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Corollary 3.4.4 If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous, then it has an absolute minimum and an absolute maximum on \( \left\lbrack {a, b}\right\rbrack \) . | Corollary 3.4.4 is sometimes referred to as the Extreme Value Theorem. It follows immediately from Theorem 3.4.2, and the fact that the interval \( \left\lbrack {a, b}\right\rbrack \) is compact (see Example 2.6.4). | Yes |
Lemma 3.4.5 Let \( f : D \rightarrow \mathbb{R} \) be continuous at \( c \in D \) . Suppose \( f\left( c\right) > 0 \) . Then there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) > 0\text{ for every }x \in B\left( {c;\delta }\right) \cap D. \] | Proof: Let \( \varepsilon = f\left( c\right) > 0 \) . By the continuity of \( f \) at \( c \), there exists \( \delta > 0 \) such that if \( x \in D \) and \( \left| {x - c}\right| < \delta \), then\n\n\[ \left| {f\left( x\right) - f\left( c\right) }\right| < \varepsilon \]\n\nThis implies, in particular, that \( f\lef... | Yes |
Theorem 3.4.7 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Suppose \( f\left( a\right) \cdot f\left( b\right) < 0 \) (this means either \( f\left( a\right) < 0 < f\left( b\right) \) or \( f\left( a\right) > 0 > f\left( b\right) ) \) . Then there exists \( c \in \left( ... | Proof: We prove only the case \( f\left( a\right) < 0 < f\left( b\right) \) (the case \( f\left( a\right) > 0 > f\left( b\right) \) is completely analogous). Define\n\n\[ A = \{ x \in \left\lbrack {a, b}\right\rbrack : f\left( x\right) \leq 0\} . \]\n\nThis set is nonempty since \( a \in A \) . This set is also bounded... | Yes |
Theorem 3.4.8 — Intermediate Value Theorem. Let \( f \) : \( \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Suppose \( f\left( a\right) < \gamma < f\left( b\right) \) . Then there exists a number \( c \in \left( {a, b}\right) \) such that \( f\left( c\right) = \gamma \) . | Proof: Define\n\n\[ \varphi \left( x\right) = f\left( x\right) - \gamma, x \in \left\lbrack {a, b}\right\rbrack . \]\n\nThen \( \varphi \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) . Moreover,\n\n\[ \varphi \left( a\right) \varphi \left( b\right) = \left\lbrack {f\left( a\right) - \gamma }\right\rbrack \l... | Yes |
Theorem 3.4.10 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be strictly increasing and continuous on \( \left\lbrack {a, b}\right\rbrack \) . Let \( c = f\left( a\right) \) and \( d = f\left( b\right) \) . Then \( f \) is one-to-one, \( f\left( \left\lbrack {a, b}\right\rbrack \right) = \left\l... | Proof: The first two assertions follow from the monotonicity of \( f \) and the Intermediate Value Theorem (see also Corollary 3.4.9). We will prove that \( {f}^{-1} \) is continuous on \( \left\lbrack {c, d}\right\rbrack \) . Fix any \( \bar{y} \in \left\lbrack {c, d}\right\rbrack \) and fix any sequence \( \left\{ {y... | Yes |
Theorem 3.5.2 If a function \( f : D \rightarrow \mathbb{R} \) is Hölder continuous, then it is uniformly continuous. | Proof: Since \( f \) is Hölder continuous, there are constants \( \ell \geq 0 \) and \( \alpha > 0 \) such that\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell {\left| u - v\right| }^{\alpha }\text{ for every }u, v \in D.\]\n\nIf \( \ell = 0 \), then \( f \) is constant and, thus, uniformly continu... | Yes |
Theorem 3.5.3 Let \( D \) be a nonempty subset of \( \mathbb{R} \) and \( f : D \rightarrow \mathbb{R} \) . Then \( f \) is uniformly continuous on \( D \) if and only if the following condition holds\n\n(C) for every two sequences \( \left\{ {u}_{n}\right\} ,\left\{ {v}_{n}\right\} \) in \( D \) such that \( \mathop{\... | Proof: Suppose first that \( f \) is uniformly continuous and let \( \left\{ {u}_{n}\right\} ,\left\{ {v}_{n}\right\} \) be sequences in \( D \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{u}_{n} - {v}_{n}}\right) = 0 \) . Let \( \varepsilon > 0 \) . Choose \( \delta > 0 \) such that \( \left| ... | Yes |
Theorem 3.5.4 Let \( f : D \rightarrow \mathbb{R} \) be a continuous function. Suppose \( D \) is compact. Then \( f \) is uniformly continuous on \( D \) . | Proof: Suppose by contradiction that \( f \) is not uniformly continuous on \( D \) . Then there exists \( {\varepsilon }_{0} > 0 \) such that for any \( \delta > 0 \), there exist \( u, v \in D \) with\n\n\[ \left| {u - v}\right| < \delta \text{ and }\left| {f\left( u\right) - f\left( v\right) }\right| \geq {\varepsil... | Yes |
Theorem 3.5.5 Let \( a, b \in \mathbb{R} \) and \( a < b \) . A function \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is uniformly continuous if and only if \( f \) can be extended to a continuous function \( \widetilde{f} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) (that is, there is a contin... | Proof: Suppose first that there exists a continuous function \( \widetilde{f} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) such that \( f = {\widetilde{f}}_{\mid \left( {a, b}\right) } \) . By Theorem 3.5.4, the function \( \widetilde{f} \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack ... | Yes |
Theorem 3.6.1 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then \( \ell = \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) if and only if the following two conditions hold:\n\n(1) For every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\... | Proof: Suppose \( \ell = \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Then\n\n\[ \ell = \mathop{\inf }\limits_{{\delta > 0}}g\left( \delta \right) \]\n\nwhere \( g \) is defined in (3.10). For any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \ell \leq g\left( \delta \r... | Yes |
Corollary 3.6.2 Suppose \( \ell = \lim \mathop{\sup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Then there exists a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) such that \( \left\{ {x}_{k}\right\} \) converges to \( \bar{x},{x}_{k} \neq \bar{x} \) for every \( k \), and\n\n\[ \mathop{\lim }\limits_{{k... | Proof: For each \( k \in \mathbb{N} \), take \( {\varepsilon }_{k} = \frac{1}{k} \) . By (1) of Theorem 3.6.1, there exists \( {\delta }_{k} > 0 \) such that\n\n\[ f\left( x\right) < \ell + {\varepsilon }_{k}\text{ for all }x \in {B}_{0}\left( {\bar{x};{\delta }_{k}}\right) \cap D. \]\n\n(3.11)\n\nLet \( {\delta }_{k}^... | Yes |
