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Theorem 5. Let \( E/K \) be a finite extension of local fields. For \( \left\lbrack A\right\rbrack \in \operatorname{Br}K \), there is equivalence between:\n\n(i) The exponent \( e\left( A\right) \) divides \( E : K \) .\n\n(ii) \( E \) is a splitting field of \( A \) .\n\n(Thus, whether \( E \) is a splitting field of... | Proof. Set \( \alpha = \left\lbrack A\right\rbrack \) . From the definition of \( e\left( A\right) = e\left( \alpha \right) \) we see that (i) is equivalent to \( {\alpha }^{E : K} = 1 \), and then again, since \( {\operatorname{inv}}_{K} \) is an isomorphism, to\n\n(61)\n\n\[ \left( {E : K}\right) {\operatorname{inv}}... | Yes |
Theorem 6. Let \( D \) be a division algebra of Schur index \( n \) over a local field \( K \). For any extension \( E/K \) of degree \( n \), the field \( E \) is isomorphic over \( K \) to a maximal subfield of \( D \). | Proof. By Theorem 5, any such \( E \) is a splitting field of \( D \), and hence \( K \) -isomorphic to a maximal subfield of \( D \) (Chapter 29, Theorem 19). | No |
Let \( n \) be a natural number and let \( L/K \) be an unramified extension of degree \( n \) . If \( E/K \) is any extension of degree \( n \) we have\n\n\[ \operatorname{Br}\left( {E/K}\right) = \operatorname{Br}\left( {L/K}\right) \] | The first statement follows immediately from Theorem 5, which says that the elements of \( \operatorname{Br}K \) that split over \( E \) are exactly the same that split over \( L \) . If \( E/K \) is Galois, we have\n\n\[ {H}^{2}\left( {G,{E}^{ \times }}\right) \simeq \operatorname{Br}\left( {E/K}\right) = \operatornam... | Yes |
Theorem 10. Let \( E/K \) be a extension of local fields. There is a unique homomorphism \( \operatorname{Br}E \rightarrow \operatorname{Br}K \) such that the diagram commutes, namely, the algebra-theoretic corestriction \( {\operatorname{cor}}_{E/K} \) (see Definition \( {4}^{ * } \) on page 221). Thus \( {\operatorna... | Proof. The existence and uniqueness of the desired map are clear. There remains to show that \( {\operatorname{cor}}_{E/K} \) preserves the Hasse invariant. Take an arbitrary \( \beta \in \operatorname{Br}E \) . By Theorem 9 there exists \( \alpha \in \operatorname{Br}K \) such that \( \beta = {\operatorname{res}}_{E/K... | Yes |
Theorem 11. Let \( L/K \) be a Galois extension of local fields with Galois group \( G \) . \( {H}^{2}\left( {G,{U}_{L}}\right) \) is taken by the natural map \( {H}^{2}\left( {G,{U}_{L}}\right) \rightarrow {H}^{2}\left( {G,{L}^{ \times }}\right) \) isomorphically onto the subgroup of the cyclic group \( {H}^{2}\left( ... | Proof. We start from the exact sequence of \( G \) -modules\n\n(67)\n\n\[ 1 \rightarrow {U}_{L} \rightarrow {L}^{ \times }\overset{{w}_{L}}{ \rightarrow }\mathbb{Z} \rightarrow 0. \]\n\nPassing to the exact sequence of cohomology groups (see F2* in Chapter 30, page 216) we obtain the exact sequence\n\n(68)\n\n\[ {H}^{1... | Yes |
Lemma 1. If \( m \in \mathbb{N} \) is not divisible by \( \operatorname{char}K \), then \( {K}^{\times m} \) is a norm group of the local field \( K \) . | Proof. By the assumption, we can take the extension \( {K}^{\prime } \) of \( K \) obtained by adjoining a primitive \( m \) -th root of unity of \( K \) . By F9, \( {K}^{\prime \times m} \) is the norm group of an abelian extension \( E/{K}^{\prime } \) . Then\n\n\[ \n{N}_{E/K}{E}^{ \times } = {N}_{{K}^{\prime }/K}\le... | Yes |
Theorem 2 (Local existence theorem). Let \( K \) be a local field. The map \[ L \mapsto {\mathcal{N}}_{L} = {N}_{L/K}{K}^{ \times } \] induces a one-to-one correspondence between finite abelian extensions \( L/K \) of \( K \) (in a fixed algebraic closure of \( K \) ) and open subgroups \( H \) of finite index in \( {K... | Proof. (a) By F10, each \( {\mathcal{N}}_{L} \) is open and has finite index in \( {K}^{ \times } \) . (b) Let \( L/K \) and \( {L}^{\prime }/K \) be finite subextensions of \( {K}_{a}/K \) . The local reciprocity law immediately implies that \[ {\mathcal{N}}_{L{L}^{\prime }} = {\mathcal{N}}_{L} \cap {\mathcal{N}}_{{L}... | Yes |
Theorem 3. Any finite abelian extension \( E/{\mathbb{Q}}_{p} \) is contained in some extension of the form \( {\mathbb{Q}}_{p}\left( \zeta \right) /{\mathbb{Q}}_{p} \), where \( \zeta \) is a root of unity. In other words: The largest abelian extension of \( {\mathbb{Q}}_{p} \) is obtained from \( {\mathbb{Q}}_{p} \) ... | Proof. In view of Theorem 2 and F13, the field \( E \) is contained in the class field \( L \) of some group of the form (46). By F14, however, \( L \) is the composite of \( {\mathbb{Q}}_{p}\left( {\zeta }_{{p}^{f} - 1}\right) \) with \( {\mathbb{Q}}_{p}\left( {\zeta }_{{p}^{m}}\right) \) ; that is, \( L = {\mathbb{Q}... | Yes |
Theorem 1 (Maschke). If \( G \) is a finite group, its group algebra \( {KG} \) over a field \( K \) is semisimple if and only if \( \operatorname{char}K \) does not divide the order of \( G \) . | Proof. The sufficiency of the condition for the semisimplicity of \( {KG} \) follows from F10 (because of F10 in Chapter 28).\n\nFor the converse, suppose that \( \operatorname{char}K = p > 0 \) and that \( n = G : 1 \) is a multiple of \( p \) . The nonzero element \( x = {N}_{G} \) of \( {KG} \) satisfies \( {xg} = x... | Yes |
Theorem 2. Let \( G \) be a finite group of order \( n \), and \( K \) a field with \( n \in {K}^{ \times } \). If \( G \) has, up to equivalence, precisely \( k = k\left( {G, K}\right) \) irreducible \( K \) -representations, and if \( h \) is the number of conjugacy classes of \( G \), then \( k \leq h \). The follow... | Proof. By F3, the center \( Z\left( {KG}\right) \) of \( {KG} \) has dimension \( h \) over \( K \). Then (17) implies that \( k \leq h \), as well as the equivalence of (i)-(iv). If \( K \) is a splitting field of \( G \), it follows from F12 that condition (iv) hold. | Yes |
Theorem 3. Let \( A = {KG} \) be the group algebra of a finite group \( G \) of order \( n \) over a field such that \( n \in {K}^{ \times } \) . For each simple component \( {A}_{i} \) of \( {KG} \) (see the notations of F11), the unit element \( {e}_{i} \) can be expressed in terms of the irreducible character \( {\c... | Proof. Write \( {e}_{i} = \mathop{\sum }\limits_{{x \in G}}{a}_{x}x \), with \( {a}_{x} \in K \) . For each \( g \in G \) we have \( {g}^{-1}{e}_{i} = \) \( \mathop{\sum }\limits_{x}{a}_{x}{g}^{-1}x \) . Applying the regular character \( \rho \) we obtain, by (7),\n\n\[ \n\rho \left( {{g}^{-1}{e}_{i}}\right) = \mathop{... | Yes |
