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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\[ {f}^{\prime }\left( c\right) = 0\text{.} \]
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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: \( f\left( a\right) = f\left( b\right) = M = m \) . In this case \( f\left( x\right) \) is constant (why?). Therefore \( {f}^{\prime }\left( x\right) = 0 \) for every \( x \in \left( {a, b}\right) \) .\n\nCase 2: \( f\left( a\right) = f\left( b\right) = M \neq m \) . In this case there is a real number \( c \in \left( {a, b}\right) \) such that \( f\left( c\right) \) is a local minimum. By Fermat’s Theorem, \( {f}^{\prime }\left( c\right) = 0 \) .\n\nCase 3: \( f\left( a\right) = f\left( b\right) = m \neq M \) . In this case there is a real number \( c \in \left( {a, b}\right) \) such that \( f\left( c\right) \) is a local maximum. By Fermat’s Theorem, \( {f}^{\prime }\left( c\right) = 0 \) .\n\nCase 4: \( f\left( a\right) = f\left( b\right) \) is neither a maximum nor a minimum. In this case there is a real number \( {c}_{1} \in \left( {a, b}\right) \) such that \( f\left( {c}_{1}\right) \) is a local maximum, and a real number \( {c}_{2} \in \left( {a, b}\right) \) such that \( f\left( {c}_{2}\right) \) is a local minimum. By Fermat’s Theorem, \( {f}^{\prime }\left( {c}_{1}\right) = {f}^{\prime }\left( {c}_{2}\right) = 0 \) .
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Yes
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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 \( \left( {a, b}\right) \) .
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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 \( {f}^{\prime }\left( \alpha \right) > 0 \) and \( d - c > 0 \) we have \( f\left( d\right) - f\left( c\right) > 0 \), or \( f\left( d\right) > f\left( c\right) \) .
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Yes
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Problem 132. Use the definition of continuity to prove that the constant function \( g\left( x\right) = c \) is continuous at any point \( a \) .
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Problem 133.
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No
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Problem 135. Let \( \left( {x}_{n}\right) ,\left( {y}_{n}\right) \) be sequences as in the NIP. Show that for all \( n, m \in \mathbb{N},{x}_{n} \leq {y}_{m} \) .
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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 \( \left. n\right\rbrack \) .
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No
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Theorem 25. Suppose \( a \in \mathbb{R}, a \geq 0 \) . There exists a real number \( c \geq 0 \) such that \( {c}^{2} = a \) .
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Notice that we can’t just say,\
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No
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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.
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(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\n\n(b) Let \( {z}_{n} = \ln \left( {n + 1}\right) - \left( {\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n + 1}}\right) \) . Use a similar diagram to show that\n\n\[ {z}_{1} \leq {z}_{2} \leq {z}_{3} \leq \cdots \]\n\n(c) Let \( {y}_{n} = 1 - {z}_{n} \) . Show that \( \left( {x}_{n}\right) \) and \( \left( {y}_{n}\right) \) satisfy the hypotheses of the nested interval property and use the NIP to conclude that there is a real number \( \gamma \) such that\n\n\( {x}_{n} \leq \gamma \leq {y}_{n} \) for all \( n \) .\n\n(d) Conclude that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}\right) - \ln \left( {n + 1}\right) }\right) = \gamma \) .
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Yes
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Theorem 26. (Intermediate Value Theorem) 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 \( f\left( c\right) = v \) .
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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 \( {x}_{1} \leq {y}_{1} \) and \( f\left( {x}_{1}\right) \leq v \leq f\left( {y}_{1}\right) \) . Let \( {m}_{1} \) be the midpoint of \( \left\lbrack {{x}_{1},{y}_{1}}\right\rbrack \) and notice that we have either \( f\left( {m}_{1}\right) \leq v \) or \( f\left( {m}_{1}\right) \geq v \) . If \( f\left( {m}_{1}\right) \leq v \), then we relabel \( {x}_{2} = {m}_{1} \) and \( {y}_{2} = {y}_{1} \) . If \( f\left( {m}_{1}\right) \geq v \), then we relabel \( {x}_{2} = {x}_{1} \) and \( {y}_{2} = {m}_{1} \) . In either case, we end up with \( {x}_{1} \leq {x}_{2} \leq {y}_{2} \leq {y}_{1},{y}_{2} - {x}_{2} = \frac{1}{2}\left( {{y}_{1} - {x}_{1}}\right) \) , \( f\left( {x}_{1}\right) \leq v \leq f\left( {y}_{1}\right) \), and \( f\left( {x}_{2}\right) \leq v \leq f\left( {y}_{2}\right) .\n\nNow play the same game with the interval \( \left\lbrack {{x}_{2},{y}_{2}}\right\rbrack \) . If we keep playing this game, we will generate two sequences \( \left( {x}_{n}\right) \) and \( \left( {y}_{n}\right) \), satisfying all of the conditions of the nested interval property. These sequences will also satisfy the following extra property: \( \forall n, f\left( {x}_{n}\right) \leq v \leq f\left( {y}_{n}\right) \) . By the NIP, there exists a \( c \) such that \( {x}_{n} \leq c \leq {y}_{n},\forall n \) . This should be the \( c \) that we seek though this is not obvious. Specifically, we need to show that \( f\left( c\right) = v \) . This should be where the continuity of \( f \) at \( c \) and the extra property on \( \left( {x}_{n}\right) \) and \( \left( {y}_{n}\right) \) come into play. QED?
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No
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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 \( x \in \left\lbrack {a, b}\right\rbrack \) .
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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 \( f \) .) IF the sequence \( \left( {x}_{n}\right) \) converged to something in \( \left\lbrack {a, b}\right\rbrack \), say \( c \), then we would have our contradiction. Indeed, we would have \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = c \) . By the continuity of \( f \) at \( c \) and Theorem 15 of Chapter 6, we would have \( \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = f\left( c\right) \) . This would say that the sequence \( \left( {f\left( {x}_{n}\right) }\right) \) converges, so by Lemma 2 of Chapter 4, it must be bounded. This would provide our contradiction, as we had \( \left| {f\left( {x}_{n}\right) }\right| > n \) , for all positive integers \( n \) . QED?
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No
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Example 14. Given the sequence \( \left( {x}_{n}\right) \), the following are subsequences.
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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.
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Yes
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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 \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = c \) .
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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 original sequence, we will need to use different letters in setting ourselves up for the NIP. With this in mind, let \( {a}_{1} = a \) and \( {b}_{1} = b \), and notice that \( {x}_{n} \in \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack \) for infinitely many \( n \) . (This is, in fact true for all \( n \), but you’ll see why we said it the way we did.) Let \( {m}_{1} \) be the midpoint of \( \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack \) and notice that either \( {x}_{n} \in \left\lbrack {{a}_{1},{m}_{1}}\right\rbrack \) for infinitely many \( n \) or \( {x}_{n} \in \left\lbrack {{m}_{1},{b}_{1}}\right\rbrack \) for infinitely many \( n \) . If \( {x}_{n} \in \left\lbrack {{a}_{1},{m}_{1}}\right\rbrack \) for infinitely many \( n \), then we relabel \( {a}_{2} = {a}_{1} \) and \( {b}_{2} = {m}_{1} \) . If \( {x}_{n} \in \left\lbrack {{m}_{1},{b}_{1}}\right\rbrack \) for infinitely many \( n \), then relabel \( {a}_{2} = {m}_{1} \) and \( {b}_{2} = {b}_{1} \) . In either case, we get \( {a}_{1} \leq {a}_{2} \leq {b}_{2} \leq {b}_{1},{b}_{2} - {a}_{2} = \frac{1}{2}\left( {{b}_{1} - {a}_{1}}\right) \), and \( {x}_{n} \in \left\lbrack {{a}_{2},{b}_{2}}\right\rbrack \) for infinitely many \( n \) .\n\nNow we consider the interval \( \left\lbrack {{a}_{2},{b}_{2}}\right\rbrack \) and let \( {m}_{2} \) be the midpoint of \( \left\lbrack {{a}_{2},{b}_{2}}\right\rbrack \) . Since \( {x}_{n} \in \left\lbrack {{a}_{2},{b}_{2}}\right\rbrack \) for infinitely many \( n \), then either \( {x}_{n} \in \left\lbrack {{a}_{2},{m}_{2}}\right\rbrack \) for infinitely many \( n \) or \( {x}_{n} \in \left\lbrack {{m}_{2},{b}_{2}}\right\rbrack \) for infinitely many \( n \) . If \( {x}_{n} \in \left\lbrack {{a}_{2},{m}_{2}}\right\rbrack \) for infinitely many \( n \), then we relabel \( {a}_{3} = {a}_{2} \) and \( {b}_{3} = {m}_{2} \) . If \( {x}_{n} \in \left\lbrack {{m}_{2},{b}_{2}}\right\rbrack \) for infinitely many \( n \), then we relabel \( {a}_{3} = {m}_{2} \) and \( {b}_{3} = {b}_{2} \) . In either case, we get \( {a}_{1} \leq {a}_{2} \leq \) \( {a}_{3}
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No
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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.
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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 \( \exists x \in S \) such that \( x \geq {x}_{1} \) (namely, \( {x}_{1} \) itself) and \( \forall x \in S,{y}_{1} \geq x \) . You probably guessed what’s coming next: let \( {m}_{1} \) be the midpoint of \( \left\lbrack {{x}_{1},{y}_{1}}\right\rbrack \) . Notice that either \( {m}_{1} \geq x,\forall x \in S \) or \( \exists x \in S \) such that \( x \geq {m}_{1} \) . In the former case, we relabel, letting \( {x}_{2} = {x}_{1} \) and \( {y}_{2} = {m}_{1} \) . In the latter case, we let \( {x}_{2} = {m}_{1} \) and \( {y}_{2} = {y}_{1} \) . In either case, we end up with \( {x}_{1} \leq {x}_{2} \leq {y}_{2} \leq {y}_{1} \) , \( {y}_{2} - {x}_{2} = \frac{1}{2}\left( {{y}_{1} - {x}_{1}}\right) \), and \( \exists x \in S \) such that \( x \geq {x}_{2} \) and \( \forall x \in S,{y}_{2} \geq x \) . If we continue this process, we end up with two sequences, \( \left( {x}_{n}\right) \) and \( \left( {y}_{n}\right) \), satisfying the following conditions:\n\n1. \( {x}_{1} \leq {x}_{2} \leq {x}_{3} \leq \ldots \)\n\n2. \( {y}_{1} \geq {y}_{2} \geq {y}_{3} \geq \ldots \)\n\n3. \( \forall n,{x}_{n} \leq {y}_{n} \)\n\n4. \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{y}_{n} - {x}_{n}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{2}^{n - 1}}\left( {{y}_{1} - {x}_{1}}\right) = 0 \)\n\n5. \( \forall n,\exists x \in S \) such that \( x \geq {x}_{n} \) and \( \forall x \in S,{y}_{n} \geq x \) ,\n\nBy properties 1-5 and the NIP there exists \( c \) such that \( {x}_{n} \leq c \leq {y}_{n},\forall n \) . We will leave it to you to use property 5 to show that \( c = \sup S \) .