Theorem 3.6.4 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then\n\n\[ \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \infty \]\n\nif and only if there exists a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) such that \( \left\{ {x}_{k}\right\} \) co... | Proof: Suppose \( \lim \mathop{\sup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \infty \) . Then\n\n\[ \mathop{\inf }\limits_{{\delta > 0}}g\left( \delta \right) = \infty \]\n\nwhere \( g \) is the extended real-valued function defined in (3.10). Thus, \( g\left( \delta \right) = \infty \) for every \( \delta ... | Yes |
Corollary 3.6.7 Suppose \( \ell = \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Then there exists a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) such that \( {x}_{k} \) converges to \( \bar{x},{x}_{k} \neq \bar{x} \) for every \( k \), and | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) = \ell \] Moreover, if \( \left\{ {y}_{k}\right\} \) is a sequence in \( D \) that converges to \( \bar{x},{y}_{k} \neq \bar{x} \) for every \( k \), and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {y}_{k}\right) = {\ell }^{\prime } ... | Yes |
Theorem 3.6.11 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \]\n\nif and only if\n\n\[ \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \mathop{\liminf }\limits_{{x \righta... | Proof: Suppose\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \]\n\nThen for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \ell - \varepsilon < f\left( x\right) < \ell + \varepsilon \text{ for all }x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D. \]\n\nSinc... | Yes |
Theorem 3.7.1 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \in D \) be a limit point of \( D \) . Then \( f \) is lower semicontinuous at \( \bar{x} \) if and only if\n\n\[ \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \geq f\left( \bar{x}\right) \] | Proof: Suppose \( f \) is lower semicontinuous at \( \bar{x} \) . Let \( \varepsilon > 0 \) . Then there exists \( {\delta }_{0} > 0 \) such that\n\n\[ f\left( \bar{x}\right) - \varepsilon < f\left( x\right) \text{ for all }x \in B\left( {\bar{x};{\delta }_{0}}\right) \cap D. \]\n\nThis implies\n\n\[ f\left( \bar{x}\ri... | Yes |
Theorem 3.7.2 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \in D \) . Then \( f \) is l.s.c. at \( \bar{x} \) if and only if for every sequence \( \left\{ {x}_{k}\right\} \) in \( D \) that converges to \( \bar{x} \) , \[ \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \geq f\left(... | Proof: Suppose \( f \) is l.s.c. at \( \bar{x} \) . Then for any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that (3.12) holds. Since \( \left\{ {x}_{k}\right\} \) converges to \( \bar{x} \), we have \( {x}_{k} \in B\left( {\bar{x};\delta }\right) \) when \( k \) is sufficiently large. Thus, \[ f\left( \b... | Yes |
Theorem 3.7.3 Suppose \( D \) is a compact set of \( \mathbb{R} \) and \( f : D \rightarrow \mathbb{R} \) is lower semicontinuous. Then \( f \) has an absolute minimum on \( D \) . That means there exists \( \bar{x} \in D \) such that\n\n\[ f\left( x\right) \geq f\left( \bar{x}\right) \text{ for all }x \in D. \] | Proof: We first prove that \( f \) is bounded below. Suppose by contradiction that for every \( k \in \mathbb{N} \), there exists \( {x}_{k} \in D \) such that\n\n\[ f\left( {x}_{k}\right) < - k.\]\n\nSince \( D \) is compact, there exists a subsequence \( \left\{ {x}_{{k}_{\ell }}\right\} \) of \( \left\{ {x}_{k}\righ... | Yes |
Theorem 3.7.5 Let \( f : D \rightarrow \mathbb{R} \) . Then \( f \) is lower semicontinuous if and only if \( {\mathcal{L}}_{a}\left( f\right) \) is closed in \( D \) for every \( a \in \mathbb{R} \) . Similarly, \( f \) is upper semicontinuous if and only if \( {\mathcal{U}}_{a}\left( f\right) \) is closed in \( D \) ... | Proof: Suppose \( f \) is lower semicontinuous. Using Corollary 2.6.10, we will prove that for every sequence \( \left\{ {x}_{k}\right\} \) in \( {\mathcal{L}}_{a}\left( f\right) \) that converges to a point \( \bar{x} \in D \), we get \( \bar{x} \in {\mathcal{L}}_{a}\left( f\right) \) . For every \( k \), since \( {x}... | Yes |
Theorem 3.7.7 Let \( f : D \rightarrow \mathbb{R} \) . Then \( f \) is continuous if and only if for every \( a, b \in \mathbb{R} \) with \( a < b \) , the set\n\n\[ \n{O}_{a, b} = \{ x \in D : a < f\left( x\right) < b\} = {f}^{-1}\left( \left( {a, b}\right) \right)\n\]\n\nis an open set in \( D \) . | Proof: Suppose \( f \) is continuous. Then \( f \) is lower semicontinuous and upper semicontinuos. Fix \( a, b \in \mathbb{R} \) with \( a < b \) . Then\n\n\[ \n{O}_{a, b} = {L}_{b} \cap {U}_{a}\n\]\n\nBy Theorem 3.7.6, the set \( {O}_{a, b} \) is open since it is the intersection of two open sets \( {L}_{a} \) and \(... | Yes |
Theorem 4.1.1 Let \( G \) be an open subset of \( \mathbb{R} \) and let \( f \) be defined on \( G \) . If \( f \) is differentiable at \( a \in G \), then \( f \) is continuous at this point. | Proof: We have the following identity for \( x \in G \smallsetminus \{ a\} \) :\n\n\[ f\left( x\right) = f\left( x\right) - f\left( a\right) + f\left( a\right) \]\n\n\[ = \frac{f\left( x\right) - f\left( a\right) }{x - a}\left( {x - a}\right) + f\left( a\right) . \]\n\nThus,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow ... | Yes |
Theorem 4.1.3 Let \( G \) be an open subset of \( \mathbb{R} \) and let \( f, g : G \rightarrow \mathbb{R} \) . Suppose both \( f \) and \( g \) are differentiable at \( a \in G \) . Then the following hold.\n\n(a) The function \( f + g \) is differentiable at \( a \) and\n\n\[{\left( f + g\right) }^{\prime }\left( a\r... | Proof: The proofs of (a) and (b) are straightforward and we leave them as exercises. Let us prove (c). For every \( x \in G \smallsetminus \{ a\} \), we can write\n\n\[ \frac{\left( {fg}\right) \left( x\right) - \left( {fg}\right) \left( a\right) }{x - a} = \frac{f\left( x\right) g\left( x\right) - f\left( a\right) g\l... | No |
Lemma 4.1.4 Let \( G \) be an open subset of \( \mathbb{R} \) and let \( f : G \rightarrow \mathbb{R} \) . Suppose \( f \) is differentiable at \( a \) . Then there exists a function \( u : G \rightarrow \mathbb{R} \) satisfying\n\n\[ f\left( x\right) - f\left( a\right) = \left\lbrack {{f}^{\prime }\left( a\right) + u\... | Proof: Define\n\n\[ u\left( x\right) = \left\{ \begin{array}{ll} \frac{f\left( x\right) - f\left( a\right) }{x - a} - {f}^{\prime }\left( a\right) , & x \in G \smallsetminus \{ a\} \\ 0, & x = a. \end{array}\right. \]\n\nSince \( f \) is differentiable at \( a \), we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}... | Yes |
