Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Corollary 13.4.7 Every finite Boolean algebra \( \mathcal{B} = \left\lbrack {B;\vee ,\land , - }\right\rbrack \) has \( {2}^{n} \) elements for some positive integer \( n \) . | Proof. Let \( A \) be the set of all atoms of \( \mathcal{B} \) and let \( \left| A\right| = n \) . Then there are exactly \( {2}^{n} \) elements (subsets) in \( \mathcal{P}\left( A\right) \), and by Theorem 13.4.6, \( \left\lbrack {B;\vee ,\land , - }\right\rbrack \) is isomorphic to \( \left\lbrack {\mathcal{P}\left(... | Yes |
All Boolean algebras of order \( {2}^{n} \) are isomorphic to one another. | Every Boolean algebra of order \( {2}^{n} \) is isomorphic to \( \left\lbrack {\mathcal{P}\left( A\right) ;\cup ,\cap ,{}^{c}}\right\rbrack \) when \( \left| A\right| = n \) . Hence, if \( {\mathcal{B}}_{1} \) and \( {\mathcal{B}}_{2} \) each have \( {2}^{n} \) elements, they each have \( n \) atoms. Suppose their sets... | No |
Consider any Boolean algebra of order 2, \(\left\lbrack {B;\vee ,\land , - }\right\rbrack\). How many functions \( f : {B}^{2} \rightarrow B \) are there? | First, all Boolean algebras of order 2 are isomorphic to \(\left\lbrack {{B}_{2};\vee ,\land , - }\right\rbrack\) so we want to determine the number of functions \( f : {B}_{2}^{2} \rightarrow {B}_{2} \). If we consider a Boolean function of two variables, \({x}_{1}\) and \({x}_{2}\), we note that each variable has two... | No |
Theorem 13.6.6 Uniqueness of Minterm Normal Form. Let \( e\left( {{x}_{1},\ldots ,{x}_{k}}\right) \) be a Boolean expression over \( B \) . There exists a unique minterm normal form \( M\left( {{x}_{1},\ldots ,{x}_{k}}\right) \) that is equivalent to \( e\left( {{x}_{1},\ldots ,{x}_{k}}\right) \) in the sense that \( e... | The uniqueness in this theorem does not include the possible ordering of the minterms in \( M \) (commonly referred to as \ | No |
Consider the Boolean expression \( f\left( {{x}_{1},{x}_{2}}\right) = {x}_{1} \vee \overline{{x}_{2}} \) . One method of determining the minterm normal form of \( f \) is to think in terms of sets. Consider the diagram with the usual translation of notation in Figure 13.6.8. | \[ f\left( {{x}_{1},{x}_{2}}\right) = \left( {\overline{{x}_{1}} \land \overline{{x}_{2}}}\right) \vee \left( {{x}_{1} \land \overline{{x}_{2}}}\right) \vee \left( {{x}_{1} \land {x}_{2}}\right) \] \[ = {M}_{00} \vee {M}_{10} \vee {M}_{11} \] | Yes |
Consider the function \( g : {B}_{2}^{3} \rightarrow {B}_{2} \) defined by Table 13.6.9. | The minterm normal form for \( g \) can be obtained by taking the join of minterms that correspond to rows that have an image value of 1 . If \( g\left( {{a}_{1},{a}_{2},{a}_{3}}\right) = \) 1, then include the minterm \( {y}_{1} \land {y}_{2} \land {y}_{3} \) where\n\n\[ \n{y}_{j} = \left\{ \begin{array}{ll} {x}_{j} &... | Yes |
Consider the circuit in Figure 13.7.15. As usual, we assume that three inputs enter on the left and the output exits on the right. If we trace the inputs through the gates we see that this circuit realizes the boolean function \[ f\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = {x}_{1} \cdot \overline{{x}_{2}} \cdot \left( {... | \[ {x}_{1} \cdot \overline{{x}_{2}} \cdot \left( {\left( {{x}_{1} + {x}_{2}}\right) + \left( {{x}_{1} + {x}_{3}}\right) }\right) = {x}_{1} \cdot \overline{{x}_{2}} \cdot \left( {{x}_{1} + {x}_{2} + {x}_{3}}\right) \] \[ = {x}_{1} \cdot \overline{{x}_{2}} \cdot {x}_{1} + {x}_{1} \cdot \overline{{x}_{2}} \cdot {x}_{2} + ... | Yes |
Consider the following table of desired outputs for the three input bits \( {x}_{1},{x}_{2},{x}_{3} \). | The first step is to write the Minterm Normal Form of \( f \). Since we are working with the two value Boolean algebra, \( {B}_{2} \), the constants in each minterm are either 0 or 1, and we simply list the minterms that have a 1. These correspond with the rows of the table above that have an output of 1.\n\n\[ f\left(... | Yes |
Example 14.1.9 \( M = \left\lbrack {\mathcal{P}\{ 1,2,3\} ; \cap }\right\rbrack \) is isomorphic to \( {M}_{2} = \left\lbrack {{\mathbb{Z}}_{2}^{3}; \cdot }\right\rbrack \), where the operation in \( {M}_{2} \) is componentwise mod 2 multiplication. A translation rule is that if \( A \subseteq \{ 1,2,3\} \), then it is... | \[ {d}_{i} = \left\{ \begin{array}{ll} 1 & \text{ if }i \in A \\ 0 & \text{ if }i \notin A \end{array}\right. \] Two cases of how this translation rule works are: \( \{ 1,2,3\} \; \) is the identity for \( {M}_{1}\;\{ 1,2\} \cap \{ 2,3\} = \{ 2\} \) \( \updownarrow \; \updownarrow \) \( \left( {1,1,1}\right) \; \) is t... | No |
Theorem 14.2.3 If \( A \) is countable, then \( {A}^{ * } \) is countable. | Proof. Case 1. Given the alphabet \( B = \{ 0,1\} \), we can define a bijection from the positive integers into \( {B}^{ * } \) . Each positive integer has a binary expansion \( {d}_{k}{d}_{k - 1}\cdots {d}_{1}{d}_{0} \), where each \( {d}_{j} \) is 0 or 1 and \( {d}_{k} = 1 \) . If \( n \) has such a binary expansion,... | Yes |
Theorem 14.2.12 Recursive implies Generating.\n\n(a) If \( A \) is countable, then there exists a generating algorithm for \( {A}^{ * } \) .\n\n(b) If \( L \) is a recursive language over a countable alphabet, then there exists a generating algorithm for \( L \) . | Proof. Part (a) follows from the fact that \( {A}^{ * } \) is countable; therefore, there exists a complete list of strings in \( {A}^{ * } \) .\n\nTo generate all strings of \( L \), start with a list of all strings in \( {A}^{ * } \) and an empty list, \( W \), of strings in \( L \) . For each string \( s \), use a r... | No |
The language over \( B \) consisting of strings of alternating 0 's and 1's is a phrase structure language. It can be defined by the following grammar: | These rules can be visualized with a graph:  Figure 14.2.17 Production rules for the language of alternating 0 's and 1 's We can verify that a string such as 10101 belongs to the language by starting with \( S \) an... | Yes |
