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Proposition 8.3.2. The sine and cosine functions are periodic with period \( {2\pi } \) . | Proof. The result follows immediately from the definitions. | No |
Proposition 8.3.3. For any \( x \in \mathbb{R},\sin \left( {-x}\right) = - \sin \left( x\right) \) and \( \cos \left( {-x}\right) = \cos \left( x\right) \) . | Proof. The result follows immediately from the definitions.\n\nQ.E.D. | No |
Proposition 8.3.5. The range of both the sine and cosine functions is \( \left\lbrack {-1,1}\right\rbrack \) . | 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. | No |
Proposition 8.3.6. For any \( x \) in the domain of the tangent function, \n\n\[ \n\tan \left( x\right) = \frac{\sin \left( x\right) }{\cos \left( x\right) }. \n\] | Proof. The result follows immediately from the definitions. \n\nQ.E.D. | No |
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) | Proof. The result follows immediately from the definitions.\n\nQ.E.D. | No |
Proposition 8.3.8. For any \( x, y \in \mathbb{R} \) ,\n\n\[ \cos \left( {x + y}\right) = \cos \left( x\right) \cos \left( y\right) - \sin \left( x\right) \sin \left( y\right) . \] | 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... | Yes |
Proposition 8.3.9. For any \( x, y \in \mathbb{R} \), \[ \sin \left( {x + y}\right) = \sin \left( x\right) \cos \left( y\right) + \sin \left( y\right) \cos \left( x\right) . \] | Exercise 8.3.1. Prove the previous proposition. | No |
Proposition 8.3.13. \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos \left( x\right) }{x} = 0 \) . | 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}}\fr... | Yes |
Proposition 8.3.14. If \( f\left( x\right) = \sin \left( x\right) \), then \( {f}^{\prime }\left( x\right) = \cos \left( x\right) \) . | 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\ri... | Yes |
Proposition 8.3.15. If \( f\\left( x\\right) = \\cos \\left( x\\right) \), then \( {f}^{\\prime }\\left( x\\right) = - \\sin \\left( x\\right) \) . | Exercise 8.3.6. Prove the previous proposition. | No |
Proposition 8.3.16. \( 2{\int }_{-1}^{1}\sqrt{1 - {x}^{2}}{dx} = \pi \) . | 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| = ... | Yes |
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 \( x > 0 \) . | 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. | Yes |
Proposition 8.4.2. For any \( x > 0 \) ,\n\n\[ \log \left( \frac{1}{x}\right) = - \log \left( x\right) \] | Proof. Using the substitution \( t = \frac{1}{u} \), we have\n\n\[ \log \left( \frac{1}{x}\right) = {\int }_{1}^{\frac{1}{x}}\frac{1}{t}{dt} = {\int }_{1}^{x}u\left( {-\frac{1}{{u}^{2}}}\right) {du} = - {\int }_{1}^{x}\frac{1}{u}{du} = - \log \left( x\right) . \] | Yes |
Proposition 8.4.3. For any positive real numbers \( x \) and \( y \) ,\n\n\[ \log \left( {xy}\right) = \log \left( x\right) + \log \left( y\right) \] | 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 \le... | Yes |
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) | 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. | Yes |
Proposition 8.4.5. \( \mathop{\lim }\limits_{{x \rightarrow + \infty }}\log \left( x\right) = + \infty \) and \( \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\log \left( x\right) = - \infty \) . | 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\nHen... | Yes |
Proposition 8.4.6. For any rational number \( \alpha > 0 \) ,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\log \left( x\right) }{{x}^{\alpha }} = 0. \] | 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 }... | Yes |
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) \... | 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. | No |
Proposition 8.5.2. For any real numbers \( x \) and \( y \) ,\n\n\[ \exp \left( {x + y}\right) = \exp \left( x\right) \exp \left( y\right) . \] | Proof. The result follows from\n\n\[ \log \left( {\exp \left( x\right) \exp \left( y\right) }\right) = \log \left( {\exp \left( x\right) }\right) + \log \left( {\exp \left( y\right) }\right) = x + y. \] | Yes |
Proposition 8.5.3. For any real number \( x \) ,\n\n\[ \exp \left( {-x}\right) = \frac{1}{\exp \left( x\right) }.\] | 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. | Yes |
Proposition 8.5.4. For any rational number \( \alpha \) , \n\n\[ \nexp \left( \alpha \right) = {e}^{\alpha }.\n\] | 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. | Yes |
Proposition 8.5.5. For any real number \( \alpha > 0 \) ,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}{x}^{\alpha }{e}^{-x} = 0. \] | 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\nLetti... | Yes |
Proposition 8.5.6. For any real number \( \alpha \) , \n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\left( 1 + \frac{\alpha }{x}\right) }^{x} = {e}^{\alpha } \] | 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}\lo... | Yes |
Lemma 1.6.2. If the square of an integer is even, then the original integer is even. | 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} =... | No |
Theorem 3.3.1. (Euclid) The set of all prime numbers is infinite. | 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 the... | Yes |
\[ \forall x \in \mathbb{Z},{x}^{2}\text{ even } \Rightarrow x\text{ even } \] | 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} +... | Yes |
Theorem 3.3.3. Among all triangles inscribed in a fixed circle, the one with maximum area is equilateral. | 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 t... | No |
Theorem 3.5.1. \( \forall n \in \mathbb{Z} \), \( n^2 \) is of the form \( 4k \) or \( 4k + 1 \) for some \( k \in \mathbb{Z} \). | 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 \( 4m^2 \)... | Yes |
Theorem 3.6.1. There is an even prime. | Proof: The number 2 is both even and prime.\n\nQ.E.D. | Yes |
