {\left( x + 1\right) }^{2} + \left( {x + 1}\right) = {x}^{2} + {3x} + 2.
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Yes
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Theorem 12.1 Composition of functions is associative. That is if \( f : A \rightarrow B \) , \( g : B \rightarrow C \) and \( h : C \rightarrow D \), then \( \left( {h \circ g}\right) \circ f = h \circ \left( {g \circ f}\right) \) .
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Proof. Suppose \( f, g, h \) are as stated. It follows from Definition 12.5 that both \( \left( {h \circ g}\right) \circ f \) and \( h \circ \left( {g \circ f}\right) \) are functions from \( A \) to \( D \) . To show that they are equal, we just need to show\n\n\[ \left( {\left( {h \circ g}\right) \circ f}\right) \left( x\right) = \left( {h \circ \left( {g \circ f}\right) }\right) \left( x\right) \]\n\nfor every \( x \in A \) . Note that Definition 12.5 yields\n\n\[ \left( {\left( {h \circ g}\right) \circ f}\right) \left( x\right) = \left( {h \circ g}\right) \left( {f\left( x\right) }\right) = h\left( {g\left( {f\left( x\right) }\right) }\right) . \]\n\nAlso\n\n\[ \left( {h \circ \left( {g \circ f}\right) }\right) \left( x\right) = h\left( {g \circ f\left( x\right) }\right) = h\left( {g\left( {f\left( x\right) }\right) }\right) . \]\n\nThus\n\n\[ \left( {\left( {h \circ g}\right) \circ f}\right) \left( x\right) = \left( {h \circ \left( {g \circ f}\right) }\right) \left( x\right) ,\]\n\nas both sides equal \( h\left( {g\left( {f\left( x\right) }\right) }\right) \) .
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Yes
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Theorem 12.2 Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) . If both \( f \) and \( g \) are injective, then \( g \circ f \) is injective. If both \( f \) and \( g \) are surjective, then \( g \circ f \) is surjective.
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Proof. First suppose both \( f \) and \( g \) are injective. To see that \( g \circ f \) is injective, we must show that \( g \circ f\left( x\right) = g \circ f\left( y\right) \) implies \( x = y \) . Suppose \( g \circ f\left( x\right) = g \circ f\left( y\right) \) . This means \( g\left( {f\left( x\right) }\right) = g\left( {f\left( y\right) }\right) \) . It follows that \( f\left( x\right) = f\left( y\right) \) . (For otherwise \( g \) wouldn’t be injective.) But since \( f\left( x\right) = f\left( y\right) \) and \( f \) is injective, it must be that \( x = y \) . Therefore \( g \circ f \) is injective.\n\nNext suppose both \( f \) and \( g \) are surjective. To see that \( g \circ f \) is surjective, we must show that for any element \( c \in C \), there is a corresponding element \( a \in A \) for which \( g \circ f\left( a\right) = c \) . Thus consider an arbitrary \( c \in C \) . Because \( g \) is surjective, there is an element \( b \in B \) for which \( g\left( b\right) = c \) . And because \( f \) is surjective, there is an element \( a \in A \) for which \( f\left( a\right) = b \) . Therefore \( g\left( {f\left( a\right) }\right) = g\left( b\right) = c \), which means \( g \circ f\left( a\right) = c \) . Thus \( g \circ f \) is surjective.
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Yes
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Theorem 12.3 Let \( f : A \rightarrow B \) be a function. Then \( f \) is bijective if and only if the inverse relation \( {f}^{-1} \) is a function from \( B \) to \( A \) .
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Suppose \( f : A \rightarrow B \) is bijective, so according to the theorem \( {f}^{-1} \) is a function. Observe that the relation \( f \) contains all the pairs \( \left( {x, f\left( x\right) }\right) \) for \( x \in A \) , so \( {f}^{-1} \) contains all the pairs \( \left( {f\left( x\right), x}\right) \) . But \( \left( {f\left( x\right), x}\right) \in {f}^{-1} \) means \( {f}^{-1}\left( {f\left( x\right) }\right) = x \) . Therefore \( {f}^{-1} \circ f\left( x\right) = x \) for every \( x \in A \) . From this we get \( {f}^{-1} \circ f = {i}_{A} \) . Similar reasoning produces \( f \circ {f}^{-1} = {i}_{B} \) .
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Yes
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The function \( f : \mathbb{R} \rightarrow \mathbb{R} \) defined as \( f\left( x\right) = {x}^{3} + 1 \) is bijective. Find its inverse.
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We begin by writing \( y = {x}^{3} + 1 \) . Now interchange variables to obtain \( x = {y}^{3} + 1 \) . Solving for \( y \) produces \( y = \sqrt[3]{x - 1} \) . Thus\n\n\[ \n{f}^{-1}\left( x\right) = \sqrt[3]{x - 1} \n\]\n\n(You can check your answer by computing\n\n\[ \n{f}^{-1}\left( {f\left( x\right) }\right) = \sqrt[3]{f\left( x\right) - 1} = \sqrt[3]{{x}^{3} + 1 - 1} = x.\n\]\n\nTherefore \( {f}^{-1}\left( {f\left( x\right) }\right) = x \) . Any answer other than \( x \) indicates a mistake.)
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Yes
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Example 12.12 Example 12.6 showed that the function \( g : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z} \) defined by the formula \( g\left( {m, n}\right) = \left( {m + n, m + {2n}}\right) \) is bijective. Find its inverse.
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The approach outlined above should work, but we need to be careful to keep track of coordinates in \( \mathbb{Z} \times \mathbb{Z} \) . We begin by writing \( \left( {x, y}\right) = g\left( {m, n}\right) \), then interchanging the variables \( \left( {x, y}\right) \) and \( \left( {m, n}\right) \) to get \( \left( {m, n}\right) = g\left( {x, y}\right) \) . This gives\n\n\[ \left( {m, n}\right) = \left( {x + y, x + {2y}}\right) \]\n\nfrom which we get the following system of equations:\n\n\[ x + y = m \]\n\n\[ x + {2y} = n\text{. } \]\n\nSolving this system, we get\n\n\[ x\; = \;{2m} - n \]\n\n\[ y = n - m. \]\n\nThen \( \left( {x, y}\right) = \left( {{2m} - n, n - m}\right) \), so \( {g}^{-1}\left( {m, n}\right) = \left( {{2m} - n, n - m}\right) \) .\n\nWe can check this by confirming \( {g}^{-1}\left( {g\left( {m, n}\right) }\right) = \left( {m, n}\right) \) . Doing the math,\n\n\[ {g}^{-1}\left( {g\left( {m, n}\right) }\right) = {g}^{-1}\left( {m + n, m + {2n}}\right) \]\n\n\[ = \;\left( {2\left( {m + n}\right) - \left( {m + {2n}}\right) ,\left( {m + {2n}}\right) - \left( {m + n}\right) }\right) \]\n\n\[ = \left( {m, n}\right) \text{.} \]
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Yes
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Theorem 13.1 (Triangle inequality) If \( x, y, z \in \mathbb{R} \), then \( \left| {x - y}\right| \leq \left| {x - z}\right| + \left| {z - y}\right| \) .
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The name triangle inequality comes from the fact that the theorem can be interpreted as asserting that for any \
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No
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Prove that \( \mathop{\lim }\limits_{{x \rightarrow 2}}\left( {{3x} + 4}\right) = {10} \) .
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Proof. Suppose \( \varepsilon > 0 \) . Note that \( \left| {\left( {{3x} + 4}\right) - {10}}\right| = \left| {{3x} - 6}\right| = \left| {3\left( {x - 2}\right) }\right| = 3\left| {x - 2}\right| \) . So if \( \delta = \frac{\varepsilon }{3} \), then \( 0 < \left| {x - 2}\right| < \delta \) yields \( \left| {\left( {{3x} + 4}\right) - {10}}\right| = 3\left| {x - 2}\right| < {3\delta } = 3\frac{\varepsilon }{3} = \varepsilon . \) \n\nIn summary, for any \( \varepsilon > 0 \), there is a \( \delta = \frac{\varepsilon }{3} \) for which \( 0 < \left| {x - 2}\right| < \delta \) implies \( \left| {\left( {{3x} + 4}\right) - {10}}\right| < \varepsilon \) . By Definition 13.2, \( \mathop{\lim }\limits_{{x \rightarrow 2}}\left( {{3x} + 4}\right) = {10} \) .
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Yes
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Prove that \( \mathop{\lim }\limits_{{x \rightarrow 2}}5{x}^{2} = {20} \) .
