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Lemma 9.4 We have that\n\n\[ \n{y}_{c}^{\prime \prime }\left( 0\right) = \frac{-{V}_{e}{c}_{4} - {g}_{m}\alpha \sqrt{r}}{\alpha \sqrt{r}}\text{ and }{x}_{c}^{\prime \prime }\left( 0\right) = \frac{-{V}_{e}{c}_{3}}{\alpha \sqrt{r}}. \n\] | Then by the L'Hospital Theorem, [53, p. 207], the angle of the spacecraft at the take-off moment is\n\n\[ \n\arctan \frac{{V}_{e}{c}_{4} + {g}_{m}\alpha \sqrt{r}}{{V}_{e}{c}_{3}}. \n\] | No |
Proposition 10.1 If two tangents to a parabola are perpendicular to each other; then they intersect on the directrix. Conversely, two tangents which intersect on the directrix are perpendicular. | Proof Let \( {x}^{2} = {2py} \) be a parabola where \( p \) is a positive constant. Suppose that two tangents contact this parabola at the points \( \left( {a,{a}^{2}/{2p}}\right) \) and \( \left( {b,{b}^{2}/{2p}}\right) \) . The equation of the first tangent is \( y - {a}^{2}/{2p} = \left( {a/p}\right) \left( {x - a}\... | Yes |
Differentiate a Known Series | We recognize that\n\n\[ \frac{1}{{\left( 1 - x\right) }^{2}} = \frac{d}{dx}\left( \frac{1}{1 - x}\right) .\n\]\n\nWe may therefore expand it as\n\n\[ \frac{1}{{\left( 1 - x\right) }^{2}} = \frac{d}{dx}\left\lbrack {1 + x + {x}^{2} + {x}^{3} + \cdots }\right\rbrack = 1 + {2x} + 3{x}^{2} + \cdots .\n\]\n\n\( \left( {2.42... | Yes |
Example 3.3.1. Cartesian-Polar Conversions | \n\( > \mathrm{C} \mathrel{\text{:=}} 4 - 3 * \mathrm{I} \) ;\n\[ 4 - {3I} \]\n\n\( > \operatorname{abs}\left( \mathrm{C}\right) \) ; 5\n\n\( > \) argument \( \left( \mathrm{C}\right) \) ;\n\[ - \arctan \left( \frac{3}{4}\right) \]\n\n> evalc(5*exp(-I*arctan(3/4))\n\[ 4 - {3I} \]\n\n\( > \operatorname{polar}\left( {5, ... | Yes |
Example 4.4.3. Multiplication, Row, and Column Matrices\n\nConsider\n\[ A = \left( \begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right) ,\;B = \left( \begin{array}{lll} 4 & 5 & 6 \end{array}\right) . \]\n\nLet us form AB and BA: | \[ {AB} = \left( \begin{array}{rrr} 4 & 5 & 6 \\ 8 & {10} & {12} \\ {12} & {15} & {18} \end{array}\right) ,\;{BA} = \left( {4 \times 1 + 5 \times 2 + 6 \times 3}\right) = \left( {32}\right) . \]\n\nThe results speak for themselves. Often when a matrix operation leads to a \( 1 \times 1 \) matrix, the parentheses are dr... | Yes |
Example 5.3.1. Transforming a Matrix Equation\n\nConsider the matrix equation \( \mathbf{g} = \mathbf{{Af}} \), with\n\n\[ A = \left( \begin{array}{lll} 1 & 2 & 3 \\ 2 & 6 & 0 \\ 3 & 0 & 4 \end{array}\right) ,\;f = \left( \begin{array}{r} 2 \\ - 1 \\ 1 \end{array}\right) ,\;g = \left( \begin{array}{lll} 1 & 2 & 3 \\ 2 ... | Applying the matrix transformation \( U = \left( \begin{matrix} {0.408248} & {0.816497} & {0.408248} \\ - {0.707107} & 0 & {0.707107} \\ {0.577350} & - {0.577350} & {0.577350} \end{matrix}\right) \), \n\n\[ {A}^{\prime } = {UA}{U}^{\mathrm{T}} = \left( \begin{array}{rrr} {7.16667} & - {0.28868} & - {0.23570} \\ - {0.28... | Yes |
Let’s obtain the eigenvalues and eigenvectors of \( \mathrm{H} = \left( \begin{array}{rrrr} 1 & 2 & 0 & 0 \\ 2 & 0 & 1 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & - 1 \end{array}\right) \) . | Using MAPLE first, we define \( \mathrm{H} \) and access the LinearAlgebra package:\n\n\( > \mathrm{H} \mathrel{\text{:=}} \operatorname{Matrix}\left( \left\lbrack {\left\lbrack {1,2,0,0}\right\rbrack ,\left\lbrack {2,0,1,2}\right\rbrack ,\left\lbrack {0,1,0,0}\right\rbrack ,\left\lbrack {0,2,0, - 1}\right\rbrack }\rig... | No |
Let \( \mathbf{F} = {xy}{\widehat{\mathbf{e}}}_{x} + {xy}{\widehat{\mathbf{e}}}_{y} + \frac{xy}{z}{\widehat{\mathbf{e}}}_{z} \). Compute \( \mathbf{\nabla } \cdot \mathbf{F} \) as follows: | \[ \frac{\partial {F}_{x}}{\partial x} = y,\;\frac{\partial {F}_{y}}{\partial y} = x,\;\frac{\partial {F}_{z}}{\partial z} = - \frac{xy}{{z}^{2}}, \] so \( \mathbf{\nabla } \cdot \mathbf{F} = y + x - \frac{xy}{{z}^{2}} \). | Yes |
Given that \( \mathbf{F} = \mathbf{r} = x{\widehat{\mathbf{e}}}_{x} + y{\widehat{\mathbf{e}}}_{y} + z{\widehat{\mathbf{e}}}_{z} \) | we have\n\[ \frac{\partial {F}_{x}}{\partial x} = \frac{\partial x}{\partial x} = 1,\;\frac{\partial {F}_{y}}{\partial y} = 1,\;\frac{\partial {F}_{z}}{\partial z} = 1, \]\n\nand therefore\n\n\[ \nabla \cdot \mathbf{r} = 3\text{.} \] | Yes |
Example 7.2.10. Curl of Radius Vector\n\nWe now evaluate \( \mathbf{\nabla } \times \mathbf{r} \) . Writing \( \mathbf{r} = x{\widehat{\mathbf{e}}}_{x} + y{\widehat{\mathbf{e}}}_{y} + z{\widehat{\mathbf{e}}}_{z} \), we note that\n\n\[ \frac{\partial {r}_{x}}{\partial y} = \frac{\partial x}{\partial y} = 0 \] | and all the other derivatives needed for the curl also vanish, so we get\n\n\[ \mathbf{\nabla } \times \mathbf{r} = 0 \] | Yes |
Example 7.3.2. Laplacian of \( r \) | We start the evaluation with\n\n\[ \frac{dr}{dx} = \frac{x}{{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{1/2}},\;\frac{{d}^{2}r}{d{x}^{2}} = \frac{1}{{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{1/2}} - \frac{{x}^{2}}{{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{3/2}}. \]\n\nCombining the above with corresponding ex... | Yes |
Example 8.4.2 (MAPLE). Our plan is to transform Levi using the two matrices\n\n\[ \n\mathrm{U} = \left( \begin{matrix} {0.8} & {0.6} & 0 \\ - {0.6} & {0.8} & 0 \\ 0 & 0 & 1 \end{matrix}\right) \;\text{ and }\;\mathrm{V} = \left( \begin{matrix} {0.8} & {0.6} & 0 \\ - {0.6} & {0.8} & 0 \\ 0 & 0 & - 1 \end{matrix}\right) ... | Using the coding that precedes this Example, we find with MATHEMATICA that the tensors obtained using transformations \( \mathrm{U} \) and \( \mathrm{V} \) have the following MatrixForm displays:\n\n\[ \n\begin{matrix} \left( \begin{matrix} \left( \begin{matrix} 0. \\ 0. \\ 0. \end{matrix}\right) & \left( \begin{matrix... | Yes |
Consider the function\n\n\\[ \n f\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1, & 0 < x < \\pi /2, \\\\ 0, & \\pi /2 < x < \\pi . \\end{array}\\right.\n\\]\n\nExpanding \\( f\\left( x\\right) \\) first as a Fourier cosine series, we have | \n\n\\[ \n {a}_{{2n} + 1} = \\frac{2}{\\pi }{\\int }_{0}^{\\pi /2}\\cos \\left\\lbrack {\\left( {{2n} + 1}\\right) x}\\right\\rbrack {dx} = \\frac{2{\\left( -1\\right) }^{n}}{\\left( {{2n} + 1}\\right) \\pi },\n\\]\n\n\\[ \n {a}_{2n} = \\frac{2}{\\pi }{\\int }_{0}^{\\pi /2}\\cos {2nxdx} = {\\delta }_{n0}.\n\\]\n\nThe c... | Yes |
Example 13.2.1.\n\n\( G = \) FourierTransform \( \left\lbrack {F, t, w}\right\rbrack \)\n\[ \n\frac{\sqrt{\frac{2}{\pi }}\operatorname{Sin}\left\lbrack w\right\rbrack }{w} \n\] | This should agree with the earlier example. It checks. | No |