Theorem 4 (Orthogonality relations). With the same assumptions and notation as in Theorem 3 we have, for each \( g \in G \) ,\n\n(25)\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{x \in G}}{\chi }_{i}\left( {xg}\right) {\chi }_{j}\left( {x}^{-1}\right) = \left\{ \begin{matrix} \frac{1}{{n}_{i}}{\chi }_{i}\left( g\right) \te... | Proof. We know that \( {e}_{i}{e}_{j} = {\delta }_{ij}{e}_{i} \) for all \( i, j \) . Multiplying together two instances of (24) yields, thanks to (4),\n\n\[ {e}_{i}{e}_{j} = \frac{{n}_{i}}{n}\frac{{n}_{j}}{n}\mathop{\sum }\limits_{g}\left( {\mathop{\sum }\limits_{x}{\chi }_{i}\left( {x{g}^{-1}}\right) {\chi }_{j}\left... | Yes |
Theorem 5. Let \( G \) be a finite group of exponent \( m \) . Every field of characteristic \( p > 0 \) containing a primitive \( m \) -th root of unity is a splitting field of \( G \) . | Proof. Clearly \( p \) does not divide \( m \), much less the order \( n \) of \( G \) . Now use F15 and F16. | No |
Theorem 6. Under the preceding assumptions, the degree \( d \) of \( \chi \) divides the order \( n \) of the group \( G \) . | Proof. For every \( g \in G \), the value \( \chi \left( g\right) \) is integral over \( \mathbb{Z} \) (see F4 again). Because of F21, this implies that\n\n(39)\n\n\[ \mathop{\sum }\limits_{C}\frac{\left| C\right| }{d}\chi \left( C\right) \chi \left( {C}^{-1}\right) \;\text{ is integral over }\mathbb{Z}, \]\n\nwhere th... | Yes |
Theorem 7 (Burnside). If a finite group \( G \) contains a conjugacy class with \( {p}^{m} \) elements, where \( p \) is prime and \( m > 0 \), then \( G \) has a nontrivial normal subgroup. | Proof. Let \( C \) be such a conjugacy class, and let \( {\chi }_{1} = {1}_{G},{\chi }_{2},\ldots ,{\chi }_{h} \) be all the irreducible \( \mathbb{C} \) -characters of \( G \) . By (30) we have\n\n\[ 1 + {\chi }_{2}\left( 1\right) {\chi }_{2}\left( C\right) + \cdots + {\chi }_{h}\left( 1\right) {\chi }_{h}\left( C\rig... | Yes |
Theorem 8 (Burnside’s Theorem). Any group \( G \) of order \( {p}^{a}{q}^{b} \), where \( p \) and \( q \) are prime, is solvable. | Proof. We can assume that \( G \) is neither abelian nor a \( p \) -group. By induction on the order of \( G \), it suffices to show that \( G \) has a nontrivial normal subgroup \( N \), since the solvability of \( N \) and \( G/N \) implies that of \( G \) . Let \( Q \) be a Sylow \( q \) -group of \( G \) . Being a ... | Yes |
Theorem 9 (Ito’s Theorem). If char \( K = 0 \), the degree of an absolutely irreducible \( K \) -representation \( T \) of a finite group \( G \) divides the index of every abelian normal subgroup \( A \) of \( G \) . | Proof. Clearly we can assume that \( K \) is algebraically closed. We then proceed by induction on \( \operatorname{ord}\left( G\right) \) . If we’re in case (b) of F28 with \( N = A \), we can assume by induction that the dimension \( {\dim }_{K}W \) divides \( \left( {H : A}\right) \), so the degree in question, \( \... | Yes |
Theorem 10. Let \( G \) be a finite group of order \( n \) and \( K \) an algebraically closed field such that \( n \in {K}^{ \times } \) . If \( G \) is supersolvable, every irreducible \( K \) -representation of \( G \) is monomial, that is, induced by a representation of degree 1 . | Proof. We can assume the given \( T : G \rightarrow {\mathrm{{GL}}}_{K}\left( V\right) \) is faithful. If \( G \) is abelian, \( T \) is already of degree 1 ; hence assume \( G \) is not abelian. Then \( G \) has an abelian normal subgroup \( A \) not contained in the center \( Z \) of \( G \) . (To see this consider t... | Yes |
Theorem 11 (Artin’s induction theorem). For every \( K \) -character \( \chi \) of \( G \) there are cyclic subgroups \( {A}_{1},\ldots ,{A}_{r} \) of \( G \) and irreducible \( K \) -characters \( {\xi }_{i} \) of the \( {A}_{i} \) such that\n\n(53)\n\n\[ \n{n\chi } = \mathop{\sum }\limits_{{i = 1}}^{r}{a}_{i}{\operat... | Proof. (a) For any cyclic subgroup \( A \) of \( G \), consider the function \( {\gamma }_{A} : A \rightarrow \mathbb{Z} \subseteq K \) defined by\n\n(54)\n\n\[ \n{\gamma }_{A}\left( x\right) = \left\{ \begin{matrix} \left| A\right| & \text{ if }\langle x\rangle = A, \\ 0 & \text{ otherwise. } \end{matrix}\right. \n\]\... | Yes |
Theorem 13. A class function \( \varphi : G \rightarrow C \) is a generalized \( K \) -character of \( G \) if and only if for every \( K \) -elementary subgroup \( H \) of \( G \) the restriction \( {\operatorname{res}}_{H}\left( \varphi \right) \) is a generalized \( K \) -character. | Proof. Only the \ | No |
Theorem 14 (Brauer). Every \( C \) -character of \( G \) is an integer linear combination of characters of monomial \( C \) -representations. | Proof. This too follows from Theorem 12, using Theorem 10 and the obvious fact that an elementary subgroup \( H = A \times P \) is supersolvable. | No |
Theorem 1. If \( \chi \) is an irreducible \( C \) -character of \( G \), then\n\n(8)\n\n\[{\chi }_{K} = {s}_{K}\left( \chi \right) {\operatorname{Tr}}_{K}\left( \chi \right) = {s}_{K}\left( \chi \right) \mathop{\sum }\limits_{\sigma }{\chi }^{\sigma }\]\n\nis the irreducible \( K \) -character of \( G \) corresponding... | Proof. (1) We show first that\n\n(9)\n\n\[B\left( {\chi, K}\right) { \otimes }_{K}C = \mathop{\prod }\limits_{\sigma }B\left( {{\chi }^{\sigma }, C}\right) ,\]\n\nwhere \( \sigma \) runs over all the elements of \( G\left( {K\left( \chi \right) /K}\right) \), as in (7) and (8). Set \( \psi \mathrel{\text{:=}} {\chi }_{... | Yes |
Theorem 4. Let \( q \) be a fixed prime number. Every element of the \( q \) -part \( S{\left( K\right) }_{q} \) of the Schur group of \( K \) has the same Schur index as a crossed product\n\n\[ \Gamma = \left( {K\left( A\right) ,\mathfrak{g}, c}\right) \]\n\nwith the following properties: (i) \( A \) is a group of roo... | Proof. By definition, \( S{\left( K\right) }_{q} \) consists of those elements of \( S\left( K\right) \) whose exponent-or, which is the same by F3 in Chapter 30, whose Schur index - is a \( q \) -power. Let \( \left\lbrack {B\left( {\chi, K}\right) }\right\rbrack \) be any element of \( S{\left( K\right) }_{q} \) . Th... | Yes |
Lemma 1. Let \( L/K \) be a Galois extension whose Galois group \( \mathfrak{g} \) has order relatively prime to \( p \) . Then the natural maps\n\n\[ \n{H}^{i}\left( {\mathfrak{g}, W\left( L\right) }\right) \rightarrow {H}^{i}\left( {\mathfrak{g}, U\left( L\right) }\right) \n\]\n\nare isomorphisms for all \( i \) . | Proof. For any arbitrary abelian torsion group \( M \), we once again denote by \( {M}_{p} \) the \( p \) -part of \( M \) and by \( {M}_{{p}^{\prime }} \) the \( p \) -regular part (that is, the subgroup containing all\n\nelements whose orders are relatively prime to \( p \) ). Because of Chapter 24, Theorem 4(ii), we... | Yes |