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Proof. Suppose \( \varepsilon > 0 \) . Notice that\n\n\[ \left| {f\left( x\right) - L}\right| = \left| {5{x}^{2} - {20}}\right| = \left| {5\left( {{x}^{2} - 4}\right) }\right| = \left| {5\left( {x - 2}\right) \left( {x + 2}\right) }\right| = 5 \cdot \left| {x - 2}\right| \cdot \left| {x + 2}\right| .\n\]\n\nNow we have a factor of \( \left| {x - 2}\right| \) in \( \left| {f\left( x\right) - L}\right| \), but it is accompanied with \( \left| {x + 2}\right| \) . But if \( \left| {x - 2}\right| \) is small, then \( x \) is close to 2, so \( \left| {x + 2}\right| \) should be close to 4 . In fact, if \( \left| {x - 2}\right| \leq 1 \), then \( \left| {x + 2}\right| = \left| {\left( {x - 2}\right) + 4}\right| \leq \left| {x - 2}\right| + \left| 4\right| \leq 1 + 4 = 5 \) . (Here we applied the inequality (13.2) from page 245.) In other words, if \( \left| {x - 2}\right| \leq 1 \) , then \( \left| {x + 2}\right| \leq 5 \), and the above equation yields\n\n\[ \left| {f\left( x\right) - L}\right| = \left| {5{x}^{2} - {20}}\right| = 5 \cdot \left| {x - 2}\right| \cdot \left| {x + 2}\right| < 5 \cdot \left| {x - 2}\right| \cdot 5 = {25}\left| {x - 2}\right| .\n\]\n\nTake \( \delta \) to be smaller than both 1 and \( \frac{\varepsilon }{25} \) . Then \( 0 < \left| {x - 2}\right| < \delta \) implies \( \left| {5{x}^{2} - {20}}\right| < {25} \cdot \left| {x - 2}\right| < {25\delta } < {25}\frac{\varepsilon }{25} = \varepsilon \) . By Definition 13.2, \( \mathop{\lim }\limits_{{x \rightarrow 2}}5{x}^{2} = {20} \) .
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Yes
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Suppose \( f\left( x\right) = \frac{x}{2} + \frac{\left| x - 2\right| }{x - 2} + 2 \) . Prove \( \mathop{\lim }\limits_{{x \rightarrow 2}}f\left( x\right) \) does not exist.
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Proof. Notice that \( f\left( 2\right) \) is not defined, as it involves division by zero. Also, \( f\left( x\right) \) behaves differently depending on whether \( x \) is to the right or left of 2 .\n\nIf \( x > 2 \), then \( x - 2 \) is positive, so \( \left| {x - 2}\right| = x - 2 \) and \( \frac{\left| x - 2\right| }{x - 2} = 1 \), so \( f\left( x\right) = \frac{x}{2} + 3 \) .\n\nIf \( x < 2 \), then \( x - 2 \) is negative, so \( \left| {x - 2}\right| = - \left( {x - 2}\right) \) and \( \frac{\left| x - 2\right| }{x - 2} = - 1 \), so \( f\left( x\right) = \frac{x}{2} + 1 \) .\n\nTherefore \( f, \) graphed below, is a piecewise function \( f\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{2}x + 3 & \text{if }x > 2 \\ \frac{1}{2}x + 1 & \text{if }x < 2. \end{array}\right. \)\n\nSuppose for the sake of contradiction that\n\n\n\n\( \mathop{\lim }\limits_{{x \rightarrow 2}}f\left( x\right) = L \), where \( L \) is a real number. Let \( \varepsilon = \frac{1}{2} \) . By Definition 13.2, there is a real number \( \delta > 0 \) for which \( \;0 < \left| {x - 2}\right| < \delta \;\mathrm{{implies}}\;\left| {f\left( x\right) - L}\right| < \frac{1}{2}. \) Put \( a = 2 - \frac{\delta }{2} \), so \( 0 < \left| {a - 2}\right| < \delta \) . Hence \( \left| {f\left( a\right) - L}\right| < \frac{1}{2} \) . Put \( b = 2 + \frac{\delta }{2} \), so \( 0 < \left| {b - 2}\right| < \delta \) . Hence \( \left| {f\left( b\right) - L}\right| < \frac{1}{2} \) . Further, \( f\left( a\right) < 2 \) and \( f\left( b\right) > 4 \), so \( 2 < \left| {f\left( b\right) - f\left( a\right) }\right| \) . With this and the help of the inequality (13.1), we get a contradiction \( 2 < 1 \), as follows:\n\n\( 2 < \left| {f\left( b\right) - f\left( a\right) }\right| = \left| {\left( {f\left( b\right) - L}\right) - \left( {f\left( a\right) - L}\right) }\right| \leq \;\left| {f\left( b\right) - L}\right| + \left| {f\left( a\right) - L}\right| < \frac{1}{2} + \frac{1}{2} = 1. \)
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Yes
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Prove that \( \mathop{\lim }\limits_{{x \rightarrow 0}}\sin \left( \frac{1}{x}\right) \) does not exist.
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Proof. Suppose for the sake of contradiction that \( \mathop{\lim }\limits_{{x \rightarrow 0}}\sin \left( \frac{1}{x}\right) = L \) for \( L \in \mathbb{R} \) . Definition 13.2 guarantees a number \( \delta \) for which \( 0 < \left| {x - 0}\right| < \delta \) implies \( \left| {\sin \left( \frac{1}{x}\right) - L}\right| < \frac{1}{4} \) . Select \( k \in \mathbb{N} \) large enough so that \( \frac{1}{k\pi } < \delta \) . As \( 0 < \left| {\frac{1}{k\pi } - 0}\right| < \delta \) , we have \( \left| {\sin \left( \frac{1}{1/{k\pi }}\right) - L}\right| < \frac{1}{4} \), and this yields \( \left| {\sin \left( {k\pi }\right) - L}\right| = \left| {0 - L}\right| = \left| L\right| < \frac{1}{4} \) .\n\nNext, take \( \ell \in \mathbb{N} \) large enough so that \( \frac{1}{\frac{\pi }{2} + 2\ell \pi } < \delta \) . Then \( 0 < \left| {\frac{1}{\frac{\pi }{2} + 2\ell \pi } - 0}\right| < \delta \), so we have \( \left| {\sin \left( \begin{matrix} \frac{1}{\frac{1}{2} + 2\ell \pi } \\ \end{matrix}\right) - L}\right| < \frac{1}{4}, \) which simplifies to \( \left| {\sin {\left( \frac{\pi }{2} + 2\ell \pi \right) }^{2} - L}\right| = \left| {1 - L}\right| < \frac{1}{4}. \)\n\nAbove we showed \( \left| L\right| \leq \frac{1}{4} \) and \( \left| {1 - L}\right| \leq \frac{1}{4} \) . Now apply the inequality (13.2) to get the contradiction \( 1 < \frac{1}{2} \), as \( 1 = \left| {L + \left( {1 - L}\right) }\right| \leq \left| L\right| + \left| {1 - L}\right| < \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \) .
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Yes
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Example 13.5 Investigate \( \mathop{\lim }\limits_{{x \rightarrow 0}}x\sin \left( \frac{1}{x}\right) \) .
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This is like the previous example, except for the extra \( \;x.\; \) Because \( \;\left| {\;\sin \left( \frac{1}{x}\right) \;}\right| \leq 1, \) we expect \( x\sin \left( \frac{1}{x}\right) \) to go to 0 as \( x \) goes to 0 . Indeed, we prove \( \mathop{\lim }\limits_{{x \rightarrow 0}}x\sin \left( \frac{1}{x}\right) = 0 \) .\n\nProof. Given \( \varepsilon > 0 \), let \( \delta = \frac{\varepsilon }{2} \) . Suppose \( 0 < \left| {x - 0}\right| < \delta \) . Simplifying, \( \left| x\right| < \delta \) , which is the same as \( \left| x\right| < \frac{\varepsilon }{2}. \) We get \( \left| {x\sin \left( \frac{1}{x}\right) - 0}\right| = \left| {x\sin \left( \frac{1}{x}\right) }\right| = \left| x\right| \cdot \left| {\sin \left( \frac{1}{x}\right) }\right| \leq \) \( \frac{\varepsilon }{2}\left| {\sin \left( \frac{1}{x}\right) }\right| \leq \frac{\varepsilon }{2} \cdot 1 < \varepsilon \) . From this, Definition 13.2 gives \( \mathop{\lim }\limits_{{x \rightarrow 0}}x\sin \left( \frac{1}{x}\right) = 0 \) .
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Yes
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Theorem 13.2 (Constant function rule) If \( a \in \mathbb{R} \), then \( \mathop{\lim }\limits_{{x \rightarrow c}}a = a \) .
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Proof. Suppose \( a \in \mathbb{R} \). According to Definition 13.2, to prove \( \mathop{\lim }\limits_{{x \rightarrow c}}a = a \), we must show that for any \( \varepsilon > 0 \), there is a \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {a - a}\right| < \varepsilon \). This is almost too easy. Just let \( \delta = 1 \) (or any other number). Then \( \left| {a - a}\right| < \varepsilon \) is automatic, because \( \left| {a - a}\right| = 0 \).
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Yes
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Theorem 13.3 (Identity function rule) If \( c \in \mathbb{R} \), then \( \mathop{\lim }\limits_{{x \rightarrow c}}x = c \) .
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Proof. Given \( \varepsilon > 0 \), let \( \delta = \varepsilon \) . Then \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {x - c}\right| < \varepsilon \) . By Definition 13.2, this means \( \mathop{\lim }\limits_{{x \rightarrow c}}x = c \) .
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Yes
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Theorem 13.4 (Constant multiple rule)\n\nIf \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) \) exists, and \( a \in \mathbb{R} \), then \( \mathop{\lim }\limits_{{x \rightarrow c}}{af}\left( x\right) = a\mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) \) .