Theorem 3.6.2. There are irrational numbers \( \alpha \) and \( \beta \) such that \( {\alpha }^{\beta } \) is rational. | 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) ... | Yes |
Theorem 3.6.3. There is a smallest 6-digit vampire number. | 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 ther... | Yes |
Theorem 3.6.4. The smallest positive number in the \( \mathbb{Z} \) -module generated by \( a \) and \( b \) is \( d = \gcd \left( {a, b}\right) \) . | 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 \( c \) i... | Yes |
Theorem 5.1.1. For all finite sets \( A,\left| A\right| = n \Rightarrow \left| {\mathcal{P}\left( A\right) }\right| = {2}^{n} \) . | 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\nIndu... | Yes |
Theorem 5.2.1.\n\n\[ \forall n \in \mathbb{N},\mathop{\sum }\limits_{{j = 1}}^{n}j = \frac{n\left( {n + 1}\right) }{2} \] | 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... | Yes |
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\\] | 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 pr... | Yes |
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}. \\] | 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 particul... | Yes |
Theorem 5.3.2. \( \forall n \in \mathbb{N},3 \mid \left( {{n}^{3} + {2n} + 6}\right) \) | 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}\righ... | Yes |
Theorem 5.3.3.\n\n\\[ \n\\forall n \\in \\mathbb{N},6 \\mid {7}^{n} - 1 \n\\] | 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\... | Yes |
\[ \forall n \geq 4 \in \mathbb{N},{2}^{n} < n! \] | 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 r... | No |
Theorem 5.3.5. For all natural numbers \( n \), if \( n \geq 4 \) then \( {n}^{2} \leq {2}^{n} \) . | 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... | No |
Theorem 5.4.1. For all natural numbers \( n, n > 1 \) implies \( n \) has a prime factor. | 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 ... | Yes |
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 \( \mathbb{N} \) . | 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 follo... | Yes |
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 \( \mathcal{O} \) . | 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 \( {... | Yes |
Theorem 7.1.2. In an undirected graph the sum of the degrees of the vertices is even. | 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) }\rig... | Yes |
Theorem 7.1.3. A magic square of order \( n \) has a magic sum equal to \( \frac{{n}^{3} + n}{2} \) . | 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 ( \( \mathop{\sum }\limits_{{i = 1}}^{k}i = \) \( \left. \frac{{k}^{2} + k}{2}\right) \) ... | Yes |
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. | 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 ... | Yes |
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. | 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. | No |
Theorem 7.3.1. For all natural numbers \( n \) and \( k \) with \( 0 < k \leq n \) , \n\n\[ \n\left( \begin{array}{l} n \\ k \end{array}\right) = \left( \begin{matrix} n - 1 \\ k \end{matrix}\right) + \left( \begin{matrix} n - 1 \\ k - 1 \end{matrix}\right) \n\] | 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... | Yes |
Theorem 7.3.2. For all natural numbers \( n \) and \( k \) with \( 0 < k \leq n \) , \[ k \cdot \left( \begin{array}{l} n \\ k \end{array}\right) = n \cdot \left( \begin{array}{l} n - 1 \\ k - 1 \end{array}\right) . \] | 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) !\l... | Yes |
Theorem 8.3.1 (Cantor). For all sets \( A, A \) is not equivalent to \( \mathcal{P}\left( A\right) \) . | 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 \( \mathcal{P}\left( A\right) \) ) th... | Yes |
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. | 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 ang... | No |
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. | 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 forma... | No |
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\[ \text{(i)}S - \mathop{\bigcup }\limits_{i}{A}_{i} = \mathop{\bigcap }\limits_{i}\left( {S - {A}_{i}}\right) \text{;} \] | 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}... | Yes |
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. | 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 he... | Yes |
Corollary 1. If a set \( A \) is countable or finite, so is any subset \( B \subseteq A \) . | For if \( A \subset {D}_{u}^{\prime } \) for a sequence \( u \), then certainly \( B \subseteq A \subseteq {D}_{u}^{\prime } \) . | No |
Corollary 2. If \( A \) is uncountable (or just infinite), so is any superset \( B \supseteq A \) . | For, if \( B \) were countable or finite, so would be \( A \subseteq B \), by Corollary 1 . | Yes |
Theorem 1. If \( A \) and \( B \) are countable, so is their cross product \( A \times B \) . | 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)... | Yes |
Corollary 3. The set \( R \) of all rational numbers \( {}^{2} \) is countable. | 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 Theo... | Yes |
Theorem 2. The union of any sequence \( \left\{ {A}_{n}\right\} \) of countable sets is countable. | 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{\bi... | Yes |
Theorem 3. The interval \( \lbrack 0,1) \) of the real axis is uncountable. | 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.... | Yes |
Corollary 1 (rule of signs).\n\n(i) \( a\left( {-b}\right) = \left( {-a}\right) b = - \left( {ab}\right) \) ;\n\n(ii) \( \left( {-a}\right) \left( {-b}\right) = {ab} \) . | 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) ... | No |
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 \( 1 = {1}^{2} > 0 \) .) | 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... | Yes |
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 \( A \) ). | Proof outline. \( {}^{2} \) Given \( \varnothing \neq A \subseteq N \), let \( P\left( n\right) \) be the proposition \ | No |