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No
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Problem 148. Complete the above ideas to provide a formal proof of Theorem 29.
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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 bound.
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No
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Problem 149. Prove Corollary 5.
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[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!]
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No
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Consider the following curious expression \( \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{\ldots }}}} \) .
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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\[ \n{x}_{1} = \sqrt{2} \n\]\n\n\[ \n{x}_{n + 1} = \sqrt{2 + {x}_{n}}. \n\]\n\n(a) Use induction to show that \( {x}_{n} < 2 \) for \( n = 1,2,3,\ldots \) .\n\n(b) Use the result from part (a) to show that \( {x}_{n} < {x}_{n + 1} \) for \( n = 1,2,3,\ldots \) .\n\n(c) From Corollary 5, we have that \( \left( {x}_{n}\right) \) must converge to some number \( c \) . Use the fact that \( \left( {\bar{x}}_{n + 1}\right) \) must converge to \( c \) as well to compute what \( c \) must be.
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No
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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 \( x \in \left\lbrack {a, b}\right\rbrack \) .
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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 upper bound which we will label \( s \), so that \( s = \sup f\left\lbrack {a, b}\right\rbrack \) . This says that \( s \geq f\left( x\right) \), for all \( x \in \left\lbrack {a, b}\right\rbrack \) . All we need to do now is find a \( c \in \left\lbrack {a, b}\right\rbrack \) with \( f\left( c\right) = s \) . With this in mind, notice that since \( s = \sup f\left\lbrack {a, b}\right\rbrack \), then for any positive integer \( n, s - \frac{1}{n} \) is not an upper bound of \( f\left\lbrack {a, b}\right\rbrack \) . Thus there exists \( {x}_{n} \in \left\lbrack {a, b}\right\rbrack \) with \( s - \frac{1}{n} < f\left( {x}_{n}\right) \leq s \) . Now, by the Bolzano-Weierstrass Theorem, \( \left( {x}_{n}\right) \) has a convergent subsequence \( \left( {x}_{{n}_{k}}\right) \) converging to some \( c \in \left\lbrack {a, b}\right\rbrack \) . Using the continuity of \( f \) at \( c \), you should be able to show that \( f\left( c\right) = s \) . To find the minimum of \( f \), find the maximum of \( - f \) .
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No
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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 \( \left\lbrack {0, b}\right\rbrack \) .
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[Hint: \( \left| {{x}^{n} - 0}\right| = {x}^{n} \leq {b}^{n} \).
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No
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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 \( I \) .
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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\[ \left| {f\left( x\right) - f\left( a\right) }\right| = \left| {f\left( x\right) - {f}_{n}\left( x\right) + {f}_{n}\left( x\right) - {f}_{n}\left( a\right) + {f}_{n}\left( a\right) - f\left( a\right) }\right| \]\n\n\[ \leq \left| {f\left( x\right) - {f}_{n}\left( x\right) }\right| + \left| {{f}_{n}\left( x\right) - {f}_{n}\left( a\right) }\right| + \left| {{f}_{n}\left( a\right) - f\left( a\right) }\right| \]\n\nIf we choose \( n \) large enough, then we can make the first and last terms as small as we wish, noting that the uniform convergence makes the first term uniformly small for all \( x \) . Once we have a specific \( n \), then we can use the continuity of \( {f}_{n} \) to find a \( \delta > 0 \) such that the middle term is small whenever \( x \) is within \( \delta \) of \( a \) . QED?
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No
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Problem 168. Prove Theorem 34.
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[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 \( x \) .]
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No
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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.
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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.
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No
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Problem 171. Prove Lemma 5.
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[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 \( B = \max \left( {\left| {s}_{1}\right| ,\left| {s}_{2}\right| ,\ldots ,\left| {s}_{\lceil N\rceil }\right| ,\left| {s}_{m}\right| + 1}\right) \) .]
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No
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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 \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = \) \( s \) .
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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 \( \left| {{s}_{n} - s}\right| = \left| {{s}_{n} - {s}_{{n}_{k}} + {s}_{{n}_{k}} - s}\right| \leq \left| {{s}_{n} - {s}_{{n}_{k}}}\right| + \left| {{s}_{{n}_{k}} - s}\right| \) . If we choose \( n \) and \( {n}_{k} \) large enough, we should be able to make each term arbitrarily small. QED?
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No
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Problem 175. Prove the Cauchy criterion.
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\( \diamond \)
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No
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Problem 178. Prove Theorem 39.
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[Hint: Use the Cauchy criterion with the fact that \( \left| {\mathop{\sum }\limits_{{k = n + 1}}^{m}{a}_{k}}\right| \leq \mathop{\sum }\limits_{{k = n + 1}}^{m}\left| {a}_{k}\right| \) .]
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No
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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 \( s \) .
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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 sequence which is bounded by \( s \) . By Corollary 5 of Chapter 7, this sequence must converge to some number \( t \) and \( t \leq s \) . Furthermore \( \sum {a}_{n} \) is also a rearrangement of \( \sum {b}_{n} \) . Thus the result holds for this special case. (Why?) For the general case, notice that \( {a}_{n} = \frac{\left| {a}_{n}\right| + {a}_{n}}{2} - \frac{\left| {a}_{n}\right| - {a}_{n}}{2} \) and that \( \sum \frac{\left| {a}_{n}\right| + {a}_{n}}{2} \) and \( \sum \frac{\left| {a}_{n}\right| - {a}_{n}}{2} \) are both convergent series with nonnegative terms. By the special case \( \sum \frac{\left| {b}_{n}\right| + {b}_{n}}{2} = \sum \frac{\left| {a}_{n}\right| + {a}_{n}}{2} \) and \( \sum \frac{\left| {b}_{n}\right| - {b}_{n}}{2} = \sum \frac{\left| {a}_{n}\right| - {a}_{n}}{2}. \)
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No
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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 \( \left| x\right| < \left| c\right| \) .
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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\[ \left| {{a}_{n}{x}^{n}}\right| = \left| {{a}_{n}{c}^{n} \cdot {\left( \frac{x}{c}\right) }^{n}}\right| \leq B{\left| \frac{x}{c}\right| }^{n}. \]\n\nWe can now use the comparison test.
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Yes
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Problem 183. Prove Corollary 9.
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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 when \( \left| x\right| < r \) and diverges when \( \left| x\right| > r \) . In the latter case, we call \( r \) the radius of convergence of the power series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \) . In the former case, we say that the radius of convergence of \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \) is \( \infty \) . Though we can say that \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \) converges absolutely when \( \left| x\right| < r \), we cannot say that the convergence is uniform. However, we can come close. We can show that the convergence is uniform for \( \left| x\right| \leq b < r \) .
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Yes
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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\[ \left| {{f}_{n}\left( x\right) }\right| \leq {M}_{n},\forall x \in S, n = 1,2,3,\ldots \]\n\nIf \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{M}_{n} \) converges then \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n}\left( x\right) \) converges uniformly on \( S \) to some function (which we will denote by \( f\left( x\right) \) ).
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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\nFirst observe that for any \( x \in S,\mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n}\left( x\right) \) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by \( f\left( x\right) \) . This actually defines the function \( f\left( x\right) \) for all \( x \in S \) . It follows that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n}\left( x\right) \) converges pointwise to \( f\left( x\right) \).\n\nNext, let \( \varepsilon > 0 \) be given. Notice that since \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{M}_{n} \) converges, say to \( M \) , then there is a real number, \( N \), such that if \( n > N \), then\n\n\[ \mathop{\sum }\limits_{{k = n + 1}}^{\infty }{M}_{k} = \left| {\mathop{\sum }\limits_{{k = n + 1}}^{\infty }{M}_{k}}\right| = \left| {M - \mathop{\sum }\limits_{{k = 1}}^{n}{M}_{k}}\right| < \varepsilon . \]\n\nYou should be able to use this to show that if \( n > N \), then\n\n\[ \left| {f\left( x\right) - \mathop{\sum }\limits_{{k = 1}}^{n}{f}_{k}\left( x\right) }\right| < \varepsilon ,\forall x \in S. \]
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No
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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) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {{x}^{n} - {x}^{n - 1}}\right) \; \) (b) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\left( {x}^{n} - {x}^{n - 1}\right) }{n}\; \) (c) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\left( {x}^{n} - {x}^{n - 1}\right) }{{n}^{2}}\;\diamond \)
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Using the Weierstrass- \( M \) test, we can prove the following result. Theorem 43. Suppose
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No
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Problem 188. Show that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }n{x}^{n - 1} \) converges for \( \left| x\right| < 1 \) .
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[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.]
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No
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Problem 189. Prove Theorem 44.
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[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.]
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No
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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 \( \left\lbrack {-r, r}\right\rbrack \) .
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[Hint: For \( \left| x\right| \leq r,\left| {{a}_{n}{x}^{n}}\right| \leq \left| {{a}_{n}{r}^{n}}\right| \).
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No
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Prove Theorem 45.
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[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 \( 0 \leq x \leq r \) . By Lemma 7,\n\n\[ \left| {\mathop{\sum }\limits_{{k = n + 1}}^{m}{a}_{k}{x}^{k}}\right| = \left| {\mathop{\sum }\limits_{{k = n + 1}}^{m}{a}_{k}{r}^{k}{\left( \frac{x}{r}\right) }^{k}}\right| \leq \left( \frac{\epsilon }{2}\right) {\left( \frac{x}{r}\right) }^{n + 1} \leq \frac{\epsilon }{2}. \]\n\nThus for \( 0 \leq x \leq r, n > N \) ,\n\n\[ \left| {\mathop{\sum }\limits_{{k = n + 1}}^{\infty }{a}_{k}{x}^{k}}\right| = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| {\mathop{\sum }\limits_{{k = n + 1}}^{m}{a}_{k}{x}^{k}}\right| \leq \frac{\epsilon }{2} < \epsilon .\rbrack
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Yes
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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 \( c \in \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \) for all \( n \in \mathbb{N} \) .