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Proof. Suppose \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) \) exists. We must show \( \mathop{\lim }\limits_{{x \rightarrow c}}{af}\left( x\right) = a\mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) \) . If \( a = 0 \) , then this reduces to \( \mathop{\lim }\limits_{{x \rightarrow c}}0 = 0 \), which is true by Theorem 13.2. Thus, for the remainder of the proof we can assume \( a \neq 0 \) .\n\nSuppose \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) = L \) . We must prove \( \mathop{\lim }\limits_{{x \rightarrow c}}{af}\left( x\right) = {aL} \) . By Definition 13.2, this means we must show that for any \( \varepsilon > 0 \), there is a \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {{af}\left( x\right) - {aL}}\right| < \epsilon \) . Let \( \varepsilon > 0 \) . Because \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) = L \), there exists a \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {f\left( x\right) - L}\right| < \frac{\epsilon }{\left| a\right| } \) . So if \( 0 < \left| {x - c}\right| < \delta \) , then \( \left| {{af}\left( x\right) - {aL}}\right| = \left| {a\left( {f\left( x\right) - L}\right) }\right| = \left| a\right| \cdot \left| {f\left( x\right) - L}\right| < \left| a\right| \frac{\varepsilon }{\left| a\right| } = \varepsilon . \n\nIn summary, we’ve shown that for any \( \varepsilon > 0 \), there is a \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {{af}\left( x\right) - {aL}}\right| < \epsilon \) . By Definition 13.2, \( \mathop{\lim }\limits_{{x \rightarrow c}}{af}\left( x\right) = {aL}. \)
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Yes
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Theorem 13.5 (Sum rule)\n\nIf both \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) \) and \( \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) \) exist, then \( \mathop{\lim }\limits_{{x \rightarrow c}}\left( {f\left( x\right) + g\left( x\right) }\right) = \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) + \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) \) .
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Proof. Let \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) = L \) and \( \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) = M \) . We must prove \( \mathop{\lim }\limits_{{x \rightarrow c}}\left( {f\left( x\right) + g\left( x\right) }\right) = \) \( L + M \) . To prove this, take \( \varepsilon > 0 \) . We need to find a corresponding \( \delta \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {\left( {f\left( x\right) + g\left( x\right) }\right) - \left( {L + M}\right) }\right| < \varepsilon \) . With this in mind, notice that\n\n\[ \left| {\left( {f\left( x\right) + g\left( x\right) }\right) - \left( {L + M}\right) }\right| = \left| {\left( {f\left( x\right) - L}\right) + \left( {g\left( x\right) - M}\right) }\right| \]\n\n\[ \leq \left| {f\left( x\right) - L}\right| + \left| {g\left( x\right) - M}\right| \text{.} \]\n\n(A)\n\nAs \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) = L \), there is a \( {\delta }^{\prime } > 0 \) such that \( 0 < \left| {x - c}\right| < {\delta }^{\prime } \) implies \( \left| {f\left( x\right) - L}\right| < \frac{\varepsilon }{2} \) . As \( \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) = M \), there is a \( {\delta }^{\prime \prime } > 0 \) such that \( 0 < \left| {x - c}\right| < {\delta }^{\prime \prime } \) implies \( \left| {g\left( x\right) - M}\right| < \frac{\varepsilon }{2} \) . Now put \( \delta = \min \left\{ {{\delta }^{\prime },{\delta }^{\prime \prime }}\right\} \), meaning that \( \delta \) equals the smaller of \( {\delta }^{\prime } \) and \( {\delta }^{\prime \prime } \) . If \( 0 < \left| {x - c}\right| < \delta \), then (A) gives \( \left| {\left( {f\left( x\right) + g\left( x\right) }\right) - \left( {L + M}\right) }\right| \leq \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon \) .\n\nWe’ve now shown that for any \( \varepsilon > 0 \), there is a \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {\left( {f\left( x\right) + g\left( x\right) }\right) - \left( {L + M}\right) }\right| < \varepsilon . \) Thus \( \mathop{\lim }\limits_{{x \rightarrow c}}\left( {f\left( x\right) + g\left( x\right) }\right) = L + M. \)
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Yes
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Example 13.6 Find \( \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{\frac{1}{x} - 1}{1 - x} \) .
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Here \( x \) approaches \( 1, \) but simply plugging in \( x = 1 \) gives \( \frac{\frac{1}{1} - 1}{1 - 1} = \frac{0}{0} \) (undefined). So we apply whatever algebra is needed to cancel the denominator \( 1 - x \) , and follow this with limit laws:\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{\frac{1}{x} - 1}{1 - x} = \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{\frac{1}{x} - 1}{1 - x}\frac{x}{x} \]\n(multiply quotient by \( 1 = \frac{x}{x} \) )\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{\left( 1 - x\right) }{\left( {1 - x}\right) x} \]\n(distribute \( x \) on top)\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{1}{x} \]\n(cancel the \( \left( {1 - x}\right) \) )\n\n\[ = \frac{\mathop{\lim }\limits_{{x \rightarrow 1}}1}{\mathop{\lim }\limits_{{x \rightarrow 1}}x} = \frac{1}{1} = 1 \]\n(apply limit laws)
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Yes
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Theorem 13.9 (Composition rule)\n\nIf \( \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) = L \) and \( f \) is continuous at \( x = L \), then \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( {g\left( x\right) }\right) = f\left( {\mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) }\right) \) .
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Proof. Suppose \( \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) = L \) and \( f \) is continuous at \( x = L \) . We need to show \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( {g\left( x\right) }\right) = f\left( L\right) \) . According to Definition 13.2, for any \( \varepsilon > 0 \) we must find a corresponding \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {f\left( {g\left( x\right) }\right) - f\left( L\right) }\right| < \varepsilon \) .\n\nSo let \( \varepsilon > 0 \) . As \( f \) is continuous at \( L \), Definition 13.3 yields \( \mathop{\lim }\limits_{{x \rightarrow L}}f\left( x\right) = f\left( L\right) \) . From this, we know there is a real number \( {\delta }^{\prime } > 0 \) for which\n\n\[ \left| {x - L}\right| < {\delta }^{\prime }\text{ implies }\left| {f\left( x\right) - f\left( L\right) }\right| < \varepsilon . \]\n\n(A)\n\nBut also, from \( \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) = L \), we know that there is a real number \( \delta > 0 \) for which \( 0 < \left| {x - c}\right| < \delta \) implies \( \left| {g\left( x\right) - L}\right| < {\delta }^{\prime } \) .\n\nIf \( 0 < \left| {x - c}\right| < \delta \), then we have \( \left| {g\left( x\right) - L}\right| < {\delta }^{\prime } \), and from this (A) yields \( \left| {f\left( {g\left( x\right) }\right) - f\left( L\right) }\right| < \varepsilon \) . Thus \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( {g\left( x\right) }\right) = f\left( L\right) \), and the proof is complete.
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Yes
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Theorem 13.10 If \( f \) is differentiable at \( c \), then \( f \) is continuous at \( c \) .
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Proof. Suppose \( f \) is differentiable at \( c \), so \( \mathop{\lim }\limits_{{x \rightarrow c}}\frac{f\left( x\right) - f\left( c\right) }{x - c} = {f}^{\prime }\left( c\right) \) . Write \( f\left( x\right) \) as\n\n\[ f\left( x\right) = \frac{f\left( x\right) - f\left( c\right) }{x - c}\left( {x - c}\right) + f\left( c\right) . \]\n\nTaking limits of both sides and using limit laws,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) = \left( {\mathop{\lim }\limits_{{x \rightarrow c}}\frac{f\left( x\right) - f\left( c\right) }{x - c}}\right) \cdot \left( {\mathop{\lim }\limits_{{x \rightarrow c}}\left( {x - c}\right) }\right) + \mathop{\lim }\limits_{{x \rightarrow c}}f\left( c\right) = {f}^{\prime }\left( c\right) \cdot 0 + f\left( c\right) = f\left( c\right) . \]\n\nThus \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) = f\left( c\right) \), which means \( f \) is continuous at \( c \) .
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Yes
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Investigate \( \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\sin \left( x\right) }{x} \) .
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For any \( x \in \mathbb{R} \), we know that \( - 1 \leq \sin \left( x\right) \leq 1 \) . Consequently we would expect \( \frac{\sin \left( x\right) }{x} \) to be very small when \( x \) is large, that is, we expect \( \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\sin \left( x\right) }{x} = 0 \) .\n\nLet us use Definition 13.4 to prove this. Given \( \varepsilon > 0 \), put \( N = \frac{1}{\varepsilon } \) . If \( x > N \) , then \( x > \frac{1}{\varepsilon } \), so \( \frac{1}{x} < \varepsilon \), and hence \( - \varepsilon < \frac{1}{x}\sin \left( x\right) < \varepsilon \), meaning \( \left| \frac{\sin \left( x\right) }{x}\right| < \varepsilon \).\n\nIn summary, given \( \varepsilon > 0 \), there is an \( N > 0 \) for which \( x > N \) implies \( \left| {\frac{\sin \left( x\right) }{x} - 0}\right| < \varepsilon \) . By Definition 13.4, \( \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\sin \left( x\right) }{x} = 0 \) .
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Yes
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Prove that the sequence \( \left\{ {1 - \frac{1}{n}}\right\} \) converges to 1 .
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Proof. Suppose \( \varepsilon > 0 \) . Choose an integer \( N > \frac{1}{\varepsilon } \), so that \( \frac{1}{N} < \varepsilon \) . Then if \( n > N \) we have \( \left| {{a}_{n} - 1}\right| = \left| {\left( {1 - \frac{1}{n}}\right) - 1}\right| = \frac{1}{n} < \frac{1}{N} < \varepsilon \) . By Definition 13.5 the sequence \( \left\{ {1 - \frac{1}{n}}\right\} \) converges to 1 .
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Yes
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Investigate the sequence \( \left\{ \frac{{\left( -1\right) }^{n + 1}\left( {n + 1}\right) }{n}\right\} \) .