Theorem 4.1.5 - Chain rule. Let \( f : {G}_{1} \rightarrow \mathbb{R} \) and let \( g : {G}_{2} \rightarrow \mathbb{R} \), where \( {G}_{1} \) and \( {G}_{2} \) are two open subsets of \( \mathbb{R} \) with \( f\left( {G}_{1}\right) \subset {G}_{2} \) . Suppose \( f \) is differentiable at \( a \) and \( g \) is differ... | Proof: Since \( f \) is differentiable at \( a \), by Lemma 4.1.4, there exists a function \( u \) defined on \( {G}_{1} \) with \[ f\left( x\right) - f\left( a\right) = \left\lbrack {{f}^{\prime }\left( a\right) + u\left( x\right) }\right\rbrack \left( {x - a}\right) \text{ for all }x \in {G}_{1}, \] and \( \mathop{\l... | Yes |
Theorem 4.2.1 — Fermat’s Rule. Let \( I \) be an open interval and \( f : I \rightarrow \mathbb{R} \) . If \( f \) has a local minimum or maximum at \( a \in I \) and \( f \) is differentiable at \( a \), then \( {f}^{\prime }\left( a\right) = 0 \) . | Proof: Suppose \( f \) has a local minimum at \( a \) . Then there exists \( \delta > 0 \) sufficiently small such that\n\n\[ f\left( x\right) \geq f\left( a\right) \text{ for all }x \in B\left( {a;\delta }\right) . \]\n\nSince \( {B}_{ + }\left( {a;\delta }\right) \) is a subset of \( B\left( {a;\delta }\right) \), we... | Yes |
Theorem 4.2.2 - Rolle’s Theorem. Let \( a, b \in \mathbb{R} \) with \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) . Suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) with \( f\left( a\right) = f\left( b\right) \) ... | Proof: Since \( f \) is continuous on the compact set \( \left\lbrack {a, b}\right\rbrack \), by the extreme value theorem (Theorem 3.4.2) there exist \( {\bar{x}}_{1} \in \left\lbrack {a, b}\right\rbrack \) and \( {\bar{x}}_{2} \in \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ \nf\left( {\bar{x}}_{1}\right) = \m... | Yes |
Theorem 4.2.3 — Mean Value Theorem. Let \( a, b \in \mathbb{R} \) with \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R}. \) Suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Then there exists \( c \in \left( {a, b}\ri... | Proof: The linear function whose graph goes through \( \left( {a, f\left( a\right) }\right) \) and \( \left( {b, f\left( b\right) }\right) \) is\n\n\[ \ng\left( x\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) + f\left( a\right) .\n\]\nDefine\n\n\[ \nh\left( x\right) = f\left( x\right)... | Yes |
Theorem 4.2.4 — Cauchy’s Theorem. Let \( a, b \in \mathbb{R} \) with \( a < b. \) Suppose \( f \) and \( g \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \left\lbrack {f\left( b\right) - f\... | Proof: Define\n\n\[ h\left( x\right) = \left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack g\left( x\right) - \left\lbrack {g\left( b\right) - g\left( a\right) }\right\rbrack f\left( x\right) \text{ for }x \in \left\lbrack {a, b}\right\rbrack . \]\n\nThen \( h\left( a\right) = f\left( b\right) g\left( a\ri... | Yes |
Theorem 4.2.5 — Intermediate Value Theorem for Derivatives. Let \( a, b \in \mathbb{R} \) with \( a < b. \) Suppose \( f \) is differentiable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ \n{f}_{ + }^{\prime }\left( a\right) < \lambda < {f}_{ - }^{\prime }\left( b\right) \n\]\n\nThen there exists \( c \in \left( ... | Proof: Define the function \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) by\n\n\[ \ng\left( x\right) = f\left( x\right) - {\lambda x} \n\]\n\nThen \( g \) is differentiable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ \n{g}_{ + }^{\prime }\left( a\right) < 0 < {g}_{ - }^{\prime }\left( b\righ... | Yes |
Proposition 4.3.1 Let \( f \) be continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . If \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \in \left( {a, b}\right) \), then \( f \) is constant on \( \left\lbrack {a, b}\right\rbrack \) . | Proof: Suppose by contradiction that \( f \) is not constant on \( \left\lbrack {a, b}\right\rbrack \) . Then there exist \( {a}_{1} \) and \( {b}_{1} \) such that \( a \leq {a}_{1} < {b}_{1} \leq b \) and \( f\left( {a}_{1}\right) \neq f\left( {b}_{1}\right) \) . By Theorem 4.2.3, there exists \( c \in \left( {{a}_{1}... | Yes |
Proposition 4.3.2 Let \( f \) be differentiable on \( \\left( {a, b}\\right) \) .\n\n(i) If \( {f}^{\\prime }\\left( x\\right) > 0 \) for all \( x \\in \\left( {a, b}\\right) \), then \( f \) is strictly increasing on \( \\left( {a, b}\\right) \) . | Proof: Let us prove (i). Fix any \( {x}_{1},{x}_{2} \\in \\left( {a, b}\\right) \) with \( {x}_{1} < {x}_{2} \) . By Theorem 4.2.3, there exists \( c \\in \\left( {{x}_{1},{x}_{2}}\\right) \) such that\n\n\[ \n\\frac{f\\left( {x}_{2}\\right) - f\\left( {x}_{1}\\right) }{{x}_{2} - {x}_{1}} = {f}^{\\prime }\\left( c\\rig... | Yes |
Theorem 4.3.3 — Inverse Function Theorem. Suppose \( f \) is differentiable on \( I = \\left( {a, b}\\right) \) and \( {f}^{\\prime }\\left( x\\right) \\neq 0 \) for all \( x \\in \\left( {a, b}\\right) \). Then \( f \) is one-to-one, \( f\\left( I\\right) \) is an open interval, and the inverse function \( {f}^{-1} : ... | Proof: It follows from Theorem 4.2.5 that\n\n\[{f}^{\\prime }\\left( x\\right) > 0\\text{for all}x \\in \\left( {a, b}\\right) \\text{, or}{f}^{\\prime }\\left( x\\right) < 0\\text{for all}x \\in \\left( {a, b}\\right) \\text{.}\]\n\nSuppose \( {f}^{\\prime }\\left( x\\right) > 0 \) for all \( x \\in \\left( {a, b}\\ri... | Yes |
Theorem 4.4.1 Suppose \( f \) and \( g \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Suppose \( f\left( \bar{x}\right) = g\left( \bar{x}\right) = 0 \), where \( \bar{x} \in \left\lbrack {a, b}\right\rbrack \) . Suppose further that there exists \( \delta ... | Proof: Let \( \left\{ {x}_{k}\right\} \) be a sequence in \( \left\lbrack {a, b}\right\rbrack \) that converges to \( \bar{x} \) and such that \( {x}_{k} \neq \bar{x} \) for every \( k \) . By Theorem 4.2.4, for each \( k \), there exists a sequence \( \left\{ {c}_{k}\right\} \), with \( {c}_{k} \) between \( {x}_{k} \... | Yes |
Theorem 4.4.5 Let \( f \) and \( g \) be differentiable on \( \left( {a,\infty }\right) \) . Suppose \( {g}^{\prime }\left( x\right) \neq 0 \) for all \( x \in \left( {a,\infty }\right) \) and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow \infty }}g\left( ... | Example 4.4.7 Consider the limit\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\ln x}{x} \]\n\nClearly the functions \( f\left( x\right) = \ln x \) and \( g\left( x\right) = x \) satisfy the conditions of Theorem 4.4.5. We have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{f}^{\prime }\left( ... | No |