Example 14.3.3 A Parity Checking Machine. The following machine is called a parity checker. It recognizes whether or not a string in \( {B}^{ * } \) contains an even number of \( 1\mathrm{\;s} \) . The memory structure of this machine reflects the fact that in order to check the parity of a string, we need only keep tr... | The input alphabet is \( B = \{ 0,1\} \) and the output alphabet is also \( B \) . The state set is \( \{ \) even, odd \( \} \) . The following table defines the output and next-state functions.\n\n<table><thead><tr><th>\( x \)</th><th>\( S \)</th><th>\( w\left( {x, s}\right) \)</th><th>\( t\left( {x, s}\right) \)</th>... | Yes |
Example 14.3.5 A Baseball Machine. Consider the following simplified version of the game of baseball. To be precise, this machine describes one half-inning of a simplified baseball game. Suppose that in addition to home plate, there is only one base instead of the usual three bases. Also, assume that there are only two... | Let's concentrate on one state. If the current state is 01,0 outs and 1 runner on base, each input results in a different combination of output and next-state. If the batter hits the ball poorly (a double play) the output is zero runs and the inning is over (the limit of two outs has been made). A simple out also resul... | No |
Recognition in Regular Languages. As we mentioned at the outset of this section, finite-state machines can recognize strings in a regular language. Consider the language \( L \) over \( \{ a, b, c\} \) that contains the strings of positive length in which each \( a \) is followed by \( b \) and each \( b \) is followed... | A finite-state machine (Figure 14.3.8) that recognizes this language can be constructed with one state for each nonterminal symbol and an additional state (Reject) that is entered if any invalid production takes place. At the end of an input tape that encodes a string in \( \{ a, b, c{\} }^{ * } \), we will know when t... | Yes |
Example 14.4.3 The Unit-time Delay Machine. A finite-state machine called the unit-time delay machine does not echo its current state, but prints its previous state. For this reason, when we find the monoid of the unit-time delay machine, we must consider both state and output. The transition diagram of this machine ap... | \n\nFigure 14.4.4\n\n<table><thead><tr><th>Input</th><th>01</th><th>00</th><th>0110</th><th>11</th><th>100 or000</th><th>101 or001</th><th></th><th>110 or101</th><th>111 or011</th></tr></thead><tr><td>0</td><td>\\( \... | Yes |
We will construct the machine of the monoid \( \left\lbrack {{\mathbb{Z}}_{2};{ + }_{2}}\right\rbrack \) . As mentioned above, the state set and the input set are both \( {\mathbb{Z}}_{2} \) . The next state function is defined by \( t\left( {s, x}\right) = s + {}_{2}x \) . | The transition diagram for \( m\left( {\mathbb{Z}}_{2}\right) \) appears in Figure 14.5.3. Note how it is identical to the transition diagram of the parity checker, which has an associated monoid that was isomorphic to \( \left\lbrack {{\mathbb{Z}}_{2};{ + }_{2}}\right\rbrack \) | No |
Example 14.5.4 The transition diagram of the monoids \( \left\lbrack {{\mathbb{Z}}_{2};{ \times }_{2}}\right\rbrack \) and \( \left\lbrack {{\mathbb{Z}}_{3};{ \times }_{3}}\right\rbrack \) appear in Figure 14.5.5. | \n\nFigure 14.5.5 The machines of \( \left\lbrack {{\mathbb{Z}}_{2};{ \times }_{2}}\right\rbrack \) and \( \left\lbrack {{\mathbb{Z}}_{3};{ \times }_{3}}\right\rbrack \) | Yes |
Example 14.5.6 Let \( U \) be the monoid that we obtained from the unit-time delay machine (Example 14.4.3). We have seen that the machine of the monoid of the parity checker is essentially the parity checker. Will we obtain a unit-time delay machine when we construct the machine of \( U \) ? | We can’t expect to get exactly the same machine because the unit-time delay machine is not a state machine and the machine of a monoid is a state machine. However, we will see that our new machine is capable of telling us what input was received in the previous time period. The operation table for the monoid serves as ... | Yes |
Example 15.1.2 A Finite Cyclic Group. \( {\mathbb{Z}}_{12} = \left\lbrack {{\mathbb{Z}}_{12};{ + }_{12}}\right\rbrack \), where \( { + }_{12} \) is addition modulo 12, is a cyclic group. To verify this statement, all we need to do is demonstrate that some element of \( {\mathbb{Z}}_{12} \) is a generator. | One such element is 5 ; that is, \( \langle 5\rangle = {\mathbb{Z}}_{12} \) . One more obvious generator is 1 . In fact,1 is a generator of every \( \left\lbrack {{\mathbb{Z}}_{n};{ + }_{n}}\right\rbrack \) . The reader is asked to prove that if an element is a generator, then its inverse is also a generator. Thus, \( ... | No |
The additive group of integers, \( \left\lbrack {\mathbb{Z}; + }\right\rbrack \), is cyclic. | \[ \mathbb{Z} = \langle 1\rangle = \{ n \cdot 1 \mid n \in \mathbb{Z}\} \] This observation does not mean that every integer is the product of an integer times 1. It means that \[ \mathbb{Z} = \{ 0\} \cup \{ \overset{n\text{ terms }}{\overbrace{1 + 1 + \cdots + 1}} \mid n \in \mathbb{P}\} \cup \{ \overset{n\text{ terms... | Yes |
Theorem 15.1.5 Cyclic Implies Abelian. If \( \left\lbrack {G; * }\right\rbrack \) is cyclic, then it is abelian. | Proof. Let \( a \) be any generator of \( G \) and let \( b, c \in G \) . By the definition of the generator of a group, there exist integers \( m \) and \( n \) such that \( b = {ma} \) and \( c = {na} \) . Thus, using Theorem 11.3.14,\n\n\[ b * c = \left( {ma}\right) * \left( {na}\right) \]\n\n\[ = \left( {m + n}\rig... | Yes |
Example 15.1.6 A Cyclic Multiplicative Group. The group of positive integers modulo 11 with modulo 11 multiplication, \( \left\lbrack {{\mathbb{Z}}_{11}^{ * };{ \times }_{11}}\right\rbrack \), is cyclic. | One of its generators is \( 6 : {6}^{1} = 6,{6}^{2} = 3,{6}^{3} = 7,\ldots ,{6}^{9} = 2 \), and \( {6}^{10} = 1 \), the identity of the group. | Yes |
A Non-cyclic Group. The real numbers with addition, \( \left\lbrack {\mathbb{R}; + }\right\rbrack \) is a noncyclic group. | The proof of this statement requires a bit more generality since we are saying that for all \( r \in \mathbb{R},\langle r\rangle \) is a proper subset of \( \mathbb{R} \) . If \( r \) is nonzero, the multiples of \( r \) are distributed over the real line, as in Figure 15.1.8. It is clear then that there are many real ... | Yes |