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Proof. Suppose for the sake of contradiction that the sequence \( \left\{ \frac{{\left( -1\right) }^{n + 1}\left( {n + 1}\right) }{n}\right\} \) converges to a real number \( L \) . Let \( \varepsilon = 1 \) . By Definition 13.5 there is an \( N \in \mathbb{N} \) for which \( n > N \) implies \( \left| {\frac{{\left( -1\right) }^{n + 1}\left( {n + 1}\right) }{n} - L}\right| < 1 \) .\n\nIf \( n \) is odd, then the \( n \) th term of the sequence is \( {a}_{n} = \frac{{\left( -1\right) }^{n + 1}\left( {n + 1}\right) }{n} = \frac{n + 1}{n} > 1 \) . For \( n \) even, the \( n \) th term of the sequence is is \( {a}_{n} = \frac{{\left( -1\right) }^{n + 1}\left( {n + 1}\right) }{n} = - \frac{n + 1}{n} < - 1 \) . Take an odd number \( m > N \) and an even number \( n > N \) . The above three lines yield\n\n\[ 2 = 1 - \left( {-1}\right) < {a}_{m} - {a}_{n}\;\left( {\text{ because }1 < {a}_{m}\text{ and }1 < - {a}_{n}}\right) \]\n\n\[ = \left| {{a}_{m} - {a}_{n}}\right| \;\left( {{a}_{m} - {a}_{n}\text{is positive}}\right) \]\n\n\[ = \left| {\left( {{a}_{m} - L}\right) - \left( {{a}_{n} - L}\right) }\right| \;\left( {\text{add }0 = L - L\text{ to }{a}_{m} - {a}_{n}}\right) \]\n\n\[ \leq \left| {{a}_{m} - L}\right| + \left| {{a}_{n} - L}\right| \;\text{(using}\left| {x - y}\right| < \left| x\right| + \left| y\right| \text{)} \]\n\n\[ < 1 + 1 = 2.\;\text{(because}\left| {{a}_{n} - L}\right| < 1\text{when}n > N\text{)} \]\n\nThus \( 2 < 2 \), which is a contradiction. Consequently the series diverges.
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Yes
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Prove that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{2}^{k}} = 1 \) .
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Consider the partial sum \( {s}_{n} = \frac{1}{{2}^{1}} + \frac{1}{{2}^{2}} + \frac{1}{{2}^{3}} + \cdots + \frac{1}{{2}^{n}} \) . We can get a neat formula for \( {s}_{n} \) by noting \( {s}_{n} = 2{s}_{n} - {s}_{n} \) . Then simplify and cancel like terms:\n\n\[ {s}_{n} = 2{s}_{n} - {s}_{n} = 2\left( {\frac{1}{{2}^{1}} + \frac{1}{{2}^{2}} + \frac{1}{{2}^{3}} + \cdots + \frac{1}{{2}^{n - 1}} + \frac{1}{{2}^{n}}}\right) - \left( {\frac{1}{{2}^{1}} + \frac{1}{{2}^{2}} + \frac{1}{{2}^{3}} + \cdots + \frac{1}{{2}^{n - 1}} + \frac{1}{{2}^{n}}}\right) \]\n\n\[ = \left( {\frac{2}{{2}^{1}} + \frac{2}{{2}^{2}} + \frac{2}{{2}^{3}} + \cdots + \frac{2}{{2}^{n - 1}} + \frac{2}{{2}^{n}}}\right) - \left( {\frac{1}{{2}^{1}} + \frac{1}{{2}^{2}} + \frac{1}{{2}^{3}} + \cdots + \frac{1}{{2}^{n - 1}} + \frac{1}{{2}^{n}}}\right) \]\n\n\[ = \left( {1 + \frac{1}{{2}^{1}} + \frac{1}{{2}^{2}} + \cdots + \frac{1}{{2}^{n - 2}} + \frac{1}{{2}^{n - 1}}}\right) - \left( {\frac{1}{{2}^{1}} + \frac{1}{{2}^{2}} + \frac{1}{{2}^{3}} + \cdots + \frac{1}{{2}^{n - 1}} + \frac{1}{{2}^{n}}}\right) = 1 - \frac{1}{{2}^{n}}. \]\n\nThus \( {s}_{n} = 1 - \frac{1}{{2}^{n}} \), so the sequence of partial sums is \( \left\{ {s}_{n}\right\} = \left\{ {1 - \frac{1}{{2}^{n}}}\right\} \), which converges to 1 by Exercise 13.7.4. Definition 13.7 yields \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{2}^{k}} = 1 \) .
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No
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Theorem 13.11 If \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k} \) converges, then the sequence \( \left\{ {a}_{n}\right\} \) converges to 0 .
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Proof. We use direct proof. Suppose \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k} \) converges, and say \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k} = S \) . Then by Definition 13.7, the sequence of partial sums \( \left\{ {s}_{n}\right\} \) converges to \( S \) . From this, Definition 13.5 says that for any \( \varepsilon > 0 \) there is an \( N \in \mathbb{N} \) for which \( n > N \) implies \( \left| {{s}_{n} - S}\right| < \varepsilon \) . Thus also \( n - 1 > N \) implies \( \left| {{s}_{n - 1} - S}\right| < \varepsilon \) .\n\nWe need to show that \( \left\{ {a}_{n}\right\} \) converges to 0 . So take \( \varepsilon > 0 \) . By the previous paragraph, there is an \( {N}^{\prime } \in \mathbb{N} \) for which \( n > {N}^{\prime } \) implies \( \left| {{s}_{n} - S}\right| < \frac{\varepsilon }{2} \) and \( \left| {{s}_{n - 1} - S}\right| < \frac{\varepsilon }{2} \) . Notice that \( {a}_{n} = {s}_{n} - {s}_{n - 1} \) for any \( n > 2 \) . So if \( n > {N}^{\prime } \) we have\n\n\[ \left| {{a}_{n} - 0}\right| = \left| {{s}_{n} - {s}_{n - 1}}\right| = \left| {\left( {{s}_{n} - S}\right) - \left( {{s}_{n - 1} - S}\right) }\right| \]\n\n\[ \leq \left| {{s}_{n} - S}\right| + \left| {{s}_{n - 1} - S}\right| < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon . \]\n\nTherefore, by Definition 13.5, the sequence \( \left\{ {a}_{n}\right\} \) converges to 0 .
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Yes
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The sets \( A = \{ n \in \mathbb{Z} : 0 \leq n \leq 5\} \) and \( B = \{ n \in \mathbb{Z} : - 5 \leq n \leq 0\} \) have the same cardinality because there is a bijective function \( f : A \rightarrow B \) given by the rule \( f\left( n\right) = - n \) .
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Several comments are in order. First, if \( \left| A\right| = \left| B\right| \), there can be lots of bijective functions from \( A \) to \( B \) . We only need to find one of them in order to conclude \( \left| A\right| = \left| B\right| \) . Second, as bijective functions play such a big role here, we use the word bijection to mean bijective function. Thus the function \( f\left( n\right) = - n \) from Example 14.1 is a bijection. Also, an injective function is called an injection and a surjective function is called a surjection.
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No
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This example shows that \( \left| \mathbb{N}\right| = \left| \mathbb{Z}\right| \) .
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To see why this is true, notice that the following table describes a bijection \( f : \mathbb{N} \rightarrow \mathbb{Z} \) . \n\n<table><tr><td>\( n \)</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>...</td></tr><tr><td>\( f\left( n\right) \)</td><td>0</td><td>1</td><td>\( - 1 \)</td><td>2</td><td>\( - 2 \)</td><td>3</td><td>\( - 3 \)</td><td>4</td><td>\( - 4 \)</td><td>5</td><td>\( - 5 \)</td><td>6</td><td>\( - 6 \)</td><td>7</td><td>\( - 7 \)</td><td>...</td></tr></table>\n\nNotice that \( f \) is described in such a way that it is both injective and surjective. Every integer appears exactly once on the infinitely long second row. Thus, according to the table, given any \( b \in \mathbb{Z} \) there is some natural number \( n \) with \( f\left( n\right) = b \), so \( f \) is surjective. It is injective because the way the table is constructed forces \( f\left( m\right) \neq f\left( n\right) \) whenever \( m \neq n \) . Because of this bijection \( f : \mathbb{N} \rightarrow \mathbb{Z} \), we must conclude from Definition 14.1 that \( \left| \mathbb{N}\right| = \left| \mathbb{Z}\right| \) .
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Yes
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Show that \( \left| \left( {0,\infty }\right) \right| = \left| \left( {0,1}\right) \right| \) .
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To accomplish this, we need to show that there is a bijection \( f : \left( {0,\infty }\right) \rightarrow \left( {0,1}\right) \) . We describe this function geometrically. Consider the interval \( \left( {0,\infty }\right) \) as the positive \( x \) -axis of \( {\mathbb{R}}^{2} \) . Let the interval \( \left( {0,1}\right) \) be on the \( y \) -axis as illustrated in Figure 14.1, so that \( \left( {0,\infty }\right) \) and \( \left( {0,1}\right) \) are perpendicular to each other.\n\nThe figure also shows a point \( P = \left( {-1,1}\right) \) . Define \( f\left( x\right) \) to be the point on \( \left( {0,1}\right) \) where the line from \( P \) to \( x \in \left( {0,\infty }\right) \) intersects the \( y \) -axis. By similar triangles, we have\n\n\[ \frac{1}{x + 1} = \frac{f\left( x\right) }{x} \]\n\nand therefore\n\n\[ f\left( x\right) = \frac{x}{x + 1} \]\n\nIf it is not clear from the figure that \( f : \left( {0,\infty }\right) \rightarrow \left( {0,1}\right) \) is bijective, then you can verify it using the techniques from Section 12.2. (Exercise 16, below.)