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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 \( n,{a}_{n} \leq c \leq {b}_{n} \) . This says \( c \in \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \) for all \( n \in \mathbb{N} \) .
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Yes
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Theorem 52. Show that \( \mathbb{Q} \) is countable.
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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 part (a) of Problem 200. Does this complete the proof or is there more to do? QEI
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No
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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 for all \( n = 1,\ldots, k \) \( {r}_{n} \in \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \) and
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\[ \mathop{\sum }\limits_{{n = 1}}^{k}{b}_{n} - {a}_{n} = \mathop{\sum }\limits_{{n = 1}}^{k}\frac{\varepsilon }{k} = \varepsilon . \]
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No
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Prove: If \( \left| S\right| = n \), then \( \left| {P\left( S\right) }\right| = {2}^{n} \) .
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[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\[ \n\left\{ {a}_{1}\right\} \leftrightarrow \{ Y, N, N,\ldots, N\} \n\] \n\n\[ \n\left\{ {a}_{2}\right\} \leftrightarrow \{ N, Y, N,\ldots, N\} \n\] \n\n\[ \n\vdots \n\] \n\n\[ \nS \leftrightarrow \{ Y, Y, Y,\ldots, Y\} \n\] \n\nHow many elements are in \( T \) ?]
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No
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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.
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[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 \( X - \left( {{Y}_{0} \cup {Y}_{1} \cup {Y}_{2}}\right) \) is an infinite set then it contains an infinite set \( {Y}_{3} \), etc. For \( n = 1,2,3,\ldots \), let \( {f}_{n} : {Y}_{n - 1} \rightarrow {Y}_{n} \) be a one-to-one correspondence and define \( f : X \rightarrow X - Y \) by\n\n\[ \left\{ {\begin{array}{ll} f\left( x\right) = {f}_{n}\left( x\right) , & \text{ if }x \in {Y}_{n}, n = 0,1,2,\ldots \\ f\left( x\right) = x, & \text{ if }x \in X - \left( {{ \cup }_{n = 0}^{\infty }{Y}_{n}}\right) \end{array}.}\right. \]\n\nShow that \( f \) is one-to-one and onto.]
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No
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Problem 210. Show that \( 0 \neq 1 \) .
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[Hint: Show that if \( x \neq 0 \), then \( 0 \cdot x \neq x \) .] \( \diamond \)
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No
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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
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\[ {\left( {s}_{n} + {t}_{n}\right) }_{n = 1}^{\infty } \equiv {\left( {\sigma }_{n} + {\tau }_{n}\right) }_{n = 1}^{\infty }. \]
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No
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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\[ \n{0}^{ * } + x = x.\n\]
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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 \( \left( {0,0,0,\ldots }\right) \) to represent \( {0}^{ * } \) and any element of \( x \), say \( \left( {{x}_{1},{x}_{2},\ldots }\right) \) , to represent \( x \) . Then\n\n\[ \n\left( {0,0,0,\ldots }\right) + \left( {{x}_{1},{x}_{2},{x}_{3},\ldots }\right) = \left( {{x}_{1},{x}_{2},{x}_{3},\ldots }\right)\n\]\n\n\[ \n= x\text{;}\n\]\n\nSince any other sequences taken from \( {0}^{ * } \) and \( x \) respectively, will yield a sum equivalent to \( x \) (see Problem 215) we conclude that\n\n\[ \n{0}^{ * } + x = x.\n\]
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Yes
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Theorem 56. (Closure with Respect to Addition) If \( \alpha \) and \( \beta \) are cuts then \( \alpha + \beta \) is a cut.
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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 number in \( \beta \) and let \( {y}_{1} \) be a rational number not in \( \beta \) . Then by Property II \( y < {y}_{1} \) .\n\nThus since \( x + y \) represents a generic element of \( \alpha + \beta \) and \( x + y < {x}_{1} + {y}_{1} \) , it follows that \( {x}_{1} + {y}_{1} \notin \alpha + \beta \) .\n\n## Proof of Property II\n\nWe will show that the contrapositive of Property II is true: If \( x \in \alpha + \beta \) and \( y < x \) then \( y \in \alpha + \beta \) .\n\nFirst, let \( x \in \alpha + \beta \) . Then there are \( {x}_{\alpha } \in \alpha \) and \( {x}_{\beta } \in \beta \) such that \( y < x = \) \( {x}_{\alpha } + {x}_{\beta } \) . Therefore \( \frac{y}{{x}_{\alpha } + {x}_{\beta }} < 1 \), so that\n\n\[ \n{x}_{\alpha }\left( \frac{y}{{x}_{\alpha } + {x}_{\beta }}\right) < {x}_{\alpha }\n\]\n\nand\n\n\[ \n{x}_{\beta }\left( \frac{y}{{x}_{\alpha } + {x}_{\beta }}\right) < {x}_{\beta }\n\]\n\nTherefore \( {x}_{\alpha }\left( \frac{y}{{x}_{\alpha } + {x}_{\beta }}\right) \in \alpha \) and \( {x}_{\beta }\left( \frac{y}{{x}_{\alpha } + {x}_{\beta }}\right) \in \beta \) . Therefore\n\n\[ \ny = {x}_{\alpha }\left( \frac{y}{{x}_{\alpha } + {x}_{\beta }}\right) + {x}_{\beta }\left( \frac{y}{{x}_{\alpha } + {x}_{\beta }}\right) \in \alpha + \beta .\n\]\n\n## Proof of Property III\n\nLet \( z \in \alpha + \beta \) . We need to find \( w > z, w \in \alpha + \beta \) . Observe that for some \( x \in \alpha \) and \( y \in \beta \)\n\n\[ \nz = x + y.\n\]\n\nSince \( \alpha \) is a cut, there is a rational number \( {x}_{1} \in \alpha \) such that \( {x}_{1} > x \) . Take \( w = {x}_{1} + y \in \alpha + \beta \) . Then\n\n\[ \nw = {x}_{1} + y > x + y = z.\n\]\n\nThis completes the proof of Theorem 56.
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Yes
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Problem 221. Prove Lemma 8.
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[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 is not in \( \beta \) then relabel it as \( {s}_{2} \) and relabel \( {r}_{1} \) as \( {r}_{2} \), etc.]
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No
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Problem 222. Let \( \alpha \) and \( \beta \) be cuts with \( \beta < \alpha \) . Prove that \( \beta + \left( {\alpha - \beta }\right) = \alpha \) .
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[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 without loss of generality that \( x < y \) . (Why?) Choose \( z \in \alpha \) with \( y < z \) . By the Lemma 8, there exists positive rational numbers \( r \) and \( s \) with \( r \in \beta \) , \( s \in \beta, s < z \), and \( s - r < z - x \) . Show that \( x < r + \left( {z - s}\right) \.]
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No
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Cauchy sequences converge
|
148
|
No
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Cantor’s first proof that \( \mathbb{R} \) is uncountable
|
160
|
No
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Theorem 1.8. If \( x \) and \( y \) are odd integers, then \( x \cdot y \) is an odd integer.
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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\[ x \cdot y = \left( {{2m} + 1}\right) \left( {{2n} + 1}\right) \]\n\n\[ = {4mn} + {2m} + {2n} + 1 \]\n\n\[ = 2\left( {{2mn} + m + n}\right) + 1\text{.}\]\n\nSince \( m \) and \( n \) are integers and the integers are closed under addition and multiplication, we conclude that \( \left( {{2mn} + m + n}\right) \) is an integer. This means that \( x \cdot y \) has been written in the form \( \left( {{2q} + 1}\right) \) for some integer \( q \), and hence, \( x \cdot y \) is an odd integer. Consequently, it has been proven that if \( x \) and \( y \) are odd integers, then \( x \cdot y \) is an odd integer.
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Yes
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The statement \( \neg \left( {P \vee Q}\right) \) is logically equivalent to \( \neg P \land \neg Q \) . This can be written as \( \neg \left( {P \vee Q}\right) \equiv \neg P \land \neg Q \) .
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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</td><td>F</td><td>F</td><td>F</td></tr><tr><td>\( \mathrm{T} \)</td><td>F</td><td>\( \mathrm{T} \)</td><td>\( \mathrm{F} \)</td><td>F</td><td>\( \mathrm{T} \)</td><td>F</td></tr><tr><td>F</td><td>\( \mathrm{T} \)</td><td>T</td><td>F</td><td>\( \mathrm{T} \)</td><td>F</td><td>F</td></tr><tr><td>F</td><td>F</td><td>F</td><td>T</td><td>\( \mathrm{T} \)</td><td>\( \mathrm{T} \)</td><td>\( \mathrm{T} \)</td></tr></table>
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Yes
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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 \\equiv \\neg Q \\rightarrow \\neg P \\; (contrapositive)\n\n\[ \nP \\rightarrow Q \\equiv \\neg P \\vee Q \n\]\n\n\[ \n\\neg \\left( {P \\rightarrow Q}\\right) \\equiv P \\land \\neg Q \n\]\n\nBiconditional Statement \\;\\left( {P \\leftrightarrow Q}\\right) \\equiv \\left( {P \\rightarrow Q}\\right) \\land \\left( {Q \\rightarrow P}\\right) \n\]\n\nDouble Negation \\;\\neg \\left( {\\neg P}\\right) \\equiv P \n\]\n\nDistributive Laws \\;P \\vee \\left( {Q \\land R}\\right) \\equiv \\left( {P \\vee Q}\\right) \\land \\left( {P \\vee R}\\right) \n\]\n\n\[ \nP \\land \\left( {Q \\vee R}\\right) \\equiv \\left( {P \\land Q}\\right) \\vee \\left( {P \\land R}\\right) \n\]\n\nConditionals with \\;P \\rightarrow \\left( {Q \\vee R}\\right) \\equiv \\left( {P\\land \\neg Q}\\right) \\rightarrow R \n\nDisjunctions
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We have already established many of these equivalencies. Others will be established in the exercises.
