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Theorem 23. (Rolle’s Theorem) Suppose \( {f}^{\prime } \) exists for every \( x \in \left( {a, b}\right), f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), and\n\n\[ f\left( a\right) = f\left( b\right) .\n\]\n\nThen there is a real number \( c \in \left( {a, b}\right) \) such that\n\n\[ {f}^{\prime }\left( ... | Proof: Since \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) we see, by the Extreme Value Theorem, 1 that \( f \) has both a maximum and a minimum on \( \left\lbrack {a, b}\right\rbrack \) . Denote the maximum by \( M \) and the minimum by \( m \) . There are several cases:\n\nCase 1: \( f\left( a\right... | Yes |
Corollary 3. If \( {f}^{\prime }\left( x\right) > 0 \) for every \( x \) in the interval \( \left( {a, b}\right) \) then for every \( c, d \in \left( {a, b}\right) \) where \( d > c \) we have\n\n\[ f\left( d\right) > f\left( c\right) \text{.} \]\n\nThat is, \( f \) is increasing on \( \left( {a, b}\right) \) . | Proof: Suppose \( c \) and \( d \) are as described in the corollary. Then by the Mean Value Theorem there is some number, say \( \alpha \in \left( {c, d}\right) \subseteq \left( {a, b}\right) \) such that\n\n\[ {f}^{\prime }\left( \alpha \right) = \frac{f\left( d\right) - f\left( c\right) }{d - c}. \]\n\nSince \( {f}^... | Yes |
Problem 132. Use the definition of continuity to prove that the constant function \( g\left( x\right) = c \) is continuous at any point \( a \) . | Problem 133. | No |
Problem 135. Let \( \left( {x}_{n}\right) ,\left( {y}_{n}\right) \) be sequences as in the NIP. Show that for all \( n, m \in \mathbb{N},{x}_{n} \leq {y}_{m} \) . | They are also coming together in the sense that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{y}_{n} - {x}_{n}}\right) = 0 \) . The NIP says that in this case there is a unique real number \( c \) in the middle of all of this \( \left\lbrack {{x}_{n} \leq c \leq {y}_{n}}\right. \) for all \( \left. n\right... | No |
Theorem 25. Suppose \( a \in \mathbb{R}, a \geq 0 \) . There exists a real number \( c \geq 0 \) such that \( {c}^{2} = a \) . | Notice that we can’t just say,\ | No |
Problem 139. The purpose of this problem is to show that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}\right) - \ln \left( {n + 1}\right) }\right) \]\n\nexists. | (a) Let \( {x}_{n} = \left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}\right) - \ln \left( {n + 1}\right) \) . Use the following diagram to\n\nshow\n\n\[ {x}_{1} \leq {x}_{2} \leq {x}_{3} \leq \cdots \]\n\n Suppose \( f\left( x\right) \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and \( v \) is any real number between \( f\left( a\right) \) and \( f\left( b\right) \) . Then there exists a real number \( c \in \left\lbrack {a, b}\right\rbrack \) such that \( f\left( c\r... | Sketch of Proof: We have two cases to consider: \( f\left( a\right) \leq v \leq f\left( b\right) \) and \( f\left( a\right) \geq v \geq f\left( b\right) \) .\n\nWe will look at the case \( f\left( a\right) \leq v \leq f\left( b\right) \) . Let \( {x}_{1} = a \) and \( {y}_{1} = b \), so we have \( {x}_{1} \leq {y}_{1} ... | No |
Theorem 27. A continuous function defined on a closed, bounded interval must be bounded. That is, let \( f \) be a continuous function defined on \( \left\lbrack {a, b}\right\rbrack \) . Then there exists a positive real number \( B \) such that \( \left| {f\left( x\right) }\right| \leq B \) for all \( x \in \left\lbra... | Sketch of Alleged Proof: Let's assume, for contradiction, that there is no such bound \( B \) . This says that for any positive integer \( n \), there must exist \( {x}_{n} \in \left\lbrack {a, b}\right\rbrack \) such that \( \left| {f\left( {x}_{n}\right) }\right| > n \) . (Otherwise \( n \) would be a bound for \( f ... | No |
Example 14. Given the sequence \( \left( {x}_{n}\right) \), the following are subsequences. | 1. \( \left( {{x}_{2},{x}_{4},{x}_{6},\ldots }\right) = {\left( {x}_{2k}\right) }_{k = 1}^{\infty } \)\n2. \( \left( {{x}_{1},{x}_{4},{x}_{9},\ldots }\right) = {\left( {x}_{{k}^{2}}\right) }_{k = 1}^{\infty } \)\n3. \( \left( {x}_{n}\right) \) itself. | Yes |
Theorem 28. The Bolzano-Weierstrass Theorem Let \( \left( {x}_{n}\right) \) be a sequence of real numbers such that \( {x}_{n} \in \left\lbrack {a, b}\right\rbrack ,\forall n \) . Then there exists \( c \in \left\lbrack {a, b}\right\rbrack \) and a subsequence \( \left( {x}_{{n}_{k}}\right) \), such that \( \mathop{\li... | Sketch of Proof: (Bolzano-Weierstrass Theorem) Suppose we have our sequence \( \left( {x}_{n}\right) \) such that \( {x}_{n} \in \left\lbrack {a, b}\right\rbrack ,\forall n \) . To find our \( c \) for the subsequence to converge to we will use the NIP. Since we are already using \( \left( {x}_{n}\right) \) as our orig... | No |
Theorem 29. (The Least Upper Bound Property (LUBP)) Let \( S \) be a non-empty subset of \( \mathbb{R} \) which is bounded above. Then \( S \) has a supremum. | Sketch of Proof: Since \( S \neq \varnothing \), then there exists \( s \in S \) . Since \( S \) is bounded above then it has an upper bound, say \( b \) . We will set ourselves up to use the Nested Interval Property. With this in mind, let \( {x}_{1} = s \) and \( {y}_{1} = b \) and notice that \( \exists x \in S \) s... | No |
Problem 148. Complete the above ideas to provide a formal proof of Theorem 29. | Notice that we really used the fact that \( S \) was non-empty and bounded above in the proof of Theorem 29. This makes sense, since a set which is not bounded above cannot possibly have a least upper bound. In fact, any real number is an upper bound of the empty set so that the empty set would not have a least upper b... | No |