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No
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Show that \( \left| \mathbb{R}\right| = \left| \left( {0,1}\right) \right| \) .
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Because of the bijection \( g : \mathbb{R} \rightarrow \left( {0,\infty }\right) \) where \( g\left( x\right) = {2}^{x} \), we have \( \left| \mathbb{R}\right| = \left| \left( {0,\infty }\right) \right| \) . Also, Example 14.3 shows that \( \left| \left( {0,\infty }\right) \right| = \left| \left( {0,1}\right) \right| \) . Therefore \( \left| \mathbb{R}\right| = \left| \left( {0,1}\right) \right| .
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Yes
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Theorem 14.5 If \( A \) and \( B \) are both countably infinite, then so is \( A \times B \) .
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Proof. Suppose \( A \) and \( B \) are both countably infinite. By Theorem 14.3, we know we can write \( A \) and \( B \) in list form as\n\n\[ A = \left\{ {{a}_{1},{a}_{2},{a}_{3},{a}_{4},\ldots }\right\} \]\n\n\[ B = \left\{ {{b}_{1},{b}_{2},{b}_{3},{b}_{4},\ldots }\right\} .\n\nFigure 14.2 shows how to form an infinite path winding through all of \( A \times B \) . Therefore \( A \times B \) can be written in list form, so it is countably infinite.
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Yes
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Corollary 14.1 Given \( n \) countably infinite sets \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \), with \( n \geq 2 \) , the Cartesian product \( {A}_{1} \times {A}_{2} \times \cdots \times {A}_{n} \) is also countably infinite.
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Proof. The proof is by induction on \( n \) . For the basis step, notice that when \( n = 2 \) the statement asserts that for countably infinite sets \( {A}_{1} \) and \( {A}_{2} \), the product \( {A}_{1} \times {A}_{2} \) is countably infinite, and this is true by Theorem 14.5.\n\nAssume that for some \( k \geq 2 \), any product \( {A}_{1} \times {A}_{2} \times \cdots \times {A}_{k} \) of countably infinite sets is countably infinite. Consider a product \( {A}_{1} \times {A}_{2} \times \cdots \times {A}_{k} \times {A}_{k + 1} \) of \( k + 1 \) countably infinite sets. It is easy to confirm that the function\n\n\[ f : {A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k} \times {A}_{k + 1} \rightarrow \left( {{A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k}}\right) \times {A}_{k + 1} \]\n\n\[ f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{k},{x}_{k + 1}}\right) \; = \;\left( {\left( {{x}_{1},{x}_{2},\ldots ,{x}_{k}}\right) ,{x}_{k + 1}}\right) \]\n\nis bijective, so \( \left| {{A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k} \times {A}_{k + 1}}\right| = \left| {\left( {{A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k}}\right) \times {A}_{k + 1}}\right| . \) By the induction hypothesis, \( \left( {{A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k}}\right) \times {A}_{k + 1} \) is a product of two countably infinite sets, so it is countably infinite by Theorem 14.5. As noted above, \( {A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k} \times {A}_{k + 1} \) has the same cardinality as the set \( \left( {{A}_{1} \times {A}_{2} \times {A}_{3} \times \cdots \times {A}_{k}}\right) \times {A}_{k + 1} \), so it too is countably infinite.
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Yes
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Theorem 14.6 If \( A \) and \( B \) are both countably infinite, then their union \( A \cup B \) is countably infinite.
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Proof. Suppose \( A \) and \( B \) are both countably infinite. By Theorem 14.3, we know we can write \( A \) and \( B \) in list form as\n\n\[ A = \left\{ {{a}_{1},{a}_{2},{a}_{3},{a}_{4},\ldots }\right\} ,\]\n\n\[ B = \left\{ {{b}_{1},{b}_{2},{b}_{3},{b}_{4},\ldots }\right\} .\n\nWe can \
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No
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Theorem 14.8 An infinite subset of a countably infinite set is countably infinite.
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Proof. Suppose \( A \) is an infinite subset of the countably infinite set \( B \) . As \( B \) is countably infinite, its elements can be written in a list \( {b}_{1},{b}_{2},{b}_{3},{b}_{4},\ldots \)\n\nThen we can also write \( A \) ’s elements in list form by proceeding through the elements of \( B \), in order, and selecting those that belong to \( A \) . Thus \( A \) can be written in list form, and since \( A \) is infinite, its list will be infinite. Consequently \( A \) is countably infinite.
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Yes
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Theorem 14.9 If \( U \subseteq A \), and \( U \) is uncountable, then \( A \) is uncountable.
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Proof. For the sake of contradiction say that \( U \subseteq A \), and \( U \) is uncountable but \( A \) is not uncountable. Then since \( U \subseteq A \) and \( U \) is infinite, then \( A \) must be infinite too. Since \( A \) is infinite, and not uncountable, it must be countably infinite. Then \( U \) is an infinite subset of a countably infinite set \( A \), so \( U \) is countably infinite by Theorem 14.8. Thus \( U \) is both uncountable and countably infinite, a contradiction.
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Yes
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Example 14.6 The intervals \( \lbrack 0,1) \) and \( \left( {0,1}\right) \) in \( \mathbb{R} \) have equal cardinalities.
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For a simpler approach, note that \( f\left( x\right) = \frac{1}{4} + \frac{1}{2}x \) is an injection \( \lbrack 0,1) \rightarrow \left( {0,1}\right) . \) Also, \( g\left( x\right) = x \) is an injection \( \left( {0,1}\right) \rightarrow \lbrack 0,1) \) . The Cantor-Bernstein-Schröder theorem guarantees a bijection \( h : \lbrack 0,1) \rightarrow \left( {0,1}\right) \), so \( \left| {\lbrack 0,1)}\right| = \left| \left( {0,1}\right) \right| \) .
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Yes
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Proposition 1.2.17. \( - 1 < 0 \)
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Proof. This will be our first exposure to a proof technique called proof by contradiction. We'll make use of this technique throughout the book. In this case, the idea goes as follows. There's no way that -1 could be positive, because if it were then \( 1 + \left( {-1}\right) \) would also have to be positive, which it isn't because we know it's 0 . There's also no way that -1 could be 0, because if it were we’d have \( - 1 = 0 \), and adding 1 to both sides gives \( 0 = 1 \), which is false because 1 is positive and 0 isn’t. Since -1 isn’t positive and it isn't equal to 0 , the only option left is that it's negative. This is the gist of the argument, but we have to write it out more carefully to satisfy those nit-picking mathematicians. Every step in our argument must have a solid reason.\n\nSo here goes the formal proof. We'll give a logical sequence of mathematical statements, followed by a reason that justifies each statement-this is called statement-reason format.\n\nFirst we show that \( - 1 > 0 \) is false:\n\n<table><thead><tr><th>Statement</th><th>Reason</th></tr></thead><tr><td>Suppose \( - 1 > 0 \) .</td><td>Proof by contradiction: supposing the opposite</td></tr><tr><td>\( 1 > 0 \)</td><td>Prop. (A) in Section 1.2.2</td></tr><tr><td>\( 1 + \left( {-1}\right) > 0 \) .</td><td>Prop. (C) in Section 1.2.2</td></tr><tr><td>\( 1 + \left( {-1}\right) = 0 \)</td><td>Prop. (B) in Section 1.2.1</td></tr><tr><td>Contradiction is acheived</td><td>The last 2 statements contradict</td></tr><tr><td>\( - 1 > 0 \) is false</td><td>The supposition must be false</td></tr></table>\n\nNext, we show that \( - 1 = 0 \) is false:\n\n<table><thead><tr><th>Statement</th><th>Reason</th></tr></thead><tr><td>Suppose \( - 1 = 0 \)</td><td>Proof by contradiction: supposing the opposite</td></tr><tr><td>\( 1 + \left( {-1}\right) = 1 + 0 \)</td><td>Follows from previous statement by substitution</td></tr><tr><td>\( 0 = 1 \)</td><td>Props. (A) and (B) in Section 1.2.1</td></tr><tr><td>\( 0 > 0 \)</td><td>Prop. (A) in Section 1.2.2</td></tr><tr><td>Contradiction is achieved</td><td>\( 0 > 0 \) contradicts Prop. (B) in Section 1.2.2</td></tr><tr><td>\( - 1 = 0 \) is false</td><td>The supposition must be false</td></tr></table>\n\nAccording to Property (B) in Section 1.2.2, there are three possibilities: either \( - 1 > 0, - 1 = 0 \), or \( - 1 < 0 \) . We have eliminated the first two possibilities. So the third possibility must be true: \( - 1 < 0 \) . This completes the proof.
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Yes
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Proposition 2.1.1. -1 has no real square root.