Let \( I = {\int }_{0}^{2\pi }\frac{\cos {2\theta d\theta }}{5 - 4\cos \theta } \) . | Using the relationships in Eqs. (17.29) and (17.30), we rewrite \( I \) as\n\n\[ I = \oint \frac{\frac{1}{2}\left( {{z}^{2} + {z}^{-2}}\right) }{5 - 2\left( {z + {z}^{-1}}\right) }\frac{dz}{iz}, \]\n\nwhere the contour is the unit circle. Equation (17.32) simplifies to\n\n\[ I = \frac{i}{4}\oint \frac{\left( {{z}^{4} +... | Yes |
Consider\n\[ I = {\int }_{0}^{\infty }\frac{dx}{{x}^{3} + 1}. \] | The integrand of \( I \) is not an even function, and in fact the entire integrand does not have definite parity. However, it does have the feature that \( {z}^{3} \) assumes the same set of values on the ray at angle \( \theta = {2\pi }/3 \) as it does on the positive real axis. We therefore look closely at the integr... | Yes |
Consider the series\n\n\[ S = \frac{1}{{1}^{3}} - \frac{1}{{3}^{3}} + \frac{1}{{5}^{3}} - \cdots = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{\left( -1\right) }^{n}}{{\left( 2n + 1\right) }^{3}}. \] | We note that\n\n\[ S = \frac{1}{2}{S}^{\prime },\;\text{ with }\;{S}^{\prime } = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\frac{{\left( -1\right) }^{n}}{{\left( 2n + 1\right) }^{3}}. \]\n\nThe summation \( {S}^{\prime } \) is of the type represented by Eq. (17.51), with \( f\left( z\right) = 1/{\left( 2z\right)... | Yes |
Theorem 2. The probability that two independent events both occur is the product of their individual probabilities. | Since it is reasonable to assume that the probability of the number on the red die is completely independent of the number on the white die and that all numbers on either die are equally probable (with probability \( 1/6 \) ), each of the above tabular entries occurs with a probability \( \left( {1/6}\right) \times \le... | Yes |
Theorem 3. The formula for the probability of \( A \cup B \) (i.e., the probability of the occurrence of any of the events in \( A \cup B \) ) is\n\n\[ p\left( {A \cup B}\right) = p\left( A\right) + p\left( B\right) - p\left( {A \cap B}\right) . \] | ## Example 18.1.4. Checking Theorem 3\n\nThe sets \( A \) and \( B \) introduced earlier in this section have respectively 6 and 5 members; we computed their individual probabilities as \( p\left( A\right) = 6/{36} \) and \( p\left( B\right) = 5/{36} \) . We also determined that the intersection \( A \cap B \) had one ... | Yes |
Theorem 2.1 (Pythagorean Triples Theorem). We will get every primitive Pythagorean triple \( \left( {a, b, c}\right) \) with \( a \) odd and \( b \) even by using the formulas\n\n\[ a = {st},\;b = \frac{{s}^{2} - {t}^{2}}{2},\;c = \frac{{s}^{2} + {t}^{2}}{2}, \]\n\nwhere \( s > t \geq 1 \) are chosen to be any odd inte... | Why did we say that we have \ | No |
Theorem 3.1. Every point on the circle\n\n\[ \n{x}^{2} + {y}^{2} = 1 \n\]\n\nwhose coordinates are rational numbers can be obtained from the formula\n\n\[ \n\\left( {x, y}\\right) = \\left( {\\frac{1 - {m}^{2}}{1 + {m}^{2}},\\frac{2m}{1 + {m}^{2}}}\\right) \n\]\n\nby substituting in rational numbers for \( m \) [except... | How is this formula for rational points on a circle related to our formula for Pythagorean triples? If we write the rational number \( m \) as a fraction \( v/u \), then our formula becomes\n\n\[ \n\\left( {x, y}\\right) = \\left( {\\frac{{u}^{2} - {v}^{2}}{{u}^{2} + {v}^{2}},\\frac{2uv}{{u}^{2} + {v}^{2}}}\\right) \n\... | Yes |
Theorem 5.1 (Euclidean Algorithm). To compute the greatest common divisor of two numbers \( a \) and \( b \), let \( {r}_{-1} = a \), let \( {r}_{0} = b \), and compute successive quotients and remainders\n\n\[{r}_{i - 1} = {q}_{i + 1} \times {r}_{i} + {r}_{i + 1}\]\n\nfor \( i = 0,1,2,\ldots \) until some remainder \(... | There remains the question of why the Euclidean algorithm always finishes. In other words, we know that the last nonzero remainder will be the desired gcd, but how do we know that we ever get a remainder that does equal 0 ? This is not a silly question, since it is easy to give algorithms that do not terminate; and the... | Yes |
Theorem 6.1 (Linear Equation Theorem). Let \( a \) and \( b \) be nonzero integers, and let \( g = \gcd \left( {a, b}\right) \) . The equation\n\n\[ \n{ax} + {by} = g \n\]\n\nalways has a solution \( \left( {{x}_{1},{y}_{1}}\right) \) in integers, and this solution can be found by the Euclidean algorithm method describ... | For example, we saw that the equation\n\n\[ \n{60x} + {22y} = \gcd \left( {{60},{22}}\right) = 2 \n\]\n\nhas the solution \( x = - 4, y = {11} \) . Then our Linear Equation Theorem says that every solution is obtained from the formula\n\n\[ \n\left( {-4 + {11k},{11} - {30k}}\right) \;\text{with}k\text{any integer.} \n\... | Yes |
Lemma 7.1. Let \( p \) be a prime number, and suppose that \( p \) divides the product \( {ab} \) . Then either \( p \) divides \( a \) or \( p \) divides \( b \) (or \( p \) divides both \( a \) and \( b \) ). \( {}^{2} \) | Proof. We are given that \( p \) divides the product \( {ab} \) . If \( p \) divides \( a \), we are done, so we may as well assume that \( p \) does not divide \( a \) . Now consider what \( \gcd \left( {p, a}\right) \) can be. It divides \( p \), so it is either 1 or \( p \) . It also divides \( a \), so it isn’t \( ... | Yes |
Theorem 7.2 (Prime Divisibility Property). Let \( p \) be a prime number; and suppose that \( p \) divides the product \( {a}_{1}{a}_{2}\cdots {a}_{r} \) . Then \( p \) divides at least one of the factors \( {a}_{1},{a}_{2},\ldots ,{a}_{r} \) . | Proof. If \( p \) divides \( {a}_{1} \), we’re done. If not, we apply the lemma to the product\n\n\[ \n{a}_{1}\left( {{a}_{2}{a}_{3}\cdots {a}_{r}}\right) \n\]\n\nto conclude that \( p \) must divide \( {a}_{2}{a}_{3}\cdots {a}_{r} \) . In other words, we are applying the lemma with \( a = {a}_{1} \) and \( b = {a}_{2}... | Yes |
Theorem 7.3 (The Fundamental Theorem of Arithmetic). Every integer \( n \geq 2 \) can be factored into a product of primes\n\n\[ n = {p}_{1}{p}_{2}\cdots {p}_{r} \]\nin exactly one way. | Proof. The Fundamental Theorem of Arithmetic really contains two assertions.\n\nAssertion 1. The number \( n \) can be factored into a product of primes in some way.\n\nAssertion 2. There is only one such factorization (aside from rearranging the factors).\n\nWe begin with Assertion 1. We are going to give a proof by i... | Yes |
Theorem 8.1 (Linear Congruence Theorem). Let \( a, c \), and \( m \) be integers with \( m \geq 1 \), and let \( g = \gcd \left( {a, m}\right) \).\n\n(a) If \( g \nmid c \), then the congruence \( {ax} \equiv c\left( {\;\operatorname{mod}\;m}\right) \) has no solutions.\n\n(b) If \( g \mid c \), then the congruence \( ... | For example, the congruence\n\n\[{943x} \equiv {381}\left( {\;\operatorname{mod}\;{2576}}\right)\]\n\nhas no solutions, since \( \gcd \left( {{943},{2576}}\right) = {23} \) does not divide 381 . On the other hand, the congruence\n\n\[{893x} \equiv {266}\left( {\;\operatorname{mod}\;{2432}}\right)\]\n\nhas 19 solutions,... | Yes |