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No
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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\[ \n\neg \left( {\exists x \in U}\right) \left\lbrack {P\left( x\right) }\right\rbrack \equiv \left( {\forall x \in U}\right) \left\lbrack {\neg P\left( x\right) }\right\rbrack . \n\]
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## 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 \( x,{x}^{3} \geq {x}^{2} \) . \n\n- If \( x \) is a real number, then \( {x}^{3} \) is greater than or equal to \( {x}^{2} \) . \n\nThe second statement shows that in a conditional statement, there is often a hidden universal quantifier. This statement is false since there are real numbers \( x \) for which \( {x}^{3} \) is not greater than or equal to \( {x}^{2} \) . For example, we could use \( x = - 1 \) or \( x = \frac{1}{2} \) . \n\nThis means that the negation must be true. We can form the negation as follows: \n\n\[ \n\neg \left( {\forall x \in \mathbb{R}}\right) \left( {{x}^{3} \geq {x}^{2}}\right) \equiv \left( {\exists x \in \mathbb{R}}\right) \neg \left( {{x}^{3} \geq {x}^{2}}\right) . \n\] \n\nIn most cases, we want to write this negation in a way that does not use the negation symbol. In this case, we can now write the open sentence \( \neg \left( {{x}^{3} \geq {x}^{2}}\right) \) as \( \left( {{x}^{3} < {x}^{2}}\right) \) . (That is, the negation of \
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Yes
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Theorem 2.8. Important Logical Equivalencies. For statements P, Q, and R,\n\n\[ \text{De Morgan’s Laws}\;\neg \left( {P \land Q}\right) \equiv \neg P \vee \neg Q \]
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\[ \neg \left( {P \vee Q}\right) \equiv \neg P \land \neg Q \]
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No
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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 \( c \) .
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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 \( s \) and \( t \) such that\n\n\[ b = a \cdot s\text{, and} \]\n\n(1)\n\n\[ c = b \cdot t\text{.} \]\n\n(2)\n\nWe can now substitute the expression for \( b \) from equation (1) into equation (2). This gives\n\n\[ c = \left( {a \cdot s}\right) \cdot t. \]\n\nUsing the associate property for multiplication, we can rearrange the right side of the last equation to obtain\n\n\[ c = a \cdot \left( {s \cdot t}\right) . \]\n\nBecause both \( s \) and \( t \) are integers, and since the integers are closed under multiplication, we know that \( s \cdot t \in \mathbb{Z} \) . Therefore, the previous equation proves that \( a \) divides \( c \) . Consequently, we have proven that whenever \( a, b \), and \( c \) are integers with \( a \neq 0 \) and \( b \neq 0 \) such that \( a \) divides \( b \) and \( b \) divides \( c \), then \( a \) divides \( c \) .
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Yes
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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 \( \left( {a + b}\right) \equiv 2\left( {\;\operatorname{mod}\;8}\right) \) .
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## Progress Check 3.6 (Proving Proposition 3.5)\n\nWe will use \
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No
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Theorem 3.7. For each integer \( n \), if \( {n}^{2} \) is an even integer, then \( n \) is an even integer.
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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 an odd integer. So using \( x = y = n \), we can conclude that if \( n \) is an odd integer, then \( n \cdot n \), or \( {n}^{2} \), is an odd integer. We have thus proved the contrapositive of the theorem, and consequently, we have proved that if \( {n}^{2} \) is an even integer, then \( n \) is an even integer.
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Yes
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For all real numbers \( a \) and \( b \), if \( a \neq 0 \) and \( b \neq 0 \), then \( {ab} \neq 0 \) .
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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 \( a \neq 0 \), then \( b = 0 \) .\n\nTo prove this, we let \( a \) and \( b \) be real numbers and assume that \( {ab} = 0 \) and \( a \neq 0 \) . We can then multiply both sides of the equation \( {ab} = 0 \) by \( \frac{1}{a} \) . This gives\n\nTherefore, \( b = 0 \) . This completes the proof of a statement that is logically equivalent to the contrapositive, and hence, we have proven the proposition.
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Yes
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Proposition 3.11. Let \( x \in \mathbb{R} \) . The real number \( x \) equals 2 if and only if \( {x}^{3} - 2{x}^{2} + x = 2 \) .
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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 assume \( x = 2 \) and prove that \( {x}^{3} - 2{x}^{2} + x = 2 \) . We can do this by substituting \( x = 2 \) into the expression \( {x}^{3} - 2{x}^{2} + x \) . This gives\n\n\[ \n{x}^{3} - 2{x}^{2} + x = {2}^{3} - 2\left( {2}^{2}\right) + 2 \n\]\n\n\[ \n= 8 - 8 + 2 \n\]\n\n\[ \n= 2\text{.} \n\]\n\nThis completes the first part of the proof.\n\nFor the second part, we assume that \( {x}^{3} - 2{x}^{2} + x = 2 \) and from this assumption, we will prove that \( x = 2 \) . We will do this by solving this equation for \( x \) . To do so, we first rewrite the equation \( {x}^{3} - 2{x}^{2} + x = 2 \) by subtracting 2 from both sides:\n\n\[ \n{x}^{3} - 2{x}^{2} + x - 2 = 0. \n\]\n\nWe can now factor the left side of this equation by factoring an \( {x}^{2} \) from the first two terms and then factoring \( \left( {x - 2}\right) \) from the resulting two terms. This is shown\n\nbelow.\n\n\[ \n{x}^{3} - 2{x}^{2} + x - 2 = 0 \n\]\n\n\[ \n{x}^{2}\left( {x - 2}\right) + \left( {x - 2}\right) = 0 \n\]\n\n\[ \n\left( {x - 2}\right) \left( {{x}^{2} + 1}\right) = 0 \n\]\n\nNow, in the real numbers, if a product of two factors is equal to zero, then one of the factors must be zero. So this last equation implies that\n\n\[ \nx - 2 = 0\text{ or }{x}^{2} + 1 = 0. \n\]\n\nThe equation \( {x}^{2} + 1 = 0 \) has no real number solution. So since \( x \) is a real number, the only possibility is that \( x - 2 = 0 \) . From this we can conclude that \( x \) must be equal to 2 .\n\nSince we have now proven both conditional statements, we have proven that \( x = 2 \) if and only if \( {x}^{3} - 2{x}^{2} + x = 2 \) .
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Yes
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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 \( x = \frac{c - b}{a} \) .
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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\nThis shows that if there is a solution, then it must be \( x = \frac{c - b}{a} \) . We also see that if \( x = \frac{c - b}{a} \), then\n\n\[ {ax} + b = a\left( \frac{c - b}{a}\right) + b \]\n\n\[ = \left( {c - b}\right) + b \]\n\n\[ = c\text{.} \]\n\nTherefore, the linear equation \( {ax} + b = c \) has exactly one real number solution and the solution is \( x = \frac{c - b}{a} \) .
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Yes
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Proposition 3.14. For each real number \( x \), if \( 0 < x < 1 \), then \( \frac{1}{x\left( {1 - x}\right) } \geq 4 \) .
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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 \( \left( {1 - x}\right) > 0 \) . Hence, \( x\left( {1 - x}\right) > 0 \) and if we multiply both sides of inequality (1) by \( x\left( {1 - x}\right) \) , we obtain\n\n\[ 1 < {4x}\left( {1 - x}\right) \text{.} \]\n\nWe can now use algebra to rewrite the last inequality as follows:\n\n\[ 1 < {4x} - 4{x}^{2} \]\n\n\[ 4{x}^{2} - {4x} + 1 < 0 \]\n\n\[ {\left( 2x - 1\right) }^{2} < 0 \]\n\nHowever, \( \left( {{2x} - 1}\right) \) is a real number and the last inequality says that a real number squared is less than zero. This is a contradiction since the square of any real number must be greater than or equal to zero. Hence, the proposition cannot be false, and we have proved that for each real number \( x \), if \( 0 < x < 1 \), then \( \frac{1}{x\left( {1 - x}\right) } \geq 4 \) . ∎
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Yes
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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 \( {x}^{2} + {y}^{2} = {z}^{2} \) .
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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 \( x = {2m} + 1 \) and \( y = {2n} + 1 \) .\n\n1. Use the assumptions that \( x \) and \( y \) are odd to prove that \( {x}^{2} + {y}^{2} \) is even and hence, \( {z}^{2} \) is even. (See Theorem 3.7 on page 105.)\n\nWe can now conclude that \( z \) is even. (See Theorem 3.7 on page 105.) So there exists an integer \( k \) such that \( z = {2k} \) . If we substitute for \( x, y \), and \( z \) in the equation \( {x}^{2} + {y}^{2} = {z}^{2} \), we obtain\n\n\[{\left( 2m + 1\right) }^{2} + {\left( 2n + 1\right) }^{2} = {\left( 2k\right) }^{2}.\n\]\n\n2. Use the previous equation to obtain a contradiction. Hint: One way is to use algebra to obtain an equation where the left side is an odd integer and the right side is an even integer.
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No
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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.
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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 rational numbers, we know that \( \frac{1}{x} \in \mathbb{Q} \) . We now know that \( x \cdot y \) and \( \frac{1}{x} \) are rational numbers and since the rational numbers are closed under multiplication, we conclude that\n\n\[ \frac{1}{x} \cdot \left( {xy}\right) \in \mathbb{Q}. \]\n\nHowever, \( \frac{1}{x} \cdot \left( {xy}\right) = y \) and hence, \( y \) must be a rational number. Since a real number cannot be both rational and irrational, this is a contradiction to the assumption that \( y \) is irrational. We have therefore proved that 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.
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Yes
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Theorem 3.20. If \( r \) is a real number such that \( {r}^{2} = 2 \), then \( r \) is an irrational number.
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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\[ r = \frac{m}{n} \]\n\nand \( m \) and \( n \) have no common factor greater than 1 . We will obtain a contradiction by showing that \( m \) and \( n \) must both be even. Squaring both sides of the last equation and using the fact that \( {r}^{2} = 2 \), we obtain\n\n\[ 2 = \frac{{m}^{2}}{{n}^{2}} \]\n\n\[ {m}^{2} = 2{n}^{2}\text{.} \]\n\n(1)\n\nEquation (1) implies that \( {m}^{2} \) is even, and hence, by Theorem 3.7, \( m \) must be an even integer. This means that there exists an integer \( p \) such that \( m = {2p} \) . We can now substitute this into equation (1), which gives\n\n\[ {\left( 2p\right) }^{2} = 2{n}^{2} \]\n\n\[ 4{p}^{2} = 2{n}^{2} \]\n\n(2)\n\nWe can divide both sides of equation (2) by 2 to obtain \( {n}^{2} = 2{p}^{2} \) . Consequently, \( {n}^{2} \) is even and we can once again use Theorem 3.7 to conclude that \( n \) is an even integer.\n\nWe have now established that both \( m \) and \( n \) are even. This means that 2 is a common factor of \( m \) and \( n \), which contradicts the assumption that \( m \) and \( n \) have no common factor greater than 1. Consequently, the statement of the theorem cannot be false, and we have proved that if \( r \) is a real number such that \( {r}^{2} = 2 \), then \( r \) is an irrational number.