Lemma 2. Let \( K \) be, as above, an extension of finite degree over \( {\mathbb{Q}}_{p} \), and let \( \Gamma = \) \( \left( {K\left( \eta \right) /K, c}\right) \) be a cyclotomic algebra over \( K \) . If \( p \neq 2 \) then \( s\left( \Gamma \right) \) divides \( p - 1 \), and if \( p = 2 \) then \( s\left( \Gamma ... | Proof. First we recall from Theorem 5 in Chapter 31 that in this situation \( s\left( \Gamma \right) \) coincides with the order of \( \left\lbrack \Gamma \right\rbrack \) in \( \operatorname{Br}K \) . Let \( {K}_{0} \) be the fixed field of \( G\left( {K\left( \eta \right) /K}\right) \) in \( {\mathbb{Q}}_{p}\left( \e... | Yes |
Theorem 7. Let \( K \) be an extension of finite degree over \( {\mathbb{Q}}_{p} \) . If \( p \neq 2 \), then \( S\left( K\right) \) is a finite cyclic group whose order equals the ramification index of \( K\left( {\zeta }_{p}\right) /K \) . | Proof. Set \( e = e\left( {K\left( {\zeta }_{p}\right) /K}\right) \) . Though we have proved (38), we still need to show that there exists \( \left\lbrack \Gamma \right\rbrack \in S\left( K\right) \) of order \( s\left( \Gamma \right) = e \) . To this end we consider \( L = K\left( {\zeta }_{p}\right) \) and \( \mathfr... | Yes |
Theorem 9. For every local field \( K \) of characteristic 0, the Schur group \( S\left( K\right) \) is isomorphic to the torsion part of the Galois group of \( K\left( W\right) /K \), where in the 2-adic case we must also assume that \( K \) is a cyclotomic field over \( {\mathbb{Q}}_{2} \) . | The proof is left to the reader; note that the torsion part of the group\n\n\[ G\left( {{\mathbb{Q}}_{p}\left( W\right) /{\mathbb{Q}}_{p}}\right) = \widehat{\mathbb{Z}} \times {\mathbb{Z}}_{p}^{ \times } \]\n\n is generated by either \( \left( {{\zeta }_{p - 1},{\mathbb{Q}}_{p}\left( W\right) /{\mathbb{Q}}_{p}}\right) ... | No |
Proposition 1.1 (Least Criminal). Let \( k \) be a natural number, and let \( S\left( k\right) \) , \( S\left( {k + 1}\right) ,\ldots, S\left( n\right) ,\ldots \) be a list of statements. If some of these statements are false, then there is a first false statement. | Proof. Let \( C \) be the set of all those natural numbers \( n \geq k \) for which \( S\left( n\right) \) is false; by hypothesis, \( C \) is a nonempty subset of \( \mathbb{N} \) . The Least Integer Axiom provides a smallest integer \( m \) in \( C \), and \( S\left( m\right) \) is the first false statement. | Yes |
Theorem 1.2. Every integer \( n \geq 2 \) is either a prime or a product of primes. | Proof. Were this not so, there would be \ | No |
Proposition 1.3. If \( m \geq 2 \) is a positive integer which is not divisible by any prime \( p \) with \( p \leq \sqrt{m} \), then \( m \) is a prime. | Proof. If \( m \) is not prime, then \( m = {ab} \), where \( a < m \) and \( b < m \) are positive integers. If \( a > \sqrt{m} \) and \( b > \sqrt{m} \), then \( m = {ab} > \sqrt{m}\sqrt{m} = m \), a contradiction. Therefore, we may assume that \( a \leq \sqrt{m} \) . By Theorem 1.2, \( a \) is either a prime or a pr... | Yes |
Theorem 1.4 (Mathematical Induction \( {}^{4} \) ). Given statements \( S\left( n\right) \), one for each natural number \( n \), suppose that:\n\n(i) Base Step : \( S\left( 0\right) \) is true;\n\n(ii) Inductive Step : if \( S\left( n\right) \) is true, then \( S\left( {n + 1}\right) \) is true.\n\nThen \( S\left( n\r... | Proof. We must show that the collection \( C \) of all those natural numbers \( n \) for which the statement \( S\left( n\right) \) is false is empty.\n\nIf, on the contrary, \( C \) is nonempty, then there is a first false statement \( S\left( m\right) \) . Since \( S\left( 0\right) \) is true, by (i), we must have \(... | Yes |
Proposition 1.5. \( {2}^{n} > n \) for all integers \( n \geq 0 \) . | Proof. The \( n \) th statement \( S\left( n\right) \) is\n\n\[ S\left( n\right) : {2}^{n} > n\text{.} \]\n\nTwo steps are required for induction, corresponding to the two hypotheses in Theorem 1.4.\n\nBase step. The initial statement\n\n\[ S\left( 0\right) : {2}^{0} > 0 \]\n\nis true, for \( {2}^{0} = 1 > 0 \) .\n\nIn... | Yes |
Proposition 1.6. \( \;1 + 2 + \cdots + n = \frac{1}{2}n\left( {n + 1}\right) \) for every integer \( n \geq 1 \) . | Proof. The proof is by induction on \( n \geq 1 \) .\n\nBase step. If \( n = 1 \), then the left side is 1 and the right side is \( \frac{1}{2}1\left( {1 + 1}\right) = 1 \) , as desired.\n\nInductive step. It is always a good idea to write the \( \left( {n + 1}\right) \) st statement \( S\left( {n + 1}\right) \) so one... | Yes |
If we assume \( {\left( fg\right) }^{\prime } = {f}^{\prime }g + f{g}^{\prime } \), the product rule for derivatives, then\n\n\[{\left( {x}^{n}\right) }^{\prime } = n{x}^{n - 1}\text{ for all integers }n \geq 1.\] | Proof. We proceed by induction on \( n \geq 1 \) .\n\nBase step. If \( n = 1 \), then we ask whether \( {\left( x\right) }^{\prime } = {x}^{0} \equiv 1 \), the constant function identically equal to 1 . By definition,\n\n\[{f}^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {x + h}\rig... | Yes |
Proposition 1.8. \( {2}^{n} > {n}^{2} \) is true for all integers \( n \geq 5 \) . | Proof. We have just checked the base step \( S\left( 5\right) \) . In proving\n\n\[ S\left( {n + 1}\right) : {2}^{n + 1} > {\left( n + 1\right) }^{2}, \]\n\nwe are allowed to assume that \( n \geq 5 \) (actually, we will need only \( n \geq 3 \) to prove the inductive step) as well as the inductive hypothesis. Multiply... | Yes |
Theorem 1.9 (Second Form of Induction). Let \( S\left( n\right) \) be a family of statements, one for each natural number \( n \), and suppose that:\n\n(i) \( S\left( 0\right) \) is true;\n\n(ii) if \( S\left( k\right) \) is true for all predecessors \( k \) of \( n \), then \( S\left( n\right) \) is itself true.\n\nTh... | Proof. It suffices to show that there are no integers \( n \) for which \( S\left( n\right) \) is false; that is, the collection \( C \) of all positive integers \( n \) for which \( S\left( n\right) \) is false is empty.\n\nIf, on the contrary, \( C \) is nonempty, then there is a least criminal \( m \) : there is a f... | Yes |
Theorem 1.10 (= Theorem 1.2). Every integer \( n \geq 2 \) is either a prime or a product of primes. | Proof. \( {}^{8} \) Base step. The statement is true when \( n = 2 \) because 2 is a prime.\n\nInductive step. If \( n \geq 2 \) is a prime, we are done. Otherwise, \( n = {ab} \), where \( 2 \leq a < n \) and \( 2 \leq b < n \) . As \( a \) and \( b \) are predecessors of \( n \), each of them is either prime or a pro... | Yes |