Theorem 4.5.1 - Taylor’s Theorem. Let \( n \) be a positive integer. Suppose \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a function such that \( {f}^{\left( n\right) } \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), and \( {f}^{\left( n + 1\right) }\left( x\right) \) exists for all ... | Proof: Let \( \bar{x} \) be as in the statement and let us fix \( x \neq \bar{x} \) . Since \( x - \bar{x} \neq 0 \), there exists a number \( \lambda \in \mathbb{R} \) such that\n\n\[ f\left( x\right) = {P}_{n}\left( x\right) + \frac{\lambda }{\left( {n + 1}\right) !}{\left( x - \bar{x}\right) }^{n + 1}. \]\n\nWe will... | Yes |
Theorem 4.5.3 Let \( n \) be an even positive integer. Suppose \( {f}^{\left( n\right) } \) exists and continuous on \( \left( {a, b}\right) \) . Let \( \bar{x} \in \left( {a, b}\right) \) satisfy\n\n\[ \n{f}^{\prime }\left( \bar{x}\right) = \ldots = {f}^{\left( n - 1\right) }\left( \bar{x}\right) = 0\text{ and }{f}^{\... | Proof: We will prove (a). Suppose \( {f}^{\left( n\right) }\left( \bar{x}\right) > 0 \) . Since \( {f}^{\left( n\right) }\left( \bar{x}\right) > 0 \) and \( {f}^{\left( n\right) } \) is continuous at \( \bar{x} \), there exists \( \delta > 0 \) such that\n\n\[ \n{f}^{\left( n\right) }\left( t\right) > 0\text{ for all }... | Yes |
Theorem 4.6.1 Let \( I \) be an interval of \( \mathbb{R} \) . A function \( f : I \rightarrow \mathbb{R} \) is convex if and only if for every \( {\lambda }_{i} \geq 0, i = 1,\ldots, n \), with \( \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i} = 1\left( {n \geq 2}\right) \) and for every \( {x}_{i} \in I, i = 1,\ld... | Proof: Since the converse holds trivially, we only need to prove the implication by induction. The conclusion holds for \( n = 2 \) by the definition of convexity. Let \( k \) be such that the conclusion holds for any \( n \) with \( 2 \leq n \leq k \) . We will show that it also holds for \( n = k + 1 \) . Fix \( {\la... | Yes |
Theorem 4.6.2 Let \( I \) be an interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Then \( f \) has a local minimum at \( \bar{x} \) if and only if \( f \) has an absolute minimum at \( \bar{x} \) . | Proof: Clearly if \( f \) has a global minimum at \( \bar{x} \), then it also has a local minimum at \( \bar{x} \). Conversely, suppose that \( f \) has a local minimum at \( \bar{x} \). Then there exists \( \delta > 0 \) such that \[ f\left( u\right) \geq f\left( \bar{x}\right) \text{ for all }u \in B\left( {\bar{x};\... | Yes |
Theorem 4.6.3 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Suppose \( f \) is differentiable at \( \bar{x} \) . Then\n\n\[ \n{f}^{\prime }\left( \bar{x}\right) \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \in I.\n\] | Proof: For any \( x \in I \) and \( t \in \left( {0,1}\right) \), we have\n\n\[ \n\frac{f\left( {\bar{x} + t\left( {x - \bar{x}}\right) }\right) - f\left( \bar{x}\right) }{t} = \frac{f\left( {{tx} + \left( {1 - t}\right) \bar{x}}\right) - f\left( \bar{x}\right) }{t}\n\]\n\n\[ \n\leq \frac{{tf}\left( x\right) + \left( {... | Yes |
Corollary 4.6.4 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Suppose \( f \) is differentiable at \( \bar{x} \) . Then \( f \) has an absolute minimum at \( \bar{x} \) if and only if \( {f}^{\prime }\left( \bar{x}\right) = 0 \) . | Proof: Suppose \( f \) has an absolute minimum at \( \bar{x} \) . By Theorem 4.2.1, \( {f}^{\prime }\left( \bar{x}\right) = 0 \) . Let us prove the converse. Suppose \( {f}^{\prime }\left( \bar{x}\right) = 0 \) . It follows from Theorem 4.6.3 that\n\n\[ 0 = {f}^{\prime }\left( \bar{x}\right) \left( {x - \bar{x}}\right)... | Yes |
Lemma 4.6.5 Let \( I \) be an open interval and suppose \( f : I \rightarrow \mathbb{R} \) is a convex function. Fix \( a, b, x \in I \) with \( a < x < b \) . Then\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq \frac{f\left( b\right) - f\left( x\ri... | Proof: Let\n\n\[ t = \frac{x - a}{b - a} \]\n\nThen \( t \in \left( {0,1}\right) \) and\n\n\[ f\left( x\right) = f\left( {a + \left( {x - a}\right) }\right) = f\left( {a + \frac{x - a}{b - a}\left( {b - a}\right) }\right) = f\left( {a + t\left( {b - a}\right) }\right) = f\left( {{tb} + \left( {1 - t}\right) a}\right) .... | Yes |
Theorem 4.6.6 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a differentiable function. Then \( f \) is convex if and only if \( {f}^{\prime } \) is increasing on \( I \) . | Proof: Suppose \( f \) is convex. Fix \( a < b \) with \( a, b \in I \) . By Lemma 4.6.5, for any \( x \in \left( {a, b}\right) \), we have\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq \frac{f\left( b\right) - f\left( a\right) }{b - a}. \]\n\nThis implies, taking limits, that\n\n\[ {f}^{\prime }\left( ... | Yes |
Corollary 4.6.7 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a function. Suppose \( f \) is twice differentiable on \( I \) . Then \( f \) is convex if and only if \( {f}^{\prime \prime }\left( x\right) \geq 0 \) for all \( x \in I \) . | Proof: It follows from Proposition 4.3.2 that \( {f}^{\prime \prime }\left( x\right) \geq 0 \) for all \( x \in I \) if and only if the derivative function \( {f}^{\prime } \) is increasing on \( I \) . The conclusion then follows directly from Theorem 4.6.6. \( ▱ \) | Yes |
Theorem 4.6.8 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Then it is locally Lipschitz continuous in the sense that for any \( \bar{x} \in I \), there exist \( \ell \geq 0 \) and \( \delta > 0 \) such that\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \l... | Proof: Fix any \( \bar{x} \in I \) . Choose four numbers \( a, b, c, d \) satisfying\n\n\[ a < b < \bar{x} < c < d\text{with}a, d \in I\text{.} \]\n\nChoose \( \delta > 0 \) such that \( B\left( {\bar{x};\delta }\right) \subset \left( {b, c}\right) \) . Let \( u, v \in B\left( {\bar{x};\delta }\right) \) with \( v < u ... | Yes |
Lemma 4.7.1 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function. Fix \( a \in \mathbb{R} \) . Define the slope function \( {\phi }_{a} \) by\n\n\[ \n{\phi }_{a}\left( x\right) = \frac{f\left( x\right) - f\left( a\right) }{x - a} \]\n\n(4.17)\n\nfor \( x \in \left( {-\infty, a}\right) \cup \left( {a,\in... | Proof: This lemma follows directly from Lemma 4.6.5. \( ▱ \) | No |
Theorem 4.7.2 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and let \( \bar{x} \in \mathbb{R} \) . Then \( f \) has left derivative and right derivative at \( \bar{x} \) . Moreover, \[ \mathop{\sup }\limits_{{x < \bar{x}}}{\phi }_{\bar{x}}\left( x\right) = {f}_{ - }^{\prime }\left( \bar{x}\right)... | Proof: By Lemma 4.7.1, the slope function \( {\phi }_{\bar{x}} \) defined by (4.17) is increasing on the interval \( \left( {\bar{x},\infty }\right) \) and bounded below by \( {\phi }_{\bar{x}}\left( {\bar{x} - 1}\right) \) . By Theorem 3.2.4, the limit \[ \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}{\phi }_... | Yes |
