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Proposition 8.3.2. The sine and cosine functions are periodic with period \( {2\pi } \) .
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Proof. The result follows immediately from the definitions.
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No
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Proposition 8.3.3. For any \( x \in \mathbb{R},\sin \left( {-x}\right) = - \sin \left( x\right) \) and \( \cos \left( {-x}\right) = \cos \left( x\right) \) .
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Proof. The result follows immediately from the definitions.\n\nQ.E.D.
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No
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Proposition 8.3.5. The range of both the sine and cosine functions is \( \left\lbrack {-1,1}\right\rbrack \) .
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Proof. The result follows immediately from the definitions along with the facts that\n\n\[ \sqrt{1 + {y}^{2}} \geq \sqrt{{y}^{2}} = \left| y\right| \]\n\n(8.3.20)\n\nand\n\n\[ \sqrt{1 + {y}^{2}} \geq 1 \]\n\n(8.3.21)\n\nfor any \( y \in \mathbb{R} \) .\n\nQ.E.D.
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No
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Proposition 8.3.6. For any \( x \) in the domain of the tangent function, \n\n\[ \n\tan \left( x\right) = \frac{\sin \left( x\right) }{\cos \left( x\right) }. \n\]
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Proof. The result follows immediately from the definitions. \n\nQ.E.D.
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No
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Proposition 8.3.7. For any \( x \) in the domain of the tangent function,\n\n\[ \n{\sin }^{2}\left( x\right) = \frac{{\tan }^{2}\left( x\right) }{1 + {\tan }^{2}\left( x\right) }\n\]\n\n\( \left( {8.3.23}\right) \)\n\nand\n\n\[ \n{\cos }^{2}\left( x\right) = \frac{1}{1 + {\tan }^{2}\left( x\right) }.\n\]\n\n(8.3.24)
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Proof. The result follows immediately from the definitions.\n\nQ.E.D.
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No
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Proposition 8.3.8. For any \( x, y \in \mathbb{R} \) ,\n\n\[ \cos \left( {x + y}\right) = \cos \left( x\right) \cos \left( y\right) - \sin \left( x\right) \sin \left( y\right) . \]
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Proof. First suppose \( x, y \), and \( x + y \) are in the domain of the tangent function.\n\nThen\n\n\[ {\cos }^{2}\left( {x + y}\right) = \frac{1}{1 + {\tan }^{2}\left( {x + y}\right) }\n\n= \frac{1}{1 + {\left( \frac{\tan \left( x\right) + \tan \left( y\right) }{1 - \tan \left( x\right) \tan \left( y\right) }\right) }^{2}}\n\n= \frac{{\left( 1 - \tan \left( x\right) \tan \left( y\right) \right) }^{2}}{{\left( 1 - \tan \left( x\right) \tan \left( y\right) \right) }^{2} + {\left( \tan \left( x\right) + \tan \left( y\right) \right) }^{2}}\n\n= \frac{{\left( 1 - \tan \left( x\right) \tan \left( y\right) \right) }^{2}}{\left( {1 + {\tan }^{2}\left( x\right) }\right) \left( {1 + {\tan }^{2}\left( y\right) }\right) }\n\n= {\left( \frac{1}{\sqrt{1 + {\tan }^{2}(x)}\sqrt{1 + {\tan }^{2}(y)}} - \frac{\tan (x)\tan (y)}{\sqrt{1 + {\tan }^{2}(x)}\sqrt{1 + {\tan }^{2}(y)}}\right) }^{2}\n\n= {\left( \cos \left( x\right) \cos \left( y\right) - \sin \left( x\right) \sin \left( y\right) \right) }^{2}. \]\n\nHence\n\n\[ \cos \left( {x + y}\right) = \pm \left( {\cos \left( x\right) \cos \left( y\right) - \sin \left( x\right) \sin \left( y\right) }\right) . \]\n\nConsider a fixed value of \( x \) . Note that the positive sign must be chosen when \( y = 0 \) . Moreover, increasing \( y \) by \( \pi \) changes the sign on both sides, so the positive sign must be chosen when \( y \) is any multiple of \( \pi \) . Since sine and cosine are continuous functions, the choice of sign could change only at points at which both sides are 0, but these points are separated by a distance of \( \pi \), so we must always choose the positive sign. Hence we have\n\n\[ \cos \left( {x + y}\right) = \cos \left( x\right) \cos \left( y\right) - \sin \left( x\right) \sin \left( y\right) \]\n\nfor all \( x, y \in \mathbb{R} \) for which \( x, y \), and \( x + y \) are in the domain of the tangent function. The identity for the other values of \( x \) and \( y \) now follows from the continuity of the sine and cosine functions. Q.E.D.
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Yes
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Proposition 8.3.9. For any \( x, y \in \mathbb{R} \), \[ \sin \left( {x + y}\right) = \sin \left( x\right) \cos \left( y\right) + \sin \left( y\right) \cos \left( x\right) . \]
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Exercise 8.3.1. Prove the previous proposition.
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No
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Proposition 8.3.13. \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos \left( x\right) }{x} = 0 \) .
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Proof. We have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos \left( x\right) }{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( \frac{1 - \cos \left( x\right) }{x}\right) \left( \frac{1 + \cos \left( x\right) }{1 + \cos \left( x\right) }\right) \]\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - {\cos }^{2}\left( x\right) }{x\left( {1 + \cos \left( x\right) }\right) } \]\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( \frac{\sin \left( x\right) }{x}\right) \left( \frac{\sin \left( x\right) }{1 + \cos \left( x\right) }\right) \]\n\n\[ = \left( 1\right) \left( 0\right) \]\n\n\[ = 0\text{.} \]
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Yes
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Proposition 8.3.14. If \( f\left( x\right) = \sin \left( x\right) \), then \( {f}^{\prime }\left( x\right) = \cos \left( x\right) \) .
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Proof. We have\n\n\[ \n{f}^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\sin \left( {x + h}\right) - \sin \left( x\right) }{h} \]\n\n\[ \n= \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\sin \left( x\right) \cos \left( h\right) + \sin \left( h\right) \cos \left( x\right) - \sin \left( x\right) }{h} \]\n\n\[ \n= \sin \left( x\right) \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\cos \left( h\right) - 1}{h} + \cos \left( x\right) \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\sin \left( h\right) }{h} \]\n\n\[ \n= \cos \left( x\right) \text{.} \]\n\n(8.3.42)\n\nQ.E.D.
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Yes
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Proposition 8.3.15. If \( f\\left( x\\right) = \\cos \\left( x\\right) \), then \( {f}^{\\prime }\\left( x\\right) = - \\sin \\left( x\\right) \) .
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Exercise 8.3.6. Prove the previous proposition.
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No
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Proposition 8.3.16. \( 2{\int }_{-1}^{1}\sqrt{1 - {x}^{2}}{dx} = \pi \) .
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Proof. Let \( x = \sin \left( u\right) \) . Then as \( u \) varies from \( - \frac{\pi }{2} \) to \( \frac{\pi }{2}, x \) varies from -1 to 1 . And, for these values, we have\n\n\[ \sqrt{1 - {x}^{2}} = \sqrt{1 - {\sin }^{2}\left( u\right) } = \sqrt{{\cos }^{2}\left( u\right) } = \left| {\cos \left( u\right) }\right| = \cos \left( u\right) . \]\n\n(8.3.50)\n\nHence\n\n\[ {\int }_{-1}^{1}\sqrt{1 - {x}^{2}}{dx} = {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^{2}\left( u\right) {du} \]\n\n\[ = {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1 + \cos \left( {2u}\right) }{2}{du} \]\n\n\[ = {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}{du} + \frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \left( {2u}\right) {du} \]\n\n\[ = \frac{\pi }{2} + \frac{1}{4}\left( {\sin \left( \pi \right) - \sin \left( {-\pi }\right) }\right) \]\n\n\[ = \frac{\pi }{2}\text{. } \]\n\n(8.3.51)
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Yes
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Proposition 8.4.1. The function \( f\left( x\right) = \log \left( x\right) \) is an increasing, differentiable function with \[ {f}^{\prime }\left( x\right) = \frac{1}{x} \] for all \( x > 0 \) .
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Proof. Using the Fundamental Theorem of Calculus, we have \[ {f}^{\prime }\left( x\right) = \frac{1}{x} > 0 \] for all \( x > 0 \), from which the result follows. Q.E.D.
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Yes
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Proposition 8.4.2. For any \( x > 0 \) ,\n\n\[ \log \left( \frac{1}{x}\right) = - \log \left( x\right) \]
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Proof. Using the substitution \( t = \frac{1}{u} \), we have\n\n\[ \log \left( \frac{1}{x}\right) = {\int }_{1}^{\frac{1}{x}}\frac{1}{t}{dt} = {\int }_{1}^{x}u\left( {-\frac{1}{{u}^{2}}}\right) {du} = - {\int }_{1}^{x}\frac{1}{u}{du} = - \log \left( x\right) . \]
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Yes
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Proposition 8.4.3. For any positive real numbers \( x \) and \( y \) ,\n\n\[ \log \left( {xy}\right) = \log \left( x\right) + \log \left( y\right) \]
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Proof. Using the substitution \( t = {xu} \), we have\n\n\[ \log \left( {xy}\right) = {\int }_{1}^{xy}\frac{1}{t}{dt} \]\n\n\[ = {\int }_{\frac{1}{x}}^{y}\frac{x}{xu}{du} \]\n\n\[ = {\int }_{\frac{1}{x}}^{1}\frac{1}{u}{du} + {\int }_{1}^{y}\frac{1}{u}{du} \]\n\n\[ = - {\int }_{1}^{\frac{1}{x}}\frac{1}{u}{du} + \log \left( y\right) \]\n\n\[ = - \log \left( \frac{1}{x}\right) + \log \left( y\right) \]\n\n\[ = \log \left( x\right) + \log \left( y\right) \]
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Yes
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Proposition 8.4.4. If \( r \in \mathbb{Q} \) and \( x \) is a positive real number, then\n\n\[ \log \left( {x}^{r}\right) = r\log \left( x\right) . \]\n\n(8.4.8)
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Proof. Using the substitution \( t = {u}^{r} \), we have\n\n\[ \log \left( {x}^{r}\right) = {\int }_{1}^{{x}^{r}}\frac{1}{t}{dt} = {\int }_{1}^{x}\frac{r{u}^{r - 1}}{{u}^{r}}{du} = r{\int }_{1}^{x}\frac{1}{u}{du} = r\log \left( x\right) . \]\n\n(8.4.9)\n\nQ.E.D.
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Yes
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Proposition 8.4.5. \( \mathop{\lim }\limits_{{x \rightarrow + \infty }}\log \left( x\right) = + \infty \) and \( \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\log \left( x\right) = - \infty \) .
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Proof. Given a real number \( M \), choose an integer \( n \) for which \( n\log \left( 2\right) > M \) (there exists such an \( n \) since \( \log \left( 2\right) > 0 \) ). Then for any \( x > {2}^{n} \), we have\n\n\[ \log \left( x\right) > \log \left( {2}^{n}\right) = n\log \left( 2\right) > M. \]\n\n(8.4.10)\n\nHence \( \mathop{\lim }\limits_{{x \rightarrow + \infty }}\log \left( x\right) = + \infty \).\n\nSimilarly, given any real number \( M \), we may choose an integer \( n \) for which \( - n\log \left( 2\right) < M \) . Then for any \( 0 < x < \frac{1}{{2}^{n}} \), we have\n\n\[ \log \left( x\right) < \log \left( \frac{1}{{2}^{n}}\right) = - n\log \left( 2\right) < M. \]\n\n(8.4.11)\n\nHence \( \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\log \left( x\right) = - \infty \).\n\nQ.E.D.
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Yes
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Proposition 8.4.6. For any rational number \( \alpha > 0 \) ,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\log \left( x\right) }{{x}^{\alpha }} = 0. \]
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Proof. Choose a rational number \( \beta \) such that \( 0 < \beta < \alpha \) . Now for any \( t > 1 \) ,\n\n\[ \frac{1}{t} < \frac{1}{t}{t}^{\beta } = \frac{1}{{t}^{1 - \beta }} \]\n\nHence\n\n\[ \log \left( x\right) = {\int }_{1}^{x}\frac{1}{t}{dt} < {\int }_{1}^{x}\frac{1}{{t}^{1 - \beta }}{dt} = \frac{{x}^{\beta } - 1}{\beta } < \frac{{x}^{\beta }}{\beta } \]\n\nwhenever \( x > 1 \) . Thus\n\n\[ 0 < \frac{\log \left( x\right) }{{x}^{\alpha }} < \frac{1}{\beta {x}^{\alpha - \beta }} \]\n\nfor \( x > 1 \) . But\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1}{\beta {x}^{\alpha - \beta }} = 0 \]\n\nso\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\log \left( x\right) }{{x}^{\alpha }} = 0. \]\n\nQ.E.D.
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Yes
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Proposition 8.5.1. The exponential function has domain \( \mathbb{R} \) and range \( \left( {0, + \infty }\right) \) . Moreover, the exponential function is increasing and differentiable on \( \mathbb{R} \) . If \( f\left( x\right) = \exp \left( x\right) \), then \( {f}^{\prime }\left( x\right) = \exp \left( x\right) \) .
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Proof. Only the final statement of the proposition requires proof. If we let \( g\left( x\right) = \log \left( x\right) \), then\n\n\[ \n{f}^{\prime }\left( x\right) = \frac{1}{{g}^{\prime }\left( {\exp \left( x\right) }\right) } = \exp \left( x\right) .\n\]\n\n(8.5.1)\n\nQ.E.D.
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No
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Proposition 8.5.2. For any real numbers \( x \) and \( y \) ,\n\n\[ \exp \left( {x + y}\right) = \exp \left( x\right) \exp \left( y\right) . \]
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Proof. The result follows from\n\n\[ \log \left( {\exp \left( x\right) \exp \left( y\right) }\right) = \log \left( {\exp \left( x\right) }\right) + \log \left( {\exp \left( y\right) }\right) = x + y. \]
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Yes
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Proposition 8.5.3. For any real number \( x \) ,\n\n\[ \exp \left( {-x}\right) = \frac{1}{\exp \left( x\right) }.\]
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Proof. The result follows from\n\n\[ \log \left( \frac{1}{\exp \left( x\right) }\right) = - \log \left( {\exp \left( x\right) }\right) = - x.\]\n\n\( \left( {8.5.5}\right) \)\n\nQ.E.D.
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Yes
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Proposition 8.5.4. For any rational number \( \alpha \) , \n\n\[ \nexp \left( \alpha \right) = {e}^{\alpha }.\n\]
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Proof. Since \( \log \left( e\right) = 1 \), we have \n\n\[ \n\log \left( {e}^{\alpha }\right) = \alpha \log \left( e\right) = \alpha .\n\] \n\nQ.E.D.
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Yes
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Proposition 8.5.5. For any real number \( \alpha > 0 \) ,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}{x}^{\alpha }{e}^{-x} = 0. \]
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Proof. We know that\n\n\[ \mathop{\lim }\limits_{{y \rightarrow + \infty }}\frac{\log \left( y\right) }{{y}^{\frac{1}{\alpha }}} = 0. \]\n\n(8.5.14)\n\nHence\n\n\[ \mathop{\lim }\limits_{{y \rightarrow + \infty }}\frac{{\left( \log \left( y\right) \right) }^{\alpha }}{y} = 0. \]\n\n\( \left( {8.5.15}\right) \)\n\nLetting \( y = {e}^{x} \), we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{\alpha }}{{e}^{x}} = 0 \]\n\n\( \left( {8.5.16}\right) \)\n\nQ.E.D.