Theorem 8.2 (Polynomial Roots Mod \( p \) Theorem). Let \( p \) be a prime number and let\n\n\[ f\left( x\right) = {a}_{0}{x}^{d} + {a}_{1}{x}^{d - 1} + \cdots + {a}_{d} \]\n\nbe a polynomial of degree \( d \geq 1 \) with integer coefficients and with \( p \nmid {a}_{0} \). Then the congruence\n\n\[ f\left( x\right) \e... | In a proof by contradiction, we begin by making a statement. We then use that statement to make deductions, eventually ending up with a conclusion that is clearly false. This allows us to deduce that the original statement was false, since it led to a false conclusion. \n\nThe particular statement with which we begin i... | Yes |
Lemma 9.2. Let \( p \) be a prime number and let \( a \) be a number with \( a ≢ 0\\left( {\\operatorname{mod}p}\\right) \) . Then the numbers\n\n\[ \na,{2a},{3a},\\ldots ,\\left( {p - 1}\\right) a\\left( {\\operatorname{mod}p}\\right)\n\]\n\nare the same as the numbers\n\n\[ \n1,2,3,\\ldots ,\\left( {p - 1}\\right) \\... | Proof. The list \( a,{2a},{3a},\\ldots ,\\left( {p - 1}\\right) a \) contains \( p - 1 \) numbers, and clearly none of them are divisible by \( p \) . Suppose that we take two numbers \( {ja} \) and \( {ka} \) in this list, and suppose that they happen to be congruent,\n\n\[ \n{ja} \equiv {ka}\\left( {\\operatorname{mo... | Yes |
Theorem 10.1 (Euler’s Formula). If \( \gcd \left( {a, m}\right) = 1 \), then\n\n\[{a}^{\phi \left( m\right) } \equiv 1\left( {\;\operatorname{mod}\;m}\right)\] | Proof. Now that we have identified the correct set of numbers to consider, the proof of Euler's formula is almost identical to the proof of Fermat's Little Theorem. So we let\n\n\[1 \leq {b}_{1} < {b}_{2} < \cdots < {b}_{\phi \left( m\right) } < m\]\n\nbe the \( \phi \left( m\right) \) numbers between 0 and \( m \) tha... | No |
Lemma 10.2. If \( \gcd \left( {a, m}\right) = 1 \), then the numbers\n\n\[ \n{b}_{1}a,{b}_{2}a,{b}_{3}a,\ldots ,{b}_{\phi \left( m\right) }a\left( {\;\operatorname{mod}\;m}\right) \n\]\n\nare the same as the numbers\n\n\[ \n{b}_{1},{b}_{2},{b}_{3},\ldots ,{b}_{\phi \left( m\right) }\left( {\;\operatorname{mod}\;m}\righ... | Proof of the lemma. We note that if \( b \) is relatively prime to \( m \), then \( {ab} \) is also relatively prime to \( m \) . Hence, each of the numbers in the list\n\n\[ \n{b}_{1}a,{b}_{2}a,{b}_{3}a,\ldots ,{b}_{\phi \left( m\right) }a\left( {\;\operatorname{mod}\;m}\right) \n\]\n\nis congruent to one number in th... | Yes |
Theorem 11.2 (Chinese Remainder Theorem). Let \( m \) and \( n \) be integers satisfying \( \gcd \left( {m, n}\right) = 1 \), and let \( b \) and \( c \) be any integers. Then the simultaneous congruences\n\n\[ x \equiv b\left( {\;\operatorname{mod}\;m}\right) \;\text{ and }\;x \equiv c\left( {\;\operatorname{mod}\;n}\... | Proof. Let's start, as usual, with an example. Suppose we want to solve\n\n\[ x \equiv 8\left( {\;\operatorname{mod}\;{11}}\right) \;\text{ and }\;x \equiv 3\left( {\;\operatorname{mod}\;{19}}\right) .\n\nThe solution to the first congruence consists of all numbers that have the form \( x = {11y} + 8 \) . We substitute... | Yes |
Theorem 12.1 (Infinitely Many Primes Theorem). There are infinitely many prime numbers. | Euclid's Proof. Suppose that you have already compiled a (finite) list of primes. I am going to show you how to find a new prime that isn't in your list. Since you can then add the new prime to the list and repeat the process, this will show that there must be infinitely many primes.\n\nSo suppose we start with some li... | Yes |
Theorem 12.2 (Primes 3 (Mod 4) Theorem). There are infinitely many primes that are congruent to 3 modulo 4. | Proof. We suppose that we have already compiled a (finite) list of primes, all of which are congruent to 3 modulo 4. Our goal is to make the list longer by finding a new 3 modulo 4 prime. Repeating this process gives a list of any desired length, thereby proving that there are infinitely many primes congruent to 3 modu... | Yes |
Theorem 12.3 (Dirichlet’s Theorem on Primes in Arithmetic Progressions \( {}^{3} \) ). Let \( a \) and \( m \) be integers with \( \gcd \left( {a, m}\right) = 1 \) . Then there are infinitely many primes that are congruent to a modulo \( m \) . That is, there are infinitely many prime numbers \( p \) satisfying\n\n\[ p... | Earlier in this chapter we proved Dirichlet’s Theorem for \( \left( {a, m}\right) = \left( {3,4}\right) \), and Exercise 12.2 asks you to do \( \left( {a, m}\right) = \left( {5,6}\right) \) . In Chapter 21, we will deal with \( \left( {a, m}\right) = \left( {1,4}\right) \) . Unfortunately, the proof of Dirichlet’s Theo... | No |
Theorem 15.1 (Euclid’s Perfect Number Formula). If \( {2}^{p} - 1 \) is a prime number, then \( {2}^{p - 1}\left( {{2}^{p} - 1}\right) \) is a perfect number. | Using the same sort of idea, we can easily verify that Euclid's Perfect Number Formula is true in general. We let \( q = {2}^{p} - 1 \), and we need to check that \( {2}^{p - 1}q \) is a perfect number. The proper divisors of \( {2}^{p - 1}q \) are\n\n\[ 1,2,4,\ldots ,{2}^{p - 1}\;\text{ and }\;q,{2q},{4q},\ldots ,{2}^... | Yes |
Theorem 15.4 (Euler’s Perfect Number Theorem). If \( n \) is an even perfect number, then \( n \) looks like\n\n\[ n = {2}^{p - 1}\left( {{2}^{p} - 1}\right) \]\n\nwhere \( {2}^{p} - 1 \) is a Mersenne prime. | Proof. Suppose that \( n \) is an even perfect number. The fact that \( n \) is even means that we can factor it as\n\n\[ n = {2}^{k}m\;\text{ with }k \geq 1\text{ and }m\text{ odd. }\]\n\nNext we use the sigma function formulas to compute \( \sigma \left( n\right) \),\n\n\[ \sigma \left( n\right) = \sigma \left( {{2}^... | Yes |
Theorem 19.1 (Korselt’s Criterion for Carmichael Numbers). Let \( n \) be a composite number. Then \( n \) is a Carmichael number if and only if it is odd and every prime \( p \) dividing \( n \) satisfies the following two conditions:\n\n(1) \( {p}^{2} \) does not divide \( n \) .\n\n(2) \( p - 1 \) divides \( n - 1 \... | Proof. Suppose first that \( n \) is a composite number, and further suppose that every prime divisor \( p \) of \( n \) satisfies conditions (1) and (2). We want to prove that \( n \) is a Carmichael number. Our proof uses the same arguments that we used to prove that 561 is a Carmichael number.\n\nWe factor \( n \) a... | No |