Theorem 15.1.9 Possible Cyclic Group Structures. If \( G \) is a cyclic group, then \( G \) is either finite or countably infinite. If \( G \) is finite and \( \left| G\right| = n \) , it is isomorphic to \( \left\lbrack {{\mathbb{Z}}_{n};{ + }_{n}}\right\rbrack \) . If \( G \) is infinite, it is isomorphic to \( \left... | Proof. Case 1: \( \left| G\right| < \infty \) . If \( a \) is a generator of \( G \) and \( \left| G\right| = n \), define \( \phi : {\mathbb{Z}}_{n} \rightarrow G \) by \( \phi \left( k\right) = {ka} \) for all \( k \in {\mathbb{Z}}_{n} \). Since \( \langle a\rangle \) is finite, we can use the fact that the elements ... | No |
Theorem 15.1.10 Subgroups of Cyclic Groups. Every subgroup of a cyclic group is cyclic. | Proof. Let \( G \) be cyclic with generator \( a \) and let \( H \leq G \) . If \( H = \{ e\}, H \) has \( e \) as a generator. We may now assume that \( \left| H\right| \geq 2 \) and \( a \neq e \) . Let \( m \) be the least positive integer such that \( {ma} \) belongs to \( H \) . This is the key step. It lets us ge... | Yes |
All subgroups of \( {\mathbb{Z}}_{10} \) | The only proper subgroups of \( {\mathbb{Z}}_{10} \) are \( {H}_{1} = \{ 0,5\} \) and \( {H}_{2} = \{ 0,2,4,6,8\} \) . They are both cyclic: \( {H}_{1} = \langle 5\rangle \) , while \( {H}_{2} = \langle 2\rangle = \langle 4\rangle = \langle 6\rangle = \langle 8\rangle \) . The generators of \( {\mathbb{Z}}_{10} \) are ... | Yes |
All subgroups of \( \mathbb{Z} \). With the exception of \( \{ 0\} \), all subgroups of \( \mathbb{Z} \) are isomorphic to \( \mathbb{Z} \). If \( H \leq \mathbb{Z} \), then \( H \) is the cyclic subgroup generated by the least positive element of \( H \). | It is infinite and so by Theorem 15.1.10 it is isomorphic to \( \mathbb{Z} \). | No |
Theorem 15.1.13 The order of elements of a finite cyclic group. If \( G \) is a cyclic group of order \( n \) and \( a \) is a generator of \( G \), the order of \( {ka} \) is \( n/d \) , where \( d \) is the greatest common divisor of \( n \) and \( k \) . | Proof. The proof of this theorem is left to the reader. | No |
Computation of an order in a cyclic group. To compute the order of \( \langle {18}\rangle \) in \( {\mathbb{Z}}_{30} \), we first observe that 1 is a generator of \( {\mathbb{Z}}_{30} \) and \( {18} = {18}\left( 1\right) \) . The greatest common divisor of 18 and 30 is 6 . Hence, the order of \( \langle {18}\rangle \) ... | The greatest common divisor of 18 and 30 is 6 . Hence, the order of \( \langle {18}\rangle \) is \( {30}/6 \), or 5 . | Yes |
Theorem 15.2.6 If \( b \in a * H \), then \( a * H = b * H \), and if \( b \in H * a \), then \( H * a = H * b. \) | Proof. In light of the remark above, we need only prove the first part of this theorem. Suppose that \( x \in a * H \) . We need only find a way of expressing \( x \) as \ | No |
In Figure 15.2.1, you can start at either 1 or 7 and obtain the same path by taking jumps of three tacks in each step. Thus, | \[ 1 + {}_{12}\{ 0,3,6,9\} = 7 + {}_{12}\{ 0,3,6,9\} = \{ 1,4,7,{10}\} .\] | Yes |
Theorem 15.2.8 Cosets Partition a Group. If \( \left\lbrack {G; * }\right\rbrack \) is a group and \( H \leq G \), the set of left cosets of \( H \) is a partition of \( G \) . In addition, all of the left cosets of \( H \) have the same cardinality. The same is true for right cosets. | Proof. That every element of \( G \) belongs to a left coset is clear because \( a \in a * H \) for all \( a \in G \) . If \( a * H \) and \( b * H \) are left cosets, we will prove that they are either equal or disjoint. If \( a * H \) and \( b * H \) are not disjoint, \( a * H \cap b * H \) is nonempty and some eleme... | Yes |
Corollary 15.2.9 A Coset Counting Formula. If \( \left| G\right| < \infty \) and \( H \leq G \) , the number of distinct left cosets of \( H \) equals \( \frac{\left| G\right| }{\left| H\right| } \) . For this reason we use \( G/H \) to denote the set of left cosets of \( H \) in \( G \) | Proof. This follows from the partitioning of \( G \) into equal sized sets, one of which is \( H \) . | No |
Consider the cosets described in Example 15.2.10. For brevity, we rename \( 0 + 4\mathbb{Z},1 + 4\mathbb{Z},2 + 4\mathbb{Z} \) , and \( 3 + 4\mathbb{Z} \) with the symbols \( \overline{0},\overline{1},\overline{2} \), and \( \overline{3} \) . Let’s do a typical calculation, \( \overline{1} + \overline{3} \) . We will s... | For example, \( 9 \in \overline{1},7 \in \overline{3} \), and \( 9 + 7 = {16} \in \overline{0} \) . Our choice of the representatives \( \overline{1} \) and \( \overline{3} \) were completely arbitrary. | Yes |
Consider the group of real numbers, \( \left\lbrack {\mathbb{R}; + }\right\rbrack \), and its subgroup of integers, \( \mathbb{Z} \). Every element of \( \mathbb{R}/\mathbb{Z} \) has the same cardinality as \( \mathbb{Z} \). Let \( s, t \in \mathbb{R} \). \( s \in t + \mathbb{Z} \) if \( s \) can be written \( t + n \)... | Now consider the coset \( {0.25} + \mathbb{Z} \). Real numbers that differ by an integer from 0.25 are \( {1.25},{2.25},{3.25},\ldots \) and \( - {0.75}, - {1.75}, - {2.75},\ldots \) If any real number is selected, there exists a representative of its coset that is greater than or equal to 0 and less than 1. We will ca... | Yes |
Consider the group \( {\mathbb{Z}}_{2}{}^{4} = {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) . Let \( H \) be \( \langle \left( {1,0,1,0}\right) \rangle \), the cyclic subgroup of \( {\mathbb{Z}}_{2}{}^{4} \) generate by \( \left( {1,0,1,0}\right) \) . | Since\n\n\[ \left( {1,0,1,0}\right) + \left( {1,0,1,0}\right) = \left( {1{ + }_{2}1,0{ + }_{2}0,1{ + }_{2}1,0{ + }_{2}0}\right) = \left( {0,0,0,0}\right) \]\n\nthe order of \( H \) is 2 and, \( {\mathbb{Z}}_{2}{}^{4}/H \) has \( \left| {{\mathbb{Z}}_{2}^{4}/H}\right| = \frac{\left| {\mathbb{Z}}_{2}^{4}\right| }{\left| ... | Yes |
Theorem 15.2.18 Coset operation is well-defined (Abelian Case). If \( G \) is an abelian group, and \( H \leq G \), the operation induced on cosets of \( H \) by the operation of \( G \) is well defined. | Proof. Suppose that \( a, b \), and \( {a}^{\prime },{b}^{\prime } \) . are two choices for representatives of cosets \( C \) and \( D \) . That is to say that \( a,{a}^{\prime } \in C, b,{b}^{\prime } \in D \) . We will show that \( a * b \) and \( {a}^{\prime } * {b}^{\prime } \) are representatives of the same coset... | Yes |