Proposition 1.11. Every integer \( n \geq 1 \) has a unique factorization \( n = {2}^{k}m \) , where \( k \geq 0 \) and \( m \geq 1 \) is odd. | Proof. We use the second form of induction on \( n \geq 1 \) to prove the existence of \( k \) and \( m \) ; the reader should see that it is more appropriate here than the first form.\n\nBase step. If \( n = 1 \), take \( k = 0 \) and \( m = 1 \) .\n\nInductive step. If \( n \geq 1 \), then \( n \) is either odd or ev... | Yes |
Theorem 1.12. If \( {F}_{n} \) denotes the nth term of the Fibonacci sequence, then for all \( n \geq 0 \) , \[ {F}_{n} = \frac{1}{\sqrt{5}}\left( {{\alpha }^{n} - {\beta }^{n}}\right) \] where \( \alpha = \frac{1}{2}\left( {1 + \sqrt{5}}\right) \) and \( \beta = \frac{1}{2}\left( {1 - \sqrt{5}}\right) \) . | Proof. We are going to use the second form of induction [the second form is the appropriate induction here, for the equation \( {F}_{n} = {F}_{n - 1} + {F}_{n - 2} \) suggests that proving \( S\left( n\right) \) will involve not only \( S\left( {n - 1}\right) \) but \( S\left( {n - 2}\right) \) as well]. Base step. The... | Yes |
Corollary 1.13. If \( \alpha = \frac{1}{2}\left( {1 + \sqrt{5}}\right) \), then \( {F}_{n} > {\alpha }^{n - 2} \) for all integers \( n \geq 3 \) . | Proof. Base step. If \( n = 3 \), then \( {F}_{3} = 2 > \alpha \), for \( \alpha \approx {1.618} \) .\n\nInductive step. We must show that \( {F}_{n + 1} > {\alpha }^{n - 1} \) . By the inductive hypothesis,\n\n\[ \n{F}_{n + 1} = {F}_{n} + {F}_{n - 1} > {\alpha }^{n - 2} + {\alpha }^{n - 3} \n\]\n\n\[ \n= {\alpha }^{n ... | Yes |
Lemma 1.14. For all integers \( n \geq 1 \) and all \( r \) with \( 0 < r < n + 1 \) ,\n\n\[ \left( \begin{matrix} n + 1 \\ r \end{matrix}\right) = \left( \begin{matrix} n \\ r - 1 \end{matrix}\right) + \left( \begin{array}{l} n \\ r \end{array}\right) \] | Proof. We must show, for all \( n \geq 1 \), that if\n\n\[ {\left( 1 + x\right) }^{n} = {c}_{0} + {c}_{1}x + {c}_{2}{x}^{2} + \cdots + {c}_{n}{x}^{n}, \]\n\nthen the coefficient of \( {x}^{r} \) in \( {\left( 1 + x\right) }^{n + 1} \) is \( {c}_{r - 1} + {c}_{r} \) . Since \( {c}_{0} = 1 \) ,\n\n\[ {\left( 1 + x\right)... | Yes |
Proposition 1.15 (Pascal). For all \( n \geq 0 \) and all \( r \) with \( 0 \leq r \leq n \) , \[ \left( \begin{array}{l} n \\ r \end{array}\right) = \frac{n!}{r!\left( {n - r}\right) !} \] | Proof. We prove the proposition by induction on \( n \geq 0 \). Base step. \( {}^{11} \) If \( n = 0 \), then \[ \left( \begin{array}{l} 0 \\ 0 \end{array}\right) = 0!/0!0! = 1 \] Inductive step. Assuming the formula for \( \left( \begin{array}{l} n \\ r \end{array}\right) \) for all \( r \), we must prove \[ \left( \b... | Yes |
Corollary 1.16. For any real number \( x \) and for all integers \( n \geq 0 \) ,\n\n\[{\left( 1 + x\right) }^{n} = \mathop{\sum }\limits_{{r = 0}}^{n}\left( \begin{array}{l} n \\ r \end{array}\right) {x}^{r} = \mathop{\sum }\limits_{{r = 0}}^{n}\frac{n!}{r!\left( {n - r}\right) !}{x}^{r}.\] | Proof. The first equation is the definition of the binomial coefficients, and the second equation replaces \( \left( \begin{array}{l} n \\ r \end{array}\right) \) by the value given in Pascal’s theorem. | No |
Corollary 1.17 (Binomial Theorem). For all real numbers a and b and for all integers \( n \geq 1 \) , \[ {\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{r = 0}}^{n}\left( \begin{array}{l} n \\ r \end{array}\right) {a}^{n - r}{b}^{r} = \mathop{\sum }\limits_{{r = 0}}^{n}\left( \frac{n!}{r!\left( {n - r}\right) !}\r... | Proof. The result is trivially true when \( a = 0 \) (if we agree that \( {0}^{0} = 1 \) ). If \( a \neq 0 \), set \( x = b/a \) in Corollary 1.16, and observe that \[ {\left( 1 + \frac{b}{a}\right) }^{n} = {\left( \frac{a + b}{a}\right) }^{n} = \frac{{\left( a + b\right) }^{n}}{{a}^{n}}. \] Therefore, \[ {\left( a + b... | Yes |
Proposition 1.18 (Polar Decomposition). Every complex number \( z \) has a factorization\n\n\[ z = r\left( {\cos \theta + i\sin \theta }\right) \]\n\nwhere \( r = \left| z\right| \geq 0 \) and \( 0 \leq \theta < {2\pi } \) . | Proof. If \( z = 0 \), then \( \left| z\right| = 0 \) and any choice of \( \theta \) works. If \( z = a + {bi} \neq 0 \), then \( \left| z\right| \neq 0 \) . Now \( z/\left| z\right| = a/\left| z\right| + {ib}/\left| z\right| \) has modulus 1, for \( {\left( a/\left| z\right| \right) }^{2} + {\left( b/\left| z\right| \... | Yes |
Proposition 1.19 (Addition Theorem). If\n\n\[ z = \cos \theta + i\sin \theta \;\text{ and }\;w = \cos \psi + i\sin \psi ,\]\n\nthen\n\n\[ {zw} = \cos \left( {\theta + \psi }\right) + i\sin \left( {\theta + \psi }\right) . \] | Proof.\n\n\[ {zw} = \left( {\cos \theta + i\sin \theta }\right) \left( {\cos \psi + i\sin \psi }\right) \]\n\n\[ = \left( {\cos \theta \cos \psi - \sin \theta \sin \psi }\right) + i\left( {\sin \theta \cos \psi + \cos \theta \sin \psi }\right) . \]\n\nThe trigonometric addition formulas show that\n\n\[ {zw} = \cos \lef... | Yes |
Corollary 1.20. If \( z \) and \( w \) are complex numbers, then\n\n\[ \left| {zw}\right| = \left| z\right| \left| w\right| \] | Proof. If the polar decompositions of \( z \) and \( w \) are \( z = r\left( {\cos \theta + i\sin \theta }\right) \) and \( w = s\left( {\cos \psi + i\sin \psi }\right) \), respectively, then we have just seen that \( \left| z\right| = r,\left| w\right| = s \) , and \( \left| {zw}\right| = {rs} \). | Yes |
Theorem 1.21 (De Moivre). For every real number \( x \) and every positive integer \( n \) ,\n\n\[ \cos \left( {nx}\right) + i\sin \left( {nx}\right) = {\left( \cos x + i\sin x\right) }^{n}. \] | Proof. We prove De Moivre’s theorem by induction on \( n \geq 1 \) . The base step\n\n\( n = 1 \) is obviously true. For the inductive step,\n\n\[ {\left( \cos x + i\sin x\right) }^{n + 1} = {\left( \cos x + i\sin x\right) }^{n}\left( {\cos x + i\sin x}\right) \]\n\n\[ = \left\lbrack {\cos \left( {nx}\right) + i\sin \l... | Yes |
Let us find the value of \( {\left( \cos {3}^{ \circ } + i\sin {3}^{ \circ }\right) }^{40} \) . | By De Moivre’s theorem,\n\n\[ \n{\left( \cos {3}^{ \circ } + i\sin {3}^{ \circ }\right) }^{40} = \cos {120}^{ \circ } + i\sin {120}^{ \circ } = - \frac{1}{2} + i\frac{\sqrt{3}}{2}.\; \blacktriangleleft \n\] | Yes |
Proposition 1.24. For all \( n \geq 1 \), there is a polynomial \( {f}_{n}\left( x\right) \) having all coefficients integers such that\n\n\[ \cos \left( {nx}\right) = {f}_{n}\left( {\cos x}\right) \] | Proof. By De Moivre's theorem,\n\n\[ \cos \left( {nx}\right) + i\sin \left( {nx}\right) = {\left( \cos x + i\sin x\right) }^{n} \]\n\n\[ = \mathop{\sum }\limits_{{r = 0}}^{n}\left( \begin{array}{l} n \\ r \end{array}\right) {\left( \cos x\right) }^{n - r}{\left( i\sin x\right) }^{r}. \]\n\nThe real part of the left sid... | Yes |