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Yes
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Proposition 8.5.6. For any real number \( \alpha \) , \n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\left( 1 + \frac{\alpha }{x}\right) }^{x} = {e}^{\alpha } \]
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Proof. First note that, letting \( x = \frac{1}{h} \) , \n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\left( 1 + \frac{\alpha }{x}\right) }^{x} = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}{\left( 1 + \alpha h\right) }^{\frac{1}{h}} = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}{e}^{\frac{1}{h}\log \left( {1 + {\alpha h}}\right) }. \] \n\nUsing l'Hôpital's rule, we have \n\n\[ \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{\log \left( {1 + {\alpha h}}\right) }{h} = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{\alpha }{1 + {\alpha h}} = \alpha , \] \n\nand the result follows from the continuity of the exponential function. Q.E.D.
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Yes
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Lemma 1.6.2. If the square of an integer is even, then the original integer is even.
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Proof: Suppose to the contrary that \( \sqrt{2} \) is a rational number. Then by the definition of the set of rational numbers, we know that there are integers \( a \) and \( b \) having the following properties: \( \sqrt{2} = \frac{a}{b} \) and \( \gcd \left( {a, b}\right) = 1. \) Consider the expression \( \sqrt{2} = \frac{a}{b} \) . By squaring both sides of this we obtain \[ 2 = \frac{{a}^{2}}{{b}^{2}} \] This last expression can be rearranged to give \[ {a}^{2} = 2{b}^{2} \] An immediate consequence of this last equation is that \( {a}^{2} \) is an even number. Using the lemma above we now know that \( a \) is an even integer and hence that there is an integer \( m \) such that \( a = \) \( {2m} \) . Substituting this last expression into the previous equation gives \[ {\left( 2m\right) }^{2} = 2{b}^{2} \] thus, \[ 4{m}^{2} = 2{b}^{2} \] so \[ 2{m}^{2} = {b}^{2}\text{.} \] This tells us that \( {b}^{2} \) is even, and hence (by the lemma), \( b \) is even. Finally, we have arrived at the desired absurdity because if \( a \) and \( b \) are both even then \( \gcd \left( {a, b}\right) \geq 2 \), but, on the other hand, one of our initial assumptions is that \( \gcd \left( {a, b}\right) = 1 \) .
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No
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Theorem 3.3.1. (Euclid) The set of all prime numbers is infinite.
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Proof: Suppose on the contrary that there are only a finite number of primes. This finite set of prime numbers could, in principle, be listed in ascending order.\n\n\\[ \n\\left\\{ {{p}_{1},{p}_{2},{p}_{3},\\ldots ,{p}_{n}}\\right\\} \n\\]\n\nConsider the number \\( N \\) formed by adding 1 to the product of all of these primes.\n\n\\[ \nN = 1 + \\mathop{\\prod }\\limits_{{k = 1}}^{n}{p}_{k} \n\\]\n\nClearly, \\( N \\) is much larger than the largest prime \\( {p}_{n} \\), so \\( N \\) cannot be a prime number itself. Thus \\( N \\) must be a product of some of the primes in the list. Suppose that \\( {p}_{j} \\) is one of the primes that divides \\( N \\) . Now notice that, by construction, \\( N \\) would leave remainder 1 upon division by \\( {p}_{j} \\) . This is a contradiction since we cannot have both \\( {p}_{j} \\mid N \\) and \\( {p}_{j} \\nmid N \\) .\n\nSince the supposition that there are only finitely many primes leads to a contradiction, there must indeed be an infinite number of primes.\n\n## Q.E.D.
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Yes
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\[ \forall x \in \mathbb{Z},{x}^{2}\text{ even } \Rightarrow x\text{ even } \]
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Proof: This statement is logically equivalent to \[ \forall x \in \mathbb{Z}, x\text{ odd } \Rightarrow {x}^{2}\text{ odd } \] so we prove that instead. Suppose that \( x \) is a particular but arbitrarily chosen integer such that \( x \) is odd. Since \( x \) is odd, there is an integer \( k \) such that \( x = {2k} + 1 \) . It follows that \( {x}^{2} = {\left( 2k + 1\right) }^{2} = 4{k}^{2} + {4k} + 1 = \) \( 2\left( {2{k}^{2} + {2k}}\right) + 1 \) . Finally, we see that \( {x}^{2} \) must be odd because it is of the form \( {2m} + 1 \), where \( m = 2{k}^{2} + {2k} \) is clearly an integer. Q.E.D.
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Yes
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Theorem 3.3.3. Among all triangles inscribed in a fixed circle, the one with maximum area is equilateral.
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Proof: We'll proceed by contradiction. Suppose to the contrary that there is a triangle, \( \bigtriangleup {ABC} \), inscribed in a circle having maximum area that is not equilateral. Since \( \bigtriangleup {ABC} \) is not equilateral, there are two sides of it that are not equal. Without loss of generality, suppose that sides \( \overline{AB} \) and \( \overline{BC} \) have different lengths. Consider the remaining side \( \left( \overline{AC}\right) \) to be the base of this triangle. We can construct another triangle \( \bigtriangleup A{B}^{\prime }C \), also inscribed in our circle, and also having \( \overline{AC} \) as its base, having a greater altitude than \( \bigtriangleup {ABC} \) - since the area of a triangle is given by the formula \( {bh}/2 \) (where \( b \) is the base, and \( h \) is the altitude), this triangle’s area is evidently greater than that of \( \bigtriangleup {ABC} \) . This is a contradiction since \( \bigtriangleup {ABC} \) was presumed to have maximal area. We leave the actual construction \( \bigtriangleup A{B}^{\prime }C \) to the following exercise. Q.E.D.
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No
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Theorem 3.5.1. \( \forall n \in \mathbb{Z} \), \( n^2 \) is of the form \( 4k \) or \( 4k + 1 \) for some \( k \in \mathbb{Z} \).
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Proof: We will consider the four cases determined by the four possible residues mod 4.\n\ncase i) If \( n \equiv 0\left( {\;\operatorname{mod}\;4}\right) \) then there is an integer \( m \) such that \( n = 4m \). It follows that \( n^2 = \left( 4m\right)^2 = 16m^2 \) is of the form \( 4k \) where \( k \) is \( 4m^2 \).\n\ncase ii) If \( n \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) then there is an integer \( m \) such that \( n = 4m + 1 \). It follows that \( n^2 = \left( 4m + 1\right)^2 = 16m^2 + 8m + 1 \) is of the form \( 4k + 1 \) where \( k \) is \( 4m^2 + 2m \).\n\ncase iii) If \( n \equiv 2\left( {\;\operatorname{mod}\;4}\right) \) then there is an integer \( m \) such that \( n = 4m + 2 \). It follows that \( n^2 = \left( 4m + 2\right)^2 = 16m^2 + 16m + 4 \) is of the form \( 4k \) where \( k \) is \( 4m^2 + 4m + 1 \).\n\ncase iv) If \( n \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) then there is an integer \( m \) such that \( n = 4m + 3 \). It follows that \( n^2 = \left( 4m + 3\right)^2 = 16m^2 + 24m + 9 \) is of the form \( 4k + 1 \) where \( k \) is \( 4m^2 + 6m + 2 \).\n\nSince these four cases exhaust the possibilities and since the desired result holds in each case, our proof is complete.\n\nQ.E.D.
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Yes
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Theorem 3.6.1. There is an even prime.
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Proof: The number 2 is both even and prime.\n\nQ.E.D.
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Yes
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Theorem 3.6.2. There are irrational numbers \( \alpha \) and \( \beta \) such that \( {\alpha }^{\beta } \) is rational.
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Proof: If \( {\sqrt{2}}^{\sqrt{2}} \) is rational then we are done. (Let \( \alpha = \beta = \sqrt{2} \) .)\n\nOtherwise, let \( \alpha = {\sqrt{2}}^{\sqrt{2}} \) and \( \beta = \sqrt{2} \) . The result follows because\n\n\( {\left( {\sqrt{2}}^{\sqrt{2}}\right) }^{\sqrt{2}} = {\sqrt{2}}^{\left( \sqrt{2}\sqrt{2}\right) } = {\sqrt{2}}^{2} = 2 \), which is clearly rational.\n\nQ.E.D.
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Yes
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Theorem 3.6.3. There is a smallest 6-digit vampire number.
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Proof: The number 125460 is a vampire number (in fact this is the smallest example of a vampire number with two sets of fangs: \( {125460} = {204} \cdot {615} = {246} \cdot {510}) \) . Since the set of 6-digit vampire numbers is non-empty, the well-ordering principle of the natural numbers allows us to deduce that there is a smallest 6- digit vampire number.\n\nQ.E.D.
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Yes
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Theorem 3.6.4. The smallest positive number in the \( \mathbb{Z} \) -module generated by \( a \) and \( b \) is \( d = \gcd \left( {a, b}\right) \) .
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Proof: Suppose that \( d \) is the smallest positive number in the \( \mathbb{Z} \) - module \( \{ {xa} + {yb} \mid x, y \in \mathbb{Z}\} \) . There are particular values of \( x \) and \( y \) (which we will distinguish with over-lines) such that \( d = \bar{x}a + \bar{y}b \) . Now, it is easy to see that if \( c \) is any common divisor of \( a \) and \( b \) then \( c \mid d \), so what remains to be proved is that \( d \) itself is a divisor of both \( a \) and \( b \) . Consider dividing \( d \) into \( a \) . By the division algorithm there are uniquely determined numbers \( q \) and \( r \) such that \( a = {qd} + r \) with \( 0 \leq r < d \) . We will show that \( r = 0 \) . Suppose, to the contrary, that \( r \) is positive. Note that we can write \( r \) as \( r = a - {qd} = a - q\left( {\bar{x}a + \bar{y}b}\right) = \left( {1 - q\bar{x}}\right) a - q\bar{y}b \) . The last equality shows that \( r \) is in the \( \mathbb{Z} \) -module under consideration, and so, since \( d \) is the smallest positive integer in this \( \mathbb{Z} \) -module it follows that \( r \geq d \) which contradicts the previously noted fact that \( r < d \) . Thus, \( r = 0 \) and so it follows that \( d \mid a \) . An entirely analogous argument can be used to show that \( d \mid b \) which completes the proof that \( d = \gcd \left( {a, b}\right) \) .
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Yes
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Theorem 5.1.1. For all finite sets \( A,\left| A\right| = n \Rightarrow \left| {\mathcal{P}\left( A\right) }\right| = {2}^{n} \) .
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Proof: Let \( n = \left| A\right| \) and proceed by induction on \( n \) .\n\nBasis: Suppose \( A \) is a finite set and \( \left| A\right| = 0 \), it follows that \( A = \varnothing \) . The power set of \( \varnothing \) is \( \{ \varnothing \} \) which is a set having 1 element. Note that \( {2}^{0} = 1 \) .\n\nInductive step: Suppose that \( A \) is a finite set with \( \left| A\right| = k + 1 \) . Choose some particular element of \( A \), say \( a \), and note that we can divide the subsets of \( A \) (i.e. elements of \( \mathcal{P}\left( A\right) \) ) into two categories, those that contain \( a \) and those that don’t. Let \( {S}_{1} = \{ X \in \mathcal{P}\left( A\right) \mid a \in X\} \) and let \( {S}_{2} = \{ X \in \mathcal{P}\left( A\right) \mid a \notin X\} \) . We have created two sets that contain all the elements of \( \mathcal{P}\left( A\right) \) , and which are disjoint from one another. In symbolic form, \( {S}_{1} \cup \) \( {S}_{2} = \mathcal{P}\left( A\right) \) and \( {S}_{1} \cap {S}_{2} = \varnothing \) . It follows that \( \left| {\mathcal{P}\left( A\right) }\right| = \left| {S}_{1}\right| + \left| {S}_{2}\right| \) . Notice that \( {S}_{2} \) is actually the power set of the \( k \) -element set \( A \smallsetminus \) \( \{ a\} \) . By the inductive hypothesis, \( \left| {S}_{2}\right| = {2}^{k} \) . Also, notice that each set in \( {S}_{1} \) corresponds uniquely to a set in \( {S}_{2} \) if we just remove the element \( a \) from it. This shows that \( \left| {S}_{1}\right| = \left| {S}_{2}\right| \) . Putting this all together we get that \( \left| {\mathcal{P}\left( A\right) }\right| = {2}^{k} + {2}^{k} = 2\left( {2}^{k}\right) = {2}^{k + 1} \) .\n\nQ.E.D.
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Yes
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Theorem 5.2.1.\n\n\[ \forall n \in \mathbb{N},\mathop{\sum }\limits_{{j = 1}}^{n}j = \frac{n\left( {n + 1}\right) }{2} \]
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Proof: We proceed by induction on \( n \) .\n\nBasis: Notice that when \( n = 0 \) the sum on the left-hand side has no terms in it! This is known as an empty sum, and by definition, an empty sum’s value is 0 . Also, when \( n = 0 \) the formula on the right-hand side becomes \( \left( {0 \cdot 1}\right) /2 \) and this is 0 as well. \( {}^{4} \)\n\nInductive step: Consider the sum on the left-hand side of the \( k + 1 \) -th version of our formula.\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{{k + 1}}j \]\n\nWe can separate out the last term of this sum.\n\n\[ = \left( {k + 1}\right) + \mathop{\sum }\limits_{{j = 1}}^{k}j \]\n\nNext, we can use the inductive hypothesis to replace the sum (the part that goes from 1 to \( k \) ) with a formula.\n\n\[ = \left( {k + 1}\right) + \frac{k\left( {k + 1}\right) }{2} \]\n\nFrom here on out it's just algebra ...\n\n\[ = \frac{2\left( {k + 1}\right) }{2} + \frac{k\left( {k + 1}\right) }{2} \]\n\n\[ = \frac{2\left( {k + 1}\right) + k\left( {k + 1}\right) }{2} \]\n\n\[ = \frac{\left( {k + 1}\right) \cdot \left( {k + 2}\right) }{2}. \]\n\nQ.E.D.