Problem 149. Prove Corollary 5. | [Hint: Let \( c = \sup \left\{ {{x}_{n} \mid n = 1,2,3,\ldots }\right\} \) . To show that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = c \), let \( \epsilon \) \( > 0 \) . Note that \( c - \epsilon \) is not an upper bound. You take it from here!] | No |
Consider the following curious expression \( \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{\ldots }}}} \) . | We will use Corollary 5 to show that this actually converges to some real number. After we know it converges we can actually compute what it is. Of course to do so, we need to define things a bit more precisely. With this in mind consider the following sequence \( \left( {x}_{n}\right) \) defined as follows:\n\n\[ \n{x... | No |
Theorem 30. (Extreme Value Theorem (EVT)) Suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then there exists \( c, d \in \left\lbrack {a, b}\right\rbrack \) such that \( f\left( d\right) \leq f\left( x\right) \leq f\left( c\right) \), for all \( x \in \left\lbrack {a, b}\right\rbrack \) . | Sketch of Proof: We will first show that \( f \) attains its maximum. To this end, recall that Theorem 27 tells us that \( f\left\lbrack {a, b}\right\rbrack = \{ f\left( x\right) \mid x \in \left\lbrack {a, b}\right\rbrack \} \) is a bounded set. By the LUBP, \( f\left\lbrack {a, b}\right\rbrack \) must have a least up... | No |
Problem 162. Let \( 0 < b < 1 \) and consider the sequence of functions \( \left( {f}_{n}\right) \) defined on \( \left\lbrack {0, b}\right\rbrack \) by \( {f}_{n}\left( x\right) = {x}^{n} \) . Use the definition to show that \( {f}_{n}\overset{unif}{ \rightarrow }0 \) on \( \left\lbrack {0, b}\right\rbrack \) . | [Hint: \( \left| {{x}^{n} - 0}\right| = {x}^{n} \leq {b}^{n} \). | No |
Theorem 32. Consider a sequence of functions \( \left( {f}_{n}\right) \) which are all continuous on an interval \( I \) . Suppose \( {f}_{n}\overset{unif}{ \rightarrow }f \) on \( I \) . Then \( f \) must be continuous on \( I \) . | Sketch of Proof: Let \( a \in I \) and let \( \varepsilon > 0 \) . The idea is to use uniform convergence to replace \( f \) with one of the known continuous functions \( {f}_{n} \) . Specifically, by uncancelling, we can write\n\n\[ \left| {f\left( x\right) - f\left( a\right) }\right| = \left| {f\left( x\right) - {f}_... | No |
Problem 168. Prove Theorem 34. | [Hint: Let a be an arbitrary fixed point in \( I \) and let \( x \in I \) . By the Fundamental Theorem of Calculus, we have\n\n\[ \n{\int }_{t = a}^{x}{f}_{n}^{\prime }\left( t\right) \mathrm{d}t = {f}_{n}\left( x\right) - {f}_{n}\left( a\right) .\n\]\n\nTake the limit of both sides and differentiate with respect to \(... | No |
Theorem 35. Suppose \( \left( {s}_{n}\right) \) is a sequence of real numbers which converges to s. Then \( \left( {s}_{n}\right) \) is a Cauchy sequence. | Intuitively, this result makes sense. If the terms in a sequence are getting arbitrarily close to \( s \), then they should be getting arbitrarily close to each other \( {}^{2} \) This is the basis of the proof. | No |
Problem 171. Prove Lemma 5. | [Hint: This is similar to problem 67 of Chapter 4. There exists \( N \) such that if \( m, n > N \) then \( \left| {{s}_{n} - {s}_{m}}\right| < 1 \) . Choose a fixed \( m > N \) and let \( B = \max \left( {\left| {s}_{1}\right| ,\left| {s}_{2}\right| ,\ldots ,\left| {s}_{\lceil N\rceil }\right| ,\left| {s}_{m}\right| +... | No |
Theorem 36. (Cauchy sequences converge) Suppose \( \left( {s}_{n}\right) \) is a Cauchy sequence of real numbers. There exists a real number \( s \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = \) \( s \) . | Sketch of Proof: We know that \( \left( {s}_{n}\right) \) is bounded, so by the Bolzano-Weierstrass Theorem, it has a convergent subsequence \( \left( {s}_{{n}_{k}}\right) \) converging to some real number \( s \) . We have \( \left| {{s}_{n} - s}\right| = \left| {{s}_{n} - {s}_{{n}_{k}} + {s}_{{n}_{k}} - s}\right| \le... | No |
Problem 175. Prove the Cauchy criterion. | \( \diamond \) | No |
Problem 178. Prove Theorem 39. | [Hint: Use the Cauchy criterion with the fact that \( \left| {\mathop{\sum }\limits_{{k = n + 1}}^{m}{a}_{k}}\right| \leq \mathop{\sum }\limits_{{k = n + 1}}^{m}\left| {a}_{k}\right| \) .] | No |
Theorem 40. Suppose \( \sum {a}_{n} \) converges absolutely and let \( s = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n} \) . Then any rearrangement of \( \sum {a}_{n} \) must converge to \( s \) . | Sketch of Proof: We will first show that this result is true in the case where \( {a}_{n} \geq 0 \) . If \( \sum {b}_{n} \) represents a rearrangement of \( \sum {a}_{n} \), then notice that the sequence of partial sums \( {\left( \mathop{\sum }\limits_{{k = 0}}^{n}{b}_{k}\right) }_{n = 0}^{\infty } \) is an increasing... | No |
Theorem 41. Suppose \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{c}^{n} \) converges for some nonzero real number \( c \) . Then \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \) converges absolutely for all \( x \) such that \( \left| x\right| < \left| c\right| \) . | To prove Theorem 41 first note that by Problem 176, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}{c}^{n} = 0 \) . Thus \( \left( {{a}_{n}{c}^{n}}\right) \) is a bounded sequence. Let \( B \) be a bound: \( \left| {a{c}^{n}}\right| \leq B \) . Then\n\n\[ \left| {{a}_{n}{x}^{n}}\right| = \left| {{a}_{n}{c}^{n... | Yes |
Problem 183. Prove Corollary 9. | As a result of Theorem 41 and Corollary 9, we have the following: either \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \) converges absolutely for all \( x \) or there exists some nonnegative real number \( r \) such that \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \) converges absolutely wh... | Yes |