Theorem 4.7.3 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and let \( \bar{x} \in \mathbb{R} \) . Then\n\n\[ \partial f\left( \bar{x}\right) = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack . \] | Proof: Suppose \( u \in \partial f\left( \bar{x}\right) \) . By the definition (4.16), we have\n\n\[ u \cdot \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x > \bar{x}. \]\n\nThis implies\n\n\[ u \leq \frac{f\left( x\right) - f\left( \bar{x}\right) }{x - \bar{x}}\text{ for al... | Yes |
Corollary 4.7.4 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and \( \bar{x} \in \mathbb{R} \) . Then \( f \) is differentiable at \( \bar{x} \) if and only if \( \partial f\left( \bar{x}\right) \) is a singleton. In this case, \[ \partial f\left( \bar{x}\right) = \left\{ {{f}^{\prime }\left( \ba... | Proof: Suppose \( f \) is differentiable at \( \bar{x} \) . Then \[ {f}_{ - }^{\prime }\left( \bar{x}\right) = {f}_{ + }^{\prime }\left( \bar{x}\right) = {f}^{\prime }\left( \bar{x}\right) . \] By Theorem 4.7.3, \[ \partial f\left( \bar{x}\right) = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\pri... | Yes |
Theorem 4.7.5 Let \( f, g : \mathbb{R} \rightarrow \mathbb{R} \) be convex functions and let \( \alpha > 0 \) . Then \( f + g \) and \( {\alpha f} \) are convex functions and\n\n\[ \partial \left( {f + g}\right) \left( \bar{x}\right) = \partial f\left( \bar{x}\right) + \partial g\left( \bar{x}\right) \]\n\n\[ \partial ... | Proof: It is not hard to see that \( f + g \) is a convex function and\n\n\[ {\left( f + g\right) }_{ + }^{\prime }\left( \bar{x}\right) = {f}_{ + }^{\prime }\left( \bar{x}\right) + {g}_{ + }^{\prime }\left( \bar{x}\right) \]\n\n\[ {\left( f + g\right) }_{ - }^{\prime }\left( \bar{x}\right) = {f}_{ - }^{\prime }\left( ... | Yes |
Theorem 4.7.6 Let \( {f}_{i} : \mathbb{R} \rightarrow \mathbb{R}, i = 1,\ldots, n \), be convex functions. Define\n\n\[ f\left( x\right) = \max \left\{ {{f}_{i}\left( x\right) : i = 1,\ldots, n}\right\} \text{and}I\left( u\right) = \left\{ {i = 1,\ldots, n : {f}_{i}\left( u\right) = f\left( u\right) }\right\} \text{.} ... | Proof: Fix \( u, v \in \mathbb{R} \) and \( \lambda \in \left( {0,1}\right) \) . For any \( i = 1,\ldots, n \), we have\n\n\[ {f}_{i}\left( {{\lambda u} + \left( {1 - \lambda }\right) v}\right) \leq \lambda {f}_{i}\left( u\right) + \left( {1 - \lambda }\right) {f}_{i}\left( v\right) \leq {\lambda f}\left( u\right) + \l... | Yes |
Theorem 4.7.8 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function. Then \( f \) has an absolute minimum at \( \bar{x} \) if and only if\n\n\[ 0 \in \partial f\left( \bar{x}\right) = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack . \]\n | Proof: Suppose \( f \) has an absolute minimum at \( \bar{x} \) . Then\n\n\[ f\left( \bar{x}\right) \leq f\left( x\right) \text{ for all }x \in \mathbb{R}. \]\n\nThis implies\n\n\[ 0 \cdot \left( {x - \bar{x}}\right) = 0 \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \in \mathbb{R}. \]\n\nIt follows f... | Yes |
Theorem 4.7.9 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and let \( a < b \) . Then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} \in \partial f\left( c\right) . \] | Proof: Define\n\n\[ g\left( x\right) = f\left( x\right) - \left\lbrack {\frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) + f\left( a\right) }\right\rbrack . \]\n\nThen \( g \) is a convex function and \( g\left( a\right) = g\left( b\right) \) . Thus, \( g \) has a local minimum at some \( c \in \... | Yes |
Corollary 4.7.10 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function. Then \( f \) is Lipschitz continuous if and only if there exists \( \ell \geq 0 \) such that\n\n\[ \partial f\left( x\right) \subset \left\lbrack {-\ell ,\ell }\right\rbrack \text{for all}x \in \mathbb{R}\text{.} \]\n | Proof: Suppose \( f \) is Lipschitz continuous on \( \mathbb{R} \) . Then there exists \( \ell \geq 0 \) such that\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell \left| {u - v}\right| \text{ for all }u, v \in \mathbb{R}. \]\n\nThen for any \( x \in \mathbb{R} \) ,\n\n\[ {f}_{ + }^{\prime }\left( x... | Yes |
Example 1.1 Here are some further illustrations of set-builder notation. | 1. \( \{ n : n \) is a prime number \( \} = \{ 2,3,5,7,{11},{13},{17},\ldots \} \)\n\n2. \( \{ n \in \mathbb{N} : n \) is prime \( \} = \{ 2,3,5,7,{11},{13},{17},\ldots \} \)\n\n3. \( \left\{ {{n}^{2} : n \in \mathbb{Z}}\right\} = \{ 0,1,4,9,{16},{25},\ldots \} \)\n\n4. \( \left\{ {x \in \mathbb{R} : {x}^{2} - 2 = 0}\r... | Yes |
Example 1.2 Describe the set \( A = \{ {7a} + {3b} : a, b \in \mathbb{Z}\} \) . | Solution: This set contains all numbers of form \( {7a} + {3b} \), where \( a \) and \( b \) are integers. Each such number \( {7a} + {3b} \) is an integer, so \( A \) contains only integers. But which integers? If \( n \) is any integer, then \( n = {7n} + 3\left( {-{2n}}\right) \), so \( n = {7a} + {3b} \) where \( a... | Yes |
This brings us to a significant fact: If \( B \) is any set whatsoever, then \( \varnothing \subseteq B \) . To see why this is true, look at the last sentence of Definition 1.3. It says that \( \varnothing \notin B \) would mean that there is at least one element of \( \varnothing \) that is not an element of \( B \) ... | Fact 1.2 The empty set is a subset of all sets, that is, \( \varphi \subseteq B \) for any set \( B \) . Here is another way to look at it. Imagine a subset of \( B \) as a thing you make by starting with braces \( \{ \} \), then filling them with selections from \( B \) . For example, to make one particular subset of ... | Yes |
1. \( 1 \in \{ 1,\{ 1\} \} \) . | 1 is the first element listed in \( \{ 1,\{ 1\} \} \) | Yes |
Example 1.7 You should examine the following statements and make sure you understand how the answers were obtained. In particular, notice that in each instance the equation \( \left| {\mathcal{P}\left( A\right) }\right| = {2}^{\left| A\right| } \) is true. | 1. \( \mathcal{P}\left( {\{ 0,1,3\} }\right) = \{ \varnothing ,\{ 0\} ,\{ 1\} ,\{ 3\} ,\{ 0,1\} ,\{ 0,3\} ,\{ 1,3\} ,\{ 0,1,3\} \} \)\n\n2. \( \mathcal{P}\left( {\{ 1,2\} }\right) = \{ \varnothing ,\{ 1\} ,\{ 2\} ,\{ 1,2\} \} \)\n\n3. \( \mathcal{P}\left( {\{ 1\} }\right) = \{ \varnothing ,\{ 1\} \} \)\n\n4. \( \mathca... | Yes |