Corollary 1.25. Every nth root of unity \( \zeta \) is equal to\n\n\[ \n{e}^{{2\pi ik}/n} = \cos \left( {{2\pi k}/n}\right) + i\sin \left( {{2\pi k}/n}\right) \n\]\n\nfor some \( k \) with \( 0 \leq k \leq n - 1 \) . | Proof. If \( \zeta = \cos \left( {{2\pi }/n}\right) + i\sin \left( {{2\pi }/n}\right) \), then De Moivre’s theorem, Theorem 1.21, gives\n\n\[ \n{\zeta }^{n} = {\left\lbrack \cos \left( 2\pi /n\right) + i\sin \left( 2\pi /n\right) \right\rbrack }^{n} \n\]\n\n\[ \n= \cos \left( {{n2\pi }/n}\right) + i\sin \left( {{n2\pi ... | Yes |
Lemma 1.26. Let \( \zeta \) be a primitive dth root of unity. If \( {\zeta }^{n} = 1 \), then \( d \) must be a divisor of \( n \) . | Proof. By long division, \( n/d = q + r/d \), where \( q \) and \( r \) are natural numbers and \( 0 \leq r/d < 1 \) ; that is, \( n = {qd} + r \), where \( 0 \leq r < d \) . But\n\n\[ 1 = {\zeta }^{n} = {\zeta }^{{qd} + r} = {\zeta }^{qd}{\zeta }^{r} = {\zeta }^{r}, \]\n\nbecause \( {\zeta }^{qd} = {\left( {\zeta }^{d... | Yes |
Proposition 1.27. For every integer \( n \geq 1 \) ,\n\n\[ \n{x}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Phi }_{d}\left( x\right) \n\] \n\nwhere \( d \) ranges over all the positive divisors \( d \) of \( n \) [in particular, \( {\Phi }_{1}\left( x\right) \) and \( \left. {{\Phi }_{n}\left( x\right) \text{occur}... | Proof. In light of Corollary 1.25, the proposition follows by collecting, for each divisor \( d \) of \( n \), all terms in the equation \( {x}^{n} - 1 = \prod \left( {x - \zeta }\right) \) with \( \zeta \) a primitive \( d \) th root of unity. | No |
Corollary 1.28. For every integer \( n \geq 1 \), we have\n\n\[ n = \mathop{\sum }\limits_{{d \mid n}}\phi \left( d\right) \] | Proof. Note that \( \phi \left( n\right) \) is the degree of \( {\Phi }_{n}\left( x\right) \), and use the fact that the degree of a product of polynomials is the sum of the degrees of the factors. - | No |
Theorem 1.29 (Division Algorithm). Given integers \( a \) and \( b \) with \( a \neq 0 \) , there exist unique integers \( q \) and \( r \) with\n\n\[ b = {qa} + r\;\text{ and }\;0 \leq r < \left| a\right| . \] | Proof. We will prove the theorem in the special case in which \( a > 0 \) and \( b \geq 0 \) ; Exercise 1.42 on page 51 asks the reader to complete the proof. Long division involves finding the largest integer \( q \) with \( {qa} \leq b \), which is the same thing as finding the smallest nonnegative integer of the for... | No |
Corollary 1.30. There are infinitely many primes. | Proof. (Euclid) Suppose, on the contrary, that there are only finitely many primes. If \( {p}_{1},{p}_{2},\ldots ,{p}_{k} \) is the complete list of all the primes, define \( M = \) \( \left( {{p}_{1}\cdots {p}_{k}}\right) + 1 \) . By Theorem 1.2, \( M \) is either a prime or a product of primes. But \( M \) is neither... | Yes |
Proposition 1.31. If \( p \) is a prime and \( b \) is any integer, then\n\n\[ \left( {p, b}\right) = \left\{ \begin{array}{l} p\text{ if }p \mid b \\ 1\text{ otherwise. } \end{array}\right. \] | Proof. A common divisor \( c \) of \( p \) and \( a \) is, of course, a divisor of \( p \) . But the only positive divisors of \( p \) are \( p \) and 1, and so \( \left( {p, a}\right) = p \) or 1 ; it is \( p \) if \( p \mid a \) , and it is 1 otherwise. - | Yes |
Theorem 1.32. If \( a \) and \( b \) are integers, then \( \gcd \left( {a, b}\right) \) is a linear combination of \( a \) and \( b \) . | Proof. We may assume that at least one of \( a \) and \( b \) is not zero (otherwise, the gcd is 0 and the result is obvious). Consider the set \( I \) of all the linear combinations:\n\n\[ I = \{ {sa} + {tb} : s, t\text{ in }\mathbb{Z}\} . \]\n\nBoth \( a \) and \( b \) are in \( I \) (take \( s = 1 \) and \( t = 0 \)... | Yes |
Corollary 1.33. Let \( a \) and \( b \) be integers. A nonnegative common divisor \( d \) is their \( \gcd \) if and only if \( c \mid d \) for every common divisor \( c \) . | Proof. Necessity (i.e., the implication \( \Rightarrow \) ) That every common divisor \( c \) of \( a \) and \( b \) is a divisor of \( d = {sa} + {tb} \), has already been proved at the end of the proof of Theorem 1.32.\n\nSufficiency (i.e., the implication \( \Leftarrow \) ) Let \( d \) denote the gcd of \( a \) and ... | Yes |
Corollary 1.34. Let \( I \) be a subset of \( \mathbb{Z} \) such that\n\n(i) 0 is in \( I \) ;\n\n(ii) if \( a \) and \( b \) are in \( I \), then \( a - b \) is in \( I \) ;\n\n(iii) if \( a \) is in \( I \) and \( q \) is in \( \mathbb{Z} \), then \( q \) a is in \( I \) .\n\nThen there is a nonnegative integer \( d ... | Proof. If \( I \) consists of only the single integer 0, take \( d = 0 \) . If \( I \) contains a nonzero integer \( a \), then \( \left( {-1}\right) a = - a \) is in \( I \), by (iii). Thus, \( I \) contains \( \pm a \), one of which is positive. By the Least Integer Axiom, \( I \) contains a smallest positive integer... | Yes |
Theorem 1.35 (Euclid’s Lemma). If \( p \) is a prime and \( p \mid {ab} \), then \( p \mid a \) or \( p \mid b \) . More generally, if a prime \( p \) divides a product \( {a}_{1}{a}_{2}\cdots {a}_{n} \), then it must divide at least one of the factors \( {a}_{i} \) . Conversely, if \( m \geq 2 \) is an integer such th... | Proof. Assume that \( p \nmid a \) ; that is, \( p \) does not divide \( a \) ; we must show that \( p \mid b \) . Now the gcd \( \left( {p, a}\right) = 1 \), by Proposition 1.31. By Theorem 1.32, there are integers \( s \) and \( t \) with \( 1 = {sp} + {ta} \), and so\n\n\[ b = {spb} + {tab}. \]\n\nSince \( p \mid {a... | Yes |
Proposition 1.36. If \( p \) is a prime, then \( p \mid \left( \begin{array}{l} p \\ j \end{array}\right) \) for \( 0 < j < p \) . | Proof. Recall that\n\n\[ \left( \begin{array}{l} p \\ j \end{array}\right) = \frac{p!}{j!\left( {p - j}\right) !} = \frac{p\left( {p - 1}\right) \cdots \left( {p - j + 1}\right) }{j!}. \]\n\nCross multiplying gives\n\n\[ j!\left( \begin{array}{l} p \\ j \end{array}\right) = p\left( {p - 1}\right) \cdots \left( {p - j +... | Yes |
Corollary 1.37. Let \( a, b \), and \( c \) be integers. If \( c \) and \( a \) are relatively prime and if \( c \mid {ab} \), then \( c \mid b \) . | Proof. By hypothesis, \( {ab} = {cd} \) for some integer \( d \) . There are integers \( s \) and \( t \) with \( 1 = {sc} + {ta} \), and so \( b = {scb} + {tab} = {scb} + {tcd} = c\left( {{sb} + {td}}\right) \) . \( \; \bullet \) | Yes |