Theorem 15.2.19 Let \( G \) be a group and \( H \leq G \) . If the operation induced on left cosets of \( H \) by the operation of \( G \) is well defined, then the set of left cosets forms a group under that operation. | Proof. Let \( {C}_{1},{C}_{2} \), and \( {C}_{3} \) be the left cosets with representatives \( {r}_{1},{r}_{2} \), and \( {r}_{3} \) , respectively. The values of \( {C}_{1} \otimes \left( {{C}_{2} \otimes {C}_{3}}\right) \) and \( \left( {{C}_{1} \otimes {C}_{2}}\right) \otimes {C}_{3} \) are determined by \( {r}_{1} ... | Yes |
The significance of \( {S}_{3} \) . Our opening example, \( {S}_{3} \), is the smallest non-abelian group. For that reason, all of its proper subgroups are abelian: in fact, they are all cyclic. | Figure 15.3.7 shows the Hasse diagram for the subgroups of \( {S}_{3} \). | No |
The only abelian symmetric groups are \( {S}_{1} \) and \( {S}_{2} \), with 1 and 2 elements, respectively. | The elements of \( {S}_{2} \) are \( i = \left( \begin{array}{ll} 1 & 2 \\ 1 & 2 \end{array}\right) \) and \( \alpha = \left( \begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right) .{S}_{2} \) is isomorphic to \( {\mathbb{Z}}_{2} \). | No |
Theorem 15.3.9 For \( n \geq 1,\left| {S}_{n}\right| = n \) ! and for \( n \geq 3,{S}_{n} \) is non-abelian. | Proof. The first part of the theorem follows from the extended rule of products (see Chapter 2). We leave the details of proof of the second part to the reader after the following hint. Consider \( f \) in \( {S}_{n} \) where \( f\left( 1\right) = 2, f\left( 2\right) = 3, f\left( 3\right) = 1 \) , and \( f\left( j\righ... | No |
Theorem 15.3.16 Decomposition into Cycles. Every cycle of length greater than 2 can be expressed as a product of transpositions. | Proof. We need only indicate how the product of transpositions can be obtained. It is easy to verify that a cycle of length \( k,\left( {{a}_{1},{a}_{2},{a}_{3},\ldots ,{a}_{k}}\right) \), is equal to the following product of \( k - 1 \) transpositions:\n\n\[ \left( {{a}_{1},{a}_{k}}\right) \cdots \left( {{a}_{1},{a}_{... | Yes |
Theorem 15.3.19 Let \( n \geq 2 \) . The alternating group is indeed a group and has order \( \frac{n!}{2} \) . | Proof. In this proof, the symbols \( {s}_{i} \) and \( {t}_{i} \) stand for transpositions and \( p, q \) are even nonnegative integers. If \( f, g \in {A}_{n} \), we can write the two permutations as products of even numbers of transpositions, \( f = {s}_{1}{s}_{2}\cdots {s}_{p} \) and \( g = {t}_{1}{t}_{2}\cdots {t}_... | Yes |
Example 15.3.20 The Sliding Tile Puzzle. Consider the sliding-tile puzzles pictured in Figure 15.3.21. Each numbered square is a tile and the dark square is a gap. Any tile that is adjacent to the gap can slide into the gap. In most versions of this puzzle, the tiles are locked into a frame so that they can be moved on... | We will associate a change in the configuration of the puzzle with an element of \( {S}_{16} \) . Imagine that a tile numbered 16 fills in the gap. For any configuration of the puzzle, the identity \( i \), is the function that leave the configurate \ | No |
Cosets of \( {A}_{3} \) . We have seen that \( {A}_{3} = \left\{ {i,{r}_{1},{r}_{2}}\right\} \) is a subgroup of \( {S}_{3} \), and its left cosets are \( {A}_{3} \) itself and \( {B}_{3} = \left\{ {{f}_{1},{f}_{2},{f}_{3}}\right\} \) . Whether \( \left\{ {{A}_{3},{B}_{3}}\right\} \) is a group boils down to determinin... | Consider the operation table for \( {S}_{3} \) in Figure 15.4.2. We have shaded in all occurrences of the elements of \( {B}_{3} \) in gray. We will call these elements the gray elements and the elements of \( {A}_{3} \) the white ones. Now consider the process of computing the coset product \( {A}_{3} \circ {B}_{3} \)... | No |
Example 15.4.3 Cosets of another subgroup of \( {S}_{3} \) . Now let’s try the left cosets of \( \left\langle {f}_{1}\right\rangle \) in \( {S}_{3} \) . There are three of them. Will we get a complicated version of \( {\mathbb{Z}}_{3} \) ? The left cosets are \( {C}_{0} = \left\langle {f}_{1}\right\rangle ,{C}_{1} = {r... | The reader might be expecting something to go wrong eventually, and here it is. To determine \( {C}_{1} \circ {C}_{2} \) we can choose from four pairs of representatives:\n\n\[ \n{r}_{1} \in {C}_{1},{r}_{2} \in {C}_{2} \rightarrow {r}_{1} \circ {r}_{2} = i \in {C}_{0} \]\n\n\[ \n{r}_{1} \in {C}_{1},{f}_{2} \in {C}_{2} ... | Yes |
Theorem 15.4.6 If \( H \leq G \), then the operation induced on left cosets of \( H \) by the operation of \( G \) is well defined if and only if any one of the following conditions is true:\n\n(a) \( H \) is a normal subgroup of \( G \) .\n\n(b) If \( h \in H, a \in G \), then there exists \( {h}^{\prime } \in H \) su... | Proof. We leave the proof of this theorem to the reader. | No |
A non-normal subgroup. The right cosets of \( \left\langle {f}_{1}\right\rangle \leq {S}_{3} \) are \( \left\{ {i,{f}_{1}}\right\} ,\left\{ {{r}_{1}{f}_{2}}\right\} \), and \( \left\{ {{r}_{2},{f}_{3}}\right\} \) . These are not the same as the left cosets of \( \left\langle {f}_{1}\right\rangle \) . | In addition, \( {f}_{2}{}^{-1}{f}_{1}{f}_{2} = {f}_{2}{f}_{1}{f}_{2} = {f}_{3} \notin \left\langle {f}_{1}\right\rangle \) . Thus, \( \left\langle {f}_{1}\right\rangle \) is not normal. | Yes |
Subgroups of \( {A}_{5} \). \( {A}_{5} \), a group in its own right with 60 elements, has many proper subgroups, but none are normal. | Although this could be done by brute force, the number of elements in the group would make the process tedious. A far more elegant way to approach the verification of this statement is to use the following fact about the cycle structure of permutations. If \( f \in {S}_{n} \) is a permutation with a certain cycle struc... | No |
Define \( \alpha : {\mathbb{Z}}_{6} \rightarrow {\mathbb{Z}}_{3} \) by \( \alpha \left( n\right) = \) \( n{\;\operatorname{mod}\;3} \). Therefore, \( \alpha \left( 0\right) = 0,\alpha \left( 1\right) = 1,\alpha \left( 2\right) = 2,\alpha \left( 3\right) = 1 + 1 + 1 = 0 \), \( \alpha \left( 4\right) = 1 \), and \( \alph... | \[ \alpha \left( {n + {6m}}\right) = \alpha \left( n\right) + {3\alpha }\left( m\right) \] (15.4.1) but we will use a line of reasoning that generalizes. We have already encountered the Chinese Remainder Theorem, which implies that the function \( \beta : {\mathbb{Z}}_{6} \rightarrow {\mathbb{Z}}_{3} \times {\mathbb{Z}... | Yes |