Theorem 42. (The Weierstrass- \( M \) Test) Let \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) be a sequence of functions defined on \( S \subseteq \mathbb{R} \) and suppose that \( {\left( {M}_{n}\right) }_{n = 1}^{\infty } \) is a sequence of nonnegative real numbers such that\n\n\[ \left| {{f}_{n}\left( x\right) }... | Sketch of Proof: Since the crucial feature of the theorem is the function \( f\left( x\right) \) that our series converges to, our plan of attack is to first define \( f\left( x\right) \) and then show that our series, \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n}\left( x\right) \), converges to it uniformly.\n\n... | No |
Problem 186. Observe that for all \( x \in \left\lbrack {-1,1}\right\rbrack \left| x\right| \leq 1 \) . Identify which of the following series converges pointwise and which converges uniformly on the interval \( \left\lbrack {-1,1}\right\rbrack \) . In every case identify the limit function. (a) \( \mathop{\sum }\limit... | Using the Weierstrass- \( M \) test, we can prove the following result. Theorem 43. Suppose | No |
Problem 188. Show that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }n{x}^{n - 1} \) converges for \( \left| x\right| < 1 \) . | [Hint: We know that \( \mathop{\sum }\limits_{{k = 0}}^{n}{x}^{k} = \frac{{x}^{n + 1} - 1}{x - 1} \) . Differentiate both sides and take the limit as \( n \) approaches infinity.] | No |
Problem 189. Prove Theorem 44. | [Hint: Let \( b \) be a number with \( \left| x\right| < b < r \) and consider \( \left| {{a}_{n}n{x}^{n - 1}}\right| = \left| {{a}_{n}{b}^{n} \cdot \frac{1}{b} \cdot n{\left( \frac{x}{b}\right) }^{n - 1}}\right| \) . You should be able to use the Comparison Test and Problem 188.] | No |
Problem 190. Suppose the power series \( \sum {a}_{n}{x}^{n} \) has radius of convergence \( r \) and the series \( \sum {a}_{n}{r}^{n} \) converges absolutely. Then \( \sum {a}_{n}{x}^{n} \) converges uniformly on \( \left\lbrack {-r, r}\right\rbrack \) . | [Hint: For \( \left| x\right| \leq r,\left| {{a}_{n}{x}^{n}}\right| \leq \left| {{a}_{n}{r}^{n}}\right| \). | No |
Prove Theorem 45. | [Hint: Let \( \epsilon > 0 \) . Since \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{r}^{n} \) converges then by the Cauchy Criterion, there exists \( N \) such that if \( m > n > N \) then \( \left| {\mathop{\sum }\limits_{{k = n + 1}}^{m}{a}_{k}{r}^{k}}\right| < \frac{\epsilon }{2} \) Let \( 0 \leq x \leq r \) .... | Yes |
Theorem 50. Let \( {\left( \left\lbrack {a}_{n},{b}_{n}\right\rbrack \right) }_{n = 1}^{\infty } \) be a sequence of nested intervals such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| {{b}_{n} - {a}_{n}}\right| > 0 \) . Then there is at least one \( c \in \mathbb{R} \) such that \( c \in \left\lbrack {... | Proof: By Corollary 5 of Chapter 7, we know that a bounded increasing sequence such as \( \left( {a}_{n}\right) \) converges, say to \( c \) . Since \( {a}_{n} \leq {a}_{m} \leq {b}_{n} \) for \( m > n \) and \( \mathop{\lim }\limits_{{m \rightarrow \infty }}{a}_{m} = c \), then for any fixed \( n,{a}_{n} \leq c \leq {... | Yes |
Theorem 52. Show that \( \mathbb{Q} \) is countable. | Sketch of Proof: First explain how you know that all of the non-negative rational numbers are in this list:\n\n\[ \n\frac{0}{1},\frac{0}{2},\frac{1}{1},\frac{0}{3},\frac{1}{2},\frac{2}{1},\frac{0}{4},\frac{1}{3},\frac{2}{2},\frac{3}{1},\cdots .\n\]\n\nHowever there is clearly some duplication. To handle this, apply par... | No |
Problem 202. Prove Corollary 11. [Hint: If we had only finitely many rationals to deal with this would be easy. Let \( \left\{ {{r}_{1},{r}_{2},\ldots ,{r}_{k}}\right\} \) be these rational numbers and take \( {a}_{n} = {r}_{n} - \frac{\varepsilon }{2k} \) and \( {b}_{n} = {r}_{n} + \frac{\varepsilon }{2k} \) . Then fo... | \[ \mathop{\sum }\limits_{{n = 1}}^{k}{b}_{n} - {a}_{n} = \mathop{\sum }\limits_{{n = 1}}^{k}\frac{\varepsilon }{k} = \varepsilon . \] | No |
Prove: If \( \left| S\right| = n \), then \( \left| {P\left( S\right) }\right| = {2}^{n} \) . | [Hint: Let \( S = {a}_{1},{a}_{2},\ldots ,{a}_{n} \) . Consider the following correspondence between the elements of \( P\left( S\right) \) and the set \( T \) of all \( n \) -tuples of yes \( \left( Y\right) \) or no \( \left( N\right) ) : \n\n\[ \n\{ \} \leftrightarrow \{ N, N, N,\ldots, N\} \n\] \n\n\[ \n\left\{ {a}... | No |
Problem 209. Suppose \( X \) is an uncountable set and \( Y \subset X \) is countably infinite. Prove that \( X \) and \( X - Y \) have the same cardinality. | [Hint: Let \( Y = {Y}_{0} \) . If \( X - {Y}_{0} \) is an infinite set, then by the previous problem it contains a countably infinite set \( {Y}_{1} \) . Likewise if \( X - \left( {{Y}_{0} \cup {Y}_{1}}\right) \) is infinite it also contains an infinite set \( {Y}_{2} \) . Again, if \( X - \left( {{Y}_{0} \cup {Y}_{1} ... | No |
Problem 210. Show that \( 0 \neq 1 \) . | [Hint: Show that if \( x \neq 0 \), then \( 0 \cdot x \neq x \) .] \( \diamond \) | No |
Problem 216. Let \( x \) and \( y \) be real numbers in \( \mathbb{Q} \) (that is, let them be sets of equivalent Cauchy sequences). If \( \left( {s}_{n}\right) \) and \( \left( {t}_{n}\right) \) are in \( x \) and \( \left( {\sigma }_{n}\right) \) and \( \left( {\tau }_{n}\right) \) are in \( y \) then | \[ {\left( {s}_{n} + {t}_{n}\right) }_{n = 1}^{\infty } \equiv {\left( {\sigma }_{n} + {\tau }_{n}\right) }_{n = 1}^{\infty }. \] | No |