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Proof. We give two proofs of this proposition. The first one explains all the details, while the second proof is more streamlined. It is the streamlined proof that you should try to imitate when you write up proofs for homework exercises.\n\n## Long drawn-out proof of Proposition 2.1.1 with all the gory details:\n\nWe will use a common proof technique called proof by contradiction. Here's how it goes:\n\nFirst we suppose that there exists a real number \( a \) such that \( {a}^{2} = \) -1. Now we know that any real number is either positive, or zero, or negative-there are no other possibilities. So we consider each of these three cases: \( a > 0 \), or \( a = 0 \), or \( a < 0 \) .\n\n- In the case that \( a > 0 \) then \( {a}^{2} = a \cdot a = \left( \text{positive}\right) \cdot \left( \text{positive}\right) = \mathrm{a} \) positive number (that is, \( {a}^{2} > 0 \) ). But this couldn’t possibly be true, because we have already supposed that \( {a}^{2} = - 1 \) : there’s no way that \( {a}^{2} > 0 \) and \( {a}^{2} = - 1 \) can both be true at the same time!\n\n---\n\n\( {}^{1} \) It’s true that the fancier graphing calculators can handle it, but that’s beside the point.\n\n---\n\n\n\n- In the case that \( a = 0 \), then \( {a}^{2} = a \cdot a = \left( 0\right) \cdot \left( 0\right) = 0 \) . But \( {a}^{2} = 0 \) also contradicts our supposition that \( {a}^{2} = - 1 \) .\n\n- In the case that \( a < 0 \), then \( {a}^{2} = a \cdot a = \left( \text{negative}\right) \cdot \left( \text{negative}\right) = \mathrm{a} \) positive number, so \( {a}^{2} > 0 \) . As in the first case, this contradicts our supposition that \( {a}^{2} = - 1 \) .\n\nSo no matter which of the three possible cases is true, we're still screwed: in every case, we always have a contradiction. We seem to have reached a dead end - a logically impossible conclusion. So what's wrong?\n\nWhat's wrong is the supposition. It must be the case that the supposition is not true. Consequently, the statement \
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Yes
|
Proposition 2.1.10. Given a right isosceles triangle where both legs have length 1 (see Figure 2.1.1) . Let \( x \) be the length of the hypotenuse. Then \( x \) is irrational-that is, it cannot be expressed as a ratio of integers.
|
Proof. The proof is by contradiction. Suppose that \( x \) is rational: that is, \( x = \frac{m}{n} \) for some integers \( m \) and \( n \) .We can always reduce a fraction to lowest terms ( as noted in Section 1.2.3), so we can assume \( m \) and \( n \) have no common factors.\n\nSince \( x \) is the hypotenuse of a right triangle, the Pythagorean Theorem gives us \( {x}^{2} = {1}^{2} + {1}^{2} = 2 \) . We can plug \( x = \frac{m}{n} \) into \( {x}^{2} = 2 \) to get \( {\left( \frac{m}{n}\right) }^{2} = 2 \) , which can be rearranged to give\n\n\[ \n{m}^{2} = 2{n}^{2} \n\]\n\nFrom this we see that \( {m}^{2} \) is divisible by 2, which means that \( {m}^{2} \) is even. Exercise 2.1.9 part (b) then tells us that \( m \) is even, so there must be an integer \( j \) such that \( m = {2j} \) . Plugging \( m = {2j} \) into \( {m}^{2} = 2{n}^{2} \) gives \( 4{j}^{2} = 2{n}^{2} \) , which simplifies to \( 2{j}^{2} = {n}^{2} \) . Hence \( {n}^{2} \) is even, and as before we conclude that \( n \) is even. So \( n = {2k} \) for some integer \( k \) .\n\nAt this point, we have \( m = {2j} \) and \( n = {2k} \), which means that \( m \) and \( n \) have a common factor of 2 . But at the beginning of the proof, we said that \( m \) and \( n \) were reduced to lowest terms, so they have no common factor. This is a contradiction. Therefore our supposition must be false, so \( x \) cannot be rational.
|
No
|
Proposition 2.1.15. Let \( a \) and \( b \) be integers, and let \( p \) be a prime number. If \( p \) divides \( {ab} \), then either \( p \) divides \( a \), or \( p \) divides \( b \) .
|
Proof. We're not ready to give a proof yet, but we'll give one later (see Exercise 3.5.23 in Section 3.5.4).
|
No
|
Example 2.2.3. Complex multiplication is commutative.
|
\[ \left( {a + {bi}}\right) \left( {c + {di}}\right) = \left( {{ac} - {bd}}\right) + \left( {{bc} + {ad}}\right) i \] (FLOI) On the other hand: \[ \left( {c + {di}}\right) \left( {a + {bi}}\right) = \left( {{ca} - {db}}\right) + \left( {{cb} + {da}}\right) i \] (FLOI) \[ = \left( {{ac} - {bd}}\right) + \left( {{bc} + {ad}}\right) i \] (commutativity of real multiplication) Since we obtain the same expression for \( \left( {a + {bi}}\right) \left( {c + {di}}\right) \) and \( \left( {c + {di}}\right) \left( {a + {bi}}\right) \) , it follows that \( \left( {a + {bi}}\right) \left( {c + {di}}\right) = \left( {c + {di}}\right) \left( {a + {bi}}\right) \) .
|
Yes
|
Proposition 2.2.11. Given that \( z = a + {bi}, w = c + {di} \), and \( z \cdot w = 0 \) . Then it must be true that either \( z = 0 \) or \( w = 0 \) .
|
The proof of Proposition 2.2.11 is outlined in the following exercise.\n\nExercise 2.2.12. Complete the proof of Proposition 2.2.11 by filling in the blanks. Note that some blanks may require an expression, and not just a single number or variable.\n\n(a) The proof is by contradiction. So we begin by supposing that \( z \neq \) \( < 1 > \) and \( w \neq < 2 > \) (which is the negation of what we’re trying to prove).\n\n(b) Since \( z \neq \_ < 3 > \), it follows that \( z \) has an inverse \( {z}^{-1} \) such that \( {z}^{-1} \cdot z = \) ___ \( < 4 > \) ___.\n\n(c) Since \( z \cdot w = 0 \), we can multiply both sides of this equation by \( < 5 > \) and obtain the equation \( w = < 6 > \) . This equation contradicts the supposition that \( < 7 > \) .\n\n(d) Since our supposition has led to a false conclusion, it follows that our supposition must be \( < 8 > \) . Therefore it cannot be true that \( < 9 > \) , so it must be true that \( < {10} > \) .\n\n\( \diamond \)
|
No
|
Let \( z = 2 + {3i} \) and \( w = 1 - {2i} \). Then
|
\[
\bar{z} = \overline{2 + {3i}} = 2 - {3i}\text{ and }\bar{w} = \overline{1 - {2i}} = 1 + {2i}.
\]
Notice also that
\[
z + w = \left( {2 + {3i}}\right) + \left( {1 - {2i}}\right) = 3 + i\text{ and }{zw} = \left( {2 + {3i}}\right) \left( {1 - {2i}}\right) = 8 - i,
\]
so that
\[
\overline{z + w} = 3 - i\text{ and }\overline{zw} = 8 + i.
\]
On the other hand, you may check that
\[
\bar{z} + \bar{w} = \left( {2 - {3i}}\right) + \left( {1 + {2i}}\right) = 3 - i\text{ and }\overline{zw} = \left( {2 - {3i}}\right) \left( {1 + {2i}}\right) = 8 + i.
\]
|
Yes
|
Proposition 2.2.21. Given \( z \) and \( w \) are complex numbers, then \( \bar{z} + \bar{w} = \overline{z + w} \).
|
Proof. We may write \( z \) as \( a + {bi} \) and \( w \) as \( c + {di} \). Then\n\n\[ \bar{z} + \bar{w} = \overline{a + {bi}} + \overline{c + {di}} \]\n\n\[ = \left( {a - {bi}}\right) + \left( {c - {di}}\right) \]\nby definition of conjugate\n\n\[ = \left( {a + c}\right) - \left( {b + d}\right) i \]\nby basic algebra\n\n\[ = \overline{\left( {a + c}\right) + \left( {b + d}\right) i} \]\nby definition of conjugate\n\n\[ = \overline{z + w} \]\nby definition of complex addition
|
Yes
|
Example 2.3.2. Let \( z = 2\operatorname{cis}\frac{\pi }{3} \) . Then
|
\[ a = 2\cos \frac{\pi }{3} = 1 \] and \[ b = 2\sin \frac{\pi }{3} = \sqrt{3} \] Hence, the rectangular representation is \( z = 1 + \sqrt{3}i \) .
|
Yes
|
Let \( z = 3\sqrt{2} - 3\sqrt{2}i \) (see Figure 2.3.3). Then the modulus of \( z \) is
|
\[ r = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{36} = 6. \] We can find the argument \( \theta \) by noticing that the tangent is equal to \( \frac{-3\sqrt{2}}{3\sqrt{2}} \) or -1 . This means that \( \theta = \arctan \left( {-1}\right) \) . Since the angle is in the fourth quadrant, this means that \( \theta = \frac{7\pi }{4} \) .