Theorem 15.4.14 Group Homomorphism Properties. If \( \theta : G \rightarrow {G}^{\prime } \) is a homomorphism, then:\n\n(a) \( \theta \left( e\right) = \theta \) (the identity of \( G \) ) \( = \) the identity of \( {G}^{\prime } = {e}^{\prime } \) .\n\n(b) \( \theta \left( {a}^{-1}\right) = \theta {\left( a\right) }^... | Proof.\n\n(a) Let \( a \) be any element of \( G \) . Then \( \theta \left( a\right) \in {G}^{\prime } \).\n\n\( \theta \left( a\right) \diamond {e}^{\prime } = \theta \left( a\right) \; \) by the definition of \( {e}^{\prime } \)\n\n\( = \theta \left( {a * e}\right) \; \) by the definition of \( e \)\n\n\( = \theta \l... | Yes |
If we define \( \pi : \mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z} \) by \( \pi \left( n\right) = n + 4\mathbb{Z} \), then \( \pi \) is a homomorphism. | The image of the subgroup \( 4\mathbb{Z} \) is the single coset \( 0 + 4\mathbb{Z} \), the identity of the factor group. Homomorphisms of this type are called natural homomorphisms. The following theorems will verify that \( \pi \) is a homomorphism and also show the connection between homomorphisms and normal subgroup... | No |
Theorem 15.4.17 If \( H \vartriangleleft G \), then the function \( \pi : G \rightarrow G/H \) defined by \( \pi \left( a\right) = {aH} \) is a homomorphism. | Proof. We leave the proof of this theorem to the reader. | No |
Theorem 15.4.20 Let \( \theta : G \rightarrow {G}^{\prime } \) be a homomorphism from \( G \) into \( {G}^{\prime } \). The kernel of \( \theta \) is a normal subgroup of \( G \). | Proof. Let \( K = \ker \theta \). We can see that \( K \) is a subgroup of \( G \) by letting \( a, b \in K \) and verify that \( a * {b}^{-1} \in K \) by computing \( \theta \left( {a * {b}^{-1}}\right) = \theta \left( a\right) * \theta {\left( b\right) }^{-1} = {e}^{\prime } * {e}^{\prime - 1} = {e}^{\prime } \). To ... | Yes |
Define \( \theta : \mathbb{Z} \rightarrow {\mathbb{Z}}_{10} \) by \( \theta \left( n\right) = n{\;\operatorname{mod}\;{10}} \) . The three previous theorems imply the following: | - \( \pi : \mathbb{Z} \rightarrow \mathbb{Z}/{10}\mathbb{Z} \) defined by \( \pi \left( n\right) = n + {10}\mathbb{Z} \) is a homomorphism.\n- \( \{ n \in \mathbb{Z} \mid \theta \left( n\right) = 0\} = \{ {10n} \mid n \in \mathbb{Z}\} = {10}\mathbb{Z} \vartriangleleft \mathbb{Z} \).\n- \( \mathbb{Z}/{10}\mathbb{Z} \) i... | Yes |
Let \( G \) be the same group of two by two invertible real matrices as in Example 15.4.11. Define \( \Phi : G \rightarrow G \) by \( \Phi \left( A\right) = \frac{A}{\sqrt{\left| \det A\right| }} \). We will let the reader verify that \( \Phi \) is a homomorphism. The theorems above imply the following. | \[ \text{-}\ker \Phi = \{ A \in G \mid \Phi \left( A\right) = I\} = \left\{ {\left. \left( \begin{array}{ll} a & 0 \\ 0 & a \end{array}\right) \right| \;a \in \mathbb{R}, a \neq 0}\right\} \vartriangleleft G\text{. This} \] verifies our statement in Example 15.4.11. As in that example, let \( \ker \Phi = \) \( {H}_{1} ... | No |
Consider \( \Phi : {\mathbb{Z}}_{2}{}^{2} \rightarrow {\mathbb{Z}}_{2}{}^{3} \) defined by \( \Phi \left( {a, b}\right) = \left( {a, b, a{ + }_{2}b}\right) \) . If \( \left( {{a}_{1},{b}_{1}}\right) ,\left( {{a}_{2},{b}_{2}}\right) \in {\mathbb{Z}}_{2}{}^{2} \) | \[ \Phi \left( {\left( {{a}_{1},{b}_{1}}\right) + \left( {{a}_{2},{b}_{2}}\right) }\right) = \Phi \left( {{a}_{1}{ + }_{2}{a}_{2},{b}_{1}{ + }_{2}{b}_{2}}\right) \] \[ = \left( {{a}_{1}{ + }_{2}{a}_{2},{b}_{1}{ + }_{2}{b}_{2},{a}_{1}{ + }_{2}{a}_{2}{ + }_{2}{b}_{1}{ + }_{2}{b}_{2}}\right) \] \[ = \left( {{a}_{1},{b}_{1... | Yes |
Theorem 15.5.3 There is a system of distinguished representatives of \( {\mathbb{Z}}_{2}{}^{6}/W \) such that each of the six-bit blocks having a single 1 is a distinguished representative of its own coset. | Now we can describe the error-correcting process. First match each of the blocks with a single 1 with its syndrome. In addition, match the identity of \( W \) with the syndrome \( \left( {0,0,0}\right) \) as in the table below. Since there are eight cosets of \( W \), select any representative of the eighth coset to be... | No |
The ring of integers. \( \left\lbrack {\mathbb{Z};+, \cdot }\right\rbrack \) is a ring, where + and - stand for regular addition and multiplication on \( \mathbb{Z} \) . | From Chapter 11, we already know that \( \left\lbrack {\mathbb{Z}; + }\right\rbrack \) is an abelian group, so we need only check parts 2 and 3 of the definition of a ring. From elementary algebra, we know that the associative law under multiplication and the distributive laws are true for \( \mathbb{Z} \) . This is ou... | No |
The ring of integers modulo \( n.\left\lbrack {{\mathbb{Z}}_{n};{ + }_{n},{ \times }_{n}}\right\rbrack \) is a ring. | The properties of modular arithmetic on \( {\mathbb{Z}}_{n} \) were described in Section 11.4, and they give us the information we need to convince ourselves that \( \left\lbrack {{\mathbb{Z}}_{n};{ + }_{n},{ \times }_{n}}\right\rbrack \) is a ring. | No |
To determine the unity in the ring \( {\mathbb{Z}}_{4} \times {\mathbb{Z}}_{3} \), we look for the element \( \left( {m, n}\right) \) such that for all elements \( \left( {a, b}\right) \in {\mathbb{Z}}_{4} \times {\mathbb{Z}}_{3},\left( {a, b}\right) = \left( {a, b}\right) \cdot \left( {m, n}\right) = \left( {m, n}\rig... | So we want \( m \) such that \( a{ \times }_{4}m = m{ \times }_{4}a = a \) in the ring \( {\mathbb{Z}}_{4} \) . The only element \( m \) in \( {\mathbb{Z}}_{4} \) that satisfies this equation is \( m = 1 \) . Similarly, we obtain value of 1 for \( n \) . So the unity of \( {\mathbb{Z}}_{4} \times {\mathbb{Z}}_{3} \), w... | No |
The equation \( {2x} = 3 \) has a solution in the ring \( \left\lbrack {\mathbb{Q};+, \cdot }\right\rbrack \) but does not have a solution in \( \left\lbrack {\mathbb{Z};+, \cdot }\right\rbrack \) | since, to solve this equation, we multiply both sides of the equation \( {2x} = 3 \) by the multiplicative inverse of 2 . This number, \( {2}^{-1} \) exists in \( \mathbb{Q} \) but does not exist in \( \mathbb{Z} \) | Yes |