Theorem 54. Let \( {0}^{ * } \) be the set of Cauchy sequences in \( \mathbb{Q} \) which are all equivalent to the sequence \( \left( {0,0,0,\ldots }\right) \) . Then\n\n\[ \n{0}^{ * } + x = x.\n\] | Proof: From Problem 216 it is clear that in forming \( {0}^{ * } + x \) we can choose any sequence in \( {0}^{ * } \) to represent \( {0}^{ * } \) and any sequence in \( x \) to represent \( x \) . (This is because any other choice will yield a sequence equivalent to \( {0}^{ * } + x \) .)\n\nThus we choose \( \left( {... | Yes |
Theorem 56. (Closure with Respect to Addition) If \( \alpha \) and \( \beta \) are cuts then \( \alpha + \beta \) is a cut. | Proof: We need to show that the set \( \alpha + \beta \) satisfies all three of the properties of a cut.\n\nProof of Property I\n\nLet \( x \) be any rational number in \( \alpha \) and let \( {x}_{1} \) be a rational number not in \( \alpha \) . Then by Property II \( x < {x}_{1} \) .\n\nLet \( y \) be any rational nu... | Yes |
Problem 221. Prove Lemma 8. | [Hint: Since \( \beta \) is a cut there exists \( {r}_{1} \in \beta \) . Let \( {s}_{1} = y \notin \beta \) . We know that \( {r}_{1} < {s}_{1} < z \) . Consider the midpoint \( \frac{{s}_{1} + {r}_{1}}{2} \) . If this is in \( \beta \) then relabel it as \( {r}_{2} \) and relabel \( {s}_{1} \) as \( {s}_{2} \) . If it... | No |
Problem 222. Let \( \alpha \) and \( \beta \) be cuts with \( \beta < \alpha \) . Prove that \( \beta + \left( {\alpha - \beta }\right) = \alpha \) . | [Hint: It is pretty straightforward to show that \( \beta + \left( {\alpha - \beta }\right) \subseteq \alpha \) . To show that \( \alpha \subseteq \beta + \left( {\alpha - \beta }\right) \), we let \( x \in \alpha \) . Since \( \beta < \alpha \), we have \( y \in \alpha \) with \( y \notin \beta \) . We can assume with... | No |
Cauchy sequences converge | 148 | No |
Cantor’s first proof that \( \mathbb{R} \) is uncountable | 160 | No |
Theorem 1.8. If \( x \) and \( y \) are odd integers, then \( x \cdot y \) is an odd integer. | Proof. We assume that \( x \) and \( y \) are odd integers and will prove that \( x \cdot y \) is an odd integer. Since \( x \) and \( y \) are odd, there exist integers \( m \) and \( n \) such that\n\n\[ x = {2m} + 1\text{ and }y = {2n} + 1.\]\n\nUsing algebra, we obtain\n\n\[ x \cdot y = \left( {{2m} + 1}\right) \le... | Yes |
The statement \( \neg \left( {P \vee Q}\right) \) is logically equivalent to \( \neg P \land \neg Q \) . This can be written as \( \neg \left( {P \vee Q}\right) \equiv \neg P \land \neg Q \) . | Table 2.3 establishes the second equivalency.\n\n<table><thead><tr><th>\( P \)</th><th>Q</th><th>\( P \vee Q \)</th><th>\( \neg \left( {P \vee Q}\right) \)</th><th>\( \neg P \)</th><th>\( \neg Q \)</th><th>\( \neg P \land \neg Q \)</th></tr></thead><tr><td>T</td><td>\( \mathrm{T} \)</td><td>\( \mathrm{T} \)</td><td>F</... | Yes |
Theorem 2.8 (Important Logical Equivalencies)\n\nFor statements \( P, Q \), and \( R \) ,\n\n\[ \n\\text{De Morgan’s Laws}\\;\\neg \\left( {P \\land Q}\\right) \\equiv \\neg P \\vee \\neg Q \n\]\n\n\[ \n\\neg \\left( {P \\vee Q}\\right) \\equiv \\neg P \\land \\neg Q \n\]\n\nConditional Statements \\;P \\rightarrow Q \... | We have already established many of these equivalencies. Others will be established in the exercises. | No |
Theorem 2.16. For any open sentence \( P\left( x\right) \) , \n\n\[ \n\neg \left( {\forall x \in U}\right) \left\lbrack {P\left( x\right) }\right\rbrack \equiv \left( {\exists x \in U}\right) \left\lbrack {\neg P\left( x\right) }\right\rbrack \text{, and} \n\] \n\n\[ \n\neg \left( {\exists x \in U}\right) \left\lbrack ... | ## Example 2.17 (Negations of Quantified Statements) \n\nConsider the following statement: \( \left( {\forall x \in \mathbb{R}}\right) \left( {{x}^{3} \geq {x}^{2}}\right) \) . \n\nWe can write this statement as an English sentence in several ways. Following are two different ways to do so. \n\n- For each real number \... | Yes |
Theorem 2.8. Important Logical Equivalencies. For statements P, Q, and R,\n\n\[ \text{De Morgan’s Laws}\;\neg \left( {P \land Q}\right) \equiv \neg P \vee \neg Q \] | \[ \neg \left( {P \vee Q}\right) \equiv \neg P \land \neg Q \] | No |
Theorem 3.1. Let \( a, b \), and \( c \) be integers with \( a \neq 0 \) and \( b \neq 0 \) . If \( a \) divides \( b \) and \( b \) divides \( c \), then \( a \) divides \( c \) . | Proof. We assume that \( a, b \), and \( c \) are integers with \( a \neq 0 \) and \( b \neq 0 \) . We further assume that \( a \) divides \( b \) and that \( b \) divides \( c \) . We will prove that \( a \) divides \( c \) .\n\nSince \( a \) divides \( b \) and \( b \) divides \( c \), there exist integers \( s \) an... | Yes |
Proposition 3.5. For all integers \( a \) and \( b \), if \( a \equiv 5\left( {\;\operatorname{mod}\;8}\right) \) and \( b \equiv 5\left( {\;\operatorname{mod}\;8}\right) \) , then \( \left( {a + b}\right) \equiv 2\left( {\;\operatorname{mod}\;8}\right) \) . | ## Progress Check 3.6 (Proving Proposition 3.5)\n\nWe will use \ | No |
Theorem 3.7. For each integer \( n \), if \( {n}^{2} \) is an even integer, then \( n \) is an even integer. | Proof. We will prove this result by proving the contrapositive of the statement, which is\n\nFor each integer \( n \), if \( n \) is an odd integer, then \( {n}^{2} \) is an odd integer.\n\nHowever, in Theorem 1.8 on page 21, we have already proven that if \( x \) and \( y \) are odd integers, then \( x \cdot y \) is a... | Yes |
For all real numbers \( a \) and \( b \), if \( a \neq 0 \) and \( b \neq 0 \), then \( {ab} \neq 0 \) . | Proof. We will prove the contrapositive of this proposition, which is\n\nFor all real numbers \( a \) and \( b \), if \( {ab} = 0 \), then \( a = 0 \) or \( b = 0 \) .\n\nThis contrapositive, however, is logically equivalent to the following:\n\nFor all real numbers \( a \) and \( b \), if \( {ab} = 0 \) and \( a \neq ... | Yes |