Lemma 1.38. Every nonzero rational number \( r \) has an expression in lowest terms. | Proof. Since \( r \) is rational, \( r = a/b \) for integers \( a \) and \( b \) . If \( d = \left( {a, b}\right) \), then \( a = {a}^{\prime }d, b = {b}^{\prime }d \), and \( a/b = {a}^{\prime }d/{b}^{\prime }d = {a}^{\prime }/{b}^{\prime } \) . But \( \left( {{a}^{\prime },{b}^{\prime }}\right) = 1 \), for if \( {d}^... | Yes |
Proposition 1.39. If \( n \geq 1 \) is an integer, then \( \phi \left( n\right) \) is the number of integers \( k \) with \( 1 \leq k \leq n \) and \( \left( {k, n}\right) = 1 \) . | Proof. It suffices to prove that \( {e}^{{2\pi ik}/n} \) is a primitive \( n \) th root of unity if and only if \( k \) and \( n \) are relatively prime.\n\nIf \( k \) and \( n \) are not relatively prime, then \( n = {dr} \) and \( k = {ds} \), where \( d, r \) , and \( s \) are integers, and \( d > 1 \) ; it follows ... | Yes |
Proposition 1.40. \( \sqrt{2} \) is irrational. | Proof. Suppose, on the contrary, that \( \sqrt{2} \) is rational; that is, \( \sqrt{2} = a/b \) . We may assume that \( a/b \) is in lowest terms; that is, \( \left( {a, b}\right) = 1 \) . Squaring, \( {a}^{2} = 2{b}^{2} \) . By Euclid’s lemma \( {}^{17},2 \mid a \), so that \( {2m} = a \), hence \( 4{m}^{2} = {a}^{2} ... | Yes |
Proposition 1.43 (Lamé’s \( {}^{19} \) Theorem). Let \( b \geq a \) be positive integers, and let \( \delta \left( a\right) \) be the number of digits in the decimal expression of \( a \) . If \( n \) is the number of steps in the Euclidean algorithm computing the \( \gcd \left( {b, a}\right) \), then\n\n\[ n \leq {5\d... | Proof. Let us denote \( b \) by \( {r}_{0} \) and \( a \) by \( {r}_{1} \) in the equations of the euclidean algorithm on page 44 , so that every equation there has the form\n\n\[ {r}_{j} = {r}_{j + 1}{q}_{j + 1} + {r}_{j + 2} \]\n\nexcept the last one, which is\n\n\[ {r}_{n - 1} = {r}_{n}{q}_{n} \]\n\nNote that \( {q}... | Yes |
Proposition 1.44. If \( b \geq 2 \) is an integer, then every positive integer \( m \) has an expression in base \( \mathbf{b} \) : there are integers \( {d}_{i} \) with \( 0 \leq {d}_{i} < b \) such that\n\n\[ m = {d}_{k}{b}^{k} + {d}_{k - 1}{b}^{k - 1} + \cdots + {d}_{0} \]\n\nmoreover, this expression is unique if \... | Proof. Let \( m \) be a positive integer; since \( b \geq 2 \), there are powers of \( b \) larger than \( m \) . We prove, by induction on \( k \geq 0 \), that if \( {b}^{k} \leq m < {b}^{k + 1} \), then \( m \) has an expression\n\n\[ m = {d}_{k}{b}^{k} + {d}_{k - 1}{b}^{k - 1} + \cdots + {d}_{0} \]\n\nin base \( b \... | Yes |
Theorem 1.48 (Fundamental Theorem of Arithmetic). Every integer \( a \geq 2 \) is a prime or a product of primes. Moreover, if a has factorizations \[ a = {p}_{1}\cdots {p}_{m}\text{ and }a = {q}_{1}\cdots {q}_{n}, \] where the \( p \) ’s and \( q \) ’s are primes, then \( n = m \) and the \( q \) ’s may be reindexed s... | Proof. We prove the theorem by induction on \( \ell \), the larger of \( m \) and \( n \) . Base step. If \( \ell = 1 \), then the given equation is \( a = {p}_{1} = {q}_{1} \), and the result is obvious. Inductive step. The equation gives \( {p}_{m} \mid {q}_{1}\cdots {q}_{n} \) . By Theorem 1.35, Euclid’s lemma, ther... | Yes |
Corollary 1.49. If \( a \geq 2 \) is an integer, then there are distinct primes \( {p}_{i} \), unique up to indexing, and unique integers \( {e}_{i} > 0 \) with\n\n\[ a = {p}_{1}^{{e}_{1}}\cdots {p}_{n}^{{e}_{n}} \] | Proof. Just collect like terms in a prime factorization. -\n\nThe uniqueness in the Fundamental Theorem of Arithmetic says that the exponents \( {e}_{1},\ldots ,{e}_{n} \) in the prime factorization \( a = {p}_{1}^{{e}_{1}}\cdots {p}_{n}^{{e}_{n}} \) are well-defined integers determined by \( a \) . | No |
Corollary 1.50. Every positive rational number \( r \neq 1 \) has a unique factorization\n\n\[ r = {p}_{1}^{{g}_{1}}\cdots {p}_{n}^{{g}_{n}} \]\n\nwhere the \( {p}_{i} \) are distinct primes and the \( {g}_{i} \) are nonzero integers. Moreover, \( r \) is an integer if and only if \( {g}_{i} > 0 \) for all \( i \) . | Proof. There are positive integers \( a \) and \( b \) with \( r = a/b \) . If \( a = {p}_{1}^{{e}_{1}}\cdots {p}_{n}^{{e}_{n}} \) and \( b = {p}_{1}^{{f}_{1}}\cdots {p}_{n}^{{f}_{n}} \), then \( r = {p}_{1}^{{g}_{1}}\cdots {p}_{n}^{{g}_{n}} \), where \( {g}_{i} = {e}_{i} - {f}_{i} \) (we may assume that the same prime... | Yes |
Lemma 1.51. Let positive integers \( a \) and \( b \) have prime factorizations\n\n\[ a = {p}_{1}^{{e}_{1}}\cdots {p}_{n}^{{e}_{n}}\text{ and }b = {p}_{1}^{{f}_{1}}\cdots {p}_{n}^{{f}_{n}}, \]\n\nwhere \( {e}_{i},{f}_{i} \geq 0 \) for all \( i \) . Then \( a \mid b \) if and only if \( {e}_{i} \leq {f}_{i} \) for all \... | Proof. If \( {e}_{i} \leq {f}_{i} \) for all \( i \), then \( b = {ac} \), where \( c = {p}_{1}^{{f}_{1} - {e}_{1}}\cdots {p}_{n}^{{f}_{n} - {e}_{n}} \) . The number \( c \) is an integer because \( {f}_{i} - {e}_{i} \geq 0 \) for all \( i \) . Therefore, \( a \mid b \) .\n\nConversely, if \( b = {ac} \), let the prime... | Yes |
Proposition 1.52. Let \( a = {p}_{1}^{{e}_{1}}\cdots {p}_{n}^{{e}_{n}} \) and let \( b = {p}_{1}^{{f}_{1}}\cdots {p}_{n}^{{f}_{n}} \), where \( {e}_{i},{f}_{i} \geq 0 \) for all \( i \) ; define\n\n\[ \n{m}_{i} = \min \left\{ {{e}_{i},{f}_{i}}\right\} \;\text{ and }\;{M}_{i} = \max \left\{ {{e}_{i},{f}_{i}}\right\} \]\... | Proof. Define \( d = {p}_{1}^{{m}_{1}}\cdots {p}_{n}^{{m}_{n}} \) . Lemma 1.51 shows that \( d \) is a (positive) common divisor of \( a \) and \( b \) ; moreover, if \( c \) is any (positive) common divisor, then \( c = {p}_{1}^{{g}_{1}}\cdots {p}_{n}^{{g}_{n}} \), where \( 0 \leq {g}_{i} \leq \min \left\{ {{e}_{i},{f... | Yes |
Proposition 1.53. If \( a \) and \( b \) are positive integers, then\n\n\[ \operatorname{lcm}\left( {a, b}\right) \gcd \left( {a, b}\right) = {ab}. \]\n | Proof. The result follows from Proposition 1.52 if one uses the identity\n\n\[ {m}_{i} + {M}_{i} = {e}_{i} + {f}_{i} \]\n\nwhere \( {m}_{i} = \min \left\{ {{e}_{i},{f}_{i}}\right\} \) and \( {M}_{i} = \max \left\{ {{e}_{i},{f}_{i}}\right\} .\n\nOf course, this proposition allows us to compute the lcm as \( {ab}/\left( ... | Yes |
Proposition 1.54. If \( m \geq 0 \) is a fixed integer, then for all integers \( a, b, c \) ,\n\n(i) \( a \equiv a{\;\operatorname{mod}\;m} \) ;\n\n(ii) if \( a \equiv b{\;\operatorname{mod}\;m} \), then \( b \equiv a{\;\operatorname{mod}\;m} \) ;\n\n(iii) if \( a \equiv b{\;\operatorname{mod}\;m} \) and \( b \equiv c{... | Proof.\n\n(i) Since \( m \mid \left( {a - a}\right) = 0 \), we have \( a \equiv a{\;\operatorname{mod}\;m} \) .\n\n(ii) If \( m \mid \left( {a - b}\right) \), then \( m \mid - \left( {a - b}\right) = b - a \) and so \( b \equiv a{\;\operatorname{mod}\;m} \) .\n\n(iii) If \( m\left| {\left( {a - b}\right) \text{and}m}\r... | Yes |