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Yes
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Theorem 5.2.2.\n\n\\[ \n\\forall n \\in {\\mathbb{Z}}^{ + },\\mathop{\\sum }\\limits_{{j = 1}}^{n}{j}^{2} = \\frac{n\\left( {n + 1}\\right) \\left( {{2n} + 1}\\right) }{6} \n\\]
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Proof: We proceed by induction on \\( n \\) .\n\nBasis: When \\( n = 1 \\) the sum has only one term, \\( {1}^{2} = 1 \\) . On the other hand, the formula is \\( \\frac{1\\left( {1 + 1}\\right) \\left( {2 \\cdot 1 + 1}\\right) }{6} = 1 \\) . Since these are equal, the basis is proved.\n\n## Inductive step:\n\nBefore proceeding with the inductive step, in this box, we will figure out what the right-hand side of our theorem looks like when \\( n \\) is replaced with \\( k + 1 \\) :\n\n\\[ \n\\frac{\\left( {k + 1}\\right) \\left( {\\left( {k + 1}\\right) + 1}\\right) \\left( {2\\left( {k + 1}\\right) + 1}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{\\left( {k + 1}\\right) \\left( {k + 2}\\right) \\left( {{2k} + 3}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{\\left( {{k}^{2} + {3k} + 2}\\right) \\left( {{2k} + 3}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{2{k}^{3} + 9{k}^{2} + {13k} + 6}{6}. \n\\]\n\nBy the inductive hypothesis,\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{k}{j}^{2} = \\frac{k\\left( {k + 1}\\right) \\left( {{2k} + 1}\\right) }{6}. \n\\]\n\nAdding \\( {\\left( k + 1\\right) }^{2} \\) to both sides of this equation gives\n\n\\[ \n{\\left( k + 1\\right) }^{2} + \\mathop{\\sum }\\limits_{{j = 1}}^{k}{j}^{2} = \\frac{k\\left( {k + 1}\\right) \\left( {{2k} + 1}\\right) }{6} + {\\left( k + 1\\right) }^{2}. \n\\]\n\nThus,\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{{k + 1}}{j}^{2} = \\frac{k\\left( {k + 1}\\right) \\left( {{2k} + 1}\\right) }{6} + \\frac{6{\\left( k + 1\\right) }^{2}}{6}. \n\\]\n\nTherefore,\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{{k + 1}}{j}^{2} = \\frac{\\left( {{k}^{2} + k}\\right) \\left( {{2k} + 1}\\right) }{6} + \\frac{6\\left( {{k}^{2} + {2k} + 1}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{\\left( {2{k}^{3} + 3{k}^{2} + k}\\right) + \\left( {6{k}^{2} + {12k} + 6}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{2{k}^{3} + 9{k}^{2} + {13k} + 6}{6} \n\\]\n\n\\[ \n= \\frac{\\left( {{k}^{2} + {3k} + 2}\\right) \\left( {{2k} + 3}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{\\left( {k + 1}\\right) \\left( {k + 2}\\right) \\left( {{2k} + 3}\\right) }{6} \n\\]\n\n\\[ \n= \\frac{\\left( {k + 1}\\right) \\left( {\\left( {k + 1}\\right) + 1}\\right) \\left( {2\\left( {k + 1}\\right) + 1}\\right) }{6}\\text{.} \n\\]\n\nThis proves the inductive step, so the result is true. Q.E.D.
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Yes
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Theorem 5.2.3.\n\n\\[ \forall n \\geq 2 \\in \\mathbb{Z},\\mathop{\\prod }\\limits_{{j = 2}}^{n}\\left( {1 - \\frac{1}{{j}^{2}}}\\right) = \\frac{n + 1}{2n}. \\]
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Proof: (Using mathematical induction on \\( n \\) .)\n\nBasis: When \\( n = 2 \\) the product has only one term, \\( 1 - 1/{2}^{2} = \\) \\( 3/4 \\) . On the other hand, the formula is \\( \\frac{2 + 1}{2 \\cdot 2} = 3/4 \\) . Since these are equal, the basis is proved.\n\nInductive step:\n\nLet \\( k \\) be a particular but arbitrarily chosen integer such that\n\n\\[ \\mathop{\\prod }\\limits_{{j = 2}}^{k}\\left( {1 - \\frac{1}{{j}^{2}}}\\right) = \\frac{k + 1}{2k} \\]\n\nMultiplying \\( {}^{5} \\) both sides by the \\( k + 1 \\) -th term of the product gives\n\n\\[ \\left( {1 - \\frac{1}{{\\left( k + 1\\right) }^{2}}}\\right) \\cdot \\mathop{\\prod }\\limits_{{j = 2}}^{k}\\left( {1 - \\frac{1}{{j}^{2}}}\\right) = \\frac{k + 1}{2k} \\cdot \\left( {1 - \\frac{1}{{\\left( k + 1\\right) }^{2}}}\\right) . \\]\n\n\\( {}^{5} \\) Really, the only reason I’m doing this silly proof is to point out to you that when you're doing the inductive step in a proof of a formula for a product, you don't add to both sides anymore, you multiply. You see that, right? Well, consider yourself to have been pointed out to or ...oh, whatever.\n\nThus\n\n\\[ \\mathop{\\prod }\\limits_{{j = 2}}^{{k + 1}}\\left( {1 - \\frac{1}{{j}^{2}}}\\right) = \\frac{k + 1}{2k} \\cdot \\left( {1 - \\frac{1}{{\\left( k + 1\\right) }^{2}}}\\right) \\]\n\n\\[ = \\frac{k + 1}{2k} - \\frac{\\left( k + 1\\right) }{{2k}{\\left( k + 1\\right) }^{2}} \\]\n\n\\[ = \\frac{k + 1}{2k} - \\frac{\\left( 1\\right) }{{2k}\\left( {k + 1}\\right) } \\]\n\n\\[ = \\frac{{\\left( k + 1\\right) }^{2} - 1}{{2k}\\left( {k + 1}\\right) } \\]\n\n\\[ = \\frac{{k}^{2} + {2k}}{{2k}\\left( {k + 1}\\right) } \\]\n\n\\[ = \\frac{k\\left( {k + 2}\\right) }{{2k}\\left( {k + 1}\\right) } \\]\n\n\\[ = \\frac{k + 2}{2\\left( {k + 1}\\right) } \\]
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Yes
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Theorem 5.3.2. \( \forall n \in \mathbb{N},3 \mid \left( {{n}^{3} + {2n} + 6}\right) \)
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Proof: (By mathematical induction)\n\nBasis: Clearly \( 3 \mid 6 \) .\n\nInductive step:\n\n(We need to show that \( 3 \mid \left( {{k}^{3} + {2k} + 6}\right) \Rightarrow 3 \mid \left( {{\left( k + 1\right) }^{3} + 2(k + }\right. \n\n1) + 6.)\n\nConsider the quantity \( {\left( k + 1\right) }^{3} + 2\left( {k + 1}\right) + 6 \) .\n\n\[ \n{\left( k + 1\right) }^{3} + 2\left( {k + 1}\right) + 6 \n\]\n\n\[ \n= \left( {{k}^{3} + 3{k}^{2} + {3k} + 1}\right) + \left( {{2k} + 2}\right) + 6 \n\]\n\n\[ \n= \left( {{k}^{3} + {2k} + 6}\right) + 3{k}^{2} + {3k} + 3 \n\]\n\n\[ \n= \left( {{k}^{3} + {2k} + 6}\right) + 3\left( {{k}^{2} + k + 1}\right) \text{.} \n\]\n\nBy the inductive hypothesis,3 is a divisor of \( {k}^{3} + {2k} + 6 \) so there\n\nis an integer \( m \) such that \( {k}^{3} + {2k} + 6 = {3m} \) . Thus,\n\n\[ \n{\left( k + 1\right) }^{3} + 2\left( {k + 1}\right) + 6 \n\]\n\n\[ \n= {3m} + 3\left( {{k}^{2} + k + 1}\right) \n\]\n\n\[ \n= 3\left( {m + {k}^{2} + k + 1}\right) \text{.} \n\]\n\nThis equation shows that 3 is a divisor of \( {\left( k + 1\right) }^{3} + 2\left( {k + 1}\right) + 6 \) ,\n\nwhich is the desired conclusion.\n\nQ.E.D.
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Yes
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Theorem 5.3.3.\n\n\\[ \n\\forall n \\in \\mathbb{N},6 \\mid {7}^{n} - 1 \n\\]
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Proof: (By PMI)\n\nBasis: Note that \\( {7}^{0} - 1 \\) is 0 and also that \\( 6 \\mid 0 \\) .\n\nInductive step:\n\n(We need to show that if \\( 6\\left| {{7}^{k} - 1\\text{then}6}\\right| {7}^{k + 1} - 1 \\) .)\n\nConsider the quantity \\( {7}^{k + 1} - 1 \\) .\n\n\\[ \n{7}^{k + 1} - 1 = 7 \\cdot {7}^{k} - 1 \n\\]\n\n\\[ \n= \\left( {6 + 1}\\right) \\cdot {7}^{k} - 1 \n\\]\n\n\\[ \n= 6 \\cdot {7}^{k} + 1 \\cdot {7}^{k} - 1 \n\\]\n\n\\[ \n= 6\\left( {7}^{k}\\right) + \\left( {{7}^{k} - 1}\\right) \n\\]\n\nBy the inductive hypothesis, \\( 6 \\mid {7}^{k} - 1 \\) so there is an integer \\( m \\) such that \\( {7}^{k} - 1 = {6m} \\) . It follows that\n\n\\[ \n{7}^{k + 1} - 1 = 6\\left( {7}^{k}\\right) + {6m}. \n\\]\n\nSo, clearly, \\( 6 \\) is a divisor of \\( {7}^{k + 1} - 1 \\) .
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Yes
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\[ \forall n \geq 4 \in \mathbb{N},{2}^{n} < n! \]
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Proof: (By mathematical induction)\n\nBasis: When \( n = 4 \) we have \( {2}^{4} < 4 \) !, which is certainly true \( \left( {{16} < {24}}\right) \). \n\nInductive step: Suppose that \( k \) is a natural number with \( k > 4 \), and that \( {2}^{k} < k \) !. Multiply the left hand side of this inequality by 2 and the right hand side by \( k + {1}^{7} \) to get\n\n\[ 2 \cdot {2}^{k} < \left( {k + 1}\right) \cdot k! \]\n\nSo\n\n\[ {2}^{k + 1} < \left( {k + 1}\right) ! \]\n\nQ.E.D.
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No
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Theorem 5.3.5. For all natural numbers \( n \), if \( n \geq 4 \) then \( {n}^{2} \leq {2}^{n} \) .
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Proof:\n\nBasis: When \( n = 4 \) we have \( {4}^{2} \leq {2}^{4} \), which is true since both numbers are 16 .\n\nInductive step: (In the inductive step we assume that \( {k}^{2} \leq {2}^{k} \) and then show that \( {\left( k + 1\right) }^{2} \leq {2}^{k + 1} \) .)\n\nThe inductive hypothesis tells us that\n\n\[ \n{k}^{2} \leq {2}^{k} \n\]\n\nIf we add \( {2k} + 1 \) to the left-hand side of this inequality and \( {2}^{k} \) to the right-hand side we will produce the desired inequality. Thus our proof will follow provided that we know that \( {2k} + 1 \leq {2}^{k} \) . Indeed, it is sufficient to show that \( {2k} + 1 \leq {k}^{2} \) since we already know (by the inductive hypothesis) that \( {k}^{2} \leq {2}^{k} \) .\n\nSo the result remains in doubt unless you can complete the exercise that follows...\n\nQ.E.D.???
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No
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Theorem 5.4.1. For all natural numbers \( n, n > 1 \) implies \( n \) has a prime factor.
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Proof: (By strong induction) Consider an arbitrary natural number \( n > 1 \) . If \( n \) is prime then \( n \) clearly has a prime factor (itself), so suppose that \( n \) is not prime. By definition, a composite natural number can be factored, so \( n = a \cdot b \) for some pair of natural numbers \( a \) and \( b \) which are both greater than 1 . Since \( a \) and \( b \) are factors of \( n \) both greater than 1, it follows that \( a < n \) (it is also true that \( b < n \) but we don’t need that \( \ldots \) ). The inductive hypothesis can now be applied to deduce that \( a \) has a prime factor \( p \) . Since \( p\left| {a\text{and}a}\right| n \), by transitivity \( p \mid n \) . Thus \( n \) has a prime factor.
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Yes
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Theorem 6.3.1. The relation \( S \) defined by\n\n\[ \forall x, y \in \mathbb{N},{xSy} \Leftrightarrow {sf}\left( x\right) = {sf}\left( y\right) \]\n\nis an equivalence relation on \( \mathbb{N} \) .
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Proof: We must show that \( \mathrm{S} \) is reflexive, symmetric and transitive.\n\nreflexive - (Here we must show that \( \forall x \in \mathbb{N}, x\mathrm{S}x \) .) Let \( x \) be an arbitrary natural number. Since \( {sf}\left( x\right) = {sf}\left( x\right) \) (this is the reflexive property of \( = \) ) it follows from the definition of \( \mathrm{S} \) that \( x\mathrm{\;S}x \) .\n\nsymmetric \( - \) (Here we must show that \( \forall x, y \in \mathbb{N}, x\mathrm{\;S}y \Rightarrow \) \( y\mathrm{\;S}x \) .) Let \( x \) and \( y \) be arbitrary natural numbers, and further suppose that \( x\mathrm{\;S}y \) . Since \( x\mathrm{\;S}y \), it follows from the definition of \( \mathrm{S} \) that \( {sf}\left( x\right) = {sf}\left( y\right) \), obviously then \( {sf}\left( y\right) = {sf}\left( x\right) \) (this is the symmetric property of \( = \) ) and so \( y\mathrm{\;S}x \) .\n\ntransitive - (Here we must show that \( \forall x, y, z \in \mathbb{N}, x\mathrm{\;S}y \land \) \( y\mathrm{\;S}z \Rightarrow x\mathrm{\;S}z \) .) Let \( x, y \) and \( z \) be arbitrary natural numbers, and further suppose that both \( x\mathrm{\;S}y \) and \( y\mathrm{\;S}z \) . From the definition of \( \mathrm{S} \) we deduce that \( {sf}\left( x\right) = {sf}\left( y\right) \) and \( {sf}\left( y\right) = {sf}\left( z\right) \) . Clearly, \( {sf}\left( x\right) = {sf}\left( z\right) \) (this deduction comes from the transitive property of \( = \) ), so \( x\mathrm{\;S}z \) .\n\nQ.E.D.
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Yes
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Theorem 6.5.1. The function \( f : \mathbb{N} \rightarrow \mathcal{O} \) defined by \( f\left( x\right) = {2x} - 1 \) is a bijection from \( \mathbb{N} \) to \( \mathcal{O} \) .
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Proof: First we will show that \( f \) is surjective. Consider an arbitrary element \( y \) of the set \( \mathcal{O} \) . Since \( y \in \mathcal{O} \) it follows that \( y \) is both positive and odd. Thus there is an integer \( k \), such that \( y = {2k} + 1 \), but also \( y > 0 \) . From this it follows that \( {2k} + 1 > 0 \) and so \( k > - 1/2 \) . Since \( k \) is also an integer, this last inequality implies that \( k \in {\mathbb{Z}}^{\text{noneg }} \) . (Recall that \( {\mathbb{Z}}^{\text{noneg }} = \{ 0,1,2,3,\ldots \} \) .) We can easily verify that a preimage for \( y \) is \( k + 1 \), since \( f\left( {k + 1}\right) = \) \( 2\left( {k + 1}\right) - 1 = {2k} + 2 - 1 = {2k} + 1 = y. \)\n\nNext we show that \( f \) is injective. Suppose that there are two input values, \( {x}_{1} \) and \( {x}_{2} \) such that \( f\left( {x}_{1}\right) = f\left( {x}_{2}\right) \) . Then \( 2{x}_{1} - 1 = 2{x}_{2} - 1 \) and simple algebra leads to \( {x}_{1} = {x}_{2} \) .\n\nQ.E.D.
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Yes
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Theorem 7.1.2. In an undirected graph the sum of the degrees of the vertices is even.