Theorem 19.2 (A Property of Prime Numbers). Let \( p \) be an odd prime and write\n\n\[ p - 1 = {2}^{k}q\;\text{ with }q\text{ odd. } \]\n\nLet \( a \) be any number not divisible by \( p \) . Then one of the following two conditions is true:\n\n(i) \( {a}^{q} \) is congruent to 1 modulo \( p \) .\n\n(ii) One of the nu... | Proof. Fermat’s Little Theorem tells us that \( {a}^{p - 1} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) . This means that, when we look at the list of numbers\n\n\[ {a}^{q},{a}^{2q},{a}^{4q},\ldots ,{a}^{{2}^{k - 1}q},{a}^{{2}^{k}q} \]\n\nwe know that the last number in the list is congruent to 1 modulo \( p \) ... | Yes |
Theorem 19.3 (Rabin-Miller Test for Composite Numbers). Let \( n \) be an odd integer and write \( n - 1 = {2}^{k}q \) with \( q \) odd. If both of the following conditions are true for some a not divisible by \( n \), then \( n \) is a composite number.\n\n(a) \( {a}^{q} ≢ 1\left( {\;\operatorname{mod}\;n}\right) \)\n... | We have already verified that the Rabin-Miller test works, since if \( n \) satisfies (a) and (b), then it does not satisfy the prime number property described in Theorem 19.2, so it must be composite. Note that the Rabin-Miller test is very fast and easy to implement on a computer, since, after computing \( {a}^{q}\le... | Yes |
Theorem 20.1. Let \( p \) be an odd prime. Then there are exactly \( \left( {p - 1}\right) /2 \) quadratic residues modulo \( p \) and exactly \( \left( {p - 1}\right) /2 \) nonresidues modulo \( p \) . | Proof. The quadratic residues are the nonzero numbers that are squares modulo \( p \) , so they are the numbers\n\n\[ \n{1}^{2},{2}^{2},\ldots ,{\left( p - 1\right) }^{2}\left( {\;\operatorname{mod}\;p}\right) .\n\]\n\nBut, as we noted above, we only need to go halfway,\n\n\[ \n{1}^{2},{2}^{2},\ldots ,{\left( \frac{p -... | Yes |
Theorem 20.2 (Quadratic Residue Multiplication Rule). (Version 1) Let \( p \) be an odd prime. Then:\n\n(i) The product of two quadratic residues modulo \( p \) is a quadratic residue.\n\n(ii) The product of a quadratic residue and a nonresidue is a nonresidue.\n\n(iii) The product of two nonresidues is a quadratic res... | Proof. We have already seen that \( \mathrm{{QR}} \times \mathrm{{QR}} = \mathrm{{QR}} \) . Suppose next that \( {a}_{1} \) is a \( \mathrm{{QR}} \) , say \( {a}_{1} \equiv {b}_{1}^{2}\left( {\;\operatorname{mod}\;p}\right) \), and that \( {a}_{2} \) is an NR. We are going to assume that \( {a}_{1}{a}_{2} \) is a QR an... | Yes |
Theorem 20.3 (Quadratic Residue Multiplication Rule). (Version 2) Let \( p \) be an odd prime. Then\n\n\[ \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) = \left( \frac{ab}{p}\right) \] | The Legendre symbol is useful for making calculations. For example, suppose that we want to know if 75 is a square modulo 97 . We can compute\n\n\[ \left( \frac{75}{97}\right) = \left( \frac{3 \cdot 5 \cdot 5}{97}\right) = \left( \frac{3}{97}\right) \left( \frac{5}{97}\right) \left( \frac{5}{97}\right) = \left( \frac{3... | No |
Theorem 21.1 (Euler’s Criterion). Let \( p \) be an odd prime. Then\n\n\[ \n{a}^{\left( {p - 1}\right) /2} \equiv \left( \frac{a}{p}\right) \left( {\;\operatorname{mod}\;p}\right) \n\] | Proof. Suppose first that \( a \) is a quadratic residue, say \( a \equiv {b}^{2}\left( {\;\operatorname{mod}\;p}\right) \) . Then Fermat's Little Theorem (Theorem 9.1) tells us that\n\n\[ \n{a}^{\left( {p - 1}\right) /2} \equiv {\left( {b}^{2}\right) }^{\left( {p - 1}\right) /2} = {b}^{p - 1} \equiv 1\left( {\;\operat... | Yes |
Theorem 21.2 (Quadratic Reciprocity). (Part I) Let \( p \) be an odd prime. Then\n\n-1 is a quadratic residue modulo \( p\; \) if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), and\n\n\[ \text{-1 is a nonresidue modulo}p\;\text{if}p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \text{.} \]\n\nIn other words... | Proof. Euler's Criterion says that\n\n\[ {\left( -1\right) }^{\left( {p - 1}\right) /2} \equiv \left( \frac{-1}{p}\right) \left( {\;\operatorname{mod}\;p}\right) \]\n\nSuppose first that \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), say \( p = {4k} + 1 \) . Then\n\n\[ {\left( -1\right) }^{\left( {p - 1}\righ... | Yes |
Theorem 21.3 (Primes 1 (Mod 4) Theorem). There are infinitely many primes that are congruent to 1 modulo 4. | Proof. Suppose we are given a list of primes \( {p}_{1},{p}_{2},\ldots ,{p}_{r} \), all of which are congruent to 1 modulo 4. We are going to find a new prime, not in our list, that is congruent to 1 modulo 4. Repeating this process gives a list of any desired length.\n\nConsider the number\n\n\[ A = {\left( 2{p}_{1}{p... | Yes |
Theorem 21.4 (Quadratic Reciprocity). (Part II) Let \( p \) be an odd prime. Then 2 is a quadratic residue modulo \( p \) if \( p \) is congruent to 1 or 7 modulo 8, and 2 is a nonresidue modulo \( p \) if \( p \) is congruent to 3 or 5 modulo 8 . In terms of the Legendre symbol,\n\n\[ \n\\left( \\frac{2}{p}\\right) = ... | Proof. There are actually four cases to consider, depending on the value of \( p \) modulo 8. We do two of them and leave the other two for you to do.\n\nWe start with the case that \( p \\equiv 3\\left( {\\;\\operatorname{mod}\\;8}\\right) \), say \( p = {8k} + 3 \) . We need to list the numbers \( 2,4,\\ldots, p - 1 ... | No |
Theorem 22.1 (Law of Quadratic Reciprocity). Let \( p \) and \( q \) be distinct odd primes. | We have proven the Law of Quadratic Reciprocity for \( \left( \frac{-1}{p}\right) \) and \( \left( \frac{2}{p}\right) \) . There are many different proofs of the relationship between \( \left( \frac{q}{p}\right) \) and \( \left( \frac{p}{q}\right) \), but none of them is easy. We will give a proof, due to Eisenstein, i... | No |
Lemma 23.2. When the numbers \( a,{2a},{3a},\ldots ,{Pa} \) are reduced modulo \( p \) into the range from \( - P \) to \( P \), the reduced values are \( \pm 1,\ldots , \pm P \) in some order, with each number appearing once with either a plus sign or a minus sign. | Proof of Lemma 23.2. We write each multiple \( {ka} \) as\n\n\[ \n{ka} = p{q}_{k} + {r}_{k}\;\text{ with } - P \leq {r}_{k} \leq P.\n\]\n\nSuppose that two of the \( {r}_{k} \) values are either the same or negatives of each another, say \( {r}_{i} = e{r}_{j} \) with \( e = \pm 1 \) . Then\n\n\[ \n{ia} - {eja} = \left(... | Yes |
Lemma 23.3. Let \( p \) be an odd prime, let \( P = \frac{p - 1}{2} \), let a be an odd integer that is not divisible by \( p \), and let \( \mu \left( {a, p}\right) \) be the quantity defined on page 171 that appears in Gauss's criterion. Then\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{P}\left\lfloor \frac{ka}{p}\right\rf... | Proof of Lemma 23.3. Just as in the proof of Lemma 23.2, we write each multiple \( {ka} \) as\n\n\[ {ka} = {q}_{k}p + {r}_{k}\;\text{ with } - P < {r}_{k} < P. \]\n\nWe next divide by \( p \) to obtain\n\n\[ \frac{ka}{p} = {q}_{k} + \frac{{r}_{k}}{p}\;\text{ with } - \frac{1}{2} < \frac{{r}_{k}}{p} < \frac{1}{2}. \]\n\... | Yes |
