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Example 8.1 Let’s investigate elements of \( A = \{ x : x \in \mathbb{N} \) and \( 7 \mid x\} \) . | 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 \( ... | Yes |
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 \( B \)? | 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 \( 5 \mid \left( {\left( {{4n} + 3}\right) - \left( {{9n} - 2}\right) }\right) \), whi... | Yes |
Example 8.5 Prove that \( \{ x \in \mathbb{Z} : {18}\;\left| {\;x\} \subseteq \{ x \in \mathbb{Z} : 6\;}\right| \;x\} . \) | 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 ... | Yes |
Prove that if \( A \) and \( B \) are sets, then \( \mathcal{P}\left( A\right) \cup \mathcal{P}\left( B\right) \subseteq \mathcal{P}\left( {A \cup B}\right) \) . | 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\... | Yes |
Example 8.9 Suppose \( A \) and \( B \) are sets. If \( \mathcal{P}\left( A\right) \subseteq \mathcal{P}\left( B\right) \), then \( A \subseteq B \) . | 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{... | Yes |
Example 8.11 Suppose \( A, B \), and \( C \) are sets, and \( C \neq \varnothing \) . Prove that if \( A \times C = B \times C \), then \( A = B \) . | 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 def... | Yes |
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) . | 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 ... | Yes |
Given sets \( A, B \), and \( C \), prove \( A \times \left( {B \cap C}\right) = \left( {A \times B}\right) \cap \left( {A \times C}\right) \) . | 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 \lef... | Yes |
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 \( A \\subseteq P \) . | 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}^{... | Yes |
Conjecture For every \( n \in \mathbb{Z} \), the integer \( f\left( n\right) = {n}^{2} - n + {11} \) is prime. | Disproof. The statement \ | No |
Conjecture If \( A, B \) and \( C \) are sets, then \( A - \left( {B \cap C}\right) = \left( {A - B}\right) \cap \left( {A - C}\right) \) . | 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(... | Yes |
Conjecture There is a real number \( x \) for which \( {x}^{4} < x < {x}^{2} \) . | 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}... | Yes |
Conjecture There exist three integers \( x, y, z \), all greater than 1 and no two equal, for which \( {x}^{y} = {y}^{z} \) . | Proof. Note that if \( x = 2, y = {16} \) and \( z = 4 \), then \( {x}^{y} = {2}^{16} = {\left( {2}^{4}\right) }^{4} = {16}^{4} = {y}^{z} \) . | Yes |
Conjecture There is a real number \( x \) for which \( {x}^{4} < x < {x}^{2} \) . | 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 \( x \) p... | Yes |
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 \( {a}_{i} \) . | 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 \(... | Yes |
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. ... | 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... | Yes |
Example 11.2 Let \( A = \{ 1,2,3,4\} \), and consider the following set:\n\n\[ S = \{ \left( {1,1}\right) ,\left( {1,3}\right) ,\left( {3,1}\right) ,\left( {3,3}\right) ,\left( {2,2}\right) ,\left( {2,4}\right) ,\left( {4,2}\right) ,\left( {4,4}\right) \} \subseteq A \times A. \] | Here we have \( {1S1},{1S3},{4S2} \), etc., but \( {3S4} \) and \( {2S1} \) . What does \( S \) mean? Think of it as meaning \ | No |
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 \( U \subseteq B \times B \). | You may be hard-pressed to invent any \ | No |
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) \} . | 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 \( ... | Yes |
Proposition Let \( n \in \mathbb{N} \) . The relation \( \equiv \left( {\;\operatorname{mod}\;n}\right) \) on the set \( \mathbb{Z} \) is reflexive, symmetric and transitive. | 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 \( x \equiv x\left( {\;\operatorname{mod}\;n}\right) \)... | Yes |
Consider the relation \( {R}_{1} \) in Figure 11.2. The equivalence class containing 2 is the set \( \left\lbrack 2\right\rbrack = \left\{ {x \in A : x{R}_{1}2}\right\} \) . | 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\... | Yes |
Consider the relation \( {R}_{2} \) in Figure 11.2. The equivalence class containing 2 is the set [2] \( = \left\{ {x \in A : x{R}_{2}2}\right\} \) . | 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}... | Yes |
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 \... | 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... | Yes |
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 \( {aRb} \) . | 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 ... | Yes |
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. | Theorem 11.2 Suppose | No |
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 \( A \) . | 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... | Yes |
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 \i... | The range of this function is the set \( \{ f\left( a\right) : a \in \mathbb{Z}\} = \{ \left| a\right| + 2 : a \in \mathbb{Z}\} = \{ 2,3,4,5,\ldots \} \). | Yes |
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 o... | 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 \( B = \{ 0,1,2\} \) . | Yes |
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... | 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 . ... | Yes |
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. | 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}^{\... | Yes |
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). | 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 \( \f... | Yes |
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. | 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... | Yes |
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. | 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 \(... | Yes |
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}\... | Then \( g \circ f = \{ \left( {a,3}\right) ,\left( {b,1}\right) ,\left( {c,3}\right) \} \). | Yes |
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) \... | 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.) | No |
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) =... | 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... | Yes |
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 \( \left( {h \circ g}\right) \circ f = h \circ \left( {g \circ f}\right) \) . | 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) \lef... | Yes |
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. | 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) = ... | Yes |
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 \( A \) . | 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 \( \l... | Yes |