Proposition 1.55. Let \( m \geq 0 \) be a fixed integer.\n\n(i) If \( a = {qm} + r \), then \( a \equiv r{\;\operatorname{mod}\;m} \).\n\n(ii) If \( 0 \leq {r}^{\prime } < r < m \), then \( r \) and \( {r}^{\prime } \) are not congruent \( {\;\operatorname{mod}\;m} \) ; in symbols, \( r ≢ {r}^{\prime }{\;\operatorname{... | Proof.\n\n(i) The equation \( a - r = {qm} \) (from the division algorithm \( a = {qm} + r \) ) shows that \( m \mid \left( {a - r}\right) \).\n\n(ii) If \( r \equiv {r}^{\prime }{\;\operatorname{mod}\;m} \), then \( m \mid \left( {r - {r}^{\prime }}\right) \) and \( m \leq r - {r}^{\prime } \) . But \( r - {r}^{\prime... | Yes |
Corollary 1.56. Given \( m \geq 2 \), every integer \( a \) is congruent \( {\;\operatorname{mod}\;m} \) to exactly one of \( 0,1,\ldots, m - 1 \) . | Proof. The division algorithm says that \( a \equiv r{\;\operatorname{mod}\;m} \), where \( 0 \leq r < m \) ; that is, \( r \) is an integer on the list \( 0,1,\ldots, m - 1 \) . If \( a \) were congruent to two integers on the list, say, \( r \) and \( {r}^{\prime } \), then \( r \equiv {r}^{\prime }{\;\operatorname{m... | Yes |
Proposition 1.57. Let \( m \geq 0 \) be a fixed integer.\n\n(i) If \( {a}_{i} \equiv {a}_{i}^{\prime }{\;\operatorname{mod}\;m} \) for \( i = 1,2,\ldots, n \), then\n\n\[ \n{a}_{1} + \cdots + {a}_{n} \equiv {a}_{1}^{\prime } + \cdots + {a}_{n}^{\prime }{\;\operatorname{mod}\;m}.\n\]\n\nIn particular, if \( a \equiv {a}... | Proof.\n\n(i) The proof is by induction on \( n \geq 2 \) . For the base step, if \( m \mid \left( {a - {a}^{\prime }}\right) \) and \( m\left| {\left( {b - {b}^{\prime }}\right) \text{, then}m}\right| \left( {a - {a}^{\prime } + b - {b}^{\prime }}\right) = \left( {a + b}\right) - \left( {{a}^{\prime } + {b}^{\prime }}... | Yes |
Proposition 1.59. There are infinitely many primes \( p \) with \( p \equiv 2{\;\operatorname{mod}\;3} \) . | Proof. We mimic Euclid's proof that there are infinitely many primes. Suppose, on the contrary, that there are only finitely many primes congruent to 2 mod 3 ; let them be \( {p}_{1},\ldots ,{p}_{s} \) . Consider the number\n\n\[ m = 1 + {p}_{1}^{2}\cdots {p}_{s}^{2} \]\n\nNow \( {p}_{i} \equiv 2{\;\operatorname{mod}\;... | Yes |
Proposition 1.60. If \( p \) is a prime and \( a \) and \( b \) are integers, then\n\n\[{\left( a + b\right) }^{p} \equiv {a}^{p} + {b}^{p}{\;\operatorname{mod}\;p}.\] | Proof. The binomial theorem gives\n\n\[{\left( a + b\right) }^{p} = {a}^{p} + {b}^{p} + \mathop{\sum }\limits_{{r = 1}}^{{p - 1}}\left( \begin{array}{l} p \\ r \end{array}\right) {a}^{p - r}{b}^{r}.\]\n\nBut Proposition 1.36 gives \( \left( \begin{array}{l} p \\ r \end{array}\right) \equiv 0{\;\operatorname{mod}\;p} \)... | Yes |
Theorem 1.61 (Fermat).\n\n(i) If \( p \) is a prime, then\n\n\[ \n{a}^{p} \equiv a{\;\operatorname{mod}\;p} \n\]\n\nfor every \( a \) in \( \mathbb{Z} \) . | Proof.\n\n(i) Assume first that \( a \geq 0 \) ; we proceed by induction on \( a \) . The base step \( a = 0 \) is plainly true. For the inductive step, observe that\n\n\[ \n{\left( a + 1\right) }^{p} \equiv {a}^{p} + 1{\;\operatorname{mod}\;p} \n\]\n\nby Proposition 1.60. The inductive hypothesis gives \( {a}^{p} \equ... | Yes |
Corollary 1.62. A positive integer \( a \) is divisible by 3 if and only if the sum of its (decimal) digits is divisible by 3. | Proof. If the decimal form of \( a \) is \( {d}_{k}\ldots {d}_{1}{d}_{0} \), then\n\n\[ a = {d}_{k}{10}^{k} + \cdots + {d}_{1}{10} + {d}_{0} \]\n\nNow \( {10} \equiv 1{\;\operatorname{mod}\;3} \), so that Proposition 1.57(iii) gives \( {10}^{i} \equiv {1}^{i} = 1{\;\operatorname{mod}\;3} \) for all \( i \) ; thus Propo... | Yes |
Corollary 1.63. Let \( p \) be a prime and let \( n \) be a positive integer. If \( m \geq 0 \) and if \( \sum \) is the sum of the p-adic digits of \( m \), then\n\n\[ \n{n}^{m} \equiv {n}^{\sum }{\;\operatorname{mod}\;p}.\n\] | Proof. Let \( m = {d}_{k}{p}^{k} + \cdots + {d}_{1}p + {d}_{0} \) be the expression of \( m \) in base \( p \) . By Fermat’s theorem, Theorem 1.61(ii), \( {n}^{{p}^{i}} \equiv n{\;\operatorname{mod}\;p} \) for all \( i \) ; thus, \( {n}^{{d}_{i}{p}^{i}} = \) \( {\left( {n}^{{d}_{i}}\right) }^{{p}^{i}} \equiv {n}^{{d}_{... | Yes |
Theorem 1.65. If \( \\left( {a, m}\\right) = 1 \), then, for every integer \( b \), the congruence\n\n\[ \n{ax} \equiv b{\\;\\operatorname{mod}\\,m} \n\]\n\ncan be solved for \( x \) ; in fact, \( x = {sb} \), where \( {sa} \equiv 1{\\;\\operatorname{mod}\\,m} \). Moreover, any two solutions are congruent \( {\\;\\oper... | Proof. Since \( \\left( {a, m}\\right) = 1 \), there is an integer \( s \) with \( {as} \equiv 1{\\;\\operatorname{mod}\\,m} \) (because there is a linear combination \( 1 = {sa} + {tm} \) ). It follows that \( b = {sab} + {tmb} \) and \( {asb} \equiv b{\\;\\operatorname{mod}\\,m} \), so that \( x = {sb} \) is a soluti... | Yes |
Corollary 1.66. If \( p \) is prime, the congruence \( {ax} \equiv b{\;\operatorname{mod}\;p} \) is always solvable if \( a \) is not divisible by \( p \) . | Proof. Since \( p \) is a prime, \( p \nmid a \) implies \( \left( {a, p}\right) = 1 \) . | No |
When \( \left( {a, m}\right) = 1 \), Theorem 1.65 says that the solutions to \( {ax} \equiv b{\;\operatorname{mod}\;m} \) are precisely those integers of the form \( {sb} + {km} \) for \( k \) in \( \mathbb{Z} \), where \( {sa} \equiv 1{\;\operatorname{mod}\;m} \) ; that is, where \( {sa} + {tm} = 1 \) . Thus, \( s \) ... | For example, let us find all the solutions to\n\n\[ \n{2x} \equiv 9{\;\operatorname{mod}\;{13}}\text{.}\n\]\n\nConsidering the products \( 2 \cdot 2,3 \cdot 2,4 \cdot 2,\ldots \) mod 13 quickly leads to \( 7 \times 2 = \) \( {14} \equiv 1{\;\operatorname{mod}\;{13}} \) ; that is, \( s = 7 \) and \( x = 7 \cdot 9 = {63}... | Yes |
Theorem 1.69 (Chinese Remainder Theorem). If \( m \) and \( {m}^{\prime } \) are relatively prime, then the two congruences\n\n\[ x \equiv b{\;\operatorname{mod}\;m} \]\n\n\[ x \equiv {b}^{\prime }{\;\operatorname{mod}\;{m}^{\prime }} \]\n\nhave a common solution, and any two solutions are congruent \( {\;\operatorname... | Proof. Every solution of the first congruence has the form \( x = b + {km} \) for some integer \( k \) ; hence, we must find \( k \) such that \( b + {km} \equiv {b}^{\prime }{\;\operatorname{mod}\;{m}^{\prime }} \) ; that is, \( {km} \equiv {b}^{\prime } - b{\;\operatorname{mod}\;{m}^{\prime }} \) . Since \( \left( {m... | Yes |