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Proof: The sum of the degrees of all the vertices in a graph \( G \) ,\n\n\[ \mathop{\sum }\limits_{{v \in V\left( G\right) }}\deg \left( v\right) \]\ncounts every edge of \( G \) exactly twice.\n\nThus,\n\n\[ \mathop{\sum }\limits_{{v \in V\left( G\right) }}\deg \left( v\right) = 2 \cdot \left| {E\left( G\right) }\right| \]\n\nIn particular we see that this sum is even.\n\nQ.E.D.
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Yes
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Theorem 7.1.3. A magic square of order \( n \) has a magic sum equal to \( \frac{{n}^{3} + n}{2} \) .
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Proof: We count the total of the entries in the magic square in\ntwo ways. The sum of all the entries in the magic square is\n\n\[ S = 1 + 2 + 3 + \ldots + {n}^{2}. \]\n\nUsing the formula for the sum of the first \( k \) naturals ( \( \mathop{\sum }\limits_{{i = 1}}^{k}i = \) \( \left. \frac{{k}^{2} + k}{2}\right) \) and evaluating at \( {n}^{2} \) gives\n\n\[ S = \frac{{n}^{4} + {n}^{2}}{2} \]\n\nOn the other hand, if the magic sum is \( M \), then each of the \( n \) rows has numbers in it which sum to \( M \) so\n\n\[ S = {nM}\text{.} \]\n\nBy equating these different expressions for \( S \) and solving for \( M \) , we prove the desired result:\n\n\[ {nM} = \frac{{n}^{4} + {n}^{2}}{2} \]\n\ntherefore\n\n\[ M = \frac{{n}^{3} + n}{2} \]
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Yes
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Theorem 7.2.1. If \( f \) is a function such that \( \left| {\operatorname{Dom}\left( f\right) }\right| > \left| {\operatorname{Rng}\left( f\right) }\right| \) then \( f \) is not injective.
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Proof: Suppose to the contrary that \( f \) is a function with \( \left| {\operatorname{Dom}\left( f\right) }\right| > \left| {\operatorname{Rng}\left( f\right) }\right| \) and that \( f \) is injective. Of course \( f \) is onto its range, so since we are presuming that \( f \) is injective it follows that \( f \) is a bijection between \( \operatorname{Dom}\left( f\right) \) and \( \operatorname{Rng}\left( f\right) \). Therefore (since \( f \) provides a one-to-one correspondence) \( \left| {\operatorname{Dom}\left( f\right) }\right| = \left| {\operatorname{Rng}\left( f\right) }\right| \). This clearly contradicts the statement that \( \left| {\operatorname{Dom}\left( f\right) }\right| > \left| {\operatorname{Rng}\left( f\right) }\right| \). Q.E.D.
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Yes
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Theorem 7.2.2. If there are \( n \) containers having capacities \( {m}_{1},{m}_{2},{m}_{3},\ldots ,{m}_{n} \) , and there are \( 1 + \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{m}_{i} - 1}\right) \) objects placed in them, then for some \( i \), container \( i \) has (at least) \( {m}_{i} \) objects in it.
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Proof: If no container holds its full capacity, then the largest the total of the objects could be is \( \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{m}_{i} - 1}\right) \). Q.E.D.
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No
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Theorem 7.3.1. For all natural numbers \( n \) and \( k \) with \( 0 < k \leq n \) , \n\n\[ \n\left( \begin{array}{l} n \\ k \end{array}\right) = \left( \begin{matrix} n - 1 \\ k \end{matrix}\right) + \left( \begin{matrix} n - 1 \\ k - 1 \end{matrix}\right) \n\]
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Proof: (The first proof is a combinatorial argument.)\n\nThere are \( \left( \begin{array}{l} n \\ k \end{array}\right) \) subsets of size \( k \) of the set \( N = \{ 1,2,3,\ldots, n\} \) . We will partition these \( k \) -subsets into two disjoint cases: those that contain the final number, \( n \), and those that do not.\n\nLet\n\n\[ \nA = \{ S \subseteq N\left| \right| S \mid = k \land n \notin S\} \n\]\n\nand, let\n\n\[ \nB = \{ S \subseteq N \mid \left| S\right| = k \land n \in S\} . \n\]\n\nSince the number \( n \) is either in a \( k \) -subset or it isn’t, these sets are disjoint and exhaustive. So the addition rule tells us that\n\n\[ \n\left( \begin{array}{l} n \\ k \end{array}\right) = \left| A\right| + \left| B\right| \n\]\n\nThe set \( A \) is really just the set of all \( k \) -subsets of the \( \left( {n - 1}\right) \) -set\n\n\( \{ 1,2,3,\ldots, n - 1\} \), so \( \left| A\right| = \left( \begin{matrix} n - 1 \\ k \end{matrix}\right) \) .\n\nAny of the sets in \( B \) can be obtained by adjoining the element \( n \) to a \( k - 1 \) subset of the \( \left( {n - 1}\right) \) -set \( \{ 1,2,3,\ldots, n - 1\} \), so \( \left| B\right| = \left( \begin{array}{l} n - 1 \\ k - 1 \end{array}\right) \) . Substituting gives us the desired result.\n\nQ.E.D.
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Yes
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Theorem 7.3.2. For all natural numbers \( n \) and \( k \) with \( 0 < k \leq n \) , \[ k \cdot \left( \begin{array}{l} n \\ k \end{array}\right) = n \cdot \left( \begin{array}{l} n - 1 \\ k - 1 \end{array}\right) . \]
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Proof: Using the formula for the value of a binomial coefficient we get \[ k \cdot \left( \begin{array}{l} n \\ k \end{array}\right) = k \cdot \frac{n!}{k!\left( {n - k}\right) !}. \] We can do some cancellation to obtain \[ k \cdot \left( \begin{array}{l} n \\ k \end{array}\right) = \frac{n!}{\left( {k - 1}\right) !\left( {n - k}\right) !}. \] Finally we factor-out an \( n \) to obtain \[ k \cdot \left( \begin{array}{l} n \\ k \end{array}\right) = n \cdot \frac{\left( {n - 1}\right) !}{\left( {k - 1}\right) !\left( {n - k}\right) !} \] since \( \left( {n - k}\right) \) is the same thing as \( \left( {\left( {n - 1}\right) - \left( {k - 1}\right) }\right) \) we have \[ k \cdot \left( \begin{array}{l} n \\ k \end{array}\right) = n \cdot \frac{\left( {n - 1}\right) !}{\left( {k - 1}\right) !\left( {\left( {n - 1}\right) - \left( {k - 1}\right) }\right) !} = n \cdot \left( \begin{array}{l} n - 1 \\ k - 1 \end{array}\right) \] Q.E.D.
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Yes
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Theorem 8.3.1 (Cantor). For all sets \( A, A \) is not equivalent to \( \mathcal{P}\left( A\right) \) .
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Proof: Suppose that there is a set \( A \) that can be placed in one-toone correspondence with its power set. Then there is a bijective function \( f : A \rightarrow \mathcal{P}\left( A\right) \) . We will deduce a contradiction by constructing a subset of \( A \) (i.e. a member of \( \mathcal{P}\left( A\right) \) ) that cannot be in the range of \( f \) .\n\n\[ \n\text{Let}S = \{ x \in A \mid x \notin f\left( x\right) \} \text{.} \n\]\n\nIf \( S \) is in the range of \( f \), there is a preimage \( y \) such that \( S = f\left( y\right) \) . But, if such a \( y \) exists then the membership question, \( y \in S \), must either be true or false. If \( y \in S \), then because \( S = f\left( y\right) \), and \( S \) consists of those elements that are not in their images, it follows that \( y \notin S \) . On the other hand, if \( y \notin S \) then \( y \notin f\left( y\right) \) so (by the definition of \( S \) ) it follows that \( y \in S \) . Either possibility leads to the other, which is a contradiction.\n\nQ.E.D.
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Yes
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Theorem 9.1.1. The points of intersection of the adjacent trisectors in an arbitrary triangle \( \bigtriangleup {ABC} \) form the vertices of an equilateral triangle.
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Let us suppose that an arbitrary triangle \( \bigtriangleup {ABC} \) is given. We want to show that the triangle whose vertices are the intersections of the adjacent tri-sectors is equilateral - this triangle will be referred to as the Morley triangle. Let’s also denote by \( A, B \) and \( C \) the measures of the angles of \( \bigtriangleup {ABC} \) . (This is what is generally known as an \
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No
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Theorem 9.3.1 (Monge's Circle Theorem). If three circles of different radii in the Euclidean plane are chosen so that no circle lies in the interior of another, the three pairs of external tangents to these circles meet in points which are collinear.
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In Figure 9.12 we see a complete example of Monge's Circle theorem in action. There are three random circles. There are three pairs of external tangents. The three points determined by the intersection of the pairs of external tangents lie on a line (shown dashed in the figure).\n\nWe won't even try to write-up a formal proof of the circle theorem. Not that it can't be done - it's just that you can probably get the point better via an informal discussion.\n\nThe main idea is simply to move to 3-dimensional space. Imagine the original flat plane containing our three random circles as being the plane \( z = 0 \) in Euclidean 3-space. Replace the three circles by three spheres of the same radius and having the same centers - clearly the intersections of these spheres with the plane \( z = 0 \) will be our original circles. While pairs of circles are encompassed by two lines (the external tangents that we've been discussing so much), when we have a pair of spheres in 3-space, they are encompassed by a cone which lies tangent to both spheres \( {}^{6} \) . Notice that the cones that lie tangent to a pair of spheres intersect the plane precisely in those infamous external tangents.\n\n\( {}^{6} \) As before, when the spheres happen to have identical radii we get a degenerate case\n- the cone becomes a cylinder.\n\n\n\nFigure 9.12: An example of Monge's circle theorem. The three pairs of external tangents to the circles intersect in points which are collinear.\n\nWell, okay, we've moved to 3-d. We've replaced our circles with spheres and our external tangents with tangent cones. The points of intersection of the external tangents are now the tips of the cones. But, what good has this all done? Is there any reason to believe that the tips of those cones lie in a line?\n\nActually, yes! There is a plane that touches all three spheres tangentially. Actually, there are two such planes, one that touches them all on their upper surfaces and one that touches them all on their lower surfaces. Oh damn! There are actually lots of planes that are tangent to all three spheres but only one that lies above the three of them. That plane intersects the plane \( z = 0 \) in a line - nothing fancy there; any pair of non-parallel planes will intersect in a line (and the only way the planes we are discussing would be parallel is if all three spheres just happened to be the same size). But that plane also lies tangent to the cones that envelope our spheres and so that plane (as well as the plane \( z = 0 \) ) contains the tips of the cones!
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No
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Theorem 2 (De Morgan’s duality laws). For any sets \( S \) and \( {A}_{i}\left( {i \in I}\right) \), the following are true:\n\n\[ \text{(i)}S - \mathop{\bigcup }\limits_{i}{A}_{i} = \mathop{\bigcap }\limits_{i}\left( {S - {A}_{i}}\right) \text{;} \]
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We now use these remarks to prove Theorem 2(i). We have to show that \( S - \bigcup {A}_{i} \) has the same elements as \( \bigcap \left( {S - {A}_{i}}\right) \), i.e., that \( x \in S - \bigcup {A}_{i} \) iff \( x \in \bigcap \left( {S - {A}_{i}}\right) \) . But, by our definitions, we have\n\n\[ x \in S - \bigcup {A}_{i} \Leftrightarrow \left\lbrack {x \in S, x \notin \bigcup {A}_{i}}\right\rbrack \]\n\n\[ \Leftrightarrow \left( {\forall i}\right) \left\lbrack {x \in S, x \notin {A}_{i}}\right\rbrack \]\n\n\[ \Leftrightarrow \left( {\forall i}\right) x \in S - {A}_{i} \]\n\n\[ \Leftrightarrow x \in \bigcap \left( {S - {A}_{i}}\right) \]\n\n as required.
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Yes
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Theorem 1. If \( R \) (also written \( \equiv \) ) is an equivalence relation on \( A \), then all \( R \) -classes are disjoint from each other, and \( A \) is their union.
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Proof. Take two \( R \) -classes, \( \left\lbrack p\right\rbrack \neq \left\lbrack q\right\rbrack \) . Seeking a contradiction, suppose they are not disjoint, so\n\n\[ \left( {\exists x}\right) \;x \in \left\lbrack p\right\rbrack \text{ and }x \in \left\lbrack q\right\rbrack \]\n\ni.e., \( p \equiv x \equiv q \) and hence \( p \equiv q \) . But then, by symmetry and transitivity,\n\n\[ y \in \left\lbrack p\right\rbrack \Leftrightarrow y \equiv p \Leftrightarrow y \equiv q \Leftrightarrow y \in \left\lbrack q\right\rbrack \]\n\ni.e., \( \left\lbrack p\right\rbrack \) and \( \left\lbrack q\right\rbrack \) consist of the same elements \( y \), contrary to assumption \( \left\lbrack p\right\rbrack \neq \left\lbrack q\right\rbrack \) . Thus, indeed, any two (distinct) \( R \) -classes are disjoint.\n\nAlso, by reflexivity,\n\n\[ \left( {\forall x \in A}\right) \;x \equiv x, \]\n\ni.e., \( x \in \left\lbrack x\right\rbrack \) . Thus each \( x \in A \) is in some \( R \) -class (namely, in \( \left\lbrack x\right\rbrack \) ); so all of \( A \) is in the union of such classes,\n\n\[ A \subseteq \mathop{\bigcup }\limits_{x}R\left\lbrack x\right\rbrack \]\n\nConversely,\n\n\[ \left( {\forall x}\right) \;R\left\lbrack x\right\rbrack \subseteq A \]\n\nsince\n\n\[ y \in R\left\lbrack x\right\rbrack \Rightarrow {xRy} \Rightarrow {yRx} \Rightarrow \left( {y, x}\right) \in R \Rightarrow y \in {D}_{R} = A, \]\n\nby definition. Thus \( A \) contains all \( R\left\lbrack x\right\rbrack \), hence their union, and so\n\n\[ A = \mathop{\bigcup }\limits_{x}R\left\lbrack x\right\rbrack \]
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Yes
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Corollary 1. If a set \( A \) is countable or finite, so is any subset \( B \subseteq A \) .
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For if \( A \subset {D}_{u}^{\prime } \) for a sequence \( u \), then certainly \( B \subseteq A \subseteq {D}_{u}^{\prime } \) .
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No
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Corollary 2. If \( A \) is uncountable (or just infinite), so is any superset \( B \supseteq A \) .
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For, if \( B \) were countable or finite, so would be \( A \subseteq B \), by Corollary 1 .
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Yes
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Theorem 1. If \( A \) and \( B \) are countable, so is their cross product \( A \times B \) .