Let us find the multiplicative inverses, when they exist, of each element of the ring \( \left\lbrack {{\mathbb{Z}}_{6};{ + }_{6},{ \times }_{6}}\right\rbrack \) . If \( u = 3 \), we want an element \( v \) such that \( u{ \times }_{6}v = 1 \) . | We do not have to check whether \( v{ \times }_{6}u = 1 \) since \( {\mathbb{Z}}_{6} \) is commutative. If we try each of the six elements, \( 0,1,2,3,4 \), and 5, of \( {\mathbb{Z}}_{6} \), we find that none of them satisfies the above equation, so 3 does not have a multiplicative inverse in \( {\mathbb{Z}}_{6} \) . | Yes |
Consider the rings \( \left\lbrack {\mathbb{Z};+, \cdot }\right\rbrack \) and \( \left\lbrack {2\mathbb{Z};+, \cdot }\right\rbrack \) . In Chapter 11 we showed that as groups, the two sets \( \mathbb{Z} \) and \( 2\mathbb{Z} \) with addition were isomorphic. The group isomorphism that proved this was the function \( f ... | We need only check whether \( f\left( {m \cdot n}\right) = \) \( f\left( m\right) \cdot f\left( n\right) \) for all \( m, n \in \mathbb{Z} \) . In fact, this condition is not satisfied:\n\n\[ f\left( {m \cdot n}\right) = 2 \cdot m \cdot n\;\text{ and }\;f\left( m\right) \cdot f\left( n\right) = {2m} \cdot {2n} = 4 \cdo... | Yes |
Next consider whether \( \left\lbrack {2\mathbb{Z};+, \cdot }\right\rbrack \) and \( \left\lbrack {3\mathbb{Z};+, \cdot }\right\rbrack \) are isomorphic. | The equation \( x + x = x \cdot x \), or \( {2x} = {x}^{2} \), makes sense in both rings. However, this equation has a nonzero solution, \( x = 2 \), in \( 2\mathbb{Z} \), but does not have a nonzero solution in \( 3\mathbb{Z} \) . Thus we have an equation solvable in one ring that cannot be solved in the other, so the... | Yes |
The set of even integers, \( 2\mathbb{Z} \), is a subring of the ring \( \left\lbrack {\mathbb{Z};+, \cdot }\right\rbrack \) since \( \left\lbrack {2\mathbb{Z}; + }\right\rbrack \) is a subgroup of the group \( \left\lbrack {\mathbb{Z}; + }\right\rbrack \) and since it is also closed with respect to multiplication: | \[ {2m},{2n} \in 2\mathbb{Z} \Rightarrow \left( {2m}\right) \cdot \left( {2n}\right) = 2\left( {2 \cdot m \cdot n}\right) \in 2\mathbb{Z} \] | Yes |
Theorem 16.1.18 Some Basic Properties. Let \( \left\lbrack {R;+, \cdot }\right\rbrack \) be a ring, with \( a, b \in R \) . Then\n\n(1) \( a \cdot 0 = 0 \cdot a = 0 \)\n\n(2) \( a \cdot \left( {-b}\right) = \left( {-a}\right) \cdot b = - \left( {a \cdot b}\right) \)\n\n(3) \( \left( {-a}\right) \cdot \left( {-b}\right)... | Proof.\n\n(1) \( a \cdot 0 = a \cdot \left( {0 + 0}\right) = a \cdot 0 + a \cdot 0 \), the last equality valid by the left distributive axiom. Hence if we add \( - \left( {a \cdot 0}\right) \) to both sides of the equality above, we obtain \( a \cdot 0 = 0 \) . Similarly, we can prove that \( 0 \cdot a = 0 \).\n\n(2) B... | No |
We will compute \( 2 \cdot \left( {-2}\right) \) in the ring \( \left\lbrack {{\mathbb{Z}}_{6};{ + }_{6},{ \times }_{6}}\right\rbrack . | 2{ \times }_{6}\left( {-2}\right) = - \left( {2{ \times }_{6}2}\right) = - 4 = 2, since the additive inverse of 4 (mod 6) is 2. Of course, we could have done the calculation directly as \( 2{ \times }_{6}\left( {-2}\right) = 2{ \times }_{6}4 = 2 \) | Yes |
Theorem 16.1.22 Multiplicative Cancellation. The multiplicative cancellation laws hold in a ring \( \\left\\lbrack {R;+, \\cdot }\\right\\rbrack \) if and only if \( R \) has no zero divisors. | Proof. We prove the theorem using the left cancellation axiom, namely that if \( a \\neq 0 \) and \( a \\cdot b = a \\cdot c \), then \( b = c \) for all \( a, b, c \\in R \). The proof using the right cancellation axiom is its mirror image.\n\n\\( \\left( \\Rightarrow \\right) \\) Assume the left cancellation law hold... | Yes |
Both \( \left\lbrack {{\mathbb{Z}}_{2};{ + }_{2},{ \times }_{2}}\right\rbrack \) and \( \left\lbrack {{\mathbb{Z}}_{3};{ + }_{3},{ \times }_{3}}\right\rbrack \) are integral domains. Consider the direct product \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \). It’s true that \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \)... | However, \( \left( {1,0}\right) \cdot \left( {0,2}\right) = \left( {0,0}\right) \), so \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \) has zero divisors and is therefore not an integral domain. | Yes |
Theorem 16.2.4 Field \( \Rightarrow \) Integral Domain. Every field is an integral domain. | Proof. The proof is fairly easy and a good exercise, so we provide a hint. Starting with the assumption that \( a \cdot b = 0 \) if we assume that \( a \neq 0 \) then the existence of \( {a}^{-1} \) makes it possible to infer that \( b = 0 \) . | No |
Theorem 16.2.5 Finite Integral Domain \( \Rightarrow \) Field. Every finite integral domain is a field. | Proof. We leave the details to the reader, but observe that if \( D \) is a finite integral domain, we can list all elements as \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \), where \( {a}_{1} = 1 \) . Now, to show that any \( {a}_{i} \) has a multiplicative inverse, consider the \( n \) products \( {a}_{i} \) . \( {a}_{1},{a}_... | No |
In \( {\mathbb{Z}}_{3}\left\lbrack x\right\rbrack \), if \( f\left( x\right) = 1 + x \) and \( g\left( x\right) = 2 + x \), then | \[ f\left( x\right) + g\left( x\right) = \left( {1 + x}\right) + \left( {2 + x}\right) \] \[ = \left( {1{ + }_{3}2}\right) + \left( {1{ + }_{3}1}\right) x \] \[ = 0 + {2x} \] \[ = {2x} \] and \[ f\left( x\right) g\left( x\right) = \left( {1 + x}\right) \cdot \left( {2 + x}\right) \] \[ = 1{ \times }_{3}2 + \left( {1{ \... | Yes |
Let \( f\left( x\right) = 2 + {x}^{2} \) and \( g\left( x\right) = - 1 + {4x} + 3{x}^{2} \) . We will compute \( f\left( x\right) \cdot g\left( x\right) \) in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . | Using the notation of the above definition, \( {a}_{0} = 2,{a}_{1} = 0 \) , \( {a}_{2} = 1,{b}_{0} = - 1,{b}_{1} = 4 \), and \( {b}_{2} = 3 \) . We want to compute the coefficients \( {d}_{0} \) , \( {d}_{1},{d}_{2},{d}_{3} \), and \( {d}_{4} \) . We will compute \( {d}_{3} \), the coefficient of the \( {x}^{3} \) term... | No |