Theorem 24.1 (Sum of Two Squares Theorem for Primes). Let \( p \) be a prime. Then \( p \) is a sum of two squares exactly when\n\n\[ p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \;\left( {\text{ or }p = 2}\right) . \] | Now that we know that statement 1 should be easy to prove, let's prove it. We are told that the prime \( p \) is a sum of two squares, say\n\n\[ p = {a}^{2} + {b}^{2} \]\n\nWe also know that \( p \) is odd, so one of \( a \) and \( b \) must be odd and the other one must be even. Switching them if necessary, we may ass... | Yes |
Theorem 25.2 (Pythagorean Hypotenuse Proposition). A number \( c \) appears as the hypotenuse of a primitive Pythagorean triple \( \left( {a, b, c}\right) \) if and only if \( c \) is a product of primes each of which is congruent to 1 modulo 4. | For example, the number \( c = {1479} \) cannot be the hypotenuse of a primitive Pythagorean triple, since \( {1479} = 3 \cdot {17} \cdot {29} \) . On the other hand, \( c = {1105} \) can be a hypotenuse, since \( {1105} = 5 \cdot {13} \cdot {17} \) . Furthermore, we can solve \( {s}^{2} + {t}^{2} = {2c} \) to find the... | Yes |
Lemma 27.1. If \( \gcd \left( {m, n}\right) = 1 \), then \( F\left( {mn}\right) = F\left( m\right) F\left( n\right) \) . | Proof. Let\n\n\[ \n{d}_{1},{d}_{2},\ldots ,{d}_{r}\text{be the divisors of}n\text{,} \n\] \n\nand \n\n\[ \n{e}_{1},{e}_{2},\ldots ,{e}_{s}\text{be the divisors of}m\text{.} \n\] \n\nThe fact that \( m \) and \( n \) are relatively prime means that the divisors of \( {mn} \) are precisely the various products \n\n\[ \n{... | Yes |
Theorem 27.2 (Euler’s Phi Function Summation Formula). Let \( {d}_{1},{d}_{2},\ldots ,{d}_{r} \) be the divisors of \( n \) . Then\n\n\[ \phi \left( {d}_{1}\right) + \phi \left( {d}_{2}\right) + \cdots + \phi \left( {d}_{r}\right) = n. \] | Proof. We let \( F\left( n\right) = \phi \left( {d}_{1}\right) + \phi \left( {d}_{2}\right) + \cdots + \phi \left( {d}_{r}\right) \), and we need to verify that \( F\left( n\right) \) always equals \( n \) . The calculation of \( \phi \left( 1\right) + \phi \left( p\right) + \phi \left( {p}^{2}\right) + \cdots + \phi \... | Yes |
Theorem 28.1 (Order Divisibility Property). Let a be an integer not divisible by the prime \( p \), and suppose that \( {a}^{n} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) . Then the order \( {e}_{p}\left( a\right) \) divides \( n \) . | Proof. The definition of the order \( {e}_{p}\left( a\right) \) tells us that\n\n\[ \n{a}^{{e}_{p}\left( a\right) } \equiv 1\left( {\;\operatorname{mod}\;p}\right) \n\] \n\nand we are assuming that \( {a}^{n} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) . We divide \( n \) by \( {e}_{p}\left( a\right) \) to get a... | Yes |
Theorem 28.2 (Primitive Root Theorem). Every prime \( p \) has a primitive root. More precisely, there are exactly \( \phi \left( {p - 1}\right) \) primitive roots modulo \( p \) . | Proof of the Primitive Root Theorem. We prove the Primitive Root Theorem using one of the most powerful techniques available in number theory: COUNTING. The use of counting was already illustrated in our proof of Theorem 11.1 on page 77. For the current proof, we will take a set of numbers and count how many numbers ar... | Yes |
Theorem 29.1 (Rules for Indices). Indices satisfy the following rules:\n\n(a) \( I\left( {ab}\right) \equiv I\left( a\right) + I\left( b\right) \;\left( {{\;\operatorname{mod}\;p} - 1}\right) \; \) [Product Rule]\n\n(b) \( I\left( {a}^{k}\right) \equiv {kI}\left( a\right) \;\left( {{\;\operatorname{mod}\;p} - 1}\right)... | Proof. These rules are nothing more than the usual laws of exponents, combined with the fact that \( g \) is a primitive root. Thus, to check (a), we compute\n\n\[ \n{g}^{I\left( {ab}\right) } \equiv {ab} \equiv {g}^{I\left( a\right) }{g}^{I\left( b\right) } \equiv {g}^{I\left( a\right) + I\left( b\right) }\left( {\;\o... | Yes |
Theorem 30.1 (Fermat’s Last Theorem for Exponent 4). The equation\n\n\\[ \n{x}^{4} + {y}^{4} = {z}^{2} \n\\]\n\nhas no solutions in positive integers \\( x, y \\), and \\( z \\) . | Proof. We use Fermat's method of descent to prove this theorem. Recall that the idea of \ | No |
Theorem 34.1 (Pell’s Equation Theorem). Let \( D \) be a positive integer that is not a perfect square. Then Pell's equation\n\n\[ \n{x}^{2} - D{y}^{2} = 1 \n\]\n\nalways has solutions in positive integers. If \( \left( {{x}_{1},{y}_{1}}\right) \) is the solution with smallest \( {x}_{1} \), then every solution \( \lef... | Proof. Our first goal is to show that Pell's equation has at least one solution. Version 1 of Dirichlet's Diophantine Approximation Theorem (Theorem 33.1) tells us that there are infinitely many pairs of positive integers \( \left( {x, y}\right) \) that satisfy the inequality\n\n\[ \n\left| {x - y\sqrt{D}}\right| < \fr... | Yes |
Theorem 35.1 (The Fundamental Theorem of Algebra). If \( {a}_{0},{a}_{1},{a}_{2},\ldots ,{a}_{d} \) are complex numbers with \( {a}_{0} \neq 0 \) and \( d \geq 1 \), then the equation\n\n\[ \n{a}_{0}{x}^{d} + {a}_{1}{x}^{d - 1} + {a}_{2}{x}^{d - 2} + \cdots + {a}_{d - 1}x + {a}_{d} = 0 \n\]\n\nhas a solution in complex... | This theorem was formulated (and used) by many mathematicians during the eighteenth century, but the first satisfactory proofs weren't discovered until the early part of the nineteenth century. Many proofs are now known, some using mostly algebra, some using analysis (calculus), and some using geometric ideas. Unfortun... | No |
Theorem 35.3 (Norm Multiplication Property). Let \( \alpha \) and \( \beta \) be any complex numbers. Then\n\n\[ \mathrm{N}\left( {\alpha \beta }\right) = \mathrm{N}\left( \alpha \right) \mathrm{N}\left( \beta \right) \] | Proof. If we write \( \alpha = a + {bi} \) and \( \beta = c + {di} \), then\n\n\[ {\alpha \beta } = \left( {{ac} - {bd}}\right) + \left( {{ad} + {bc}}\right) i \]\n\nso we need to check that\n\n\[ {\left( ac - bd\right) }^{2} + {\left( ad + bc\right) }^{2} = \left( {{a}^{2} + {b}^{2}}\right) \left( {{c}^{2} + {d}^{2}}\... | No |