Proposition 3.11. Let \( x \in \mathbb{R} \) . The real number \( x \) equals 2 if and only if \( {x}^{3} - 2{x}^{2} + x = 2 \) . | Proof. We will prove this biconditional statement by proving the following two conditional statements:\n\n- For each real number \( x \), if \( x \) equals 2, then \( {x}^{3} - 2{x}^{2} + x = 2 \) .\n\n- For each real number \( x \), if \( {x}^{3} - 2{x}^{2} + x = 2 \), then \( x \) equals 2 .\n\nFor the first part, we... | Yes |
Proposition 3.12. If \( a, b \), and \( c \) are real numbers with \( a \neq 0 \), then the linear equation \( {ax} + b = c \) has exactly one real number solution, which is \( x = \frac{c - b}{a} \) . | Proof. Assume that \( a, b \), and \( c \) are real numbers with \( a \neq 0 \) . We can solve the linear equation \( {ax} + b = c \) by adding \( - b \) to both sides of the equation and then dividing both sides of the resulting equation by \( a \) (since \( a \neq 0 \) ), to obtain\n\n\[ x = \frac{c - b}{a}. \]\n\nTh... | Yes |
Proposition 3.14. For each real number \( x \), if \( 0 < x < 1 \), then \( \frac{1}{x\left( {1 - x}\right) } \geq 4 \) . | Proof. We will use a proof by contradiction. So we assume that the proposition is false, or that there exists a real number \( x \) such that \( 0 < x < 1 \) and\n\n\[ \frac{1}{x\left( {1 - x}\right) } < 4 \]\n\n(1)\n\nWe note that since \( 0 < x < 1 \), we can conclude that \( x > 0 \) and that \( \left( {1 - x}\right... | Yes |
Proposition 3.17. For all integers \( x \) and \( y \), if \( x \) and \( y \) are odd integers, then there does not exist an integer \( z \) such that \( {x}^{2} + {y}^{2} = {z}^{2} \) . | Proof. We will use a proof by contradiction. So we assume that there exist integers \( x \) and \( y \) such that \( x \) and \( y \) are odd and there exists an integer \( z \) such that \( {x}^{2} + {y}^{2} = \) \( {z}^{2} \) . Since \( x \) and \( y \) are odd, there exist integers \( m \) and \( n \) such that \( x... | No |
Proposition 3.19. For all real numbers \( x \) and \( y \), if \( x \) is rational and \( x \neq 0 \) and \( y \) is irrational, then \( x \cdot y \) is irrational. | Proof. We will use a proof by contradiction. So we assume that there exist real numbers \( x \) and \( y \) such that \( x \) is rational, \( x \neq 0, y \) is irrational, and \( x \cdot y \) is rational. Since \( x \neq 0 \), we can divide by \( x \), and since the rational numbers are closed under division by nonzero... | Yes |
Theorem 3.20. If \( r \) is a real number such that \( {r}^{2} = 2 \), then \( r \) is an irrational number. | Proof. We will use a proof by contradiction. So we assume that the statement of the theorem is false. That is, we assume that\n\n\( r \) is a real number, \( {r}^{2} = 2 \), and \( r \) is a rational number.\n\nSince \( r \) is a rational number, there exist integers \( m \) and \( n \) with \( n > 0 \) such that\n\n\[... | Yes |
Proposition 1. If \( n \) is an integer, then \( \left( {{n}^{2} + n}\right) \) is an even integer. | If we were trying to write a direct proof of this proposition, the only thing we could assume is that \( n \) is an integer. This is not much help. In a situation such as this, we will sometimes use cases to provide additional assumptions for the forward process of the proof. Cases are usually based on some common prop... | No |
For all real numbers \( a \) and \( b \), if \( {ab} = 0 \), then \( a = 0 \) or \( b = 0 \) . | We let \( a \) and \( b \) be real numbers and assume that \( {ab} = 0 \) . We will prove that \( a = 0 \) or \( b = 0 \) by considering two cases: (1) \( a = 0 \), and (2) \( a \neq 0 \) .\n\nIn the case where \( a = 0 \), the conclusion of the proposition is true and so there is nothing to prove.\n\nIn the case where... | Yes |
Let a be a positive real number. For each real number \( x \) , 1. \( \left| x\right| = a \) if and only if \( x = a \) or \( x = - a \) . | We let \( a \) be a positive real number and let \( x \in \mathbb{R} \) . We will first prove that if \( \left| x\right| = a \), then \( x = a \) or \( x = - a \) . So we assume that \( \left| x\right| = a \) . In the case where \( x \geq 0 \), we see that \( \left| x\right| = x \), and since \( \left| x\right| = a \),... | Yes |
Theorem 3.25. Let a be a positive real number. For all real numbers \( x \) and \( y \) ,\n\n1. \( \left| x\right| < a \) if and only if \( - a < x < a \) .\n\n2. \( \left| {xy}\right| = \left| x\right| \left| y\right| \) .\n\n3. \( \left| {x + y}\right| \leq \left| x\right| + \left| y\right| \) . This is known as the ... | Proof. We will prove Part (1). The proof of Part (2) is included in Exercise (10), and the proof of Part (3) is Exercise (14). For Part (1), we will prove the biconditional proposition by proving the two associated conditional propositions.\n\nSo we let \( a \) be a positive real number and let \( x \in \mathbb{R} \) a... | No |
Proposition 3.27. If \( n \) is an integer, then 3 divides \( {n}^{3} - n \) . | Proof. Let \( n \) be an integer. We will show that 3 divides \( {n}^{3} - n \) by examining the three cases for the remainder when \( n \) is divided by 3 . By the Division Algorithm, there exist unique integers \( q \) and \( r \) such that\n\n\[ n = {3q} + r\text{, and }0 \leq r < 3.\]\n\nThis means that we can cons... | No |
Theorem 3.30 (Properties of Congruence Modulo n). Let \( n \in \mathbb{N} \), and let \( a, b \), and \( c \) be integers. 1. For every integer \( a, a \equiv a\left( {\;\operatorname{mod}\;n}\right) \). This is called the reflexive property of congruence modulo \( n \). 2. If \( a \equiv b\left( {\;\operatorname{mod}\... | Proof. We will prove the reflexive property and the transitive property. The proof of the symmetric property is Exercise (3). Let \( n \in \mathbb{N} \), and let \( a \in \mathbb{Z} \) . We will show that \( a \equiv a\left( {\;\operatorname{mod}\;n}\right) \) . Notice that \[ a - a = 0 = n \cdot 0. \] This proves that... | No |