Let \( f\left( x\right) = 1 + x + {x}^{3} \) and \( g\left( x\right) = 1 + x \) be two polynomials in \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack \). Let us divide \( f\left( x\right) \) by \( g\left( x\right) \). | \[ \frac{{x}^{3} + x + 1}{x + 1} = {x}^{2} + x + \frac{1}{x + 1} \] or equivalently, \[ {x}^{3} + x + 2 = \left( {{x}^{2} + x}\right) \cdot \left( {x + 1}\right) + 1 \] That is, \( f\left( x\right) = g\left( x\right) \cdot q\left( x\right) + r\left( x\right) \) where \( q\left( x\right) = {x}^{2} + x \) and \( r\left( ... | Yes |
Let \( f\left( x\right) = 1 + {x}^{4} \) and \( g\left( x\right) = 1 + x \) be polynomials in \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack \) . Let us divide \( f\left( x\right) \) by \( g\left( x\right) \) : | \[
\begin{matrix} {x}^{3} + {x}^{2} + x + 1 \\ x + 1){x}^{4} + 0{x}^{3} + 0{x}^{2} + {0x} + 1 \end{matrix}
\]
\[
\begin{array}{r} {x}^{4} + {x}^{3} \\ {x}^{3} \end{array}
\]
\[
\begin{array}{r} \frac{{x}^{3} + {x}^{2}}{{x}^{2}} + 1 \end{array}
\]
Thus \( {x}^{4} + 1 = \left( {{x}^{3} + {x}^{2} + x + 1}\right) \left(... | Yes |
Theorem 16.3.13 Division Property for Polynomials. Let \( \left\lbrack {F;+, \cdot }\right\rbrack \) be a field and let \( f\left( x\right) \) and \( g\left( x\right) \) be two elements of \( F\left\lbrack x\right\rbrack \) with \( g\left( x\right) \neq 0 \) . Then there exist unique polynomials \( q\left( x\right) \) ... | Proof. This theorem can be proven by induction on \( \deg f\left( x\right) \) . | No |
Theorem 16.3.14 The Factor Theorem. Let \( \left\lbrack {F;+, \cdot }\right\rbrack \) be a field. An element \( a \in F \) is a zero of \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) if and only if \( x - a \) is a factor of \( f\left( x\right) \) in \( F\left\lbrack x\right\rbrack \) . | Proof.\n\n\( \left( \Rightarrow \right) \) Assume that \( a \in F \) is a zero of \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . We wish to show that \( x - a \) is a factor of \( f\left( x\right) \) . To do so, apply the division property to \( f\left( x\right) \) and \( g\left( x\right) = \) \( x - a \) . ... | No |
Theorem 16.3.15 A nonzero polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) of degree \( n \) can have at most \( n \) zeros. | Proof. Let \( a \in F \) be a zero of \( f\left( x\right) \) . Then \( f\left( x\right) = \left( {x - a}\right) \cdot {q}_{1}\left( x\right) ,{q}_{1}\left( x\right) \in F\left\lbrack x\right\rbrack \), by the Factor Theorem. If \( b \in F \) is a zero of \( {q}_{1}\left( x\right) \), then again by Factor Theorem, \( f\... | Yes |
The polynomial \( f\left( x\right) = {x}^{4} + 1 \) is reducible over \( {\mathbb{Z}}_{2} \) | since \( {x}^{4} + 1 = \left( {x + 1}\right) \left( {{x}^{3} + {x}^{2} + x - 1}\right) \) | No |
Is the polynomial \( f\left( x\right) = {x}^{3} + x + 1 \) reducible over \( {\mathbb{Z}}_{2} \) ? | Since a factorization of a cubic polynomial can only be as a product of linear and quadratic factors, or as a product of three linear factors, \( f\left( x\right) \) is reducible if and only if it has at least one linear factor. From the Factor Theorem, \( x - a \) is a factor of \( {x}^{3} + x + 1 \) over \( {\mathbb{... | Yes |
Example 16.4.1 Extending the Rational Numbers. Let \( f\left( x\right) = {x}^{2} - 2 \in \) \( \mathbb{Q}\left\lbrack x\right\rbrack \) . It is important to remember that we are considering \( {x}^{2} - 2 \) over \( \mathbb{Q} \), no other field. We would like to find all zeros of \( f\left( x\right) \) and the smalles... | By the last condition \( \sqrt{2} \) must be an element of \( S \), and, if \( S \) is to be a field, the sum, product, difference, and quotient of elements in \( S \) must be in \( S \) . So operations involving this number, such as \( \sqrt{2},{\left( \sqrt{2}\right) }^{2},{\left( \sqrt{2}\right) }^{3},\sqrt{2} + \sq... | No |
Extending \( {\mathbb{Z}}_{2} \) . Consider the polynomial \( g\left( x\right) = {x}^{2} + x + 1 \in \) \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack \) . Let’s repeat the steps from the previous example to factor \( g\left( x\right) \) . First, \( g\left( 0\right) = 1 \) and \( g\left( 1\right) = 1 \), so none of the... | \[ {a}^{3} = {a}^{2} \cdot a \] \[ = \left( {a + 1}\right) \cdot a \] \[ = {a}^{2} + a \] \[ = \left( {a + 1}\right) + a \] \[ = 1 \] Therefore, \( {a}^{-1} = a + 1 = {a}^{2} \) and \( {\left( a + 1\right) }^{-1} = a \) . It is not difficult to see that \( {a}^{n} \) is in \( S \) for all positive \( n \) . Does \( S \... | Yes |
An Error Correcting Polynomial Code. An important observation regarding the previous example is that the nonzero elements of \( {GF}\left( 4\right) \) can be represented two ways. First as a linear combination of 1 and \( a \) . There are four such linear combinations, one of which is zero. Second, as powers of \( a \)... | \[ \n{a}^{0} = 1 \cdot 1 + 0 \cdot a \n\] \n\[ \n{a}^{1} = 0 \cdot 1 + 1 \cdot a \n\] \n\[ \n{a}^{2} = 1 \cdot 1 + 1 \cdot a \n\] \n\nNext, we briefly describe the field \( {GF}\left( 8\right) \) and how an error correcting code can be build on a the same observation about that field. \n\nFirst, we start with the irred... | No |
Let\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }i{x}^{i} = 0 + {1x} + 2{x}^{2} + 3{x}^{3} + \cdots \;\text{ and }\n\]\n\n\[ g\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }{2}^{i}{x}^{i} = 1 + {2x} + 4{x}^{2} + 8{x}^{3} + \cdots \n\]\n\nbe elements in \( \mathbb{Z}\left\lbrack \left\lb... | First the sum:\n\n\[ f\left( x\right) + g\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }i{x}^{i} + \mathop{\sum }\limits_{{i = 0}}^{\infty }{2}^{i}{x}^{i} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{\infty }\left( {i + {2}^{i}}\right) {x}^{i} \]\n\n\[ = 1 + {3x} + 6{x}^{2} + {11}{x}^{3} + \cdots \]\n\nThe p... | Yes |
Theorem 16.5.6 Polynomial Units. Let \( \\left\\lbrack {F;+, \\cdot }\\right\\rbrack \) be a field. Polynomial \( f\\left( x\\right) \) is a unit in \( F\\left\\lbrack x\\right\\rbrack \) if and only if it is a nonzero constant polynomial. | Proof.\n\n\\( \\left( \\Rightarrow \\right) \\) Let \\( f\\left( x\\right) \\) be a unit in \\( F\\left\\lbrack x\\right\\rbrack \\) . Then \\( f\\left( x\\right) \\) has a multiplicative inverse, call it \\( g\\left( x\\right) \\), such that \\( f\\left( x\\right) \\cdot g\\left( x\\right) = 1 \\) . Hence, the \\( \\d... | Yes |