Theorem 35.4 (Gaussian Prime Theorem). The Gaussian primes can be described as follows:\n\n(i) \( 1 + i \) is a Gaussian prime.\n\n(ii) Let \( p \) be an ordinary prime with \( p \equiv 3\left( {\\operatorname{mod}4}\right) \) . Then \( p \) is a Gaussian prime.\n\n(iii) Let \( p \) be an ordinary prime with \( p \equi... | Proof. As we observed previously, if \( \\mathrm{N}\\left( \\alpha \\right) \) is an ordinary prime, then \( \\alpha \) must be a Gaussian prime. The number \( 1 + i \) in category (i) has norm 2, so it is a Gaussian prime. Similarly, the numbers \( u + {vi} \) in category (iii) have norm \( {u}^{2} + {v}^{2} = p \), s... | Yes |
Lemma 35.5 (Gaussian Divisibility Lemma). Let \( \alpha = a + {bi} \) be a Gaussian integer.\n\n(a) If 2 divides \( \mathrm{N}\left( \alpha \right) \), then \( 1 + i \) divides \( \alpha \) . | Proof of Lemma. (a) We are given that 2 divides \( \mathrm{N}\left( \alpha \right) = {a}^{2} + {b}^{2} \), so \( a \) and \( b \) are either both odd or both even. It follows that \( a + b \) and \( - a + b \) are both divisible by 2 , so the quotient\n\n\[ \frac{a + {bi}}{1 + i} = \frac{\left( {a + b}\right) + \left( ... | Yes |
Theorem 36.2 (Gaussian Integer Division with Remainder). Let \( \\alpha \) and \( \\beta \) be Gaussian integers with \( \\beta \\neq 0 \) . Then there are Gaussian integers \( \\gamma \) and \( \\rho \) such that\n\n\[ \n\\alpha = {\\beta \\gamma } + \\rho \\;\\text{ and }\\;\\mathrm{N}\\left( \\rho \\right) < \\mathr... | Proof. If we divide the equation we’re trying to prove by \( \\beta \), it becomes\n\n\[ \n\\frac{\\alpha }{\\beta } = \\gamma + \\frac{\\rho }{\\beta }\\;\\text{ with }\\;\\mathrm{N}\\left( \\frac{\\rho }{\\beta }\\right) < 1.\n\]\n\nThis means that we should choose \( \\gamma \) to be as close to \( \\alpha /\\beta \... | Yes |
Theorem 36.3 (Gaussian Integer Common Divisor Property). Let \( \alpha \) and \( \beta \) be Gaussian integers, and let \( S \) be the collection of Gaussian integers \[ {A\alpha } + {B\beta },\;\text{where}A\text{and}B\text{are any Gaussian integers.} \] Among all the Gaussian integers in \( S \), choose an element \[... | Proof. We use Gaussian Integer Division with Remainder to divide \( \alpha \) by \( g \) , \[ \alpha = {g\gamma } + \rho \;\text{ with }\;0 \leq \mathrm{N}\left( \rho \right) < \mathrm{N}\left( g\right) . \] Our goal is to show that the remainder \( \rho \) is zero. Substituting \( g = {a\alpha } + {b\beta } \) into \(... | Yes |
Theorem 36.4 (Gaussian Prime Divisibility Property). Let \( \pi \) be a Gaussian prime, let \( \alpha \) and \( \beta \) be Gaussian integers, and suppose that \( \pi \) divides the product \( {\alpha \beta } \) . Then either \( \pi \) divides \( \alpha \) or \( \pi \) divides \( \beta \) (or both). | Proof. We apply the Gaussian Integer Common Divisor Property to the two numbers \( \alpha \) and \( \pi \) . This tells us that we can find Gaussian integers \( a \) and \( b \) such that the number\n\n\[ g = {a\alpha } + {b\pi }\;\text{divides both}\alpha \text{and}\pi \text{.} \]\n\nBut \( \pi \) is a prime, so the f... | Yes |
Theorem 36.5 (Sum of Two Squares Theorem (Legendre)). For a given positive integer \( N \), let\n\n\( {D}_{1} = \left( {\text{the number of positive integers }d\text{ dividing }N\text{ that satisfy }d \equiv 1\;\left( {\;\operatorname{mod}\;4}\right) }\right) ,\n\n\( {D}_{3} = \left( {\text{the number of positive integ... | Proof of Legendre's Sum of Two Squares Theorem. The proof has two steps. First we find a formula for \( R\left( N\right) \) . Next we find a formula for \( {D}_{1} - {D}_{3} \) . Comparing the two formulas completes the proof.\n\nAlthough the proof is not very hard, it may seem complicated because of the notation. So w... | Yes |
Theorem 36.6 (Difference of \( {D}_{1} - {D}_{3} \) Theorem). Factor the integer \( N \) into a product of ordinary primes as\n\n\[ N = {2}^{t}\underset{\left( 1{\;\operatorname{mod}\;4}\text{ primes }\right) }{\underbrace{{p}_{1}^{{e}_{1}}{p}_{2}^{{e}_{2}}\cdots {p}_{r}^{{e}_{r}}}} \cdot \underset{\left( 3{\;\operator... | Proof. We give a proof by induction on \( s \) . First, if \( s = 0 \), then \( N = {2}^{t}{p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \) , so every odd divisor of \( N \) is congruent to 1 modulo 4 . In other words, \( {D}_{3} = 0 \) , and \( {D}_{1} \) is the number of odd divisors of \( N \) . The odd divisors of \( N... | Yes |
Theorem 37.1 (Irrationality of \( \sqrt{2} \) Theorem). The square root of 2 is irrational. That is, there is no rational number \( r \) satisfying \( {r}^{2} = 2 \) . | Proof. We assume that there does exist a rational number \( r \) satisfying \( {r}^{2} = 2 \), and we use the supposed existence of \( r \) to end up with a contradictory statement, that is, with a statement that is clearly false. This contradiction shows that such an \( r \) does not exist. As noted in Chapter 36, thi... | Yes |
Theorem 37.2 (Liouville’s Inequality). Let \( \\alpha \) be an algebraic number; say \( \\alpha \) is a root of the polynomial\n\n\[ f\\left( X\\right) = {c}_{0}{X}^{d} + {c}_{1}{X}^{d - 1} + {c}_{2}{X}^{d - 2} + \\cdots + {c}_{d - 1}X + {c}_{d} \]\n\nhaving integer coefficients. Let \( D \) be any number with \( D > d... | Proof. The fact that \( X = \\alpha \) is a root of \( f\\left( X\\right) \) means that when we divide \( f\\left( X\\right) \) by \( X - \\alpha \), we get a remainder of 0 . In other words, \( f\\left( X\\right) \) factors as\n\n\[ f\\left( X\\right) = \\left( {X - \\alpha }\\right) g\\left( X\\right) \]\n\nfor some ... | Yes |
Lemma 37.3 (Lemma on Good Approximations to \( \mathbf{\beta } \) ). Let \( \beta \) be Liouville’s number\n\n\[ \beta = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{10}^{n!}} \]\n\nas described above. Then for every number \( D \geq 1 \) we can find infinitely many different rational numbers \( a/b \) that satis... | Proof. Intuitively, Lemma 37.3 says that we can find rational numbers that are very, very close to \( \beta \) . How might we find such good approximations? The defini-\n\ntion of\n\n\[ \beta = \frac{1}{{10}^{1!}} + \frac{1}{{10}^{2!}} + \frac{1}{{10}^{3!}} + \frac{1}{{10}^{4!}} + \frac{1}{{10}^{5!}} + \frac{1}{{10}^{6... | No |
Theorem 37.4 (Transcendence of \( \beta \) Theorem). Liouville’s number\n\n\[ \beta = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{10}^{n!}} \]\n\nis transcendental. | Proof. We give a proof by contradiction, so we start by assuming that \( \beta \) is actually algebraic and try to derive a false statement. The assumption that \( \beta \) is algebraic means that it is a root of a polynomial\n\n\[ f\left( X\right) = {c}_{0}{X}^{d} + {c}_{1}{X}^{d - 1} + {c}_{2}{X}^{d - 2} + \cdots + {... | Yes |