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Proof. If \( A \) or \( B \) is \( \varnothing \), then \( A \times B = \varnothing \), and there is nothing to prove.\n\nThus let \( A \) and \( B \) be nonvoid and countable. We may assume that they fill two infinite sequences, \( A = \left\{ {a}_{n}\right\}, B = \left\{ {b}_{n}\right\} \) (repeat terms if necessary). Then, by definition, \( A \times B \) is the set of all ordered pairs of the form\n\n\[ \left( {{a}_{n},{b}_{m}}\right) ,\;n, m \in N. \]\n\nCall \( n + m \) the rank of the pair \( \left( {{a}_{n},{b}_{m}}\right) \) . For each \( r \in N \), there are \( r - 1 \) pairs of rank \( r \) :\n\n\[ \left( {{a}_{1},{b}_{r - 1}}\right) ,\left( {{a}_{2},{b}_{r - 2}}\right) ,\ldots ,\left( {{a}_{r - 1},{b}_{1}}\right) . \]\n\n(1)\n\nWe now put all pairs \( \left( {{a}_{n},{b}_{m}}\right) \) in one sequence as follows. We start with\n\n\[ \left( {{a}_{1},{b}_{1}}\right) \]\n\nas the first term; then take the two pairs of rank three,\n\n\[ \left( {{a}_{1},{b}_{2}}\right) ,\left( {{a}_{2},{b}_{1}}\right) \]\n\nthen the three pairs of rank four, and so on. At the \( \left( {r - 1}\right) \) st step, we take all pairs of rank \( r \), in the order indicated in (1).\n\nRepeating this process for all ranks ad infinitum, we obtain the sequence of pairs\n\n\[ \left( {{a}_{1},{b}_{1}}\right) ,\left( {{a}_{1},{b}_{2}}\right) ,\left( {{a}_{2},{b}_{1}}\right) ,\left( {{a}_{1},{b}_{3}}\right) ,\left( {{a}_{2},{b}_{2}}\right) ,\left( {{a}_{3},{b}_{1}}\right) ,\ldots ,\]\n\nin which \( {u}_{1} = \left( {{a}_{1},{b}_{1}}\right) ,{u}_{2} = \left( {{a}_{1},{b}_{2}}\right) \), etc.\n\nBy construction, this sequence contains all pairs of all ranks \( r \), hence all pairs that form the set \( A \times B \) (for every such pair has some rank \( r \) and so it must eventually occur in the sequence). Thus \( A \times B \) can be put in a sequence.
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Yes
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Corollary 3. The set \( R \) of all rational numbers \( {}^{2} \) is countable.
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Proof. Consider first the set \( Q \) of all positive rationals, i.e.,\n\n\[ \text{fractions}\frac{n}{m}\text{, with}n, m \in N\text{.} \]\n\nWe may formally identify them with ordered pairs \( \left( {n, m}\right) \), i.e., with \( N \times N \) . We call \( n + m \) the rank of \( \left( {n, m}\right) \) . As in Theorem 1, we obtain the sequence\n\n\[ \frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\frac{1}{4},\frac{2}{3},\frac{3}{2},\frac{4}{1},\ldots \]\n\nBy dropping reducible fractions and inserting also 0 and the negative rationals, we put \( R \) into the sequence\n\n\[ 0,1, - 1,\frac{1}{2}, - \frac{1}{2},2, - 2,\frac{1}{3}, - \frac{1}{3},3, - 3,\ldots \text{, as required.} \]
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Yes
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Theorem 2. The union of any sequence \( \left\{ {A}_{n}\right\} \) of countable sets is countable.
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Proof. As each \( {A}_{n} \) is countable, we may put\n\n\[ \n{A}_{n} = \left\{ {{a}_{n1},{a}_{n2},\ldots ,{a}_{nm},\ldots }\right\} .\n\]\n\n(The double subscripts are to distinguish the sequences representing different sets \( {A}_{n} \) .) As before, we may assume that all sequences are infinite. Now, \( \mathop{\bigcup }\limits_{n}{A}_{n} \) obviously consists of the elements of all \( {A}_{n} \) combined, i.e., all \( {a}_{nm}\left( {n, m \in N}\right) \) . We call \( n + m \) the rank of \( {a}_{nm} \) and proceed as in Theorem 1, thus obtaining\n\n\[ \n\mathop{\bigcup }\limits_{n}{A}_{n} = \left\{ {{a}_{11},{a}_{12},{a}_{21},{a}_{13},{a}_{22},{a}_{31},\ldots }\right\} .\n\]\n\nThus \( \mathop{\bigcup }\limits_{n}{A}_{n} \) can be put in a sequence.
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Yes
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Theorem 3. The interval \( \lbrack 0,1) \) of the real axis is uncountable.
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Proof. We must show that no sequence can comprise all of \( \lbrack 0,1) \) . Indeed, given any \( \left\{ {u}_{n}\right\} \), write each term \( {u}_{n} \) as an infinite decimal fraction; say,\n\n\[ \n{u}_{n} = 0.{a}_{n1},{a}_{n2},\ldots ,{a}_{nm},\ldots \n\] \n\nNext, construct a new decimal fraction \n\n\[ \nz = 0.{x}_{1},{x}_{2},\ldots ,{x}_{n},\ldots , \n\] \n\nchoosing its digits \( {x}_{n} \) as follows.\n\nIf \( {a}_{nn} \) (i.e., the \( n \) th digit of \( {u}_{n} \) ) is 0, put \( {x}_{n} = 1 \) ; if, however, \( {a}_{nn} \neq 0 \), put \( {x}_{n} = 0 \) . Thus, in all cases, \( {x}_{n} \neq {a}_{nn} \), i.e., \( z \) differs from each \( {u}_{n} \) in at least one decimal digit (namely, the \( n \) th digit). It follows that \( z \) is different from all \( {u}_{n} \) and hence is not in \( \left\{ {u}_{n}\right\} \), even though \( z \in \lbrack 0,1) \) .\n\nThus, no matter what the choice of \( \left\{ {u}_{n}\right\} \) was, we found some \( z \in \lbrack 0,1) \) not in the range of that sequence. Hence no \( \left\{ {u}_{n}\right\} \) contains all of \( \lbrack 0,1) \) .
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Yes
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Corollary 1 (rule of signs).\n\n(i) \( a\left( {-b}\right) = \left( {-a}\right) b = - \left( {ab}\right) \) ;\n\n(ii) \( \left( {-a}\right) \left( {-b}\right) = {ab} \) .
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Proof. By Axiom VI,\n\n\[ a\left( {-b}\right) + {ab} = a\left\lbrack {\left( {-b}\right) + b}\right\rbrack = a \cdot 0 = 0. \]\n\nThus\n\n\[ a\left( {-b}\right) + {ab} = 0. \]\n\nBy definition, then, \( a\left( {-b}\right) \) is the additive inverse of \( {ab} \), i.e.,\n\n\[ a\left( {-b}\right) = - \left( {ab}\right) . \]\n\nSimilarly, we show that\n\n\[ \left( {-a}\right) b = - \left( {ab}\right) \]\n\nand that\n\n\[ - \left( {-a}\right) = a\text{.} \]\n\nFinally,(ii) is obtained from (i) when \( a \) is replaced by \( - a \) .
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No
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Corollary 2. In an ordered field, \( a \neq 0 \) implies\n\n\[ \n{a}^{2} = \left( {a \cdot a}\right) > 0.\n\]\n\n(Hence \( 1 = {1}^{2} > 0 \) .)
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Proof. If \( a > 0 \), we may multiply by \( a \) (Axiom IX(b)) to obtain\n\n\[ \na \cdot a > 0 \cdot a = 0\text{, i.e.,}{a}^{2} > 0\text{.}\n\]\n\nIf \( a < 0 \), then \( - a > 0 \) ; so we may multiply the inequality \( a < 0 \) by \( - a \) and\n\nobtain\n\n\[ \na\left( {-a}\right) < 0\left( {-a}\right) = 0\n\]\n\ni.e., by Corollary 1,\n\n\[ \n- {a}^{2} < 0\n\]\n\nwhence\n\n\[ \n{a}^{2} > 0\text{.}\n\]
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Yes
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Theorem 2 (well-ordering of \( N \) ). In an ordered field, each nonvoid set \( A \subseteq N \) has a least member (i.e., one that exceeds no other element of \( A \) ).
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Proof outline. \( {}^{2} \) Given \( \varnothing \neq A \subseteq N \), let \( P\left( n\right) \) be the proposition \
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No
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Theorem 1. If \( a \) and \( b \) are integers (or rationals) in \( F \), so are \( a + b \) and \( {ab} \) .
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Proof. For integers, this follows from Examples (a) and (d) in §§5-6; one only has to distinguish three cases:\n\n(i) \( a, b \in N \) ;\n\n(ii) \( - a \in N, b \in N \) ;\n\n(iii) \( a \in N, - b \in N \) .\n\nThe details are left to the reader (see Basic Concepts of Mathematics, Chapter \( 2,§7 \), Theorem 1).\n\nNow let \( a \) and \( b \) be rationals, say,\n\n\[ a = \frac{p}{q}\text{ and }b = \frac{r}{s} \]\n\nwhere \( p, q, r, s \in J \) and \( q, s \neq 0 \) . Then, as is easily seen,\n\n\[ a \pm b = \frac{{ps} \pm {qr}}{qs}\text{ and }{ab} = \frac{pr}{qs}, \]\n\nwhere \( {qs} \neq 0 \) ; and \( {qs} \) and \( {pr} \) are integers by the first part of the proof (since \( p, q, r, s \in J) \) .\n\nThus \( a \pm b \) and \( {ab} \) are fractions with integral numerators and denominators. Hence, by definition, \( a \pm b \in R \) and \( {ab} \in R \) .
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No
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Theorem 2. In any field \( F \), the set \( R \) of all rationals is a field itself, under the operations defined in \( F \), with the same neutral elements 0 and 1 . Moreover, \( R \) is an ordered field if \( F \) is. (We call \( R \) the rational subfield of \( F \) .)
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Proof. We have to check that \( R \) satisfies the field axioms.\n\nThe closure law I follows from Theorem 1.\n\nAxioms II, III, and VI hold for rationals because they hold for all elements of \( F \) ; similarly for Axioms VII to IX if \( F \) is ordered.\n\nAxiom IV holds in \( R \) because the neutral elements 0 and 1 belong to \( R \) ; indeed, they are integers, hence certainly rationals.\n\nTo verify Axiom V, we must show that \( - x \) and \( {x}^{-1} \) belong to \( R \) if \( x \) does. If, however,\n\n\[ x = \frac{p}{q}\;\left( {p, q \in J, q \neq 0}\right) \]\n\nthen\n\n\[ - x = \frac{-p}{q} \]\n\nwhere again \( - p \in J \) by the definition of \( J \) ; thus \( - x \in R \) .\n\nIf, in addition, \( x \neq 0 \), then \( p \neq 0 \), and\n\n\[ x = \frac{p}{q}\text{ implies }{x}^{-1} = \frac{q}{p}\text{. (Why?) } \]\n\nThus \( {x}^{-1} \in R \) .\n\nNote. The representation\n\n\[ x = \frac{p}{q}\;\left( {p, q \in J}\right) \]\n\nis not unique in general; in an ordered field, however, we can always choose \( q > 0 \), i.e., \( q \in N \) (take \( p \leq 0 \) if \( x \leq 0 \) ).\n\nAmong all such \( q \) there is a least one by Theorem 2 of §§5-6. If \( x = p/q \) , with this minimal \( q \in N \), we say that the rational \( x \) is given in lowest terms.
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Yes
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Theorem 1. In a complete field \( F \) (such as \( {E}^{1} \) ), every nonvoid left-bounded subset \( A \subset F \) has an infimum (i.e., a glb).
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Proof. Let \( B \) be the (nonvoid) set of all lower bounds of \( A \) (such bounds exist since \( A \) is left bounded). Then, clearly, no member of \( B \) exceeds any member of \( A \), and so \( B \) is right bounded by an element of \( A \) . Hence, by the assumed completeness of \( F, B \) has a supremum in \( F \), call it \( p \) .\n\nWe shall show that \( p \) is also the required infimum of \( A \), thus completing the proof.\n\nIndeed, we have\n\n(i) \( p \) is a lower bound of \( A \) . For, by definition, \( p \) is the least upper bound of\n\n\( B \) . But, as shown above, each \( x \in A \) is an upper bound of \( B \) . Thus\n\n\[ \left( {\forall x \in A}\right) \;p \leq x. \]\n\n(ii) \( p \) is the greatest lower bound of \( A \) . For \( p = \sup B \) is not exceeded by any member of \( B \) . But, by definition, \( B \) contains all lower bounds of \( A \) ; so \( p \) is not exceeded by any of them, i.e.,\n\n\[ p = \operatorname{glb}A = \inf A. \]
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Yes
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Theorem 2. In an ordered field \( F \), we have \( q = \sup A\left( {A \subset F}\right) \) iff\n\n(i) \( \left( {\forall x \in A}\right) \;x \leq q \) and\n\n(ii) each field element \( p < q \) is exceeded by some \( x \in A \) ; i.e.,\n\n\[ \left( {\forall p < q}\right) \left( {\exists x \in A}\right) \;p < x. \]\n\nEquivalently,\n\n\( \left( {\mathrm{{ii}}}^{\prime }\right) \)\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists x \in A}\right) \;q - \varepsilon < x;\;\left( {\varepsilon \in F}\right) . \]
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Proof. Condition (i) states that \( q \) is an upper bound of \( A \), while (ii) implies that no smaller element \( p \) is such a bound (since it is exceeded by some \( x \) in \( A) \) . When combined,(i) and (ii) state that \( q \) is the least upper bound.\n\nMoreover, any element \( p < q \) can be written as \( q - \varepsilon \left( {\varepsilon > 0}\right) \) . Hence (ii) can be rephrased as \( \left( {\mathrm{{ii}}}^{\prime }\right) \) .
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Yes
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Corollary 1. Let \( b \in F \) and \( A \subset F \) in an ordered field \( F \) . If each element \( x \) of \( A \) satisfies \( x \leq b\left( {x \geq b}\right) \), so does \( \sup A \) (inf \( A \), respectively), provided it exists in \( F \) .
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In fact, the condition\n\n\[ \left( {\forall x \in A}\right) \;x \leq b \]\n\nmeans that \( b \) is a right bound of \( A \) . However, \( \sup A \) is the least right bound, so \( \sup A \leq b \) ; similarly for inf \( A \) .
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Yes
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Corollary 2. In any ordered field, \( \varnothing \neq A \subseteq B \) implies\n\n\[ \sup A \leq \sup B\text{and}\inf A \geq \inf B\text{,}\]\n\n as well as\n\n\[ \inf A \leq \sup A \]\n\nprovided the suprema and infima involved exist.
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Proof. Let \( p = \inf B \) and \( q = \sup B \) .\n\nAs \( q \) is a right bound of \( B \),\n\n\[ x \leq q\text{for all}x \in B\text{.}\]\n\nBut \( A \subseteq B \), so \( B \) contains all elements of \( A \) . Thus\n\n\[ x \in A \Rightarrow x \in B \Rightarrow x \leq q \]\n\nso, by Corollary 1, also\n\n\[ \sup A \leq q = \sup B \]\n\n as claimed.\n\nSimilarly, one gets \( \inf A \geq \inf B \) .\n\nFinally, if \( A \neq \varnothing \), we can fix some \( x \in A \) . Then\n\n\[ \inf A \leq x \leq \sup A \]\nand all is proved.
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Yes
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Theorem 1. Any complete field \( F \) (e.g., \( {E}^{1} \) ) is Archimedean. \( {}^{1} \n\nThat is, given any \( x, y \in F\left( {x > 0}\right) \) in such a field, there is a natural \( n \in F \) such that \( {nx} > y \) .