Solve the simultaneous congruences\n\n\\[ \nx \\equiv 2{\\;\\operatorname{mod}\\;5} \n\\]\n\n\\[ \n{3x} \\equiv 5{\\;\\operatorname{mod}\\;{13}}\\text{.} \n\\] | Every solution to the first congruence has the form \\( x = {5k} + 2 \\) for \\( k \\) in \\( \\mathbb{Z} \\) . Substituting into the second congruence, we have\n\n\\[ \n3\\left( {{5k} + 2}\\right) \\equiv 5{\\;\\operatorname{mod}\\;{13}}\\text{.} \n\\]\n\nTherefore,\n\n\\[ \n{15k} + 6 \\equiv 5{\\;\\operatorname{mod}\... | Yes |
Corollary 1.76. The date with month \( m \), day \( d \), year \( y = {100}\mathrm{C} + N \), where \( 0 \leq N \leq {99} \), has number | Proof. If we write a year \( y = {100C} + N \), where \( 0 \leq N \leq {99} \), then\n\n\[ y = {100C} + N \equiv {2C} + N{\;\operatorname{mod}\;7}, \]\n\n\[ \lfloor y/4\rfloor = {25C} + \lfloor N/4\rfloor \equiv {4C} + \lfloor N/4\rfloor {\;\operatorname{mod}\;7}, \]\n\n\[ \lfloor y/{100}\rfloor = C\text{, and}\lfloor ... | Yes |
Proposition 1.78 (Conway). Let \( D \) be doomsday \( {100C} \), and let \( 0 \leq N \leq {99} \) . If \( N = {12q} + r \), where \( 0 \leq r < {12} \), then the formula for doomsday \( {100}\mathrm{C} + N \) is\n\n\[ D + q + r + \lfloor r/4\rfloor {\;\operatorname{mod}\;7}. \] | Proof.\n\n\[ \text{ Doomsday }\left( {{100C} + N}\right) \equiv D + N + \lfloor N/4\rfloor \]\n\n\[ \equiv D + {12q} + r + \lfloor \left( {{12q} + r}\right) /4\rfloor \]\n\n\[ \equiv D + {15q} + r + \lfloor r/4\rfloor \]\n\n\[ \equiv D + q + r + \lfloor r/4\rfloor {\;\operatorname{mod}\;7}.\; \bullet \] | Yes |
Proposition 2.2. Let \( f : X \rightarrow Y \) and \( g : {X}^{\prime } \rightarrow {Y}^{\prime } \) be functions. Then \( f = g \) if and only if \( X = {X}^{\prime }, Y = {Y}^{\prime } \), and \( f\left( a\right) = g\left( a\right) \) for every \( a \in X \) . | Proof. Assume that \( f = g \) . Functions are subsets of \( X \times Y \), and so \( f = g \) means that each of \( f \) and \( g \) is a subset of the other (informally, we are saying that \( f \) and \( g \) have the same graph). If \( a \in X \) and \( \left( {a, f\left( a\right) }\right) \in f = g \), then \( \lef... | Yes |
Lemma 2.6. Composition of functions is associative: if\n\n\\[ \nf : X \\rightarrow Y,\\;g : Y \\rightarrow Z,\\;\\text{ and }\\;h : Z \\rightarrow W \n\\]\n\nare functions, then\n\n\\[ \nh \\circ \\left( {g \\circ f}\\right) = \\left( {h \\circ g}\\right) \\circ f. \n\\] | Proof. We show that the value of either composite on an element \\( a \\in X \\) is just \\( w = h\\left( {g\\left( {f\\left( a\\right) }\\right) }\\right) \\) . If \\( x \\in X \\), then\n\n\\[ \nh \\circ \\left( {g \\circ f}\\right) : x \\mapsto \\left( {g \\circ f}\\right) \\left( x\\right) = g\\left( {f\\left( x\\r... | Yes |
Lemma 2.7. If \( f : X \rightarrow Y \), then \( {1}_{Y} \circ f = f = f \circ {1}_{X} \) . | Proof. If \( x \in X \), then\n\n\[ \n{1}_{Y} \circ f : x \mapsto f\left( x\right) \mapsto f\left( x\right) \n\] \n\nand \n\n\[ \nf \circ {1}_{X} : x \mapsto x \mapsto f\left( x\right) .\; \bullet \n\] | Yes |
Lemma 2.9. If \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) are functions such that \( g \circ f = {1}_{X} \) , then \( f \) is injective and \( g \) is surjective. | Proof. Suppose that \( f\left( a\right) = f\left( {a}^{\prime }\right) \) ; apply \( g \) to obtain \( g\left( {f\left( a\right) }\right) = g\left( {f\left( {a}^{\prime }\right) }\right) \) ; that is, \( a = {a}^{\prime } \) [because \( g\left( {f\left( a\right) }\right) = a \) ], and so \( f \) is injective. If \( x \... | Yes |
Lemma 2.10. A function \( f : X \rightarrow Y \) has an inverse \( g : Y \rightarrow X \) if and only if it is a bijection. | Proof. If \( f \) has an inverse \( g \), then Lemma 2.9 shows that \( f \) is injective and surjective, for both composites \( g \circ f \) and \( f \circ g \) are identities.\n\nAssume that \( f \) is a bijection. Let \( y \in Y \) . Since \( f \) is surjective, there is some \( a \in X \) with \( f\left( a\right) = ... | No |
Proposition 2.13. If the set of all the bijections from a set \( X \) to itself is denoted by \( {S}_{X} \), then composition of functions satisfies the following properties:\n\n(i) if \( f, g \in {S}_{X} \), then \( f \circ g \in {S}_{X} \) ;\n\n(ii) \( h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \ci... | Proof. We have merely restated results of Exercise 2.13(ii) on page 102, Lemmas 2.6, 2.7, and 2.10. - | No |
Proposition 2.14. Let \( X \) and \( Y \) be sets, and let \( f : X \rightarrow Y \) be a function.\n\n(i) If \( T \subseteq S \) are subsets of \( X \), then \( f\left( T\right) \subseteq f\left( S\right) \), and if \( U \subseteq V \) are subsets of \( Y \), then \( {f}^{-1}\left( U\right) \subseteq {f}^{-1}\left( V\... | Proof.\n\n(i) If \( y \in f\left( T\right) \), then \( y = f\left( t\right) \) for some \( t \in T \) . But \( t \in S \), because \( T \subseteq S \), and so \( f\left( t\right) \in f\left( S\right) \) . Therefore, \( f\left( T\right) \subseteq f\left( S\right) \) . The other inclusion is proved just as easily.\n\n(ii... | Yes |
Corollary 2.15. If \( f : X \rightarrow Y \) is a surjection, then \( B \mapsto {f}^{-1}\left( B\right) \) is an injection \( \mathcal{P}\left( Y\right) \rightarrow \mathcal{P}\left( X\right) \), where \( \mathcal{P}\left( Y\right) \) denotes the family of all the subsets of \( Y \) . | Proof. If \( B, C \subseteq Y \) and \( {f}^{-1}\left( B\right) = {f}^{-1}\left( C\right) \), then Proposition 2.14(ii) gives\n\n\[ B = f{f}^{-1}\left( B\right) = f{f}^{-1}\left( C\right) = C.\; \bullet \] | Yes |
Lemma 2.19. If \( \equiv \) is an equivalence relation on a set \( X \), then \( x \equiv y \) if and only if \( \left\lbrack x\right\rbrack = \left\lbrack y\right\rbrack \) . | Proof. Assume that \( x \equiv y \) . If \( z \in \left\lbrack x\right\rbrack \), then \( z \equiv x \), and so transitivity gives \( z \equiv y \) ; hence \( \left\lbrack x\right\rbrack \subseteq \left\lbrack y\right\rbrack \) . By symmetry, \( y \equiv x \), and this gives the reverse inclusion \( \left\lbrack y\righ... | Yes |
If \( \equiv \) is an equivalence relation on a set \( X \), then the equivalence classes form a partition of \( X \) . Conversely, given a partition \( \mathcal{P} \) of \( X \), there is an equivalence relation on \( X \) | Proof. Assume that an equivalence relation \( \equiv \) on \( X \) is given. Each \( x \in X \) lies in the equivalence class \( \left\lbrack x\right\rbrack \) because \( \equiv \) is reflexive; it follows that the equivalence classes are nonempty subsets whose union is \( X \) . To prove pairwise disjointness, assume ... | Yes |
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