|
Yes
|
Proposition 2.3.8. Let \( z = r\operatorname{cis}\theta \) and \( w = s\operatorname{cis}\phi \) be two nonzero complex numbers. Then\n\n\[ z \cdot w = {rs}\operatorname{cis}\left( {\theta + \phi }\right) \]\n\nAlternatively, we may write\n\n\[ r\operatorname{cis}\theta \cdot s\operatorname{cis}\phi = {rs}\operatorname{cis}\left( {\theta + \phi }\right) . \]
|
Proof. The proof uses the following trigonometric formulas (surely you remember them!):\n\n\[ \cos \left( {\theta + \phi }\right) = \cos \theta \cos \phi - \sin \theta \sin \phi \]\n\n\[ \sin \left( {\theta + \phi }\right) = \cos \theta \cdot \sin \phi + \sin \theta \cdot \cos \phi \]
|
No
|
Proposition 2.3.12. Let \( z = r\operatorname{cis}\theta \) and \( w = s\operatorname{cis}\phi \) be two nonzero complex numbers. Then \[ \frac{z}{w} = \frac{r}{s}\operatorname{cis}\left( {\theta - \phi }\right) \]
|
Alternatively, we may write \[ \frac{r\operatorname{cis}\theta }{s\operatorname{cis}\phi } = \frac{r}{s}\operatorname{cis}\left( {\theta - \phi }\right) \]
|
Yes
|
Example 2.3.13. If \( z = 3\operatorname{cis}\left( {\pi /3}\right) \) and \( w = 2\operatorname{cis}\left( {\pi /6}\right) \), then
|
\[ {zw} = \left( {2 \cdot 3}\right) \operatorname{cis}\left( {\pi /3 + \pi /6}\right) = 6\operatorname{cis}\left( {\pi /2}\right) = {6i}. \]
|
Yes
|
We will compute \( {z}^{10} \) where \( z = 1 + i \) .
|
Rather than computing \( {\left( 1 + i\right) }^{10} \) directly, it is much easier to switch to polar coordinates and calculate \( {z}^{10} \) using de Moivre’s Theorem:\n\n\[ \n{z}^{10} = {\left( 1 + i\right) }^{10} \]\n\n\[ \n= {\left( \sqrt{2}\operatorname{cis}\left( \frac{\pi }{4}\right) \right) }^{10} \]\n\n\[ \n= {\left( \sqrt{2}\right) }^{10}\operatorname{cis}\left( \frac{5\pi }{2}\right) \]\n\n\[ \n= {32}\operatorname{cis}\left( \frac{\pi }{2}\right) \]\n\n\[ \n= {32i}\text{.} \]\n
|
Yes
|
Example 2.3.22. Let \( z = 2\operatorname{cis}\left( {\pi /4}\right) \) . What is \( {z}^{-3} \) ?
|
\[ {z}^{-3} = {\left( {z}^{3}\right) }^{-1} \] \[ = {\left( {\left\lbrack 2\operatorname{cis}\left( \pi /4\right) \right\rbrack }^{3}\right) }^{-1} \] \[ = {\left( 8\operatorname{cis}\left( 3\pi /4\right) \right) }^{-1}\;\text{ (by de Moivre’s Theorem) } \] \[ = \frac{1}{8}\operatorname{cis}\left( {{5\pi }/4}\right) \;\text{ (by Exercise 2.3.11) } \]
|
No
|
The complex number \( z \) is an \( n \) th root of unity if and only if \( z \) satisfies the following condition:\n\n\[ z = \operatorname{cis}\left( \frac{2k\pi }{n}\right) \text{, where}k\text{is an integer between 0 and}n - 1\text{.} \]
|
Proof. The proposition is an \
|
No
|
Proposition 2.4.28. Given any equation of the form \( {z}^{n} + {a}_{n - 1}{z}^{n - 1} + \) \( {a}_{n - 2}{z}^{n - 2} + \ldots + {a}_{1}z = {a}_{0} \), where \( n > 0 \) and \( {a}_{0},{a}_{1},\ldots {a}_{n - 1} \) are real numbers. Then there exists at least one and at most \( n \) distinct complex numbers which solve the given equation.
|
The Fundamental Theorem of Algebra actually has two parts. The easy part is the \
|
No
|
Proposition 3.2.3. (The division algorithm) Given any integer \( a \) and any positive integer \( m \), then there exists a unique number \( r \) between 0 and \( m - 1 \) such that \( a = q \cdot m + r \) for some integer \( q \) . In this expression, \( q \) is called the quotient, and \( r \) is called the remainder.
|
Proof. It turns out that proving this \
|
No
|
Proposition 3.2.8. If \( a \equiv r\left( {\;\operatorname{mod}\;m}\right) \) and \( 0 \leq r \leq m - 1 \), then \( r = \) \( {\;\operatorname{mod}\;\left( {a, m}\right) } \) .
|
Proof. Given that \( a \equiv r\left( {\;\operatorname{mod}\;m}\right) \), by the definition of modular equivalence it follows that \( a \) and \( r \) have the same remainder mod \( m \) . But since \( 0 \leq r \leq m - 1 \), the remainder of \( r \) is \( r \) itself. It follows that the remainder of \( a \) is also \( r \) : so \( r = {\;\operatorname{mod}\;\left( {a, m}\right) } \) .
|
Yes
|
Suppose Dusty drives around the 5-mile track 112 miles in a positive direction, then \( {49} \) miles in a negative direction, then \( {322} \) miles in a positive direction. To find Dusty’s net displacement we may take 112 - \( {49} + {322} = {385} \) and then take the remainder mod 5 (which turns out to be 0).
|
\[ {\;\operatorname{mod}\;\left( {{112},5}\right) } = 2 \] \[ {\;\operatorname{mod}\;\left( {-{49},5}\right) } = 1 \] \[ {\;\operatorname{mod}\;\left( {{322},5}\right) } = 2, \] and we compute \[ 2 + 1 + 2 = 5 \equiv 0\;\left( {\;\operatorname{mod}\;5}\right) . \] We have obtained the same answer with much less work. How did we do it? By replacing each number with its remainder.
|
Yes
|
Suppose I travel on my racetrack at a 113 miles per hour in the positive direction for 17 hours. We may compute:
|
Net displacement : \( {113} \cdot {17} = {1921} \) miles\n\nFinal position : \( {1921} = {384} \cdot 5 + 1 \Rightarrow \) final position \( = 1 \) .\n\nOn the other hand, we may reach the same conclusion by a somewhat easier route:\n\n\[ \n{\;\operatorname{mod}\;\left( {{113},5}\right) } = 3 \]\n\n\[ \n{\;\operatorname{mod}\;\left( {{17},5}\right) } = 2 \]\n\nand we compute\n\n\[ \n3 \cdot 2 = 6 \equiv 1\;\left( {\;\operatorname{mod}\;5}\right) \]\n\nAgain, we have obtained the correct answer by replacing each number with its remainder.
|
Yes
|
\[ 8 + x \equiv 6\;\left( {\;\operatorname{mod}\;{11}}\right) \]
|
From algebra we understand how to solve an equation with an \( = \) sign, but what do we do with this \( \equiv \) sign? In fact, we can turn it in to an \( = \) sign by using Proposition 3.2.10, which says that \( 8 + x \equiv 6\left( {\;\operatorname{mod}\;{11}}\right) \) means the same as:\n\n\[ 8 + x = k \cdot {11} + 6 \]\n\nAnd then we can solve for \( x \) like any other equation. The result is\n\n\[ x = k \cdot {11} - 2 \]\n\nSo we solved for \( x \), but what numbers does \( x \) actually equal? What does \( k \cdot {11} - 2 \) mean? \( k \) is an integer, therefore \( x \) can equal -2 (if \( k = 0 \) ), or -13 (if \( k = - 1 \) ), or 9 (if \( k = 1 \) ), and so on. In other words \( x \) equals -2 plus any integer multiple of 11 , which, by the definition of modular equivalence, means\n\n\[ x \equiv - 2\;\left( {\;\operatorname{mod}\;{11}}\right) \]\n\nThis is a correct solution: but it's not the only way to write it. It would be just as valid to write\n\n\[ \text{-}x \equiv - {13}\left( {\;\operatorname{mod}\;{11}}\right) \]\n\n\[ \text{-}x \equiv {20}\left( {\;\operatorname{mod}\;{11}}\right) \]\n\n\[ \text{-}x \equiv {130}\left( {\;\operatorname{mod}\;{11}}\right) \]\n\n- . . .\n\nNotice however that there is only one way to write the solution in terms of a number in \( {\mathbb{Z}}_{11} \), namely:\n\n\[ {\;\operatorname{mod}\;\left( {x,{11}}\right) } = 9 \]\n\nIn order to avoid ambiguity, mathematicians and textbooks always write solutions mod \( n \) in terms of numbers in \( {\mathbb{Z}}_{n} \). In our current example, it’s easy enough to obtain the standard solution \( \left( {x \equiv 9\left( {\;\operatorname{mod}\;{11}}\right) }\right) \) directly from the equation \( x = k \cdot {11} - 2 \) ? by taking one of the 11 ’s and adding it to the -2 to get\n\n\[ x = \left( {k - 1}\right) \cdot {11} + {11} - 2 = \left( {k - 1}\right) \cdot {11} + 9. \]\n\nSince \( k \) is an arbitrary integer, \( k - 1 \) is also an arbitrary integer. So we get \( x \equiv 9\left( {\;\operatorname{mod}\;{11}}\right) \).
|
Yes
|
Given the equation\n\n\[ \n{5x} + 3 \equiv 9\;\left( {\;\operatorname{mod}\;{11}}\right) \n\]
|
Using the definition of modular equivalence, this becomes\n\n\[ \n{5x} + 3 = {11k} + 9. \n\]\n\nSolving this equality using basic algebra gives us\n\n\[ \nx = \frac{{11k} + 6}{5}. \n\]\n\nNow remember that \( x \) must be an integer. In order for the right side to be an integer, we need to find a \( k \) that makes \( \frac{{11k} + 6}{5} \) an integer. At this point we may use trial and error to find a \( k \) in \( {\mathbb{Z}}_{5} \) such that \( {11} \cdot k + 6 \) is a multiple of 5 . We get \( k = 4 \) ; and in fact adding \( 5 \cdot n \) to 4 also works for any \( n \in \mathbb{Z} \) , since \( {5n} \) is always divisible by 5 . Now we can solve for \( x \) by substituting \( k = 4 + {5n} \) back in to the previous equation:\n\n\[ \nx = \frac{{11}\left( {4 + 5 \cdot n}\right) + 6}{5} \n\]\n\n\[ \n= \frac{{11} \cdot 4 + 6}{5} + \frac{{11} \cdot \left( {5n}\right) }{5} \n\]\n\n\[ \n= {10} + {11n} \n\]\n\nTherefore \( x \equiv {10}\left( {\;\operatorname{mod}\;{11}}\right) \) is the general solution. You may check (which is always a good idea!) by plugging \( {10} + {11n} \) for a couple values of \( n \) back into the original equation, and you'll see these numbers work.