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Proof by contradiction. Suppose this fails. Thus, given \( y, x \in F\left( {x > 0}\right) \) , assume that there is no \( n \in N \) with \( {nx} > y \) .\n\nThen\n\n\[ \n\left( {\forall n \in N}\right) \;{nx} \leq y \n\] \n\ni.e., \( y \) is an upper bound of the set of all products \( {nx}\left( {n \in N}\right) \) . Let \n\n\[ \nA = \{ {nx} \mid n \in N\} . \n\] \n\nClearly, \( A \) is bounded above (by \( y \) ) and \( A \neq \varnothing \) ; so, by the assumed completeness of \( F, A \) has a supremum, say, \( q = \sup A \) .\n\nAs \( q \) is an upper bound, we have (by the definition of \( A \) ) that \( {nx} \leq q \) for all \( n \in N \), hence also \( \left( {n + 1}\right) x \leq q \) ; i.e., \n\n\[ \n{nx} \leq q - x \n\] \n\nfor all \( n \in N \) (since \( n \in N \Rightarrow n + 1 \in N \) ).\n\n--- \n\nThus \( q - x \) (which is less than \( q \) for \( x > 0 \) ) is another upper bound of all \( {nx} \) , i.e., of the set \( A \) .\n\nThis is impossible, however, since \( q = \sup A \) is the least upper bound of \( A \) .\n\nThis contradiction completes the proof.
|
Yes
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Corollary 1. In any Archimedean (hence also in any complete) field \( F \), the set \( N \) of all natural elements has no upper bounds, and the set \( J \) of all integers has neither upper nor lower bounds. Thus\n\n\[ \left( {\forall y \in F}\right) \left( {\exists m, n \in N}\right) \; - m < y < n. \]
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Proof. Given any \( y \in F \), one can use the Archimedean property (with \( x = 1 \) ) to find an \( n \in N \) such that\n\n\[ n \cdot 1 > y\text{, i.e.,}n > y\text{.} \]\n\nSimilarly, there is an \( m \in N \) such that\n\n\[ m > - y\text{, i.e.,} - m < y\text{.} \]\n\nThis proves our last assertion and shows that no \( y \in F \) can be a right bound of \( N \) (for \( y < n \in N \) ), or a left bound of \( J \) (for \( y > - m \in J \) ).
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Yes
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In any Archimedean (hence also in any complete) field \( F \), each left (right) bounded set \( A \) of integers \( \left( {\varnothing \neq A \subset J}\right) \) has a minimum (maximum, respectively).
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Proof. Suppose \( \varnothing \neq A \subseteq J \), and \( A \) has a lower bound \( y \) .\n\nThen Corollary 1 (last part) yields a natural \( m \), with \( - m < y \), so that\n\n\[ \left( {\forall x \in A}\right) \; - m < x \]\n\nand so \( x + m > 0 \) .\n\nThus, by adding \( m \) to each \( x \in A \), we obtain a set (call it \( A + m \) ) of naturals. \( {}^{2} \)\n\nNow, by Theorem 2 of \( §§5 - 6, A + m \) has a minimum; call it \( p \) . As \( p \) is the least of all sums \( x + m, p - m \) is the least of all \( x \in A \) ; so \( p - m = \min A \) exists, as claimed.\n\nNext, let \( A \) have a right bound \( z \) . Then look at the set of all additive inverses \( - x \) of points \( x \in A \) ; call it \( B \) .\n\nClearly, \( B \) is left bounded (by \( - z \) ), so it has a minimum, say, \( u = \min B \) . Then \( - u = \max A \) . (Verify!)
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Yes
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Corollary 2. Any element \( x \) of an Archimedean field \( F \) has an integral part \( \left\lbrack x\right\rbrack \) . It is the unique integer \( n \) such that\n\n\[ n \leq x < n + 1 \]\n
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(It exists, by Theorem 2.)
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No
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Theorem 3 (density of rationals). Between any elements \( a \) and \( b\\left( {a < b}\\right) \) of an Archimedean field \( F \) (such as \( {E}^{1} \) ), there is a rational \( r \in F \) with\n\n\[ a < r < b.\\text{.} \]
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Proof. Let \( p = \\left\\lbrack a\\right\\rbrack \) (the integral part of \( a \) ). The idea of the proof is to start with \( p \) and to mark off a small \
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No
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Theorem 1. Given \( a \geq 0 \) in a complete field \( F \), and a natural number \( n \in {E}^{1} \) , there always is a unique element \( p \in F, p \geq 0 \), such that\n\n\[ {p}^{n} = a\text{.} \]\n\nIt is called the \( n \) th root of \( a \), denoted\n\n\[ \sqrt[n]{a}\text{or}{a}^{1/n}\text{.} \]\n\n(Note that \( \sqrt[n]{a} \geq 0 \), by definition.)
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A direct proof, from the completeness axiom, is sketched in Problems 1 and 2 below. We shall give a simpler proof in Chapter 4, §9, Example (a). At present, we omit it and temporarily take Theorem 1 for granted.
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No
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Theorem 2. Every complete field \( F \) (such as \( {E}^{1} \) ) has irrational elements, i.e., elements that are not rational.
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Proof. By Theorem 1, \( F \) has the element\n\n\[ p = \sqrt{2}\text{with}{p}^{2} = 2\text{.} \]\n\nSeeking a contradiction, suppose \( \sqrt{2} \) is rational, i.e.,\n\n\[ \sqrt{2} = \frac{m}{n} \]\n\nfor some \( m, n \in N \) in lowest terms (see \( §7 \), final note).\n\nThen \( m \) and \( n \) are not both even (otherwise, reduction by 2 would yield a smaller \( n \) ). From \( m/n = \sqrt{2} \), we obtain\n\n\[ {m}^{2} = 2{n}^{2} \]\n\nso \( {m}^{2} \) is even.\n\nOnly even elements have even squares, however. \( {}^{2} \) Thus \( m \) itself must be even; i.e., \( m = {2r} \) for some \( r \in N \). It follows that\n\n\[ 4{r}^{2} = {m}^{2} = 2{n}^{2}\text{, i.e.,}2{r}^{2} = {n}^{2} \]\n\nand, by the same argument, \( n \) must be even.\n\nThis contradicts the fact that \( m \) and \( n \) are not both even, and this contradiction shows that \( \sqrt{2} \) must be irrational.
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Yes
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Theorem 1.\n\n(i) If \( {x}_{n} \geq b \) for infinitely many \( n \), then\n\n\[ \overline{\lim }{x}_{n} \geq b\text{ as well. } \]\n\n(ii) If \( {x}_{n} \leq a \) for all but finitely many \( n,{}^{4} \) then\n\n\[ \overline{\lim }{x}_{n} \leq a\text{ as well. } \]\n\nSimilarly for lower limits (with all inequalities reversed).
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Proof.\n\n(i) If \( {x}_{n} \geq b \) for infinitely many \( n \), then such \( n \) must occur in each set\n\n\[ {A}_{m} = \left\{ {{x}_{m},{x}_{m + 1},\ldots }\right\} .\n\nHence\n\n\[ \left( {\forall m}\right) \;{q}_{m} = \sup {A}_{m} \geq b \]\n\nso \( \bar{L} = \mathop{\inf }\limits_{m}{q}_{m} \geq b \), by Corollary 1 of \( §§8 - 9 \) .\n\n(ii) If \( {x}_{n} \leq a \) except finitely many \( n \), let \( {n}_{0} \) be the last of these \
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Yes
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(i) If \( \overline{\lim }{x}_{n} > a \), then \( {x}_{n} > a \) for infinitely many \( n \) .\n\n(ii) If \( \overline{\lim }{x}_{n} < b \), then \( {x}_{n} < b \) for all but finitely many \( n \) .
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Proof. Assume the opposite and find a contradiction to Theorem 1.
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No
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Theorem 2. We have \( q = \overline{\lim }{x}_{n} \) in \( {E}^{ * } \) iff\n\n\( \left( {\mathrm{i}}^{\prime }\right) \) each neighborhood \( {G}_{q} \) contains \( {x}_{n} \) for infinitely many \( n \), and\n\n(ii’) if \( q < b \), then \( {x}_{n} \geq b \) for at most finitely many \( n.{}^{6} \)
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Proof. If \( q = \overline{\lim }{x}_{n} \), Corollary 2 yields \( \left( {\mathrm{{ii}}}^{\prime }\right) \).\n\nIt also shows that any interval \( \left( {a, b}\right) \), with \( a < q < b \), contains infinitely many \( {x}_{n} \) (for there are infinitely many \( {x}_{n} > a \), and only finitely many \( {x}_{n} \geq b \) , by \( \left. \left( {\mathrm{{ii}}}^{\prime }\right) \right) \).\n\nNow if \( q \in {E}^{1} \),\n\n\[{G}_{q} = \left( {q - \varepsilon, q + \varepsilon }\right)\]\n\nis such an interval, so we obtain \( \left( {\mathrm{i}}^{\prime }\right) \). The cases \( q = \pm \infty \) are analogous; we leave them to the reader.\n\nConversely, assume \( \left( {\mathrm{i}}^{\prime }\right) \) and \( \left( {\mathrm{{ii}}}^{\prime }\right) \).\n\nSeeking a contradiction, let \( q < \bar{L} \) ; say,\n\n\[q < b < \overline{\lim }{x}_{n}\]\n\nThen Corollary 2(i) yields \( {x}_{n} > b \) for infinitely many \( n \), contrary to our assumption \( \left( {\mathrm{{ii}}}^{\prime }\right) \).\n\nSimilarly, \( q > \overline{\lim }{x}_{n} \) would contradict \( \left( {\mathrm{i}}^{\prime }\right) \).\n\nThus necessarily \( q = \overline{\lim }{x}_{n} \).
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No
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Theorem 3. We have \( q = \lim {x}_{n} \) in \( {E}^{ * } \) iff\n\n\[ \underline{\lim }{x}_{n} = \overline{\lim }{x}_{n} = q. \]
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Proof. Suppose\n\n\[ \underline{\lim }{x}_{n} = \overline{\lim }{x}_{n} = q \]\n\nIf \( q \in {E}^{1} \), then every \( {G}_{q} \) is an interval \( \left( {a, b}\right), a < q < b \) ; therefore, Corollary 2(ii) and its analogue for \( \lim {x}_{n} \) imply (with \( q \) treated as both \( \overline{\lim }{x}_{n} \) and \( \left. {\underline{\lim }{x}_{n}}\right) \) that\n\n\[ a < {x}_{n} < b\text{ for all but finitely many }n. \]\n\nThus by Definition \( 2, q = \lim {x}_{n} \), as claimed.\n\nConversely, if so, then any \( {G}_{q} \) (no matter how small) contains all but finitely many \( {x}_{n} \) . Hence so does any interval \( \left( {a, b}\right) \) with \( a < q < b \), for it contains some small \( {G}_{q} \).\n\nNow, exactly as in the proof of Theorem 2, one excludes\n\n\[ q \neq \underline{\lim }{x}_{n}\text{ and }q \neq \overline{\lim }{x}_{n} \]\n\nThis settles the case \( q \in {E}^{1} \) . The cases \( q = \pm \infty \) are quite analogous.
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Yes
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Theorem 1. For any vectors \( \bar{x},\bar{y} \), and \( \bar{z} \in {E}^{n} \) and any \( a, b \in {E}^{1} \), we have\n\n(a) \( \bar{x} + \bar{y} \) and \( a\bar{x} \) are vectors in \( {E}^{n} \) (closure laws);\n\n(b) \( \bar{x} + \bar{y} = \bar{y} + \bar{x} \) (commutativity of vector addition);\n\n(c) \( \left( {\bar{x} + \bar{y}}\right) + \bar{z} = \bar{x} + \left( {\bar{y} + \bar{z}}\right) \) (associativity of vector addition);\n\n(d) \( \bar{x} + \overline{0} = \overline{0} + \bar{x} = \bar{x} \), i.e., \( \overline{0} \) is the neutral element of addition;\n\n(e) \( \bar{x} + \left( {-\bar{x}}\right) = \overline{0} \), i.e., \( - \bar{x} \) is the additive inverse of \( \bar{x} \) ;\n\n(f) \( a\left( {\bar{x} + \bar{y}}\right) = a\bar{x} + a\bar{y} \) and \( \left( {a + b}\right) \bar{x} = a\bar{x} + b\bar{x} \) (distributive laws);\n\n(g) \( \left( {ab}\right) \bar{x} = a\left( {b\bar{x}}\right) \) ;\n\n(h) \( 1\bar{x} = \bar{x} \) .
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Proof. Assertion (a) is immediate from Definitions 1 and 6. The rest follows from corresponding properties of real numbers.\n\nFor example, to prove (b), let \( \bar{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) ,\bar{y} = \left( {{y}_{1},\ldots ,{y}_{n}}\right) \) . Then by definition, we have\n\n\[ \bar{x} + \bar{y} = \left( {{x}_{1} + {y}_{1},\ldots ,{x}_{n} + {y}_{n}}\right) \text{and}\bar{y} + \bar{x} = \left( {{y}_{1} + {x}_{1},\ldots ,{y}_{n} + {x}_{n}}\right) \text{.} \]\n\nThe right sides in both expressions, however, coincide since addition is commutative in \( {E}^{1} \) . Thus \( \bar{x} + \bar{y} = \bar{y} + \bar{x} \), as claimed; similarly for the rest, which we leave to the reader.
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No
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Theorem 2. If \( \bar{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is a vector in \( {E}^{n} \), then, with \( {\bar{e}}_{k} \) as above,\n\n\[ \bar{x} = {x}_{1}{\bar{e}}_{1} + {x}_{2}{\bar{e}}_{2} + \cdots + {x}_{n}{\bar{e}}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{\bar{e}}_{k}. \]\n\nMoreover, if \( \bar{x} = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\bar{e}}_{k} \) for some \( {a}_{k} \in {E}^{1} \), then necessarily \( {a}_{k} = {x}_{k} \) , \( k = 1,\ldots, n \) .
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Proof. By definition,\n\n\[ {\bar{e}}_{1} = \left( {1,0,\ldots ,0}\right) ,{\bar{e}}_{2} = \left( {0,1,\ldots ,0}\right) ,\ldots ,{\bar{e}}_{n} = \left( {0,0,\ldots ,1}\right) . \]\n\nThus\n\n\[ {x}_{1}{\bar{e}}_{1} = \left( {{x}_{1},0,\ldots ,0}\right) ,{x}_{2}{\bar{e}}_{2} = \left( {0,{x}_{2},\ldots ,0}\right) ,\ldots ,{x}_{n}{\bar{e}}_{n} = \left( {0,0,\ldots ,{x}_{n}}\right) . \]\n\nAdding up componentwise, we obtain\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{\bar{e}}_{k} = \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = \bar{x}, \]\n\nas asserted.\n\nMoreover, if the \( {x}_{k} \) are replaced by any other \( {a}_{k} \in {E}^{1} \), the same process yields\n\n\[ \left( {{a}_{1},\ldots ,{a}_{n}}\right) = \bar{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \]\n\ni.e., the two n-tuples coincide, whence \( {a}_{k} = {x}_{k}, k = 1,\ldots, n \) .