Theorem 38.2 (Binomial Theorem). The binomial coefficients in the expansion\n\n\[ \n{\\left( A + B\\right) }^{n} = \\left( \\begin{array}{l} n \\\\ 0 \\end{array}\\right) {A}^{n} + \\left( \\begin{array}{l} n \\\\ 1 \\end{array}\\right) {A}^{n - 1}B + \\left( \\begin{array}{l} n \\\\ 2 \\end{array}\\right) {A}^{n - 2}{... | Proof. We have already done the hard work of proving the first equality. To get the second formula, we simply multiply the numerator and denominator of the first fraction by \( \\left( {n - k}\\right) \) ! to get\n\n\[ \n\\frac{n\\left( {n - 1}\\right) \\left( {n - 2}\\right) \\cdots \\left( {n - k + 1}\\right) }{k!} \... | Yes |
Theorem 38.3 (Binomial Theorem Modulo \( p \) ). Let \( p \) be a prime number.\n\n(a) The binomial coefficient \( \left( \begin{array}{l} p \\ k \end{array}\right) \) is congruent to\n\n\[ \left( \begin{array}{l} p \\ k \end{array}\right) \equiv \left\{ \begin{array}{ll} 0\left( {\;\operatorname{mod}\;p}\right) & \tex... | Proof. (a) If \( k = 0 \) or \( k = p \), then we know that \( \left( \begin{array}{l} p \\ k \end{array}\right) = 1 \) . So the interesting problem is to find out what happens when \( k \) is between 1 and \( p - 1 \) . Let’s take a particular example, say \( \left( \begin{array}{l} 7 \\ 5 \end{array}\right) \), and t... | Yes |
Theorem 38.4 (Fermat’s Little Theorem). Let \( p \) be a prime number, and let a be any number with \( a ≢ 0\\left( {\\;\\operatorname{mod}\\;p}\\right) \) . Then\n\n\[ \n{a}^{p - 1} \equiv 1\\left( {\\;\\operatorname{mod}\\;p}\\right) \n\] | Proof by Induction. We start by using induction to prove that the formula\n\n\[ \n{a}^{p} \equiv a\\left( {\\;\\operatorname{mod}\\;p}\\right) \n\]\n\nis true for all numbers \( a \) . This formula is clearly true for \( a = 0 \), which gets our induction started. Next suppose that we know it is true for some particula... | Yes |
Theorem 39.1 (Binet’s Formula). The Fibonacci sequence \( {F}_{n} \) is the sequence described by the recursion\n\n\[ \n{F}_{1} = {F}_{2} = 1\;\text{ and }\;{F}_{n} = {F}_{n - 1} + {F}_{n - 2}\;\text{ for }n = 3,4,5,\ldots \n\]\n\nThen the \( {n}^{\text{th }} \) term of the Fibonacci sequence is given by the formula\n\... | Proof. For each number \( n = 1,2,3,\ldots \), let \( {H}_{n} \) be the number\n\n\[ \n{H}_{n} = \frac{1}{\sqrt{5}}\left\{ {{\left( \frac{1 + \sqrt{5}}{2}\right) }^{n} - {\left( \frac{1 - \sqrt{5}}{2}\right) }^{n}}\right\} .\n\]\n\nWe will prove by induction on \( n \) that \( {H}_{n} = {F}_{n} \) for every number \( n... | Yes |
Theorem 41.1 (Mordell’s Theorem). (L.J. Mordell, 1922) Let \( E \) be an elliptic curve given by the equation\n\n\[ E : {y}^{2} = {x}^{3} + a{x}^{2} + {bx} + c, \]\n\nwhere \( a, b, c \) are integers such that the discriminant\n\n\[ \Delta \left( E\right) = - 4{a}^{3}c + {a}^{2}{b}^{2} - 4{b}^{3} - {27}{c}^{2} + {18abc... | Mordell proved his theorem in 1922. Unfortunately, the proof is too complicated for us to give in detail, but the following outline of Mordell's proof shows that it is nothing more than a fancy version of Fermat's descent method:\n\n\( {}^{3} \) If \( \Delta \left( E\right) = 0 \), then the cubic polynomial \( {x}^{3} ... | No |
Theorem 42.3 (Siegel’s Theorem). (C.L. Siegel, 1926) Let \( E \) be an elliptic curve\n\n\[ E : {y}^{2} = {x}^{3} + a{x}^{2} + {bx} + c \]\n\ngiven by an equation whose coefficients \( a, b \), and \( c \) are integers and with discriminant \( \Delta \left( E\right) \neq 0 \). Then there are only finitely many solution... | Siegel actually gave two very different proofs of his theorem. The first, published in 1926 in the Journal of the London Mathematical Society, \( {}^{1} \) works directly with the equation for \( E \) and uses factorization methods. The second proof, published in 1929, begins with Mordell's theorem and uses the geometr... | Yes |
Theorem 43.1 (The Number of Points Modulo \( p \) on \( {E}_{2} : {y}^{2} = {x}^{3} + x \) ).\n\nLet \( p \) be an odd prime, and let \( {N}_{p} \) denote the number of points modulo \( p \) on the elliptic curve \( {E}_{2} : {y}^{2} = {x}^{3} + x \) .\n\n(a) If \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), ... | The first part is comparatively easy to verify, but we omit the proof because we will be proving a similar result later. The second part is considerably more difficult, so we are content to illustrate it with one more example. The prime \( p = {130657} \) is congruent to 1 modulo 4. Using trial and error, a computer, o... | No |
Theorem 45.1 (Hasse’s Theorem). (H. Hasse,1933) Let \( {N}_{p} \) be the number of points modulo \( p \) on an elliptic curve, and let \( {a}_{p} = p - {N}_{p} \) be the \( p \) -defect. Then | \[ \left| {a}_{p}\right| < 2\sqrt{p} \] | Yes |
減正猶如加負，減負猶如加正。 | 說明 1. 設有人存銀 1000 元,而負債 500 元,記作 \( + {1000} \) 元 \( - {500} \) 元\n\n(A)\n\n若此人以存款內100元購物， 即常於(A)之左邊減去(+100 元）故其財產之狀況為\n\n\( + {900} \) 元 -500 元 -(b)\n\n若此人購物,未用現金,係轉欠某店者，則 \( \left( \mathrm{A}\right) \) 之右邊當加入 \( \left( {-{100}\text{元}}\right) \) ,故其財產之狀况為  be a ring \( \neq 0 \) . Then the following are equivalent:\n\ni) \( A \) is a field;\n\nii) the only ideals in \( A \) are 0 and (1);\n\niii) every homomorphism of \( A \) into a non-zero ring \( B \) is injective. | Proof. i) \( \Rightarrow \) ii). Let \( \mathfrak{a} \neq 0 \) be an ideal in \( A \) . Then \( \mathfrak{a} \) contains a non-zero element \( x;x \) is a unit, hence \( a \supseteq \left( x\right) = \left( 1\right) \), hence \( a = \left( 1\right) \).\n\nii) \( \Rightarrow \) iii). Let \( \phi : A \rightarrow B \) . b... | Yes |
Corollary 1.4. If \( \mathfrak{a} \neq \left( 1\right) \) is an ideal of \( A \), there exists a maximal ideal of \( A \) containing \( \mathfrak{a} \) . | Proof. Apply (1.3) to \( A/\mathfrak{a} \), bearing in mind (1.1). Alternatively, modify the proof of (1.3). | No |
Corollary 1.5. Every non-unit of \( A \) is contained in a maximal ideal. | ∎ | No |
Proposition 1.6. i) Let \( A \) be a ring and \( \mathfrak{m} \neq \left( 1\right) \) an ideal of \( A \) such that every \( x \in A - m \) is a unit in \( A \) . Then \( A \) is a local ring and \( m \) its maximal ideal. | Proof. i) Every ideal \( \neq \left( 1\right) \) consists of non-units, hence is contained in \( m \) . Hence \( m \) is the only maximal ideal of \( A \) . | Yes |
Proposition 1.7. The set \( \mathfrak{R} \) of all nilpotent elements in a ring \( A \) is an ideal, and \( A/\mathfrak{N} \) has no nilpotent element \( \neq 0 \) . | Proof. If \( x \in \mathfrak{N} \), clearly \( {ax} \in \mathfrak{N} \) for all \( a \in A \) . Let \( x, y \in \mathfrak{N} \) : say \( {x}^{m} = 0,{y}^{n} = 0 \) . By the binomial theorem (which is valid in any commutative ring), \( {\left( x + y\right) }^{m + n - 1} \) is a sum of integer multiples of products \( {x... | Yes |