Theorem 3.31. Let \( n \in \mathbb{N} \) and let \( a \in \mathbb{Z} \) . If \( a = {nq} + r \) and \( 0 \leq r < n \) for some integers \( q \) and \( r \), then \( a \equiv r\left( {\;\operatorname{mod}\;n}\right) \) . | This theorem says that an integer is congruent \( \left( {\;\operatorname{mod}\;n}\right) \) to its remainder when it is divided by \( n \) . Since this remainder is unique and since the only possible remainders for division by \( n \) are \( 0,1,2,\ldots, n - 1 \), we can state the following result. | No |
Proposition 3.33. For each integer \( a \), if \( a ≢ 0\left( {\;\operatorname{mod}\;5}\right) \), then \( {a}^{2} \equiv 1\left( {\;\operatorname{mod}\;5}\right) \) or \( {a}^{2} \equiv 4\left( {\;\operatorname{mod}\;5}\right) \) . | Proof. We will prove this proposition using cases for \( a \) based on congruence modulo 5. In doing so, we will use the results in Theorem 3.28 and Theorem 3.30. Because the hypothesis is \( a ≢ 0\left( {\;\operatorname{mod}\;5}\right) \), we can use four cases, which are: (1) \( a \equiv 1\left( {\;\operatorname{mod}... | No |
Proposition 4.2. For each natural number \( n \) , \n\n\[ \n{1}^{2} + {2}^{2} + \cdots + {n}^{2} = \frac{n\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}. \n\] | Proof. We will use a proof by mathematical induction. For each natural number \( n \), we let \( P\left( n\right) \) be \n\n\[ \n{1}^{2} + {2}^{2} + \cdots + {n}^{2} = \frac{n\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}. \n\]\n\nWe first prove that \( P\left( 1\right) \) is true. Notice that \( \frac{1\left( {1 ... | Yes |
Proposition 4.4. For every natural number \( n,4 \) divides \( \left( {{5}^{n} - 1}\right) \) . | Proof. (Proof by Mathematical Induction) For each natural number \( n \), let \( P\left( n\right) \) be \ | No |
Proposition 4.7. For each natural number \( n \) with \( n \geq 4, n! > {2}^{n} \) . | Proof. We will use a proof by mathematical induction. For this proof, we let\n\n\[ P\left( n\right) \text{be \ | No |
Theorem 4.9. Each natural number greater than 1 either is a prime number or is a product of prime numbers. | Proof. We will use the Second Principle of Mathematical Induction. We let \( P\\left( n\\right) \) be\n\n\( n \) is a prime number or \( n \) is a product of prime numbers.\n\nFor the basis step, \( P\\left( 2\\right) \) is true since 2 is a prime number.\n\nTo prove the inductive step, we let \( k \) be a natural numb... | Yes |
Proposition 4.13. For each natural number \( n \), the Fibonacci number \( {f}_{3n} \) is an even natural number. | Hint: We have already defined the predicate \( P\left( n\right) \) to be used in an induction proof and have proved the basis step. Use the information in and after the preceding know-show table to help prove that if \( {f}_{3k} \) is even, then \( {f}_{3\left( {k + 1}\right) } \) is even. | No |
Theorem 4.14. Let \( a, r \in \mathbb{R} \) . If a geometric sequence is defined by \( {a}_{1} = a \) and for each \( n \in \mathbb{N},{a}_{n + 1} = r \cdot {a}_{n} \), then for each \( n \in \mathbb{N},{a}_{n} = a \cdot {r}^{n - 1} \) . | The proof of this proposition is Exercise (6). | No |
Theorem 4.16. Let \( a, r \in \mathbb{R} \) and \( r \neq 1 \) . If the sequence \( {S}_{1},{S}_{2},\ldots ,{S}_{n},\ldots \) is defined by \( {S}_{1} = a \) and for each \( n \in \mathbb{N},{S}_{n + 1} = a + r \cdot {S}_{n} \), then for each \( n \in \mathbb{N} \) , \( {S}_{n} = a\left( \frac{1 - {r}^{n}}{1 - r}\right... | The proof of Proposition 4.16 is Exercise (8). | No |
Lemma 5.6. Let \( A \) and \( B \) be subsets of some universal set. If \( A = B \cup \{ x\} \) , where \( x \notin B \), then any subset of \( A \) is either a subset of \( B \) or a set of the form \( C \cup \{ x\} \), where \( C \) is a subset of \( B \) . | Proof. Let \( A \) and \( B \) be subsets of some universal set, and assume that \( A = \) \( B \cup \{ x\} \) where \( x \notin B \) . Let \( Y \) be a subset of \( A \) . We need to show that \( Y \) is a subset of \( B \) or that \( Y = C \cup \{ x\} \), where \( C \) is some subset of \( B \) . There are two cases ... | Yes |
Proposition 5.7. Let \( S \) be the set of all integers that are multiples of 6, and let \( T \) be the set of all even integers. Then \( S \) is a subset of \( T \) . | Proof. Let \( S \) be the set of all integers that are multiples of 6, and let \( T \) be the set of all even integers. We will show that \( S \) is a subset of \( T \) by showing that if an integer \( x \) is an element of \( S \), then it is also an element of \( T \) .\n\nLet \( x \in S \) . (Note: The use of the wo... | Yes |
Proposition 5.10. Let \( A \) and \( B \) be subsets of the universal set \( U \) . If \( A \subseteq B \), then \( {B}^{c} \subseteq {A}^{c} \) | 1. The conclusion of the conditional statement is \( {B}^{c} \subseteq {A}^{c} \) . Explain why we should try the choose-an-element method to prove this proposition.\n\n2. Complete the following know-show table for this proposition and explain\n\nexactly where the choose-an-element method is used.\n\n<table><thead><tr>... | No |
Proposition 5.11. Let \( A \) and \( B \) be subsets of some universal set. Then \( A - \left( {A - B}\right) = A \cap B. \) | Proof. Let \( A \) and \( B \) be subsets of some universal set. We will prove that \( A - \left( {A - B}\right) = A \cap B \) by proving that \( A - \left( {A - B}\right) \subseteq A \cap B \) and that \( A \cap B \subseteq A - \left( {A - B}\right) \).