The function \( f : \mathbb{R} \rightarrow \mathbb{R} \) defined as \( f\left( x\right) = {x}^{3} + 1 \) is bijective. Find its inverse. | 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) = \sqr... | Yes |
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. | 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, ... | Yes |
Theorem 13.1 (Triangle inequality) If \( x, y, z \in \mathbb{R} \), then \( \left| {x - y}\right| \leq \left| {x - z}\right| + \left| {z - y}\right| \) . | The name triangle inequality comes from the fact that the theorem can be interpreted as asserting that for any \ | No |
Prove that \( \mathop{\lim }\limits_{{x \rightarrow 2}}\left( {{3x} + 4}\right) = {10} \) . | 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}... | Yes |
Prove that \( \mathop{\lim }\limits_{{x \rightarrow 2}}5{x}^{2} = {20} \) . | 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... | Yes |
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. | 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|... | Yes |
Prove that \( \mathop{\lim }\limits_{{x \rightarrow 0}}\sin \left( \frac{1}{x}\right) \) does not exist. | 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}\righ... | Yes |
Example 13.5 Investigate \( \mathop{\lim }\limits_{{x \rightarrow 0}}x\sin \left( \frac{1}{x}\right) \) . | 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) ... | Yes |
Theorem 13.2 (Constant function rule) If \( a \in \mathbb{R} \), then \( \mathop{\lim }\limits_{{x \rightarrow c}}a = a \) . | 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 a... | Yes |
Theorem 13.3 (Identity function rule) If \( c \in \mathbb{R} \), then \( \mathop{\lim }\limits_{{x \rightarrow c}}x = c \) . | 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 \( \mathop{\lim }\limits_{{x \rightarrow c}}x = c \) . | Yes |
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 \( \mathop{\lim }\limits_{{x \rightarrow c}}{af}\left( x\right) = a\mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) \) . | 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 \( \mathop{\lim }\limits_{{x \rightarrow c}}0 = 0 \)... | Yes |
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... | 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 > ... | Yes |
Example 13.6 Find \( \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{\frac{1}{x} - 1}{1 - x} \) . | 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... | Yes |
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 \( \mathop{\lim }\limits_{{x \rightarrow c}}f\left( {g\left( x\right) }\right) = f\left( {\mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\right) }\right) \) . | 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 ... | Yes |
Theorem 13.10 If \( f \) is differentiable at \( c \), then \( f \) is continuous at \( c \) . | 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\... | Yes |
Investigate \( \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\sin \left( x\right) }{x} \) . | 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... | Yes |
Prove that the sequence \( \left\{ {1 - \frac{1}{n}}\right\} \) converges to 1 . | 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 sequ... | Yes |
Investigate the sequence \( \left\{ \frac{{\left( -1\right) }^{n + 1}\left( {n + 1}\right) }{n}\right\} \) . | 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( -... | Yes |
Prove that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{2}^{k}} = 1 \) . | 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}... | No |
Theorem 13.11 If \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k} \) converges, then the sequence \( \left\{ {a}_{n}\right\} \) converges to 0 . | 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 a... | Yes |
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 \( f\left( n\right) = - n \) . | 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 b... | No |
This example shows that \( \left| \mathbb{N}\right| = \left| \mathbb{Z}\right| \) . | 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>...<... | Yes |
Show that \( \left| \left( {0,\infty }\right) \right| = \left| \left( {0,1}\right) \right| \) . | 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}\ri... | No |
Show that \( \left| \mathbb{R}\right| = \left| \left( {0,1}\right) \right| \) . | 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| \... | Yes |
Theorem 14.5 If \( A \) and \( B \) are both countably infinite, then so is \( A \times B \) . | 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 infin... | Yes |
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. | 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 \( k \geq 2 \),... | Yes |
Theorem 14.6 If \( A \) and \( B \) are both countably infinite, then their union \( A \cup B \) is countably infinite. | 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 \ | No |
Theorem 14.8 An infinite subset of a countably infinite set is countably infinite. | 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, an... | Yes |
Theorem 14.9 If \( U \subseteq A \), and \( U \) is uncountable, then \( A \) is uncountable. | 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 infin... | Yes |
Example 14.6 The intervals \( \lbrack 0,1) \) and \( \left( {0,1}\right) \) in \( \mathbb{R} \) have equal cardinalities. | 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 \(... | Yes |
Proposition 1.2.17. \( - 1 < 0 \) | 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... | Yes |
Proposition 2.1.1. -1 has no real square root. | 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 ... | 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... | 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} + {... | 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 \( ... | 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 }... | 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... | 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... | 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... | 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) \;... | 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 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 ... | 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. H... | 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... | 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}... | 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 \( ... | 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... | 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( {\;\operator... | 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... | 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... | 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 =... | 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 \( {a}^{\prime } \) s... | 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 m... | 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... | No |
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