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Yes
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Proposition 1. If \( n \) is an integer, then \( \left( {{n}^{2} + n}\right) \) is an even integer.
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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 properties that the element \( x \) may or may not possess. The cases must be chosen so that they exhaust all possibilities for the object \( x \) in the hypothesis of the original proposition. For Proposition 1, we know that an integer must be even or it must be odd. We can thus use the following two cases for the integer \( n \) :\n\n- The integer \( n \) is an even integer;\n\n- The integer \( n \) is an odd integer.
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No
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For all real numbers \( a \) and \( b \), if \( {ab} = 0 \), then \( a = 0 \) or \( b = 0 \) .
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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 \( a \neq 0 \), we can multiply both sides of the equation \( {ab} = 0 \)\n\nby \( \frac{1}{a} \) and obtain\n\n\[ \frac{1}{a} \cdot {ab} = \frac{1}{a} \cdot 0 \]\n\n\[ b = 0\text{.} \]\n\nSo in both cases, \( a = 0 \) or \( b = 0 \), and this proves that for all real numbers \( a \) and \( b \) , if \( {ab} = 0 \), then \( a = 0 \) or \( b = 0 \) .
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Yes
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Let a be a positive real number. For each real number \( x \) , 1. \( \left| x\right| = a \) if and only if \( x = a \) or \( x = - a \) .
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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 \( \left| x\right| = a \), we can conclude that \( x = a \) . In the case where \( x < 0 \), we see that \( \left| x\right| = - x \) . Since \( \left| x\right| = a \), we can conclude that \( - x = a \) and hence that \( x = - a \) . These two cases prove that if \( \left| x\right| = a \), then \( x = a \) or \( x = - a \) . We will now prove that if \( x = a \) or \( x = - a \), then \( \left| x\right| = a \) . We start by assuming that \( x = a \) or \( x = - a \) . Since the hypothesis of this conditional statement is a disjunction, we use two cases. When \( x = a \), we see that \[ \left| x\right| = \left| a\right| = a\;\text{ since }a > 0. \] When \( x = - a \), we conclude that \[ \left| x\right| = \left| {-a}\right| = - \left( {-a}\right) \;\text{ since } - a < 0, \] and hence, \( \left| x\right| = a \) . This proves that if \( x = a \) or \( x = - a \), then \( \left| x\right| = a \) . Because we have proven both conditional statements, we have proven that \( \left| x\right| = a \) if and only if \( x = a \) or \( x = - a \) .
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Yes
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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 Triangle Inequality.
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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 \( x \in \mathbb{R} \) and first assume that \( \left| x\right| < a \) . We will use two cases: either \( x \geq 0 \) or \( x < 0 \) .\n\n- In the case where \( x \geq 0 \), we know that \( \left| x\right| = x \) and so the inequality \( \left| x\right| < a \) implies that \( x < a \) . However, we also know that \( - a < 0 \) and that \( x > 0 \) . Therefore, we conclude that \( - a < x \) and, hence, \( - a < x < a \) .\n\n- When \( x < 0 \), we see that \( \left| x\right| = - x \) . Therefore, the inequality \( \left| x\right| < a \) implies that \( - x < a \), which in turn implies that \( - a < x \) . In this case, we also know that \( x < a \) since \( x \) is negative and \( a \) is positive. Hence, \( - a < x < a \) .\n\nSo in both cases, we have proven that \( - a < x < a \) and this proves that if \( \left| x\right| < a \), then \( - a < x < a \) . We now assume that \( - a < x < a \) .\n\n- If \( x \geq 0 \), then \( \left| x\right| = x \) and hence, \( \left| x\right| < a \) .\n\n- If \( x < 0 \), then \( \left| x\right| = - x \) and so \( x = - \left| x\right| \) . Thus, \( - a < - \left| x\right| \) . By multiplying both sides of the last inequality by -1, we conclude that \( \left| x\right| < a \) .\n\nThese two cases prove that if \( - a < x < a \), then \( \left| x\right| < a \) . Hence, we have proven that \( \left| x\right| < a \) if and only if \( - a < x < a \) .
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No
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Proposition 3.27. If \( n \) is an integer, then 3 divides \( {n}^{3} - n \) .
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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 consider the following three cases: (1) \( r = 0 \) ; (2) \( r = 1 \) ; and (3) \( r = 2 \) .\n\nIn the case where \( r = 0 \), we have \( n = {3q} \) . By substituting this into the expression \( {n}^{3} - n \), we get\n\n\[ {n}^{3} - n = {\left( 3q\right) }^{3} - \left( {3q}\right) \]\n\n\[ = {27}{q}^{3} - {3q} \]\n\n\[ = 3\left( {9{q}^{3} - q}\right) \text{.}\]\n\nSince \( \left( {9{q}^{3} - q}\right) \) is an integer, the last equation proves that \( 3 \mid \left( {{n}^{3} - n}\right) \) .\n\nIn the second case, \( r = 1 \) and \( n = {3q} + 1 \) . When we substitute this into \( \left( {{n}^{3} - n}\right) \), we obtain\n\n\[ {n}^{3} - n = {\left( 3q + 1\right) }^{3} - \left( {{3q} + 1}\right) \]\n\n\[ = \left( {{27}{q}^{3} + {27}{q}^{2} + {9q} + 1}\right) - \left( {{3q} + 1}\right) \]\n\n\[ = {27}{q}^{3} + {27}{q}^{2} + {6q} \]\n\n\[ = 3\left( {9{q}^{3} + 9{q}^{2} + {2q}}\right) \text{.}\]\n\nSince \( \left( {9{q}^{3} + 9{q}^{2} + {2q}}\right) \) is an integer, the last equation proves that \( 3 \mid \left( {{n}^{3} - n}\right) \) .\n\nThe last case is when \( r = 2 \) . The details for this case are part of Exercise (1). Once this case is completed, we will have proved that 3 divides \( {n}^{3} - n \) in all three cases. Hence, we may conclude that if \( n \) is an integer, then 3 divides \( {n}^{3} - n \) .
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No
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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 \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \), then \( b \equiv a\left( {\;\operatorname{mod}\;n}\right) \). This is called the symmetric property of congruence modulo \( n \). 3. If \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) and \( b \equiv c\left( {\;\operatorname{mod}\;n}\right) \), then \( a \equiv c\left( {\;\operatorname{mod}\;n}\right) \). This is called the transitive property of congruence modulo \( n \).
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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 \( n \) divides \( \left( {a - a}\right) \) and hence, by the definition of congruence modulo \( n \), we have proven that \( a \equiv a\left( {\;\operatorname{mod}\;n}\right) \) . To prove the transitive property, we let \( n \in \mathbb{N} \), and let \( a, b \), and \( c \) be integers. We assume that \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) and that \( b \equiv c\left( {\;\operatorname{mod}\;n}\right) \) . We will use the definition of congruence modulo \( n \) to prove that \( a \equiv c\left( {\;\operatorname{mod}\;n}\right) \) . Since \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) and \( b \equiv c\left( {\;\operatorname{mod}\;n}\right) \), we know that \( n \mid \left( {a - b}\right) \) and \( n \mid \left( {b - c}\right) \) . Hence, there exist integers \( k \) and \( q \) such that \[ a - b = {nk} \] \[ b - c = {nq}\text{.} \] By adding the corresponding sides of these two equations, we obtain \[ \left( {a - b}\right) + \left( {b - c}\right) = {nk} + {nq}. \] If we simplify the left side of the last equation and factor the right side, we get \[ a - c = n\left( {k + q}\right) . \] By the closure property of the integers, \( \left( {k + q}\right) \in \mathbb{Z} \), and so this equation proves that \( n \mid \left( {a - c}\right) \) and hence that \( a \equiv c\left( {\;\operatorname{mod}\;n}\right) \) . This completes the proof of the transitive property of congruence modulo \( n \) .
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No
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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 \( a \equiv r\left( {\;\operatorname{mod}\;n}\right) \) .
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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.
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No
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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 \( {a}^{2} \equiv 4\left( {\;\operatorname{mod}\;5}\right) \) .
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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) \( a \equiv 1\left( {\;\operatorname{mod}\;5}\right) ,\left( 2\right) a \equiv 2\left( {\;\operatorname{mod}\;5}\right) ,\left( 3\right) a \equiv 3\left( {\;\operatorname{mod}\;5}\right) \), and (4) \( a \equiv 4\left( {\;\operatorname{mod}\;5}\right) \) . Following are proofs for the first and fourth cases.\n\nCase 1. \( \left( {a \equiv 1\left( {\;\operatorname{mod}\;5}\right) }\right) \) . In this case, we use Theorem 3.28 to conclude that\n\n\[ \n{a}^{2} \equiv {1}^{2}\left( {\;\operatorname{mod}\;5}\right) \;\text{ or }\;{a}^{2} \equiv 1\left( {\;\operatorname{mod}\;5}\right) .\n\]\n\nThis proves that if \( a \equiv 1\left( {\;\operatorname{mod}\;5}\right) \), then \( {a}^{2} \equiv 1\left( {\;\operatorname{mod}\;5}\right) .\n\nCase 4. \( \left( {a \equiv 4\left( {\;\operatorname{mod}\;5}\right) }\right) \) . In this case, we use Theorem 3.28 to conclude that\n\n\[ \n{a}^{2} \equiv {4}^{2}\left( {\;\operatorname{mod}\;5}\right) \;\text{ or }\;{a}^{2} \equiv {16}\left( {\;\operatorname{mod}\;5}\right) .\n\]\n\nWe also know that \( {16} \equiv 1\left( {\;\operatorname{mod}\;5}\right) \) . So we have \( {a}^{2} \equiv {16}\left( {\;\operatorname{mod}\;5}\right) \) and \( {16} \equiv \) 1 (mod 5), and we can now use the transitive property of congruence (Theorem 3.30) to conclude that \( {a}^{2} \equiv 1\left( {\;\operatorname{mod}\;5}\right) \) . This proves that if \( a \equiv 4\left( {\;\operatorname{mod}\;5}\right) \) , then \( {a}^{2} \equiv 1\left( {\;\operatorname{mod}\;5}\right) \) .