First, let \( x \in A - \left( {A - B}\right) \). This means tha... | Yes |
Proposition 5.14. Let \( A \) and \( B \) be subsets of some universal set. Then \( A \subseteq B \) if and only if \( A \cap {B}^{c} = \varnothing \) . | Proof. Let \( A \) and \( B \) be subsets of some universal set. We will first prove that if \( A \subseteq B \), then \( A \cap {B}^{c} = \varnothing \), by proving its contrapositive. That is, we will prove\n\nIf \( A \cap {B}^{c} \neq \varnothing \), then \( A \nsubseteq B \) .\n\nSo assume that \( A \cap {B}^{c} \n... | Yes |
Proposition 5.16. Let \( a, b \), and \( t \) be integers with \( t \neq 0 \) . If \( t \) divides \( a \) and \( t \) divides \( b \), then for all integers \( x \) and \( y \) , \( t \) divides \( \left( {{ax} + {by}}\right) \) . | Proof. Let \( a, b \), and \( t \) be integers with \( t \neq 0 \), and assume that \( t \) divides \( a \) and \( t \) divides \( b \) . We will prove that for all integers \( x \) and \( y, t \) divides \( \left( {{ax} + {by}}\right) \) . So let \( x \in \mathbb{Z} \) and let \( y \in \mathbb{Z} \) . Since \( t \) di... | Yes |
Theorem 5.17. Let \( A, B \), and \( C \) be subsets of some universal set \( U \) . Then\n\n\[ \text{-}A \cap B \subseteq A\text{and}A \subseteq A \cup B\text{.} \]\n\n\[ \text{- If}A \subseteq B\text{, then}A \cap C \subseteq B \cap C\text{and}A \cup C \subseteq B \cup C\text{.} \] | Proof. The first part of this theorem was included in Exercise (7) from Section 5.2. The second part of the theorem was Exercise (12) from Section 5.2. | No |
Theorem 5.18 (Algebra of Set Operations). Let \( A, B \), and \( C \) be subsets of some universal set \( U \) . Then all of the following equalities hold.\n\n\[ \n\\text{Properties of the Empty Set}\\;A \\cap \\varnothing = \\varnothing \\;A \\cap U = A\n\]\n\n\[ \n\\text{and the Universal Set}\\;A \\cup \\varnothing ... | ## Proof of One of the Commutative Laws in Theorem 5.18\n\nProof. We will prove that \( A \\cap B = B \\cap A \) . Let \( x \\in U \) . Then\n\n\[ \nx \\in A \\cap B\\text{if and only if}x \\in A\\text{and}x \\in B\\text{.\n\]\n\n(1)\n\nHowever, we know that if \( P \) and \( Q \) are statements, then \( P \\land Q \) ... | Yes |
Theorem 5.20. Let \( A \) and \( B \) be subsets of some universal set \( U \) . Then the following are true:\n\nDe Morgan's Laws\n\[{\left( A \cup B\right) }^{c} = {A}^{c} \cap {B}^{c}\] | Proof. We will only prove one of De Morgan's Laws, namely, the one that was explored in Preview Activity 1. The proofs of the other parts are left as exercises. Let \( A \) and \( B \) be subsets of some universal set \( U \) . We will prove that \( {\left( A \cup B\right) }^{c} = \) \( {A}^{c} \cap {B}^{c} \) by provi... | No |
Theorem 5.22. Let \( A \) and \( B \) be subsets of some universal set \( U \) . The following are equivalent:\n\n1. \( A \subseteq B \) 2. \( A \cap {B}^{c} = \varnothing \) 3. \( {A}^{c} \cup B = U \) | Proof. To prove that these are equivalent conditions, we will prove that (1) implies (2), that (2) implies (3), and that (3) implies (1).\n\nLet \( A \) and \( B \) be subsets of some universal set \( U \) . We have proved that (1) implies (2) in Proposition 5.14.\n\nTo prove that (2) implies (3), we will assume that \... | Yes |
\[ \text{2.}A \times \left( {B \cup C}\right) = \left( {A \times B}\right) \cup \left( {A \times C}\right) \] | We will not prove all these results; rather, we will prove Part (2) of Theorem 5.25 and leave some of the rest to the exercises. In constructing these proofs, we need to keep in mind that Cartesian products are sets, and so we follow many of the same principles to prove set relationships that were introduced in Section... | No |
Theorem 5.25 (Part (2)). Let \( A, B \), and \( C \) be sets. Then\n\n\[ A \times \left( {B \cup C}\right) = \left( {A \times B}\right) \cup \left( {A \times C}\right) . \] | Proof. Let \( A, B \), and \( C \) be sets. We will prove that \( A \times \left( {B \cup C}\right) \) is equal to \( \left( {A \times B}\right) \cup \left( {A \times C}\right) \) by proving that each set is a subset of the other set.\n\nTo prove that \( A \times \left( {B \cup C}\right) \subseteq \left( {A \times B}\r... | Yes |
Example 5.28 (A Family of Sets Indexed by the Positive Real Numbers) For each positive real number \( \alpha \), let \( {A}_{\alpha } \) be the interval \( ( - 1,\alpha \rbrack \) . That is,\n\n\[ \n{A}_{\alpha } = \{ x \in \mathbb{R} \mid - 1 < x \leq \alpha \} .\n\]\n\nIf we let \( {\mathbb{R}}^{ + } \) be the set of... | To determine the intersection of this family, notice that\n\n- if \( y \in \mathbb{R} \) and \( y < - 1 \), then for each \( \alpha \in {\mathbb{R}}^{ + }, y \notin {A}_{\alpha } \) ;\n\n- if \( y \in \mathbb{R} \) and \( - 1 < y \leq 0 \), then \( y \in {A}_{\alpha } \) for each \( \alpha \in {\mathbb{R}}^{ + } \) ; a... | Yes |
Theorem 5.30. Let \( \Lambda \) be a nonempty indexing set and let \( \mathcal{A} = \left\{ {{A}_{\alpha } \mid \alpha \in \Lambda }\right\} \) be an indexed family of sets, each of which is a subset of some universal set \( U \) . Then\n\n1. For each \( \beta \in \Lambda ,\mathop{\bigcap }\limits_{{\alpha \in \Lambda ... | Proof. We will prove Parts (1) and (3). The proofs of Parts (2) and (4) are included in Exercise (4). So we let \( \Lambda \) be a nonempty indexing set and let \( \mathcal{A} = \left\{ {{A}_{\alpha } \mid \alpha \in \Lambda }\right\} \) be an indexed family of sets. To prove Part (1), we let \( \beta \in \Lambda \) an... | No |