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Yes
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Theorem 3. For any vectors \( \bar{x},\bar{y} \), and \( \bar{z} \in {E}^{n} \) and any \( a, b \in {E}^{1} \), we have\n\n(a) \( \bar{x} \cdot \bar{x} \geq 0 \), and \( \bar{x} \cdot \bar{x} > 0 \) iff \( \bar{x} \neq \overline{0} \) ;\n\n(b) \( \left( {a\bar{x}}\right) \cdot \left( {b\bar{y}}\right) = \left( {ab}\right) \left( {\bar{x} \cdot \bar{y}}\right) \) ;\n\n(c) \( \bar{x} \cdot \bar{y} = \bar{y} \cdot \bar{x} \) (commutativity of inner products);\n\n(d) \( \left( {\bar{x} + \bar{y}}\right) \cdot \bar{z} = \bar{x} \cdot \bar{z} + \bar{y} \cdot \bar{z} \) (distributive law).
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Proof. To prove these properties, express all in terms of the components of \( \bar{x} \) , \( \bar{y} \), and \( \bar{z} \), and proceed as in Theorem 1.\n\nNote that (b) implies \( \bar{x} \cdot \overline{0} = 0 \) (put \( a = 1, b = 0 \) ).
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Yes
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Theorem 4. For any vectors \( \bar{x} \) and \( \bar{y} \in {E}^{n} \) and any \( a \in {E}^{1} \), we have the following properties:\n\n\( \left( {\mathrm{a}}^{\prime }\right) \left| \bar{x}\right| \geq 0 \), and \( \left| \bar{x}\right| > 0 \) iff \( \bar{x} \neq \overline{0} \) .\n\n\( \left( {\mathrm{b}}^{\prime }\right) \left| {a\bar{x}}\right| = \left| a\right| \left| \bar{x}\right| .\n\n\( \left( {\mathrm{c}}^{\prime }\right) \left| {\bar{x} \cdot \bar{y}}\right| \leq \left| \bar{x}\right| \left| \bar{y}\right| \), or, in components,\n\n\[ \n{\left( \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{y}_{k}\right) }^{2} \leq \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}^{2}}\right) \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{y}_{k}^{2}}\right) \;\text{ (Cauchy-Schwarz inequality). }\n\]\n\nEquality, \( \left| {\bar{x} \cdot \bar{y}}\right| = \left| \bar{x}\right| \left| \bar{y}\right| \), holds iff \( \bar{x}\parallel \bar{y} \) .\n\n\( \left( {\mathrm{d}}^{\prime }\right) \left| {\bar{x} + \bar{y}}\right| \leq \left| \bar{x}\right| + \left| \bar{y}\right| \) and \( \left| \right| \bar{x}\left| -\right| \bar{y}\left| \right| \leq \left| {\bar{x} - \bar{y}}\right| \) (triangle inequalities).
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Proof. Property \( \left( {\mathrm{a}}^{\prime }\right) \) follows from Theorem 3(a) since\n\n\[ \n{\left| \bar{x}\right| }^{2} = \bar{x} \cdot \bar{x}\text{ (see Definition 4). }\n\]\n\nFor \( \left( {\mathrm{b}}^{\prime }\right) \), use Theorem 3(b), to obtain\n\n\[ \n\left( {a\bar{x}}\right) \cdot \left( {a\bar{x}}\right) = {a}^{2}\left( {\bar{x} \cdot \bar{x}}\right) = {a}^{2}{\left| \bar{x}\right| }^{2}.\n\]\n\nBy Definition 4, however,\n\n\[ \n\left( {a\bar{x}}\right) \cdot \left( {a\bar{x}}\right) = {\left| a\bar{x}\right| }^{2}.\n\]\n\nThus\n\n\[ \n{\left| a\bar{x}\right| }^{2} = {a}^{2}{\left| x\right| }^{2}\n\]\n\nso that \( \left| {a\bar{x}}\right| = \left| a\right| \left| \bar{x}\right| \), as claimed.\n\nNow we prove \( \left( {\mathrm{c}}^{\prime }\right) \) . If \( \bar{x}\parallel \bar{y} \) then \( \bar{x} = t\bar{y} \) or \( \bar{y} = t\bar{x} \) ; so \( \left| {\bar{x} \cdot \bar{y}}\right| = \left| \bar{x}\right| \left| \bar{y}\right| \) follows by \( \left( {\mathrm{b}}^{\prime }\right) \) . (Verify!)\n\nOtherwise, \( \bar{x} \neq t\bar{y} \) and \( \bar{y} \neq t\bar{x} \) for all \( t \in {E}^{1} \) . Then we obtain, for all \( t \in {E}^{1} \),\n\n\[ \n0 \neq {\left| t\bar{x} - \bar{y}\right| }^{2} = \mathop{\sum }\limits_{{k = 1}}^{n}{\left( t{x}_{k} - {y}_{k}\right) }^{2} = {t}^{2}\mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}^{2} - {2t}\mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{y}_{k} + \mathop{\sum }\limits_{{k = 1}}^{n}{y}_{k}^{2}.\n\]\n\nThus, setting\n\n\[ \nA = \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}^{2}, B = 2\mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{y}_{k},\text{ and }C = \mathop{\sum }\limits_{{k = 1}}^{n}{y}_{k}^{2},\n\]\n\nwe see that the quadratic equation\n\n\[ \n0 = A{t}^{2} - {Bt} + C\n\]\n\nhas no real solutions in \( t \), so its discriminant, \( {B}^{2} - {4AC} \), must be negative; i.e.,\n\n\[ \n4{\left( \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{y}_{k}\right) }^{2} - 4\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}^{2}}\right) \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{y}_{k}^{2}}\right) < 0\n\]\n\nproving \( \left( {\mathrm{c}}^{\prime }\right) \).\n\nTo prove \( \left( {\mathrm{d}}^{\prime }\right) \), use Definition 2 and Theorem 3(d), to obtain\n\n\[ \n{\left| \bar{x} + \bar{y}\right| }^{2} = \left( {\bar{x} + \bar{y}}\right) \cdot \left( {\bar{x} + \bar{y}}\right) = \bar{x} \cdot \bar{x} + \bar{y} \cdot \bar{y} + 2\bar{x} \cdot \bar{y} = {\left| \bar{x}\right| }^{2} + {\left| \bar{y}\right| }^{2} + 2\bar{x} \cdot \bar{y}.\n\]\n\nBut \( \bar{x} \cdot \bar{y} \leq \left| \bar{x}\right| \left| \bar{y}\right| \) by \( \left( {\mathrm{c}}^{\prime }\right) \) . Thus we have\n\n\[ \n{\left| \bar{x} + \bar{y}\right| }^{2} \leq {\left| \bar{x}\right| }^{2} + {\left| \bar{y}\right| }^{2} + 2\left| \bar{x}\right| \left| \bar{y}\right| = {\left( \left| \bar{x}\right| + \lef
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Yes
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Theorem 5. For any points \( \bar{x},\bar{y} \), and \( \bar{z} \in {E}^{n} \), we have\n\n(i) \( \rho \left( {\bar{x},\bar{y}}\right) \geq 0 \), and \( \rho \left( {\bar{x},\bar{y}}\right) = 0 \) iff \( \bar{x} = \bar{y} \) ;\n\n(ii) \( \rho \left( {\bar{x},\bar{y}}\right) = \rho \left( {\bar{y},\bar{x}}\right) \) ;\n\n(iii) \( \rho \left( {\bar{x},\bar{z}}\right) \leq \rho \left( {\bar{x},\bar{y}}\right) + \rho \left( {\bar{y},\bar{z}}\right) \) (triangle inequality).
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Proof.\n\n(i) By Definition 3 and Note \( 3,\rho \left( {\bar{x},\bar{y}}\right) = \left| {\bar{x} - \bar{y}}\right| \) ; therefore, by Theorem \( 4\left( {\mathrm{a}}^{\prime }\right) \) , \( \rho \left( {\bar{x},\bar{y}}\right) = \left| {\bar{x} - \bar{y}}\right| \geq 0 \)\n\nAlso, \( \left| {\bar{x} - \bar{y}}\right| > 0 \) iff \( \bar{x} - \bar{y} \neq 0 \), i.e., iff \( \bar{x} \neq \bar{y} \) . Hence \( \rho \left( {\bar{x},\bar{y}}\right) \neq 0 \) iff \( \bar{x} \neq \bar{y} \), and assertion (i) follows.\n\n(ii) By Theorem \( 4\left( {\mathrm{\;b}}^{\prime }\right) ,\left| {\bar{x} - \bar{y}}\right| = \left| {\left( {-1}\right) \left( {\bar{y} - \bar{x}}\right) }\right| = \left| {\bar{y} - \bar{x}}\right| \), so (ii) follows.\n\n(iii) By Theorem \( 4\left( {\mathrm{\;d}}^{\prime }\right) \),\n\n\[ \rho \left( {\bar{x},\bar{y}}\right) + \rho \left( {\bar{y},\bar{z}}\right) = \left| {\bar{x} - \bar{y}}\right| + \left| {\bar{y} - \bar{z}}\right| \geq \left| {\bar{x} - \bar{y} + \bar{y} - \bar{z}}\right| = \rho \left( {\bar{x},\bar{z}}\right) .\n\]
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Yes
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Theorem 1. A set \( A \subseteq {E}^{n} \) is a plane (hyperplane) iff \( A \) is exactly the set of all \( \bar{x} \in {E}^{n} \) satisfying (4) for some fixed \( c \in {E}^{1} \) and \( \overrightarrow{u} = \left( {{u}_{1},\ldots ,{u}_{n}}\right) \neq \overline{0} \) .
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Proof. Indeed, as we saw above, each plane has an equation of the form (4). Conversely, any equation of that form (with, say, \( {u}_{1} \neq 0 \) ) can be written as\n\n\[ \n{u}_{1}\left( {{x}_{1} - \frac{c}{{u}_{1}}}\right) + {u}_{2}{x}_{2} + {u}_{3}{x}_{3} + \cdots + {u}_{n}{x}_{n} = 0.\n\]\n\nThen, setting \( {a}_{1} = c/{u}_{1} \) and \( {a}_{k} = 0 \) for \( k \geq 2 \), we transform it into (3), which is, by definition, the equation of a plane through \( \bar{a} = \left( {c/{u}_{1},0,\ldots ,0}\right) \), orthogonal to \( \overrightarrow{u} = \left( {{u}_{1},\ldots ,{u}_{n}}\right) \). \n\nThus, briefly, planes are exactly all sets with linear equations (4). In this connection, equation (4) is called the general equation of a plane. The vector \( \overrightarrow{u} \) is said to be normal to the plane. Clearly, if both sides of (4) are multiplied by a nonzero scalar \( q \), one obtains an equivalent equation (representing the same set). Thus we may replace \( {u}_{k} \) by \( q{u}_{k} \), i.e., \( \overrightarrow{u} \) by \( q\overrightarrow{u} \), without affecting the plane. In particular, we may replace \( \overrightarrow{u} \) by the unit vector \( \overrightarrow{u}/\left| \overrightarrow{u}\right| \), as in lines (this is called the normalization of the equation). Thus\n\n\[ \n\frac{\overrightarrow{u}}{\left| \overrightarrow{u}\right| } \cdot \left( {\bar{x} - \bar{a}}\right) = 0\n\]\n\n(5)\n\nand\n\n\[ \n\bar{x} = \bar{a} + t\frac{\overrightarrow{u}}{\left| \overrightarrow{u}\right| }\n\]\n\n(6)\n\nare the normalized (or normal) equations of the plane (3) and line (1), respectively.
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Yes
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Theorem 1. \( {E}^{2} = C \) is a field, with zero element \( 0 = \left( {0,0}\right) \) and unity \( 1 = \) \( \left( {1,0}\right) \), under addition and multiplication as defined above.
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Proof. We only must show that multiplication obeys Axioms I-VI of the field axioms. Note that for addition, all is proved in Theorem 1 of §§1-3.\n\nAxiom I (closure) is obvious from our definition, for if \( z \) and \( {z}^{\prime } \) are in \( C \), so is \( z{z}^{\prime } \) .\n\nTo prove commutativity, take any complex numbers\n\n\[ z = \left( {x, y}\right) \text{ and }{z}^{\prime } = \left( {{x}^{\prime },{y}^{\prime }}\right) \]\n\nand verify that \( z{z}^{\prime } = {z}^{\prime }z \) . Indeed, by definition,\n\n\[ z{z}^{\prime } = \left( {x{x}^{\prime } - y{y}^{\prime }, x{y}^{\prime } + y{x}^{\prime }}\right) \text{and}{z}^{\prime }z = \left( {{x}^{\prime }x - {y}^{\prime }y,{x}^{\prime }y + {y}^{\prime }x}\right) \text{;}\n\nbut the two expressions coincide by the commutative laws for real numbers. Associativity and distributivity are proved in a similar manner.\n\nNext, we show that \( 1 = \left( {1,0}\right) \) satisfies Axiom IV(b), i.e., that \( {1z} = z \) for any complex number \( z = \left( {x, y}\right) \) . In fact, by definition, and by axioms for \( {E}^{1} \),\n\n\[ {1z} = \left( {1,0}\right) \left( {x, y}\right) = \left( {{1x} - {0y},{1y} + {0x}}\right) = \left( {x - 0, y + 0}\right) = \left( {x, y}\right) = z.\]\n\nIt remains to verify Axiom V(b), i.e., to show that each complex number \( z = \left( {x, y}\right) \neq \left( {0,0}\right) \) has an inverse \( {z}^{-1} \) such that \( z{z}^{-1} = 1 \) . It turns out that the inverse is obtained by setting\n\n\[ {z}^{-1} = \left( {\frac{x}{{\left| z\right| }^{2}}, - \frac{y}{{\left| z\right| }^{2}}}\right) .\n\nIn fact, we then get\n\n\[ z{z}^{-1} = \left( {\frac{{x}^{2}}{{\left| z\right| }^{2}} + \frac{{y}^{2}}{{\left| z\right| }^{2}}, - \frac{xy}{{\left| z\right| }^{2}} + \frac{yx}{{\left| z\right| }^{2}}}\right) = \left( {\frac{{x}^{2} + {y}^{2}}{{\left| z\right| }^{2}},0}\right) = \left( {1,0}\right) = 1 \]\n\nsince \( {x}^{2} + {y}^{2} = {\left| z\right| }^{2} \), by definition. This completes the proof.
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Yes
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Corollary 1. \( {i}^{2} = - 1 \) ; i.e., \( \left( {0,1}\right) \left( {0,1}\right) = \left( {-1,0}\right) \) .
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Proof. By definition, \( \left( {0,1}\right) \left( {0,1}\right) = \left( {0 \cdot 0 - 1 \cdot 1,0 \cdot 1 + 1 \cdot 0}\right) = \left( {-1,0}\right) \) .