Example 1.8 Suppose \( A = \{ a, b, c, d, e\}, B = \{ d, e, f\} \) and \( C = \{ 1,2,3\} \). | 1. \( A \cup B = \{ a, b, c, d, e, f\} \)\n2. \( A \cap B = \{ d, e\} \)\n3. \( A - B = \{ a, b, c\} \)\n4. \( B - A = \{ f\} \)\n5. \( \;\left( {A - B}\right) \cup \left( {B - A}\right) = \{ a, b, c, f\} \)\n6. \( A \cup C = \{ a, b, c, d, e,1,2,3\} \)\n7. \( A \cap C = \varnothing \)\n8. \( A - C = \{ a, b, c, d, e\}... | Yes |
Example 1.12 Suppose \( {A}_{1} = \{ 0,2,5\} ,{A}_{2} = \{ 1,2,5\} \) and \( {A}_{3} = \{ 2,5,7\} \) . Then | \[ \mathop{\bigcup }\limits_{{i = 1}}^{3}{A}_{i} = {A}_{1} \cup {A}_{2} \cup {A}_{3} = \{ 0,1,2,5,7\} \;\text{ and }\;\mathop{\bigcap }\limits_{{i = 1}}^{3}{A}_{i} = {A}_{1} \cap {A}_{2} \cap {A}_{3} = \{ 2,5\} . \] | Yes |
This example involves the following infinite list of sets. \( \begin{matrix} {A}_{1} = \{ - 1,0,1\} , & {A}_{2} = \{ - 2,0,2\} , & {A}_{3} = \{ - 3,0,3\} , & \cdots , & {A}_{i} = \{ - i,0, i\} , & \cdots \end{matrix} \) | Observe that \( \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} = \mathbb{Z} \), and \( \mathop{\bigcap }\limits_{{i = 1}}^{\infty }{A}_{i} = \{ 0\} \) | Yes |
In this example, all sets \( {A}_{\alpha } \) are all subsets of the plane \( {\mathbb{R}}^{2} \) . Each \( \alpha \) belongs to the index set \( I = \left\lbrack {0,2}\right\rbrack = \{ x \in \mathbb{R} : 0 \leq x \leq 2\} \), which is the set of all real numbers between 0 and 2 . For each number \( \alpha \in I \), d... | Now consider the infinite union \( \mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha } \) . It is the shaded triangle shown below, because any point \( \left( {x, y}\right) \) on this triangle belongs to the set \( {A}_{x} \), and is therefore in the union. (And any point not on the triangle is not in any \( {A}_{x}... | Yes |
Example 1.15 Here our sets are indexed by \( {\mathbb{R}}^{2} \) . For any \( \left( {a, b}\right) \in {\mathbb{R}}^{2} \), let \( {P}_{\left( a, b\right) } \) be the following subset of \( {\mathbb{R}}^{3} \) :\n\n\[ \n{P}_{\left( a, b\right) } = \left\{ {\left( {x, y, z}\right) \in {\mathbb{R}}^{3} : {ax} + {by} = 0}... | For any point \( \left( {a, b}\right) \in {\mathbb{R}}^{2} \) with \( \left( {a, b}\right) \neq \left( {0,0}\right) \), we can visualize \( {P}_{\left( a, b\right) } \) as the vertical plane that cuts the \( {xy} \) -plane at the line \( {ax} + {by} = 0 \) . Figure 1.11 (right) shows a few of the \( {P}_{\left( a, b\ri... | Yes |
Every integer that is not odd is even. | \( \forall n \in \mathbb{Z}, \sim \left( {n\text{ is odd }}\right) \Rightarrow \left( {n\text{ is even }}\right) ,\; \) or \( \;\forall n \in \mathbb{Z}, \sim O\left( n\right) \Rightarrow E\left( n\right) . \) | Yes |
Every integer is even. | \( \forall n \in \mathbb{Z}, E\left( n\right) \) | No |
\[ \forall x \in \mathbb{R},\exists y \in \mathbb{R},{y}^{3} = x. \] | This statement is true, for no matter what number \( x \) is there exists a number \( y = \sqrt[3]{x} \) for which \( {y}^{3} = x \) . | Yes |
Example 2.8 Consider the Mean Value Theorem from Calculus:\n\nIf \( f \) is continuous on the interval \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \), then there is a number \( c \in \left( {a, b}\right) \) for which \( {f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\l... | Here is a translation to symbolic form:\n\n\( \left( {\left( {f\text{ cont. on }\left\lbrack {a, b}\right\rbrack }\right) \land \left( {f\text{ is diff. on }\left( {a, b}\right) }\right) }\right) \Rightarrow \left( {\exists \;c \in \left( {a, b}\right) ,{f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\... | Yes |
Example 2.10 Consider negating the following statement.\n\n\( R \) : You can solve it by factoring or with the quadratic formula. | Now, \( R \) means (You can solve it by factoring) \( \vee \) (You can solve it with Q.F.), which we will denote as \( P \vee Q \) . The negation of this is\n\n\[ \sim \left( {P \vee Q}\right) = \left( { \sim P}\right) \land \left( { \sim Q}\right) . \]\n\nTherefore, in words, the negation of \( R \) is\n\n\( \sim R \)... | Yes |
Example 2.11 We will negate the following sentence.\n\n\( R \) : The numbers \( x \) and \( y \) are both odd. | This statement means \( \left( {x\text{is odd}}\right) \land \left( {y\text{is odd}}\right) \), so its negation is\n\n\[ \sim \left( {\left( {x\text{ is odd}}\right) \land \left( {y\text{ is odd}}\right) }\right) \; = \; \sim \left( {x\text{ is odd}}\right) \vee \sim \left( {y\text{ is odd}}\right) \]\n\n\[ = \;\left( ... | Yes |
Example 2.13 If a statement has multiple quantifiers, negating it involves several iterations of Equations (2.8) and (2.9). Consider the following:\n\n\( S \) : For every real number \( x \) there is a real number \( y \) for which \( {y}^{3} = x \) . | This statement asserts any real number \( x \) has a cube root \( y \), so it’s true. Symbolically \( S \) can be expressed as\n\n\[ \forall x \in \mathbb{R},\exists y \in \mathbb{R},{y}^{3} = x. \]\n\nLet's work out the negation of this statement.\n\n\[ \sim \left( {\forall x \in \mathbb{R},\exists y \in \mathbb{R},{y... | Yes |
Example 2.14 Negate the following statement about a particular (i.e., constant) number \( a \) .\n\n\( R \) : If \( a \) is odd then \( {a}^{2} \) is odd. | Using Equation (2.10), we get the following negation.\n\n\( \sim R : a \) is odd and \( {a}^{2} \) is not odd. | Yes |
Example 2.15 This example is like the previous one, but the constant \( a \) is replaced by a variable \( x \) . We will negate the following statement.\n\n\( R \) : If \( x \) is odd then \( {x}^{2} \) is odd. | As in Section 2.8, we interpret this as the universally quantified statement\n\n\[ R : \forall x \in \mathbb{Z},\left( {x\text{ odd }}\right) \Rightarrow \left( {{x}^{2}\text{ odd }}\right) .\n\]\n\nBy Equations (2.8) and (2.10), we get the following negation for \( R \) .\n\n\[ \sim \left( {\forall x \in \mathbb{Z},\l... | Yes |
In ordering a café latte, you have a choice of whole, skim or soy milk; small, medium or large; and either one or two shots of espresso. How many choices do you have in ordering one drink? | Solution: Your choice is modeled by a list of form (milk, size, shots). There are 3 choices for the first entry, 3 for the second and 2 for the third. By the multiplication principle, the number of choices is \( 3 \cdot 3 \cdot 2 = {18} \). | Yes |