Theorem 1. If \( a \) and \( b \) are integers (or rationals) in \( F \), so are \( a + b \) and \( {ab} \) . | 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... | No |
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 \( F \) .) | 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... | Yes |
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). | 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 ... | Yes |
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\n... | 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 \( ... | Yes |
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 \( F \) . | 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 \( A \) . | Yes |
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. | 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 \l... | Yes |
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 \( {nx} > y \) . | 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 \( {nx}\left( {n \in N}\right) \) ... | Yes |
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\[ \left( {\forall y \in F}\right) \left( {\exists m, n \in N}\right) \; - m < y < n. \] | 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 ... | Yes |
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). | 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... | Yes |
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 | (It exists, by Theorem 2.) | No |
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\[ a < r < b.\\text{.} \] | 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 \ | No |
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 \(... | 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. | No |
Theorem 2. Every complete field \( F \) (such as \( {E}^{1} \) ) has irrational elements, i.e., elements that are not rational. | 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 (o... | Yes |
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... | 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 Coroll... | Yes |
(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 \( n \) . | Proof. Assume the opposite and find a contradiction to Theorem 1. | No |
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 \( n.{}^{6} \) | 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} \... | No |
Theorem 3. We have \( q = \lim {x}_{n} \) in \( {E}^{ * } \) iff\n\n\[ \underline{\lim }{x}_{n} = \overline{\lim }{x}_{n} = q. \] | 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 \( \overline{\lim }{x}_{n} \) a... | Yes |
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( {... | 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} =... | No |
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... | 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) ,\ldot... | Yes |
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}\rig... | 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 \( a = 1, b = 0 \) ). | Yes |
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 }\... | 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}}\... | Yes |
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... | 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... | Yes |
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 \( \overrightarrow{u} = \left( {{u}_{1},\ldots ,{u}_{n}}\right) \neq \overline{0} \) . | 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}_{... | Yes |
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. | 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 co... | Yes |
Corollary 1. \( {i}^{2} = - 1 \) ; i.e., \( \left( {0,1}\right) \left( {0,1}\right) = \left( {-1,0}\right) \) . | Proof. By definition, \( \left( {0,1}\right) \left( {0,1}\right) = \left( {0 \cdot 0 - 1 \cdot 1,0 \cdot 1 + 1 \cdot 0}\right) = \left( {-1,0}\right) \) . | Yes |
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\[ z = x + {yi}\text{ iff }z = \left( {x, y}\right) . \] | 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}\righ... | Yes |
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. | 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( \va... | Yes |
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 se... | 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\[... | Yes |
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). | 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... | Yes |
Corollary 1. A nonempty set \( A \subseteq \left( {S,\rho }\right) \) is open iff \( A \) is a union of open globes. | 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... | Yes |
Corollary 2. Every finite set \( F \) in a metric space \( \left( {S,\rho }\right) \) is closed. | 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. | Yes |
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 \( S \). | 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 ... | No |
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. | 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 ... | Yes |
Theorem 2. A set \( A \subseteq {E}^{n} \) is bounded iff there is a real \( K > 0 \) such that\n\n\[ \left( {\forall \bar{x} \in A}\right) \;\left| \bar{x}\right| < K \] | 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{ o... | Yes |
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.) | 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... | Yes |
(i) We have \( {x}_{m} \rightarrow p \) in \( \left( {S,\rho }\right) \) iff \( \rho \left( {{x}_{m}, p}\right) \rightarrow 0 \) in \( {E}^{1} \) . | 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), howev... | Yes |
Corollary 3. If \( {x}_{m} \) tends to \( p \), then so does each subsequence \( {x}_{{m}_{k}} \) . | 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 \( \left\{ {x}_{{m}_{k}}\right\} \) . | Yes |
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