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No
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Proposition 4.2. For each natural number \( n \) , \n\n\[ \n{1}^{2} + {2}^{2} + \cdots + {n}^{2} = \frac{n\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}. \n\]
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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 \( \frac{1\left( {1 + 1}\right) \left( {2 \cdot 1 + 1}\right) }{6} = 1 \) . This shows that \n\n\[ \n{1}^{2} = \frac{1\left( {1 + 1}\right) \left( {2 \cdot 1 + 1}\right) }{6} \n\]\n\nwhich proves that \( P\left( 1\right) \) is true.\n\nFor the inductive step, we prove that for each \( k \in \mathbb{N} \), if \( P\left( k\right) \) is true, then \( P\left( {k + 1}\right) \) is true. So let \( k \) be a natural number and assume that \( P\left( k\right) \) is true. That is, assume that \n\n\[ \n{1}^{2} + {2}^{2} + \cdots + {k}^{2} = \frac{k\left( {k + 1}\right) \left( {{2k} + 1}\right) }{6}. \n\]\n\n(1)\n\nThe goal now is to prove that \( P\left( {k + 1}\right) \) is true. That is, it must be proved that \n\n\[ \n{1}^{2} + {2}^{2} + \cdots + {k}^{2} + {\left( k + 1\right) }^{2} = \frac{\left( {k + 1}\right) \left\lbrack {\left( {k + 1}\right) + 1}\right\rbrack \left\lbrack {2\left( {k + 1}\right) + 1}\right\rbrack }{6} \n\]\n\n\[ \n= \frac{\left( {k + 1}\right) \left( {k + 2}\right) \left( {{2k} + 3}\right) }{6}\text{.} \n\]\n\n(2)\n\nTo do this, we add \( {\left( k + 1\right) }^{2} \) to both sides of equation (1) and algebraically rewrite the right side of the resulting equation. This gives \n\n\[ \n{1}^{2} + {2}^{2} + \cdots + {k}^{2} + {\left( k + 1\right) }^{2} = \frac{k\left( {k + 1}\right) \left( {{2k} + 1}\right) }{6} + {\left( k + 1\right) }^{2} \n\]\n\n\[ \n= \frac{k\left( {k + 1}\right) \left( {{2k} + 1}\right) + 6{\left( k + 1\right) }^{2}}{6} \n\]\n\n\[ \n= \frac{\left( {k + 1}\right) \left\lbrack {k\left( {{2k} + 1}\right) + 6\left( {k + 1}\right) }\right\rbrack }{6} \n\]\n\n\[ \n= \frac{\left( {k + 1}\right) \left( {2{k}^{2} + {7k} + 6}\right) }{6} \n\]\n\n\[ \n= \frac{\left( {k + 1}\right) \left( {k + 2}\right) \left( {{2k} + 3}\right) }{6}\text{. } \n\]\n\nComparing this result to equation (2), we see that if \( P\left( k\right) \) is true, then \( P\left( {k + 1}\right) \) is true. Hence, the inductive step has been established, and by the Principle of Mathematical Induction, we have proved that for each natural number \( n \) , \( {1}^{2} + {2}^{2} + \cdots + {n}^{2} = \frac{n\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}. \)
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Yes
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Proposition 4.4. For every natural number \( n,4 \) divides \( \left( {{5}^{n} - 1}\right) \) .
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Proof. (Proof by Mathematical Induction) For each natural number \( n \), let \( P\left( n\right) \) be \
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No
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Proposition 4.7. For each natural number \( n \) with \( n \geq 4, n! > {2}^{n} \) .
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Proof. We will use a proof by mathematical induction. For this proof, we let\n\n\[ P\left( n\right) \text{be \
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No
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Theorem 4.9. Each natural number greater than 1 either is a prime number or is a product of prime numbers.
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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 number with \( k \\geq 2 \) . We assume that \( P\\left( 2\\right), P\\left( 3\\right) ,\\ldots, P\\left( k\\right) \) are true. That is, we assume that each of the natural numbers \( 2,3,\\ldots, k \) is a prime number or a product of prime numbers. The goal is to prove that \( P\\left( {k + 1}\\right) \) is true or that \( \\left( {k + 1}\\right) \) is a prime number or a product of prime numbers.\n\nCase 1: If \( \\left( {k + 1}\\right) \) is a prime number, then \( P\\left( {k + 1}\\right) \) is true.\n\nCase 2: If \( \\left( {k + 1}\\right) \) is not a prime number, then \( \\left( {k + 1}\\right) \) can be factored into a product of natural numbers with each one being less than \( \\left( {k + 1}\\right) \) . That is, there exist natural numbers \( a \) and \( b \) with\n\n\\[ k + 1 = a \\cdot b,\\;\\text{ and }\\;1 < a \\leq k\\text{ and }1 < b \\leq k.\n\\]\nUsing the inductive assumption, this means that \( P\\left( a\\right) \) and \( P\\left( b\\right) \) are both true. Consequently, \( a \) and \( b \) are prime numbers or are products of prime numbers. Since \( k + 1 = a \\cdot b \), we conclude that \( \\left( {k + 1}\\right) \) is a product of prime numbers. That is, we conclude that \( P\\left( {k + 1}\\right) \) is true. This proves the inductive step.\n\nHence, by the Second Principle of Mathematical Induction, we conclude that \( P\\left( n\\right) \) is true for all \( n \\in \\mathbb{N} \) with \( n \\geq 2 \), and this means that each natural number greater than 1 is either a prime number or is a product of prime numbers.
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Yes
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Proposition 4.13. For each natural number \( n \), the Fibonacci number \( {f}_{3n} \) is an even natural number.
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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.
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No
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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 \( n \in \mathbb{N},{a}_{n} = a \cdot {r}^{n - 1} \) .
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The proof of this proposition is Exercise (6).
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No
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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) .
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The proof of Proposition 4.16 is Exercise (8).
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No
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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 \( B \) .
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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 to consider: (1) \( x \) is not an element of \( Y \), and (2) \( x \) is an element of \( Y \) .\n\nCase 1: Assume that \( x \notin Y \) . Let \( y \in Y \) . Then \( y \in A \) and \( y \neq x \) . Since\n\n\[ A = B \cup \{ x\} \]\n\nthis means that \( y \) must be in \( B \) . Therefore, \( Y \subseteq B \) .\n\nCase 2: Assume that \( x \in Y \) . In this case, let \( C = Y - \{ x\} \) . Then every element of \( C \) is an element of \( B \) . Hence, we can conclude that \( C \subseteq B \) and that \( Y = C \cup \{ x\} \) .\n\nCases (1) and (2) show that if \( Y \subseteq A \), then \( Y \subseteq B \) or \( Y = C \cup \{ x\} \), where \( C \subseteq B \) .
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Yes
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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 \( T \) .
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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 word \
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Yes
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Proposition 5.10. Let \( A \) and \( B \) be subsets of the universal set \( U \) . If \( A \subseteq B \), then \( {B}^{c} \subseteq {A}^{c} \)
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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><th>Step</th><th>Know</th><th>Reason</th></tr></thead><tr><td>\( P \)</td><td>\( A \subseteq B \)</td><td>Hypothesis</td></tr><tr><td>\( {P1} \)</td><td>Let \( x \in {B}^{c} \) .</td><td>Choose an arbitrary element of \( {B}^{c} \) .</td></tr><tr><td>\( {P2} \)</td><td>If \( x \in A \), then \( x \in B \) .</td><td>Definition of \
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No
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Proposition 5.11. Let \( A \) and \( B \) be subsets of some universal set. Then \( A - \left( {A - B}\right) = A \cap B. \)
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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 that
\[
x \in A\text{and}x \notin \left( {A - B}\right) \text{.}
\]
We know that an element is in \( \left( {A - B}\right) \) if and only if it is in \( A \) and not in \( B \). Since \( x \notin \left( {A - B}\right) \), we conclude that \( x \notin A \) or \( x \in B \). However, we also know that \( x \in A \) and so we conclude that \( x \in B \). This proves that
\[
x \in A\text{and}x \in B\text{.}
\]
This means that \( x \in A \cap B \), and hence we have proved that \( A - \left( {A - B}\right) \subseteq A \cap B \).
Now choose \( y \in A \cap B \). This means that
\[
y \in A\text{and}y \in B\text{.}
\]
We note that \( y \in \left( {A - B}\right) \) if and only if \( y \in A \) and \( y \notin B \) and hence, \( y \notin \left( {A - B}\right) \) if and only if \( y \notin A \) or \( y \in B \). Since we have proved that \( y \in B \), we conclude that \( y \notin \left( {A - B}\right) \), and hence, we have established that \( y \in A \) and \( y \notin \left( {A - B}\right) \). This proves that if \( y \in A \cap B \), then \( y \in A - \left( {A - B}\right) \) and hence, \( A \cap B \subseteq A - \left( {A - B}\right) \).
Since we have proved that \( A - \left( {A - B}\right) \subseteq A \cap B \) and \( A \cap B \subseteq A - \left( {A - B}\right) \), we conclude that \( A - \left( {A - B}\right) = A \cap B \).
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Yes
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Proposition 5.14. Let \( A \) and \( B \) be subsets of some universal set. Then \( A \subseteq B \) if and only if \( A \cap {B}^{c} = \varnothing \) .
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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 \( A \cap {B}^{c} \neq \varnothing \) . We will prove that \( A \nsubseteq B \) by proving that there must exist an element \( x \) such that \( x \in A \) and \( x \notin B \) .\n\nSince \( A \cap {B}^{c} \neq \varnothing \), there exists an element \( x \) that is in \( A \cap {B}^{c} \) . This means that\n\n\[ x \in A\text{and}x \in {B}^{c}\text{.} \]\n\nNow, the fact that \( x \in {B}^{c} \) means that \( x \notin B \) . Hence, we can conclude that\n\n\[ x \in A\text{and}x \notin B\text{.} \]\n\nThis means that \( A \nsubseteq B \), and hence, we have proved that if \( A \cap {B}^{c} \neq \varnothing \), then \( A \nsubseteq B \), and therefore, we have proved that if \( A \subseteq B \), then \( A \cap {B}^{c} = \varnothing \) .