|
Yes
|
To solve the equation \( {4x} + 5 \equiv 7\left( {\;\operatorname{mod}\;{11}}\right) \)
|
\[ {4x} + 5 \equiv 7\;\left( {\;\operatorname{mod}\;{11}}\right) \] \[ \Rightarrow {4x} + 5 = {11k} + 7 \] (by modular equivalence) \[ \Rightarrow x = \frac{{11k} + 2}{4} \] (basic algebra) Now, \( {11k} + 2 \) is a multiple of 4 when \( k = 2 \), as well as when \( k \) equals 2 plus any multiple of 4 . Therefore \( k = 2 + {4n} \), hence we may continue from the previous equation: \[ x = \frac{2 + {11k}}{4} \] \[ \Rightarrow x = \frac{2 + {11} \cdot \left( {2 + {4n}}\right) }{4}\;\text{(substitution)} \] \[ \Rightarrow x = 6 + {11n}.\;\text{(simplification)} \] Therefore \( x \equiv 6\left( {\;\operatorname{mod}\;{11}}\right) \) is the general solution.
|
Yes
|
Consider the equation \( {79x} \equiv 9\left( {\;\operatorname{mod}\;{15}}\right) \)
|
In Section 3.2 we mentioned that when we’re doing arithmetic mod \( n \), we can replace any number with its remainder \( {\;\operatorname{mod}\;n} \) without changing the answer. In this example then, we can replace the 79 with its remainder mod 15 , which is 4 . Thus we have\n\n\[ \n{4x} \equiv 9\;\left( {\;\operatorname{mod}\;{15}}\right) \n\]\n\nwhich leads to\n\[ \nx = \frac{{15k} + 9}{4}. \n\]\n\nBy rewriting the numerator, we can simplify the right-hand side:\n\n\[ \nx = \frac{\left( {{12k} + {3k}}\right) + \left( {8 + 1}\right) }{4} = {3k} + 2 + \frac{{3k} + 1}{4}. \n\]\n\nand we readily discover that \( k = 1 + {4n} \) makes the right-hand side an integer, so that\n\n\[ \nx = \frac{{15} \cdot \left( {1 + {4n}}\right) + 9}{4} = 6 + {15n}\text{, or }x \equiv 6\left( {\;\operatorname{mod}\;{15}}\right) . \n\]
|
Yes
|
To solve the equation \( {447x} + {53} \equiv {712}\left( {\;\operatorname{mod}\;{111}}\right) \) we proceed as follows:
|
\[ \Rightarrow {447x} \equiv {659}\;\left( {\;\operatorname{mod}\;{111}}\right) \;\text{(subtract 53 from both sides)} \] \[ \Rightarrow {3x} \equiv {104}\;\left( {\;\operatorname{modular}\text{equivalence}}\right) \] \[ \Rightarrow {3x} = {104} + {111k}\;\text{(basic algebra)} \] \[ \Rightarrow x = \frac{{104} + {111k}}{3}\;\text{(basic algebra)} \] \[ \Rightarrow x = {34} + \frac{2}{3} + {37k} \] It should be clear that no value of \( k \) makes the right side an integer. Hence \( x \) has no solution.
|
Yes
|
Proposition 3.4.4. Given \( \ell, m \in \mathbb{Z} \) .\n\n(a) \( {\;\operatorname{mod}\;\left( {\ell + m, n}\right) } = {\;\operatorname{mod}\;\left( {\ell, n}\right) } \oplus {\;\operatorname{mod}\;\left( {m, n}\right) } \)
|
Proof. For simplicity we let \( a \mathrel{\text{:=}} {\;\operatorname{mod}\;\left( {\ell, n}\right) } \) and \( b \mathrel{\text{:=}} {\;\operatorname{mod}\;\left( {m, n}\right) } \) . Then according to the definition of remainder \( {\;\operatorname{mod}\;n} \) we have\n\n\[ \ell = a + {sn}\;\text{ and }\;m = b + {tn}. \]\n\nAdding these two equations (which is basically substitution) and basic algebra we find\n\n\[ \ell + m = a + b + \left( {s + t}\right) n \]\n\nNow by the definition of \( \oplus \), there is some \( p \in \mathbb{Z} \) such that \( a + b = \left( {a \oplus b}\right) + {pn} \) ; therefore\n\n\[ \ell + m = \left( {a \oplus b}\right) + {pn} + \left( {s + t}\right) n = \left( {a \oplus b}\right) + \left( {p + s + t}\right) n.\;\text{ (subs. and basic algebra) } \]\n\nHence by the definition of modular equivalence,\n\n\[ \ell + m \equiv a \oplus b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nNow since \( a \oplus b \) is between 0 and \( n - 1 \) by definition, it follows from Proposition 3.2.8 that\n\n\[ {\;\operatorname{mod}\;\left( {\ell + m, n}\right) } = a \oplus b. \]\n\nRecalling the definitions of \( a \) and \( b \) above we get finally:\n\n\[ {\;\operatorname{mod}\;\left( {\ell + m, n}\right) } = {\;\operatorname{mod}\;\left( {\ell, n}\right) } \oplus {\;\operatorname{mod}\;\left( {m, n}\right) }, \]\n\nand we're finished!
|
Yes
|
Proposition 3.4.13. \( {\mathbb{Z}}_{n} \) is closed under modular addition and multiplication, for all positive integers \( n \) .
|
Exercise 3.4.14. Prove Proposition 3.4.13. That is, show that the modular sum and modular product of two elements of \( {\mathbb{Z}}_{n} \) are also in \( {\mathbb{Z}}_{n} \) . (*Hint*) \( \diamond \)
|
No
|
Proposition 3.4.17. \( 0 \in {\mathbb{Z}}_{n} \) is the additive identity of \( {\mathbb{Z}}_{n} \) .
|
Proof. Given any \( a \in {\mathbb{Z}}_{n} \), then \( a \oplus 0 \) is computed by taking the remainder of \( a + 0{\;\operatorname{mod}\;n} \) . Since \( a + 0 = a \), and \( 0 \leq a < n \), it follows that the remainder of \( a \) is still \( a \) . Hence \( a \oplus 0 = a \) . Similarly we can show \( 0 \oplus a = a \) . Thus 0 satisfies the definition of identity for \( {\mathbb{Z}}_{n} \) .
|
Yes
|
Proposition 3.4.19. Let \( {\mathbb{Z}}_{n} \) be the integers \( {\;\operatorname{mod}\;n} \) and \( a \in {\mathbb{Z}}_{n} \) . Then for every \( a \) there is an additive inverse \( {a}^{\prime } \in {\mathbb{Z}}_{n} \) .\n\nIn other words: for any \( a \in {\mathbb{Z}}_{n} \) in we can find an \( {a}^{\prime } \) such that:\n\n\[ a \oplus {a}^{\prime } = {a}^{\prime } \oplus a = 0. \]
|
We structure the proof of Proposition 3.4.19 as an exercise. We prove the two cases \( a = 0 \) and \( a \neq 0 \) separately.\n\nExercise 3.4.20.\n\n(a) Show that \( 0 \in {\mathbb{Z}}_{n} \) has an additive inverse in \( {\mathbb{Z}}_{n} \) .\n\n(b) Suppose \( a \) is a nonzero element of \( {\mathbb{Z}}_{n} \) (in mathematical shorthand, we write this as: \( a \in {\mathbb{Z}}_{n} \smallsetminus \{ 0\} \) ), and let \( {a}^{\prime } = n - a \) .\n\n(i) Show that \( {a}^{\prime } \) is in \( {\mathbb{Z}}_{n} \) . (*Hint*)\n\n(ii) Show that \( a \oplus {a}^{\prime } = {a}^{\prime } \oplus a = 0\left( {\;\operatorname{mod}\;n}\right) \) : that is, \( {a}^{\prime } \) is the additive inverse of \( a \) .
|
No
|
Modular addition and multiplication are associative:
|
Modular addition is associative: Given \( a, b, c \) are elements of \( {\mathbb{Z}}_{n} \) , we may apply part (a) of Exercise 3.4.8 and get\n\n\[{\;\operatorname{mod}\;\left( {\left( {a + b}\right) + c}\right), n) = \left( {a \oplus b}\right) \oplus c}.\n\nSimilarly, we may apply part (b) of Exercise 3.4.8 to get\n\n\[{\;\operatorname{mod}\;\left( {a + \left( {b + c}\right), n}\right) } = a \oplus \left( {b \oplus c}\right) .\n\nNow here's where we use regular arithmetic. The associative property of integer addition tells us that \( \left( {a + b}\right) + c = a + \left( {b + c}\right) \), so the left-hand sides are equal. So \( \left( {a \oplus b}\right) \oplus c = a \oplus \left( {b \oplus c}\right) \), and the proof is complete.
|
No
|
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