Theorem 5.31. Let \( \Lambda \) be a nonempty indexing set, let \( \mathcal{A} = \left\{ {{A}_{\alpha } \mid \alpha \in \Lambda }\right\} \) be an indexed family of sets, and let \( B \) be a set. Then\n\n\[ \text{1.}B \cap \left( {\mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }}\right) = \mathop{\bigcup ... | The proof of Theorem 5.31 is Exercise (5). | No |
Proposition 1. For all \( a, b \in \mathbb{R} \), if \( g\left( a\right) = g\left( b\right) \), then \( a = b \) . | Proof. We let \( a, b \in \mathbb{R} \), and we assume that \( g\left( a\right) = g\left( b\right) \) and will prove that \( a = b \) . Since \( g\left( a\right) = g\left( b\right) \), we know that\n\n\[ \n{5a} + 3 = {5b} + 3.\n\] | Yes |
Proposition 2. For all \( b \in \mathbb{R} \), there exists an \( a \in \mathbb{R} \) such that \( g\left( a\right) = b \) . | Proof. We let \( b \in \mathbb{R} \) . We will prove that there exists an \( a \in \mathbb{R} \) such that \( g\left( a\right) = b \) by constructing such an \( a \) in \( \mathbb{R} \) . In order for this to happen, we need \( g\left( a\right) = {5a} + 3 = b \) . | Yes |
Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be defined by \( f\left( x\right) = {x}^{2} + 1 \). Notice that\n\n\[ f\left( 2\right) = 5\text{and}f\left( {-2}\right) = 5\text{.} \]\n\nThis is enough to prove that the function \( f \) is not an injection since this shows that there exist two different inputs that prod... | Since \( f\left( x\right) = {x}^{2} + 1 \), we know that \( f\left( x\right) \geq 1 \) for all \( x \in \mathbb{R} \). This implies that the function \( f \) is not a surjection. For example,-2 is in the codomain of \( f \) and \( f\left( x\right) \neq - 2 \) for all \( x \) in the domain of \( f \) . | No |
Is the function \( F \) a surjection? That is, does \( F \) map \( \mathbb{R} \) onto \( T \) ? | To see if it is a surjection, we must determine if it is true that for every \( y \in T \) , there exists an \( x \in \mathbb{R} \) such that \( F\left( x\right) = y \) . So we choose \( y \in T \) . The goal is to determine if there exists an \( x \in \mathbb{R} \) such that\n\n\[ F\left( x\right) = y\text{, or} \]\n\... | Yes |
Example 6.14 (A Function that Is an Injection but Is Not a Surjection)\n\nLet \( {\mathbb{Z}}^{ * } = \{ x \in \mathbb{Z} \mid x \geq 0\} = \mathbb{N} \cup \{ 0\} \) . Define \( g : {\mathbb{Z}}^{ * } \rightarrow \mathbb{N} \) by \( g\left( x\right) = {x}^{2} + 1 \) . (Notice that this is the same formula used in Examp... | To prove that \( g \) is not a surjection, pick an element of \( \mathbb{N} \) that does not appear to be in the range. We will use 3 , and we will use a proof by contradiction to prove that there is no \( x \) in the domain \( \left( {\mathbb{Z}}^{ * }\right) \) such that \( g\left( x\right) = 3 \) . So we assume that... | Yes |
Theorem 6.22. Let \( A \) and \( B \) be nonempty sets and let \( f \) be a subset of \( A \times B \) that satisfies the following two properties:\n\n- For every \( a \in A \), there exists \( b \in B \) such that \( \left( {a, b}\right) \in f \) ; and\n\n- For every \( a \in A \) and every \( b, c \in B \), if \( \le... | A Note about Theorem 6.22. The first condition in Theorem 6.22 means that every element of \( A \) is an input, and the second condition ensures that every input has exactly one output. Many texts will use Theorem 6.22 as the definition of a function. Many mathematicians believe that this ordered pair representation of... | No |
Theorem 6.26. Let \( A \) and \( B \) be nonempty sets and let \( f : A \rightarrow B \) be a bijection. Then \( {f}^{-1} : B \rightarrow A \) is a function, and for every \( a \in A \) and \( b \in B \) , | \[ f\left( a\right) = b\text{ if and only if }{f}^{-1}\left( b\right) = a. \] | No |
For an example of the use of the notation in Theorem 6.26, let \( {\mathbb{R}}^{ + } = \{ x \in \mathbb{R} \mid x > 0\} \) . Define \[ f : \mathbb{R} \rightarrow \mathbb{R}\text{by}f\left( x\right) = {x}^{3}\text{; and}g : \mathbb{R} \rightarrow {\mathbb{R}}^{ + }\text{by}g\left( x\right) = {e}^{x}\text{.} \] Notice th... | In fact, \[ {f}^{-1} : \mathbb{R} \rightarrow \mathbb{R}\text{by}{f}^{-1}\left( y\right) = \sqrt[3]{y}\text{; and}{g}^{-1} : {\mathbb{R}}^{ + } \rightarrow \mathbb{R}\text{by}{g}^{-1}\left( y\right) = \ln y\text{.} \] For each function (and its inverse), we can write the result of Theorem 6.26 as follows: <table><thead... | Yes |
Corollary 6.28. Let \( A \) and \( B \) be nonempty sets and let \( f : A \rightarrow B \) be a bijection. Then\n\n1. For every \( x \) in \( A,\left( {{f}^{-1} \circ f}\right) \left( x\right) = x \) .\n\n2. For every \( y \) in \( B,\left( {f \circ {f}^{-1}}\right) \left( y\right) = y \) . | Proof. Let \( A \) and \( B \) be nonempty sets and assume that \( f : A \rightarrow B \) is a bijection. So let \( x \in A \) and let \( f\left( x\right) = y \) . By Theorem 6.26, we can conclude that \( {f}^{-1}\left( y\right) = x \) . Therefore,\n\n\[ \left( {{f}^{-1} \circ f}\right) \left( x\right) = {f}^{-1}\left(... | No |
Theorem 6.29. Let \( f : A \rightarrow B \) and \( g : B \rightarrow C \) be bijections. Then \( g \circ f \) is a bijection and \( {\left( g \circ f\right) }^{-1} = {f}^{-1} \circ {g}^{-1} \) . | Proof. Let \( f : A \rightarrow B \) and \( g : B \rightarrow C \) be bijections. Then \( {f}^{-1} : B \rightarrow A \) and \( {g}^{-1} : C \rightarrow B \) . Hence, \( {f}^{-1} \circ {g}^{-1} : C \rightarrow A \) . Also, by Theorem 6.20, \( g \circ f : A \rightarrow C \) is a bijection, and hence \( {\left( g \circ f\... | Yes |
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