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Yes
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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 \( \left( {{ax} + {by}}\right) \) .
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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 \( t \) divides \( a \), there exists an integer \( m \) such that . . .
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Yes
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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\[ \text{- If}A \subseteq B\text{, then}A \cap C \subseteq B \cap C\text{and}A \cup C \subseteq B \cup C\text{.} \]
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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.
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No
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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\[ \n\\text{and the Universal Set}\\;A \\cup \\varnothing = A\\;A \\cup U = U\n\]\n\n\[ \n\\text{Idempotent Laws}\\;A \\cap A = A\\;A \\cup A = A\n\]\n\n\[ \n\\text{Commutative Laws}\\;A \\cap B = B \\cap A\\;A \\cup B = B \\cup A\n\]\n\n\[ \n\\left( {A \\cap B}\\right) \\cap C = A \\cap \\left( {B \\cap C}\\right)\n\]\n\n\[ \n\\left( {A \\cup B}\\right) \\cup C = A \\cup \\left( {B \\cup C}\\right)\n\]\n\n\[ \nA \\cap \\left( {B \\cup C}\\right) = \\left( {A \\cap B}\\right) \\cup \\left( {A \\cap C}\\right)\n\]\n\n\[ \nA \\cup \\left( {B \\cap C}\\right) = \\left( {A \\cup B}\\right) \\cap \\left( {A \\cup C}\\right)\n\]
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## 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 \( P \\land Q \) is logically equivalent to \( Q \\land P \) . Consequently, we can conclude that\n\n\[ \nx \\in A\\text{and}x \\in B\\text{if and only if}x \\in B\\text{and}x \\in A\\text{.\n\]\n\n(2)\n\nNow we know that\n\n\[ \nx \\in B\\text{and}x \\in A\\text{if and only if}x \\in B \\cap A\\text{.\n\]\n\n(3)\n\nThis means that we can use (1), (2), and (3) to conclude that\n\n\[ \nx \\in A \\cap B\\text{if and only if}x \\in B \\cap A\\text{,\n\]\n\nand, hence, we have proved that \( A \\cap B = B \\cap A \) .
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Yes
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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\[{\left( A \cup B\right) }^{c} = {A}^{c} \cap {B}^{c}\]
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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 proving that an element is in \( {\left( A \cup B\right) }^{c} \) if and only if it is in \( {A}^{c} \cap {B}^{c} \) . So let \( x \) be in the universal set \( U \) .\n\n\[x \in {\left( A \cup B\right) }^{c}\text{ if and only if }x \notin A \cup B,\]\n\n(1)\n\nand\n\n\[x \notin A \cup B\text{if and only if}x \notin A\text{and}x \notin B.\]\n\n(2)\n\nCombining (1) and (2), we see that\n\n\[x \in {\left( A \cup B\right) }^{c}\text{ if and only if }x \notin A\text{ and }x \notin B.\]\n\n(3)\n\n\n\nIn addition, we know that\n\n\[x \notin A\text{and}x \notin B\text{if and only if}x \in {A}^{c}\text{and}x \in {B}^{c},\]\n\n(4)\n\nand this is true if and only if \( x \in {A}^{c} \cap {B}^{c} \) . So we can use (3) and (4) to conclude that\n\n\[x \in {\left( A \cup B\right) }^{c}\text{ if and only if }x \in {A}^{c} \cap {B}^{c},\]\n\nand, hence, that \( {\left( A \cup B\right) }^{c} = {A}^{c} \cap {B}^{c} \) .
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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. \( {A}^{c} \cup B = U \)
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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 \( A \cap {B}^{c} = \varnothing \) and use the fact that \( {\varnothing }^{c} = U \) . We then see that\n\n\[ \n{\left( A \cap {B}^{c}\right) }^{c} = {\varnothing }^{c} \n\]\n\nThen, using one of De Morgan's Laws, we obtain\n\n\[ \n{A}^{c} \cup {\left( {B}^{c}\right) }^{c} = U \n\]\n\n\[ \n{A}^{c} \cup B = U.\text{.} \n\]\nThis completes the proof that (2) implies (3).\n\nWe now need to prove that (3) implies (1). We assume that \( {A}^{c} \cup B = U \) and will prove that \( A \subseteq B \) by proving that every element of \( A \) must be in \( B \) .\n\nSo let \( x \in A \) . Then we know that \( x \notin {A}^{c} \) . However, \( x \in U \) and since \( {A}^{c} \cup B = U \), we can conclude that \( x \in {A}^{c} \cup B \) . Since \( x \notin {A}^{c} \), we conclude that \( x \in B \) . This proves that \( A \subseteq B \) and hence that (3) implies (1).\n\nSince we have now proved that (1) implies (2), that (2) implies (3), and that (3) implies (1), we have proved that the three conditions are equivalent.
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Yes
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\[ \text{2.}A \times \left( {B \cup C}\right) = \left( {A \times B}\right) \cup \left( {A \times C}\right) \]
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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 Sections 5.2 and 5.3.\n\nThe other thing to remember is that the elements of a Cartesian product are ordered pairs. So when we start a proof of a result such as Part (2) of Theorem 5.25, the primary goal is to prove that the two sets are equal. We will do this by proving that each one is a subset of the other one. So if we want to prove that \( A \times \left( {B \cup C}\right) \subseteq \) \( \left( {A \times B}\right) \cup \left( {A \times C}\right) \), we can start by choosing an arbitrary element of \( A \times \left( {B \cup C}\right) \) . The goal is then to show that this element must be in \( \left( {A \times B}\right) \cup \left( {A \times C}\right) \) . When we start by choosing an arbitrary element of \( A \times \left( {B \cup C}\right) \), we could give that element a name. For example, we could start by letting\n\n\[ u\text{be an element of}A \times \left( {B \cup C}\right) \text{.} \]\n\n(1)\n\nWe can then use the definition of \
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No
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Theorem 5.25 (Part (2)). Let \( A, B \), and \( C \) be sets. Then\n\n\[ A \times \left( {B \cup C}\right) = \left( {A \times B}\right) \cup \left( {A \times C}\right) . \]
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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 \( A \times \left( {B \cup C}\right) \subseteq \left( {A \times B}\right) \cup \left( {A \times C}\right) \), we let \( u \in A \times \left( {B \cup C}\right) \) . Then there exists \( x \in A \) and there exists \( y \in B \cup C \) such that \( u = \left( {x, y}\right) \) . Since \( y \in B \cup C \), we know that \( y \in B \) or \( y \in C \) .\n\nIn the case where \( y \in B \), we have \( u = \left( {x, y}\right) \), where \( x \in A \) and \( y \in B \) . So in this case, \( u \in A \times B \), and hence \( u \in \left( {A \times B}\right) \cup \left( {A \times C}\right) \) . Similarly, in the case where \( y \in C \), we have \( u = \left( {x, y}\right) \), where \( x \in A \) and \( y \in C \) . So in this case, \( u \in A \times C \) and, hence, \( u \in \left( {A \times B}\right) \cup \left( {A \times C}\right) \) .\n\nIn both cases, \( u \in \left( {A \times B}\right) \cup \left( {A \times C}\right) \) . Hence, we may conclude that if \( u \) is an element of \( A \times \left( {B \cup C}\right) \), then \( u \in \left( {A \times B}\right) \cup \left( {A \times C}\right) \), and this proves that\n\n\[ A \times \left( {B \cup C}\right) \subseteq \left( {A \times B}\right) \cup \left( {A \times C}\right) . \]\n\n(1)\n\nWe must now prove that \( \left( {A \times B}\right) \cup \left( {A \times C}\right) \subseteq A \times \left( {B \cup C}\right) \) . So we let \( v \in \left( {A \times B}\right) \cup \left( {A \times C}\right) \) . Then \( v \in \left( {A \times B}\right) \) or \( v \in \left( {A \times C}\right) \) .\n\nIn the case where \( v \in \left( {A \times B}\right) \), we know that there exists \( s \in A \) and there exists \( t \in B \) such that \( v = \left( {s, t}\right) \) . But since \( t \in B \), we know that \( t \in B \cup C \), and hence \( v \in A \times \left( {B \cup C}\right) \) . Similarly, in the case where \( v \in \left( {A \times C}\right) \), we know that there exists \( s \in A \) and there exists \( t \in C \) such that \( v = \left( {s, t}\right) \) . But because \( t \in C \) , we can conclude that \( t \in B \cup C \) and, hence, \( v \in A \times \left( {B \cup C}\right) \) .\n\nIn both cases, \( v \in A \times \left( {B \cup C}\right) \) . Hence, we may conclude that if \( v \in \left( {A \times B}\right) \cup \left( {A \times C}\right) \), then \( v \in A \times \left( {B \cup C}\right) \), and this proves that\n\n\[ \left( {A \times B}\right) \cup \left( {A \times C}\right) \subseteq A \times \left( {B \cup C}\right) . \]\n\n(2)\n\nThe relationships in (1) and (2) prove that \( A \times \left( {B \cup C}\right) = \left( {A \times B}\right) \cup \left( {A \times C}\right) \) .