Example 3.3 Consider lists of length 4 made with symbols \( A, B, C, D, E, F, G \) .\n\n(a) How many such lists are possible if repetition is allowed? | (a) Imagine the list as containing four boxes that we fill with selections from the letters \( A, B, C, D, E, F \) and \( G \), as illustrated below.\n\n\n\nWe have 7 choices in filling each box. The multiplication pri... | Yes |
Example 3.4 A non-repetitive list of length 5 is to be made from the symbols \( A, B, C, D, E, F, G \) . The first entry must be either a \( B, C \) or \( D \), and the last entry must be a vowel. How many such lists are possible? | Solution: Start by making a list of five boxes. The first box must contain either \( B, C \) or \( D \), so there are three choices for it.\n\n\n\nNow there are 6 letters left for the remaining 4 boxes. The knee-jerk a... | Yes |
Example 3.5 How many length-4 non-repetitive lists can be made from the symbols \( A, B, C, D, E, F, G \), if the list must contain an \( E \) ? | In Example 3.3 (c) our approach was to divide these lists into four types, depending on whether the \( E \) is in the first, second, third or fourth position. Then we used the multiplication principle to count the lists of type 1. There are 6 choices for the second entry, 5 for the third, and 4 for the fourth. This is ... | Yes |
Example 3.6 How many even 5-digit numbers are there for which no digit is 0, and the digit 6 appears exactly once? For instance, 55634 and 16118 are such numbers, but not 63304 (has a 0), nor 63364 (too many 6’s), nor 55637 (not even). | Solution: Let \( X \) be the set of all such numbers. The answer will be \( \left| X\right| \), so our task is to find \( \left| X\right| \) . Put \( X = {X}_{1} \cup {X}_{2} \cup {X}_{3} \cup {X}_{4} \cup {X}_{5} \), where \( {X}_{i} \) is the set of those numbers in \( X \) whose \( i \) th digit is 6, as diagramed b... | Yes |
Example 3.7 How many length-4 lists can be made from the symbols \( A, B, C, D, E, F, G \) if the list has at least one \( E \), and repetition is allowed? | Solution: Such a list might contain one, two, three or four \( E \) ’s, which could occur in various positions. This is a fairly complex situation.\n\nBut it is very easy to count the set \( U \) of all lists of length 4 made from \( A, B, C, D, E, F, G \) if we don’t care whether or not the lists have any \( E \) ’s. ... | Yes |
How many such lists are there if repetition is not allowed? | To answer the first question, note that there are seven letters, so the number of lists is \( 7! = \mathbf{5040} \) . | Yes |
Example 3.9 Ten contestants run a marathon. All finish, and there are no ties. How many different possible rankings are there for first-, second-and third-place? | Solution: Call the contestants \( A, B, C, D, E, F, G, H, I \) and \( J \) . A ranking of winners can be regarded as a 3-permutation of the set of 10 contestants. For example, \( E{CH} \) means \( E \) in first-place, \( C \) in second-place and \( H \) in third. Thus there are \( P\left( {{10},3}\right) = {10} \cdot 9... | Yes |
Example 3.10 You deal five cards off of a standard 52-card deck, and line them up in a row. How many such lineups are there that either consist of all red cards, or all clubs? | Solution: There are 26 red cards. The number of ways to line up five of them is \( P\left( {{26},5}\right) = {26} \cdot {25} \cdot {24} \cdot {23} \cdot {22} = 7,{893},{600} \) . \n\nThere are 13 club cards (which are black). The number of ways to line up five of them is \( P\left( {{13},5}\right) = {13} \cdot {12} \cd... | Yes |
Example 3.11 How many size-4 subsets does \( \{ 1,2,3,4,5,6,7,8,9\} \) have? | The answer is \( \begin{aligned} \left( \begin{aligned} 9 \\ 4 \end{aligned}\right) = \frac{9!}{4!\left( {9 - 4}\right) !} & = \frac{9!}{4!5!} = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{4!5!} \\ & = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{24} = \mathbf{{126}}. \end{aligned} \) | Yes |
Example 3.13 A single 5-card hand is dealt off of a standard 52-card deck. How many different 5-card hands are possible? | Solution: Think of the deck as a set \( D \) of 52 cards. Then a 5-card hand is just a 5-element subset of \( D \) . There are many such subsets, such as \n\nThus the number of 5-card hands is the number of 5-element s... | Yes |
Example 3.14 This problem concerns 5-card hands that can be dealt off of a 52-card deck. How many such hands are there in which two of the cards are clubs and three are hearts? | Solution: Such a hand is described by a list of length two of the form\n\n\[ \left( {\left\{ {\left\lbrack \begin{matrix} * \\ * \\ * \end{matrix}\right\rbrack ,\left\lbrack \begin{matrix} * \\ * \\ * \end{matrix}\right\rbrack }\right\} ,\left\{ {\left\lbrack \begin{matrix} * \\ \heartsuit \end{matrix}\right\rbrack ,\l... | Yes |
Example 3.15 A lottery features a bucket of 36 balls numbered 1 through 36. Six balls will be drawn randomly. For \$1 you buy a ticket with six blanks: \( ▱▱▱▱▱▱▱ \) . You fill in the blanks with six different numbers between 1 and 36. You win \( \$ 1,{000},{000} \) if you chose the same numbers that are drawn, regardl... | Solution: In filling out the ticket you are choosing six numbers from a set of 36 numbers. Thus there are \( \left( \begin{matrix} {36} \\ 6 \end{matrix}\right) = \frac{{36}!}{6!\left( {{36} - 6}\right) !} = 1,{947},{792} \) different combinations of numbers you might write. Only one of these will be a winner. Your cha... | Yes |
Example 3.16 How many 7-digit binary strings (0010100,1101011, etc.) have an odd number of 1’s? | Solution: Let \( A \) be the set of all 7-digit binary strings with an odd number of 1’s, so the answer will be \( \left| A\right| \) . To find \( \left| A\right| \), we break \( A \) into smaller parts. Notice any string in \( A \) will have either one, three, five or seven 1’s. Let \( {A}_{1} \) be the set of 7-digit... | Yes |
Theorem 3.1 (Binomial Theorem) If \( n \) is a non-negative integer, then \( {\left( x + y\right) }^{n} = \left( \begin{aligned} n \\ 0 \end{aligned}\right) {x}^{n} + \left( \begin{aligned} n \\ 1 \end{aligned}\right) {x}^{n - 1}y + \left( \begin{aligned} n \\ 2 \end{aligned}\right) {x}^{n - 2}{y}^{2} + \left( \begin{a... | For now we will be content to accept the binomial theorem without proof. (You will be asked to prove it in an exercise in Chapter 10.) | No |