Proposition 1.8. The nilradical of \( A \) is the intersection of all the prime ideals of \( A \) . | Proof. Let \( {\mathfrak{R}}^{\prime } \) denote the intersection of all the prime ideals of \( A \) . If \( f \in A \) is nilpotent and if \( \mathfrak{p} \) is a prime ideal, then \( {f}^{n} = 0 \in \mathfrak{p} \) for some \( n > 0 \), hence \( f \in \mathfrak{p} \) (because \( \mathfrak{p} \) is prime). Hence \( f ... | Yes |
Proposition 1.9. \( x \in \Re \Leftrightarrow 1 - {xy} \) is a unit in \( A \) for all \( y \in A \) . | Proof. \( \Rightarrow \) : Suppose \( 1 - {xy} \) is not a unit. By (1.5) it belongs to some maximal ideal \( \mathfrak{m} \) ; but \( x \in \mathfrak{R} \subseteq \mathfrak{m} \), hence \( {xy} \in \mathfrak{m} \) and therefore \( 1 \in \mathfrak{m} \), which is absurd.\n\n\( \Leftarrow \) : Suppose \( x \notin \mathf... | Yes |
Proposition 1.10. i) If \( {\mathfrak{a}}_{i},{\mathfrak{a}}_{j} \) are coprime whenever \( i \neq j \), then \( \Pi {\mathfrak{a}}_{i} = \bigcap {\mathfrak{a}}_{i} \) . | Proof. i) by induction on \( n \) . The case \( n = 2 \) is dealt with above. Suppose \( n > 2 \) and the result true for \( {a}_{1},\ldots ,{a}_{n - 1} \), and let \( \mathfrak{b} = \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}{a}_{i} = \mathop{\bigcap }\limits_{{i = 1}}^{{n - 1}}{a}_{i} \) . Since \( {a}_{i} + {a}_{n} =... | Yes |
Proposition 1.11. i) Let \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{n} \) be prime ideals and let \( \mathfrak{a} \) be an ideal contained in \( \mathop{\bigcup }\limits_{{i = 1}}^{n}{\mathfrak{p}}_{i} \) . Then \( \mathfrak{a} \subseteq {\mathfrak{p}}_{1} \) for some \( i \) . | Proof. i) is proved by induction on \( n \) in the form\n\n\[ \mathfrak{a} \nsubseteq {\mathfrak{p}}_{i}\left( {1 \leq i \leq n}\right) \Rightarrow \mathfrak{a} \nsubseteq \mathop{\bigcup }\limits_{{i = 1}}^{n}{\mathfrak{p}}_{i}. \]\n\nIt is certainly true for \( n = 1 \) . If \( n > 1 \) and the result is true for \( ... | Yes |
Proposition 1.15. \( D = \) set of zero-divisors of \( A = \mathop{\bigcup }\limits_{{x \neq 0}}r\left( {\operatorname{Ann}\left( x\right) }\right) \) . | Proof. \( D = r\left( D\right) = r\left( {\mathop{\bigcup }\limits_{{x \neq 0}}\operatorname{Ann}\left( x\right) }\right) = \mathop{\bigcup }\limits_{{x \neq 0}}r\left( {\operatorname{Ann}\left( x\right) }\right) \) . | Yes |
Proposition 1.16. Let \( \\mathfrak{a},\\mathfrak{b} \) be ideals in a ring \( A \) such that \( r\\left( \\mathfrak{a}\\right), r\\left( \\mathfrak{b}\\right) \) are coprime. Then \( \\mathfrak{a},\\mathfrak{b} \) are coprime. | Proof. \( r\\left( {\\mathfrak{a} + \\mathfrak{b}}\\right) = r\\left( {r\\left( \\mathfrak{a}\\right) + r\\left( \\mathfrak{b}\\right) }\\right) = r\\left( 1\\right) = \\left( 1\\right) \), hence \( \\mathfrak{a} + \\mathfrak{b} = \\left( 1\\right) \) by (1.13). | Yes |
Proposition 1.17. i) \( \mathfrak{a} \subseteq {\mathfrak{a}}^{ec},\mathfrak{b} \supseteq {\mathfrak{b}}^{ce} \) ;\n\nii) \( {b}^{c} = {b}^{cec},{a}^{e} = {a}^{ece} \) ;\n\niii) If \( C \) is the set of contracted ideals in \( A \) and if \( E \) is the set of extended ideals\nin \( B \), then \( C = \left\{ {\mathfrak... | Proof. i) is trivial, and ii) follows from i).\n\niii) If \( a \in C \), then \( a = {b}^{c} \cdot = {b}^{cec} = {a}^{ec} \) ; conversely if \( a = {a}^{ec} \) then \( a \) is the contraction of \( {\mathfrak{a}}^{e} \) . Similarly for \( E \) . ∎ | No |
Proposition 2.1. i) If \( L \supseteq M \supseteq N \) are \( A \) -modules, then\n\n\[ \left( {L/N}\right) /\left( {M/N}\right) \cong L/M \] | Proof. i) Define \( \theta : L/N \rightarrow L/M \) by \( \theta \left( {x + N}\right) = x + M \) . Then \( \theta \) is a well-defined \( A \) -module homomorphism of \( L/N \) onto \( L/M \), and its kernel is \( M/N \) ; hence (i). | Yes |
Proposition 2.3. \( M \) is a finitely generated \( A \) -module \( \Leftrightarrow M \) is isomorphic to a quotient of \( {A}^{n} \) for some integer \( n > 0 \) . | Proof. \( \Rightarrow : \) Let \( {x}_{1},\ldots ,{x}_{n} \) generate \( M \) . Define \( \phi : {A}^{n} \rightarrow M \) by \( \phi \left( {{a}_{1},\ldots ,{a}_{n}}\right) = \) \( {a}_{1}{x}_{1} + \cdots + {a}_{n}{x}_{n} \) . Then \( \phi \) is an \( A \) -module homomorphism onto \( M \), and therefore \( M \cong {A}... | Yes |
Proposition 2.4. Let \( M \) be a finitely generated \( A \) -module, let \( \mathfrak{a} \) be an ideal of \( A \), and let \( \phi \) be an \( A \) -module endomorphism of \( M \) such that \( \phi \left( M\right) \subseteq \mathfrak{a}M \) . Then \( \phi \) satisfies an equation of the form\n\n\[{\phi }^{n} + {a}_{1... | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be a set of generators of \( M \) . Then each \( \phi \left( {x}_{i}\right) \in \mathfrak{a}M \), so that we have say \( \phi \left( {x}_{i}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j}\left( {1 \leq i \leq n;{a}_{ij} \in \mathfrak{a}}\right) \), i.e.,\n\n\[ \math... | Yes |
Corollary 2.5. Let \( M \) be a finitely generated \( A \) -module and let \( \mathfrak{a} \) be an ideal of \( A \) such that \( \mathfrak{a}M = M \) . Then there exists \( x \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \) such that \( {xM} = 0 \) . | Proof. Take \( \phi = \) identity, \( x = 1 + {a}_{1} + \cdots + {a}_{n} \) in (2.4). | No |
Proposition 2.6. (Nakayama's lemma). Let \( M \) be a finitely generated \( A \) -module and \( \mathfrak{a} \) an ideal of \( A \) contained in the Jacobson radical \( \mathfrak{R} \) of \( A \) . Then \( \mathfrak{a}M = M \) implies \( M = 0 \) . | First Proof. By (2.5) we have \( {xM} = 0 \) for some \( x \equiv 1\left( {\;\operatorname{mod}\;\Re }\right) \) . By (1.9) \( x \) is a unit in \( A \), hence \( M = {x}^{-1}{xM} = 0 \) . | Yes |
Corollary 2.7. Let \( M \) be a finitely generated \( A \) -module, \( N \) a submodule of \( M \), \( \mathfrak{a} \varsubsetneq \mathfrak{R} \) an ideal. Then \( M = \mathfrak{a}M + N \Rightarrow M = N \) . | Proof. Apply (2.6) to \( M/N \), observing that \( \mathfrak{a}\left( {M/N}\right) = \left( {\mathfrak{a}M + N}\right) /N \) . ∎ | Yes |
Proposition 2.8. Let \( {x}_{i}\left( {1 \leq i \leq n}\right) \) be elements of \( M \) whose images in \( M/\mathfrak{m}M \) form a basis of this vector space. Then the \( {x}_{t} \) generate \( M \) . | Proof. Let \( N \) be the submodule of \( M \) generated by the \( {x}_{i} \) . Then the composite map \( N \rightarrow M \rightarrow M/\mathfrak{m}M \) maps \( N \) onto \( M/\mathfrak{m}M \), hence \( N + \mathfrak{m}M = M \), hence \( N = M \) by (2.7). | Yes |
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