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Yes
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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 positive real numbers, then we have a family of sets indexed by \( {\mathbb{R}}^{ + } \) . We will first determine the union of this family of sets. Notice that for each \( \alpha \in {\mathbb{R}}^{ + },\alpha \in {A}_{\alpha } \), and if \( y \) is a real number with \( - 1 < y \leq 0 \), then \( y \in {A}_{\alpha } \) . Also notice that if \( y \in \mathbb{R} \) and \( y < - 1 \), then for each \( \alpha \in {\mathbb{R}}^{ + }, y \notin {A}_{\alpha } \) . With these observations, we conclude that\n\n\[ \n\mathop{\bigcup }\limits_{{\alpha \in {\mathbb{R}}^{ + }}}{A}_{\alpha } = \left( {-1,\infty }\right) = \{ x \in \mathbb{R} \mid - 1 < x\} .\n\]
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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 \( \alpha \in {\mathbb{R}}^{ + } \) ; and\n\n- if \( y \in \mathbb{R} \) and \( y > 0 \), then if we let \( \beta = \frac{y}{2}, y > \beta \) and \( y \notin {A}_{\beta } \).\n\nFrom these observations, we conclude that\n\n\[ \n\mathop{\bigcap }\limits_{{\alpha \in {\mathbb{R}}^{ + }}}{A}_{\alpha } = ( - 1,0\rbrack = \{ x \in \mathbb{R} \mid - 1 < x \leq 0\} .\n\]
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Yes
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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 \( \beta \in \Lambda ,\mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha } \subseteq {A}_{\beta } \).\n\n2. For each \( \beta \in \Lambda ,{A}_{\beta } \subseteq \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha } \).\n\n\[ \text{3.}{\left( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} = \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \]\n\n\[ \text{4.}{\left( \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} = \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \]
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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 \( \beta \in \Lambda \) and note that if \( x \in \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha } \), then \( x \in {A}_{\alpha } \), for all \( \alpha \in \Lambda \) . Since \( \beta \) is one element in \( \Lambda \), we may\n\nconclude that \( x \in {A}_{\beta } \) . This proves that \( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha } \subseteq {A}_{\beta } \).\n\nTo prove Part (3), we will prove that each set is a subset of the other set. We first let \( x \in {\left( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} \) . This means that \( x \notin \left( {\mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }}\right) \), and this means that\n\nthere exists a \( \beta \in \Lambda \) such that \( x \notin {A}_{\beta } \).\n\nHence, \( x \in {A}_{\beta }^{c} \), which implies that \( x \in \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \) . Therefore, we have proved that\n\n\[ {\left( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} \subseteq \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \]\n\n(1)\n\nWe now let \( y \in \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \) . This means that there exists a \( \beta \in \Lambda \) such that \( y \in {A}_{\beta }^{c} \) or \( y \notin {A}_{\beta } \) . However, since \( y \notin {A}_{\beta } \), we may conclude that \( y \notin \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha } \) and, hence, \( y \in {\left( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} \) . This proves that\n\n\[ \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \subseteq {\left( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} \]\n\n(2)\n\nUsing the results in (1) and (2), we have proved that \( {\left( \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }\right) }^{c} = \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }^{c} \) .
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No
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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\[ \text{1.}B \cap \left( {\mathop{\bigcup }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }}\right) = \mathop{\bigcup }\limits_{{\alpha \in \Lambda }}\left( {B \cap {A}_{\alpha }}\right) \text{, and} \]\n\n\[ \text{2.}B \cup \left( {\mathop{\bigcap }\limits_{{\alpha \in \Lambda }}{A}_{\alpha }}\right) = \mathop{\bigcap }\limits_{{\alpha \in \Lambda }}\left( {B \cup {A}_{\alpha }}\right) \text{.} \]
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The proof of Theorem 5.31 is Exercise (5).
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No
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Proposition 1. For all \( a, b \in \mathbb{R} \), if \( g\left( a\right) = g\left( b\right) \), then \( a = b \) .
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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\[ \n{5a} + 3 = {5b} + 3.\n\]
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Yes
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Proposition 2. For all \( b \in \mathbb{R} \), there exists an \( a \in \mathbb{R} \) such that \( g\left( a\right) = b \) .
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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 \( g\left( a\right) = {5a} + 3 = b \) .
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Yes
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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 produce the same output.
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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 \( f \) .
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No
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Is the function \( F \) a surjection? That is, does \( F \) map \( \mathbb{R} \) onto \( T \) ?
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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\n\[ {x}^{2} + 1 = y\text{.} \]\n\nOne way to proceed is to work backward and solve the last equation (if possible) for \( x \) . Doing so, we get\n\n\[ {x}^{2} = y - 1 \]\n\n\[ x = \sqrt{y - 1}\text{ or }x = - \sqrt{y - 1}. \]\n\nNow, since \( y \in T \), we know that \( y \geq 1 \) and hence that \( y - 1 \geq 0 \) . This means that \( \sqrt{y - 1} \in \mathbb{R} \) . Hence, if we use \( x = \sqrt{y - 1} \), then \( x \in \mathbb{R} \), and\n\n\[ F\left( x\right) = F\left( \sqrt{y - 1}\right) \]\n\n\[ = {\left( \sqrt{y - 1}\right) }^{2} + 1 \]\n\n\[ = \left( {y - 1}\right) + 1 \]\n\n\[ = y\text{.} \]\n\nThis proves that \( F \) is a surjection since we have shown that for all \( y \in T \), there exists an \( x \in \mathbb{R} \) such that \( F\left( x\right) = y \) .
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Yes
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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 Examples 6.12 and 6.13.) Following is a table of values for some inputs for the function \( g \) .\n\n<table><thead><tr><th>\( x \)</th><th>\( g\left( x\right) \)</th><th>\( x \)</th><th>\( g\left( x\right) \)</th></tr></thead><tr><td>0</td><td>1</td><td>3</td><td>10</td></tr><tr><td>1</td><td>2</td><td>4</td><td>17</td></tr><tr><td>2</td><td>5</td><td>5</td><td>26</td></tr></table>\n\nNotice that the codomain is \( \mathbb{N} \), and the table of values suggests that some natural numbers are not outputs of this function. So it appears that the function \( g \) is not a surjection.
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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 there exists an \( x \in {\mathbb{Z}}^{ * } \) with \( g\left( x\right) = 3 \) . Then\n\n\[ \n{x}^{2} + 1 = 3 \n\]\n\n\[ \n{x}^{2} = 2 \n\]\n\n\[ \nx = \pm \sqrt{2}\text{.} \n\]\n\nBut this is not possible since \( \sqrt{2} \notin {\mathbb{Z}}^{ * } \) . Therefore, there is no \( x \in {\mathbb{Z}}^{ * } \) with \( g\left( x\right) = 3 \) . This means that for every \( x \in {\mathbb{Z}}^{ * }, g\left( x\right) \neq 3 \) . Therefore,3 is not in the range of \( g \), and hence \( g \) is not a surjection.\n\nThe table of values suggests that different inputs produce different outputs, and hence that \( g \) is an injection. To prove that \( g \) is an injection, assume that \( s, t \in {\mathbb{Z}}^{ * } \)\n\n(the domain) with \( g\left( s\right) = g\left( t\right) \) . Then\n\n\[ \n{s}^{2} + 1 = {t}^{2} + 1 \n\]\n\n\[ \n{s}^{2} = {t}^{2} \n\]\n\nSince \( s, t \in {\mathbb{Z}}^{ * } \), we know that \( s \geq 0 \) and \( t \geq 0 \) . So the preceding equation implies that \( s = t \) . Hence, \( g \) is an injection.
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Yes
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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 \( \left( {a, b}\right) \in f \) and \( \left( {a, c}\right) \in f \), then \( b = c \) .\n\nIf we use \( f\left( a\right) = b \) whenever \( \left( {a, b}\right) \in f \), then \( f \) is a function from \( A \) to \( B \) .
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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 a function is the most rigorous definition of a function. It allows us to use set theory to work with and compare functions. For example, equality of functions becomes a question of equality of sets. Therefore, many textbooks will use the ordered pair representation of a function as the definition of a function.
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No
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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 \( b \in B \) ,
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\[ f\left( a\right) = b\text{ if and only if }{f}^{-1}\left( b\right) = a. \]
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No
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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 that \( {\mathbb{R}}^{ + } \) is the codomain of \( g \) . We can then say that both \( f \) and \( g \) are bijections. Consequently, the inverses of these functions are also functions.
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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><tr><th>Theorem 6.26</th><th>Translates to:</th></tr></thead><tr><td>For \( x, y \in \mathbb{R}, f\left( x\right) = y \) if and only if \( {f}^{-1}\left( y\right) = x \) .</td><td>For \( x, y \in \mathbb{R},{x}^{3} = y \) if and only if \( \sqrt[3]{y} = x \) .</td></tr><tr><td>For \( x \in \mathbb{R}, y \in {\mathbb{R}}^{ + }, g\left( x\right) = y \) if and only if \( {g}^{-1}\left( y\right) = x \) .</td><td>For \( x \in \mathbb{R}, y \in {\mathbb{R}}^{ + },{e}^{x} = y \) if and only if \( \ln y = x \) .</td></tr></table>
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Yes
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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 \( B,\left( {f \circ {f}^{-1}}\right) \left( y\right) = y \) .
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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\[ \left( {{f}^{-1} \circ f}\right) \left( x\right) = {f}^{-1}\left( {f\left( x\right) }\right) \]\n\n\[ = {f}^{-1}\left( y\right) \]\n\n\[ = x\text{.} \]\n\nHence, for each \( x \in A,\left( {{f}^{-1} \circ f}\right) \left( x\right) = x \) .
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No
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Theorem 6.29. Let \( f : A \rightarrow B \) and \( g : B \rightarrow C \) be bijections. Then \( g \circ f \) is a bijection and \( {\left( g \circ f\right) }^{-1} = {f}^{-1} \circ {g}^{-1} \) .
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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 \( {\left( g \circ f\right) }^{-1} : C \rightarrow A \) . We will now prove that for each \( z \in C,{\left( g \circ f\right) }^{-1}\left( z\right) = \left( {{f}^{-1} \circ {g}^{-1}}\right) \left( z\right) \) .\n\nLet \( z \in C \) . Since the function \( g \) is a surjection, there exists a \( y \in B \) such that\n\n\[ g\left( y\right) = z. \]\n\nAlso, since \( f \) is a surjection, there exists an \( x \in A \) such that\n\n\[ f\left( x\right) = y. \]\n\nNow these two equations can be written in terms of the respective inverse functions as\n\n\[ {g}^{-1}\left( z\right) = y\text{; and} \]\n\n\[ {f}^{-1}\left( y\right) = x\text{.} \]\n\nUsing equations (3) and (4), we see that\n\n\[ \left( {{f}^{-1} \circ {g}^{-1}}\right) \left( z\right) = {f}^{-1}\left( {{g}^{-1}\left( z\right) }\right) \]\n\n\[ = {f}^{-1}\left( y\right) \]\n\n\[ = x\text{.} \]\n\nUsing equations (1) and (2) again, we see that \( \left( {g \circ f}\right) \left( x\right) = z \) . However, in terms of the inverse function, this means that\n\n\[ {\left( g \circ f\right) }^{-1}\left( z\right) = x. \]\n\nComparing equations (5) and (6), we have shown that for all \( z \in C \) , \( {\left( g \circ f\right) }^{-1}\left( z\right) = \left( {{f}^{-1} \circ {g}^{-1}}\right) \left( z\right) \) . This proves that \( {\left( g \circ f\right) }^{-1} = {f}^{-1} \circ {g}^{-1} \) .
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Yes
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