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Yes
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Theorem 2. Every \( z \in C \) has a unique representation as\n\n\[ z = x + {yi} \]\n\nwhere \( x \) and \( y \) are real and \( i = \left( {0,1}\right) \). Specifically,\n\n\[ z = x + {yi}\text{ iff }z = \left( {x, y}\right) . \]
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Proof. By our conventions, \( x = \left( {x,0}\right) \) and \( y = \left( {y,0}\right) \), so\n\n\[ x + {yi} = \left( {x,0}\right) + \left( {y,0}\right) \left( {0,1}\right) . \]\n\nComputing the right-hand expression from definitions, we have for any \( x, y \in \) \( {E}^{1} \) that\n\n\[ x + {yi} = \left( {x,0}\right) + \left( {y \cdot 0 - 0 \cdot 1, y \cdot 1 + 0 \cdot 1}\right) = \left( {x,0}\right) + \left( {0, y}\right) = \left( {x, y}\right) . \]\n\nThus \( \left( {x, y}\right) = x + {yi} \) for any \( x, y \in {E}^{1} \). In particular, if \( \left( {x, y}\right) \) is the given number \( z \in C \) of the theorem, we obtain \( z = \left( {x, y}\right) = x + {yi} \), as required.\n\nTo prove uniqueness, suppose that we also have\n\n\[ z = {x}^{\prime } + {y}^{\prime }i\text{ with }{x}^{\prime } = \left( {{x}^{\prime },0}\right) \text{ and }{y}^{\prime } = \left( {{y}^{\prime },0}\right) . \]\n\nThen, as shown above, \( z = \left( {{x}^{\prime },{y}^{\prime }}\right) \). Since also \( z = \left( {x, y}\right) \), we have \( \left( {x, y}\right) = \) \( \left( {{x}^{\prime },{y}^{\prime }}\right) \), i.e., the two ordered pairs coincide, and so \( x = {x}^{\prime } \) and \( y = {y}^{\prime } \) after all.
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Yes
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Theorem 1 (Hausdorff property \( {}^{2} \) ). Any two points \( p \) and \( q\left( {p \neq q}\right) \) in \( \left( {S,\rho }\right) \) are centers of two disjoint globes.
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Proof. As \( p \neq q \), we have \( \rho \left( {p, q}\right) > 0 \) by metric axiom \( \left( {\mathrm{i}}^{\prime }\right) \) . Thus we may put\n\n\[ \varepsilon = \frac{1}{2}\rho \left( {p, q}\right) > 0 \]\n\nIt remains to show that with this \( \varepsilon ,{G}_{p}\left( \varepsilon \right) \cap {G}_{q}\left( \varepsilon \right) = \varnothing \) .\n\nSeeking a contradiction, suppose this fails. Then there is \( x \in {G}_{p}\left( \varepsilon \right) \cap {G}_{q}\left( \varepsilon \right) \) so that \( \rho \left( {p, x}\right) < \varepsilon \) and \( \rho \left( {x, q}\right) < \varepsilon \) . By the triangle law,\n\n\[ \rho \left( {p, q}\right) \leq \rho \left( {p, x}\right) + \rho \left( {x, q}\right) < \varepsilon + \varepsilon = {2\varepsilon }\text{; i.e.,}\rho \left( {p, q}\right) < {2\varepsilon }\text{,} \]\n\nwhich is impossible since \( \rho \left( {p, q}\right) = {2\varepsilon } \) .
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Yes
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Theorem 2. The union of any finite or infinite family of open sets \( {A}_{i}\left( {i \in I}\right) \) , denoted\n\n\[ \mathop{\bigcup }\limits_{{i \in I}}{A}_{i} \]\n\n is open itself. So also is\n\n\[ \mathop{\bigcap }\limits_{{i = 1}}^{n}{A}_{i} \]\n\n for finitely many open sets. (This fails for infinitely many sets \( {A}_{i} \) ; see Problem 11 below.)
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Proof. We must show that any point \( p \) of \( A = \mathop{\bigcup }\limits_{i}{A}_{i} \) is interior to \( A \) .\n\nNow if \( p \in \mathop{\bigcup }\limits_{i}{A}_{i}, p \) is in some \( {A}_{i} \), and it is an interior point of \( {A}_{i} \) (for \( {A}_{i} \) is open, by assumption). Thus there is a globe\n\n\[ {G}_{p} \subseteq {A}_{i} \subseteq A \]\n\n as required.\n\nFor finite intersections, it suffices to consider two open sets \( A \) and \( B \) (for \( n \) sets, all then follows by induction). We must show that each \( p \in A \cap B \) is interior to \( A \cap B \) .\n\nNow as \( p \in A \) and \( A \) is open, we have some \( {G}_{p}\left( {\delta }^{\prime }\right) \subseteq A \) . Similarly, there is \( {G}_{p}\left( {\delta }^{\prime \prime }\right) \subseteq B \) . Then the smaller of the two globes, call it \( {G}_{p} \), is in both \( A \) and \( B \), so\n\n\[ {G}_{p} \subseteq A \cap B \]\n\n and \( p \) is interior to \( A \cap B \), indeed.
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Yes
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Theorem 3. If the sets \( {A}_{i}\left( {i \in I}\right) \) are closed, so is\n\n\[ \mathop{\bigcap }\limits_{{i \in I}}{A}_{i} \]\n\n(even for infinitely many sets).
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Proof. Let \( A = \mathop{\bigcap }\limits_{{i \in I}}{A}_{i} \) . To prove that \( A \) is closed, we show that \( - A \) is open. Now by set theory (see Chapter 1, §§1-3, Theorem 2),\n\n\[ - A = - \mathop{\bigcap }\limits_{i}{A}_{i} = \mathop{\bigcup }\limits_{i}\left( {-{A}_{i}}\right) \]\n\nwhere the \( \left( {-{A}_{i}}\right) \) are open (for the \( {A}_{i} \) are closed). Thus by Theorem \( 2, - A \) is open, as required.
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Yes
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Corollary 1. A nonempty set \( A \subseteq \left( {S,\rho }\right) \) is open iff \( A \) is a union of open globes.
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For if \( A \) is such a union, it is open by Theorem 2. Conversely, if \( A \) is open, then each \( p \in A \) is in some \( {G}_{p} \subseteq A \) . All such \( {G}_{p}\left( {p \in A}\right) \) cover all of \( A \), so \( A \subseteq \mathop{\bigcup }\limits_{{p \in A}}{G}_{p} \) . Also, \( \mathop{\bigcup }\limits_{{p \in A}}{G}_{p} \subseteq A \) since all \( {G}_{p} \) are in \( A \) . Thus\n\n\[ A = \mathop{\bigcup }\limits_{{p \in A}}{G}_{p} \]
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Yes
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Corollary 2. Every finite set \( F \) in a metric space \( \left( {S,\rho }\right) \) is closed.
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Proof. If \( F = \varnothing, F \) is closed by Example (5). If \( F \neq \varnothing \), let\n\n\[ F = \left\{ {{p}_{1},\ldots ,{p}_{n}}\right\} = \mathop{\bigcup }\limits_{{k = 1}}^{n}\left\{ {p}_{k}\right\} \]\n\nNow by Example (7), each \( \left\{ {p}_{k}\right\} \) is closed; hence so is \( F \) by Theorem 3.
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Yes
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Theorem 4. Let \( \\left( {A,\\rho }\\right) \) be a subspace of \( \\left( {S,\\rho }\\right) \). Then the open (closed) sets in \( \\left( {A,\\rho }\\right) \) are exactly all sets of the form \( A \\cap U \), with \( U \) open (closed) in \( S \).
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Proof. Let \( G \) be open in \( \\left( {A,\\rho }\\right) \). By Corollary \( 1, G \) is the union of some open globes \( {G}_{i}^{ * }\\left( {i \\in I}\\right) \) in \( \\left( {A,\\rho }\\right) \). (For brevity, we omit the centers and radii; we also omit the trivial case \( G = \\varnothing \).)\n\nAs was shown in \( §{11} \), however, \( {G}_{i}^{ * } = A \\cap {G}_{i} \), where \( {G}_{i} \) is an open globe in \( \\left( {S,\\rho }\\right) \). Thus\n\n\[ G = \\mathop{\\bigcup }\\limits_{i}{G}_{i}^{ * } = \\mathop{\\bigcup }\\limits_{i}\\left( {A \\cap {G}_{i}}\\right) = A \\cap \\mathop{\\bigcup }\\limits_{i}{G}_{i} \]\n\nby set theory (see Chapter 1, §§1-3, Problem 9).\n\nAgain by Corollary \( 1, U = \\mathop{\\bigcup }\\limits_{i}{G}_{i} \) is an open set in \( \\left( {S,\\rho }\\right) \). Thus \( G \) has the form\n\n\[ A \\cap \\mathop{\\bigcup }\\limits_{i}{G}_{i} = A \\cap U \]\n\nwith \( U \) open in \( S \), as asserted.\n\nConversely, assume the latter, and let \( p \\in G \). Then \( p \\in A \) and \( p \\in U \). As \( U \) is open in \( \\left( {S,\\rho }\\right) \), there is a globe \( {G}_{p} \) in \( \\left( {S,\\rho }\\right) \) such that \( p \\in {G}_{p} \\subseteq U \). As \( p \\in A \), we have\n\n\[ p \\in A \\cap {G}_{p} \\subseteq A \\cap U. \]\n\nHowever, \( A \\cap {G}_{p} \) is a globe in \( \\left( {A,\\rho }\\right) \), call it \( {G}_{p}^{ * } \). Thus\n\n\[ p \\in {G}_{p}^{ * } \\subseteq A \\cap U = G \]\n\ni.e., \( p \) is an interior point of \( G \) in \( \\left( {A,\\rho }\\right) \). We see that each \( p \\in G \) is interior to \( G \), as a set in \( \\left( {A,\\rho }\\right) \), so \( G \) is open in \( \\left( {A,\\rho }\\right) \).\n\nThis proves the theorem for open sets. Now let \( F \) be closed in \( \\left( {A,\\rho }\\right) \). Then by Definition \( 3, A - F \) is open in \( \\left( {A,\\rho }\\right) \). (Of course, when working in \( \\left( {A,\\rho }\\right) \), we replace \( S \) by \( A \) in taking complements.) Let \( G = A - F \), so \( F = A - G \), and \( G \) is open in \( \\left( {A,\\rho }\\right) \). By what was shown above, \( G = A \\cap U \) with \( U \) open in \( S \). Thus\n\n\[ F = A - G = A - \\left( {A \\cap U}\\right) = A - U = A \\cap \\left( {-U}\\right) \]\n\nby set theory. Here \( - U = S - U \) is closed in \( \\left( {S,\\rho }\\right) \) since \( U \) is open there. Thus \( F = A \\cap \\left( {-U}\\right) \), as required.\n\nThe proof of the converse (for closed sets) is left as an exercise.
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No
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Theorem 1. A set \( A \subseteq \left( {S,\rho }\right) \) is bounded iff \( A \) is contained in some globe. If so, the center \( p \) of this globe can be chosen at will.
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Proof. If \( A = \varnothing \), all is trivial.\n\nThus let \( A \neq \varnothing \) ; let \( q \in A \), and choose\n\n\n\nFIGURE 11\n\nany \( p \in S \) . Now if \( A \) is bounded, then \( {dA} < + \infty \), so we can choose a real \( \varepsilon > \) \( \rho \left( {p, q}\right) + {dA} \) as a suitable radius for a globe \( {G}_{p}\left( \varepsilon \right) \supseteq A \) (see Figure 11 for motivation). Now if \( x \in A \), then by the definition of \( {dA} \) , \( \rho \left( {q, x}\right) \leq {dA} \) ; so by the triangle law,\n\n\[ \rho \left( {p, x}\right) \leq \rho \left( {p, q}\right) + \rho \left( {q, x}\right) \]\n\n\[ \leq \rho \left( {p, q}\right) + {dA} < \varepsilon \]\n\ni.e., \( x \in {G}_{p}\left( \varepsilon \right) \) . Thus \( \left( {\forall x \in A}\right) x \in {G}_{p}\left( \varepsilon \right) \),\n\nas required.\n\nConversely, if \( A \subseteq {G}_{p}\left( \varepsilon \right) \), then any \( x, y \in A \) are also in \( {G}_{p}\left( \varepsilon \right) \) ; so \( \rho \left( {x, p}\right) < \varepsilon \) and \( \rho \left( {p, y}\right) < \varepsilon \), whence\n\n\[ \rho \left( {x, y}\right) \leq \rho \left( {x, p}\right) + \rho \left( {p, y}\right) < \varepsilon + \varepsilon = {2\varepsilon }. \]\n\nThus \( {2\varepsilon } \) is an upper bound of all \( \rho \left( {x, y}\right) \) with \( x, y \in A \) . Therefore,\n\n\[ {dA} = \sup \rho \left( {x, y}\right) \leq {2\varepsilon } < + \infty \]\n\ni.e., \( A \) is bounded, and all is proved.
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Yes
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Theorem 2. A set \( A \subseteq {E}^{n} \) is bounded iff there is a real \( K > 0 \) such that\n\n\[ \left( {\forall \bar{x} \in A}\right) \;\left| \bar{x}\right| < K \]
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Proof. By Theorem 1 (choosing \( \overline{0} \) for \( p \) ), \( A \) is bounded iff \( A \) is contained in some globe \( {G}_{\overline{0}}\left( \varepsilon \right) \) about \( \overline{0} \) . That is,\n\n\[ \left( {\forall \bar{x} \in A}\right) \;\bar{x} \in {G}_{\overline{0}}\left( \varepsilon \right) \text{ or }\rho \left( {\bar{x},\overline{0}}\right) = \left| \bar{x}\right| < \varepsilon . \]\n\nThus \( \varepsilon \) is the required \( K \) . (*The proof for normed spaces is the same.)
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Yes
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Corollary 1. If \( {x}_{m} \rightarrow p \), then \( p \) is the unique cluster point of \( \left\{ {x}_{m}\right\} \) . (Thus a sequence with two or more cluster points, or none at all, diverges.)
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For if \( p \neq q \), the Hausdorff property (Theorem 1 of \( §{12} \) ) yields an \( \varepsilon \) such that\n\n\[
{G}_{p}\left( \varepsilon \right) \cap {G}_{q}\left( \varepsilon \right) = \varnothing .
\]\n\nAs \( {x}_{m} \rightarrow p,{G}_{p}\left( \varepsilon \right) \) leaves out at most finitely many \( {x}_{m} \), and only these can possibly be in \( {\widehat{G}}_{q}\left( \varepsilon \right) \) . (Why?) Thus \( q \) fails to satisfy Definition 1 and hence is no cluster point. Hence \( \lim {x}_{m} \) (if it exists) is unique.
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Yes
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(i) We have \( {x}_{m} \rightarrow p \) in \( \left( {S,\rho }\right) \) iff \( \rho \left( {{x}_{m}, p}\right) \rightarrow 0 \) in \( {E}^{1} \) .
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Proof. By (2), we have \( \rho \left( {{x}_{m}, p}\right) \rightarrow 0 \) in \( {E}^{1} \) if\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall m > k}\right) \;\left| {\rho \left( {{x}_{m}, p}\right) - 0}\right| = \rho \left( {{x}_{m}, p}\right) < \varepsilon . \]\n\nBy (1), however, this means that \( {x}_{m} \rightarrow p \), proving our first assertion. The rest easily follows from it, since \( \rho \left( {{\bar{x}}_{m},\bar{p}}\right) = \left| {{\bar{x}}_{m} - \bar{p}}\right| \) in \( {E}^{n} \) .
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Yes
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Corollary 3. If \( {x}_{m} \) tends to \( p \), then so does each subsequence \( {x}_{{m}_{k}} \) .
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For \( {x}_{m} \rightarrow p \) means that each \( {G}_{p} \) leaves out at most finitely many \( {x}_{m} \) . This certainly still holds if we drop some terms, passing to \( \left\{ {x}_{{m}_{k}}\right\} \) .
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Yes
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