Example 3.17 A 3-card hand is dealt off of a standard 52-card deck. How many different such hands are there for which all three cards are red or all three cards are face cards? | Solution: Let \( A \) be the set of 3-card hands where all three cards are red (i.e., either \( \circ \) or \( \diamondsuit \) ). Let \( B \) be the set of 3-card hands in which all three cards are face cards (i.e., \( J, K \) or \( Q \) of any suit). These sets are illustrated below.\n\n\[ A = \left\{ {\left\{ {\left\... | Yes |
Example 3.18 A 3-card hand is dealt off of a standard 52-card deck. How many different such hands are there for which it is not the case that all 3 cards are red or all three cards are face cards? | Solution: We will use the subtraction principle combined with our answer to Example 3.17, above. The total number of 3-card hands is \( \left( \begin{matrix} {52} \\ 3 \end{matrix}\right) = \frac{{52}!}{3!\left( {{52} - 3}\right) !} = \) \( \frac{{52}!}{3!{49}!} = \frac{{52} \cdot {51} \cdot {50}}{3!} = {26} \cdot {17}... | Yes |
A bag contains 20 identical red marbles, 20 identical green marbles, and 20 identical blue marbles. You reach in and grab 20 marbles. There are many possible outcomes. You could have 11 reds, 4 greens and 5 blues. Or you could have 20 reds, 0 greens and 0 blues, etc. All together, how many outcomes are possible? | Each outcome can be thought of as a 20-element multiset made from the elements of the 3-element set \( X = \{ \mathrm{R},\mathrm{G},\mathrm{B}\} \) . For example,11 reds,4 greens and 5 blues would correspond to the multiset\n\n\[ \left\lbrack {R, R, R, R, R, R, R, R, R, R, R, G, G, G, G, B, B, B, B, B}\right\rbrack . \... | Yes |
Example 3.20 How many non-negative integer solutions does the equation \( w + x + y + z = {20} \) have? | Solution: We can model a solution with stars and bars. For example, encode the solution \( \left( {w, x, y, z}\right) = \left( {3,4,5,8}\right) \) as\n\n\n\nIn general, any solution \( \left( {w, x, y, z}\right) = \l... | Yes |
Example 3.21 This problem concerns the lists \( \left( {w, x, y, z}\right) \) of integers with the property that \( 0 \leq w \leq x \leq y \leq z \leq {10} \) . That is, each entry is an integer between 0 and 10, and the entries are ordered from smallest to largest. For example, \( \left( {0,3,3,7}\right) ,\left( {1,1,... | Solution: We can encode such a list with 10 stars and 4 bars, where \( w \) is the number of stars to the left of the first bar, \( x \) is the number of stars to the left of the second bar, \( y \) is the number of stars to the left of the third bar, and \( z \) is the number of stars to the left of the fourth bar.\n\... | Yes |
Example 3.22 Count the permutations of the letters in MISSISSIPPI. | Solution: Think of this word as an 11-element multiset with one M, four I’s, four S’s and two P’s. By Fact 3.8, it has \( \frac{{11}!}{1!4!4!2!} = {34},{650} \) permutations. | Yes |
Example 3.23 Determine the number of permutations of the multiset \( \left\lbrack {1,1,1,1,5,5,{10},{25},{25}}\right\rbrack \) . | Solution: By Fact 3.8 the answer is \( \frac{9!}{4!2!1!2!} = {3780} \) . | Yes |
Example 3.24 Pick six integers between 0 and 9 (inclusive). Show that two of them must add up to 9. | Solution: Pick six numbers between 0 and 9. Here's why two of them sum to 9: Imagine five boxes, each marked with two numbers, as shown below. Each box is labeled so that the two numbers written on it sum to 9.\n\n\n... | No |
A store has a gumball machine containing a large number of red, green, blue and white gumballs. You get one gumball for each nickel you put into the machine. The store offers the following deal: You agree to buy some number of gumballs, and if 13 or more of them have the same color you get $5. What is the fewest number... | Let \( n \) be the number of gumballs that you buy. Imagine sorting your \( n \) gumballs into four boxes labeled RED, GREEN, BLUE, and WHITE. (That is, red balls go in the red box, green balls go in the green box, etc.)\n\nThe division principle says that one box contains \( \lceil \frac{n}{4}\rceil \) or more gumball... | Yes |
Example 3.26 Nine points are randomly placed on the right triangle shown below. Show that three of these points form a triangle whose area is \( \frac{1}{8} \) square unit or less. (We allow triangles with zero area, in which case the three points lie on a line.) | Solution: Divide the triangle into four smaller triangles, as indicated by the dashed lines below. Each of these four triangles has an area of \( \frac{1}{2}{bh} = \frac{1}{2}\frac{1}{2}\frac{1}{2} = \frac{1}{8} \) square units. Think of these smaller triangles as \ | No |
Example 3.27 Use combinatorial proof to show \( \left( \begin{array}{l} n \\ k \end{array}\right) = \left( \begin{matrix} n \\ n - k \end{matrix}\right) \) . | Solution: First, by definition, if \( k < 0 \) or \( k > n \), then both sides are 0, and thus equal. Therefore for the rest of the proof we can assume \( 0 \leq k \leq n \). The left-hand side \( \left( \begin{array}{l} n \\ k \end{array}\right) \) is the number of \( k \)-element subsets of \( S = \{ 1,2,\ldots, n\} ... | Yes |
Example 3.28 Use a combinatorial proof to show that \( \mathop{\sum }\limits_{{k = 0}}^{n}{\left( \begin{array}{l} n \\ k \end{array}\right) }^{2} = \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) . | Solution: First, the right-hand side \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) is the number of ways to select \( n \) things from a set \( S \) that has \( {2n} \) elements.\n\nNow let’s count this a different way. Divide \( S \) into two equal-sized parts, \( S = A \cup B \), where \( \left| A\right| ... | Yes |
Proposition Let \( a, b \in \mathbb{Z} \) and \( n \in \mathbb{N} \) . If \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \), then \( {a}^{2} \equiv {b}^{2}\left( {\;\operatorname{mod}\;n}\right) \) . | Proof. We will use direct proof. Suppose \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \).\n\nBy definition of congruence of integers, this means \( n \mid \left( {a - b}\right) \).\n\nThen by definition of divisibility, there is an integer \( c \) for which \( a - b = {nc} \).\n\nNow multiply both sides of this... | Yes |
Proposition 7.1 If \( a, b \in \mathbb{N} \), then there exist integers \( k \) and \( \ell \) for which \( \gcd \left( {a, b}\right) = {ak} + b\ell \) . | Proof. (Direct) Suppose \( a, b \in \mathbb{N} \) . Consider the set \( A = \{ {ax} + {by} : x, y \in \mathbb{Z}\} \) . This set contains both positive and negative integers, as well as 0 . (Reason: Let \( y = 0 \) and let \( x \) range over all integers. Then \( {ax} + {by} = {ax} \) ranges over all multiples of \( a ... | Yes |
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