Theorem 16.5.7 Power Series Units. Let \( \\left\\lbrack {F;+, \\cdot }\\right\\rbrack \) be a field. Then \( f\\left( x\\right) = \\) \( \\mathop{\\sum }\\limits_{{i = 0}}^{\\infty }{a}_{i}{x}^{i} \) is a unit of \( F\\left\\lbrack \\left\\lbrack x\\right\\rbrack \\right\\rbrack \) if and only if \( {a}_{0} \\neq 0 \)... | Proof.\n\n\\( \\left( \\Rightarrow \\right) \\) If \( f\\left( x\\right) \\) is a unit of \( F\\left\\lbrack \\left\\lbrack x\\right\\rbrack \\right\\rbrack \\), then there exists \( g\\left( x\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{\\infty }{b}_{i}{x}^{i} \\) in \( F\\left\\lbrack \\left\\lbrack x\\right\\rbrac... | Yes |
Let \( f\left( x\right) = 1 + {2x} + 3{x}^{2} + 4{x}^{3} + \cdots = \mathop{\sum }\limits_{{i = 0}}^{\infty }\left( {i + 1}\right) {x}^{i} \) be an element of \( \mathbb{Q}\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \) . Then, by Theorem 16.5.7, since \( {a}_{0} = 1 \neq 0, f\left( x\right) \) is a unit and ... | Using the formulas for the \( {b}_{i}^{\prime }\mathrm{s} \), we obtain\n\n\[ \n{b}_{0} = 1 \n\]\n\n\[ \n{b}_{1} = - 1\left( {2 \cdot 1}\right) = - 2 \n\]\n\n\[ \n{b}_{2} = - 1\left( {2 \cdot \left( {-2}\right) + 3 \cdot 1}\right) = 1 \n\]\n\n\[ \n{b}_{3} = - 1\left( {2 \cdot 1 + 3 \cdot \left( {-2}\right) + 4 \cdot 1}... | Yes |
Consider a circle of radius 5 in 2-space centered at the origin. We know that we can parameterize this circle as\n\n\[ \mathbf{r}\left( t\right) = \langle 5\cos \left( t\right) ,5\sin \left( t\right) \rangle \]\n\nwhere \( t \) runs from 0 to \( {2\pi } \) . | We see that \( {\mathbf{r}}^{\prime }\left( t\right) = \langle - 5\sin \left( t\right) ,5\cos \left( t\right) \rangle \), and hence \( \left| {{\mathbf{r}}^{\prime }\left( t\right) }\right| = 5 \) . It then follows that\n\n\[ s = L\left( t\right) = {\int }_{0}^{t}\left| {{\mathbf{r}}^{\prime }\left( w\right) }\right| {... | Yes |
Let us parameterize the curve defined by\n\n\\[ \n\\mathbf{r}\\left( t\\right) = \\left\\langle {{t}^{2},\\frac{8}{3}{t}^{3/2},{4t}}\\right\\rangle \n\\]\n\nfor \\( t \\geq 0 \\) in terms of arc length. | To write \\( t \\) in terms of \\( s \\) we find \\( s \\) in terms of \\( t \\) :\n\n\\[ \ns\\left( t\\right) = {\\int }_{0}^{t}\\sqrt{{\\left( {x}^{\\prime }\\left( w\\right) \\right) }^{2} + {\\left( {y}^{\\prime }\\left( w\\right) \\right) }^{2} + {\\left( {z}^{\\prime }\\left( w\\right) \\right) }^{2}}{dw} \n\\]\n... | Yes |
Consider the function \( f \) defined by\n\n\[ f\left( {x, y}\right) = \frac{{x}^{2}{y}^{2}}{{x}^{2} + {y}^{2}}.\]\n\nWe want to know whether \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {0,0}\right) }}f\left( {x, y}\right) \) exists. | Note that if either \( x \) or \( y \) is 0, then \( f\left( {x, y}\right) = 0 \) . Therefore, if \( f \) has a limit at \( \left( {0,0}\right) \), it must be 0 . We will therefore argue that\n\n\[ \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {0,0}\right) }}f\left( {x, y}\right) = 0 \]\n\nby showing ... | Yes |
Suppose we have a machine that manufactures rectangles of width \( x = {20}\mathrm{\;{cm}} \) and height \( y = {10}\mathrm{\;{cm}} \). However, the machine isn’t perfect, and therefore the width could be off by \( {dx} = {\Delta x} = {0.2}\mathrm{\;{cm}} \) and the height could be off by \( {dy} = {\Delta y} = {0.4}\m... | We will estimate the uncertainty in the area using (10.4.2), and find that \( {\Delta A} \approx {dA} = {A}_{x}\left( {{20},{10}}\right) {dx} + {A}_{y}\left( {{20},{10}}\right) {dy}. \) Since \( {A}_{x} = y \) and \( {A}_{y} = x \), we have \( {\Delta A} \approx {dA} = {10dx} + {20dy} = {10} \cdot {0.2} + {20} \cdot {0... | Yes |
Let \( f\left( {x, y}\right) = {x}^{2}y \) be defined on the triangle \( D \) with vertices \( \left( {0,0}\right) ,\left( {2,0}\right) \), and \( \left( {2,3}\right) \). To evaluate \( {\iint }_{D}f\left( {x, y}\right) {dA} \), we must first describe the region \( D \) in terms of the variables \( x \) and \( y \). | Approach 1: Integrate first with respect to \( y \). In this case we choose to evaluate the double integral as an iterated integral in the form\n\n\[ \n{\iint }_{D}{x}^{2}{ydA} = {\int }_{x = a}^{x = b}{\int }_{y = {g}_{1}\left( x\right) }^{y = {g}_{2}\left( x\right) }{x}^{2}{ydydx} \n\]\n\nand therefore we need to des... | Yes |
Example 11.5.3 Let \( f\left( {x, y}\right) = {e}^{{x}^{2} + {y}^{2}} \) on the disk \( D = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq 1}\right\} \) . We will evaluate \( {\iint }_{D}f\left( {x, y}\right) {dA} \) . | In rectangular coordinates the double integral \( {\iint }_{D}f\left( {x, y}\right) {dA} \) can be written as the iterated integral\n\n\[ \n{\iint }_{D}f\left( {x, y}\right) {dA} = {\int }_{x = - 1}^{x = 1}{\int }_{y = - \sqrt{1 - {x}^{2}}}^{y = \sqrt{1 - {x}^{2}}}{e}^{{x}^{2} + {y}^{2}}{dydx}. \n\]\n\nWe cannot evalua... | Yes |
Example 11.6.1 Consider the torus (or doughnut) shown in Figure 11.6.2. | To find a parametrization of this torus, we recall our work in Preview Activity 11.6.1. There, we saw that a circle of radius \( r \) that has its center at the point \( \left( {0,0,{z}_{0}}\right) \) and is contained in the horizontal plane \( z = {z}_{0} \), as shown in Figure 11.6.3, can be parametrized using the ve... | Yes |
Find the mass of the tetrahedron in the first octant bounded by the coordinate planes and the plane \( x + {2y} + {3z} = 6 \) if the density at point \( \left( {x, y, z}\right) \) is given by \( \delta \left( {x, y, z}\right) = x + y + z \) . | We find the mass, \( M \), of the tetrahedron by the triple integral\n\n\[ M = {\iiint }_{S}\delta \left( {x, y, z}\right) {dV} \]\n\nwhere \( S \) is the solid tetrahedron described above. In this example, we choose to integrate with respect to \( z \) first for the innermost integral. The top of the tetrahedron is gi... | Yes |
Say we have an object, and 5 measurements of its length from the same ruler but from different people, \[ {5.1}\left\lbrack \mathrm{\;{cm}}\right\rbrack ,{4.9}\left\lbrack \mathrm{\;{cm}}\right\rbrack ,{4.7}\left\lbrack \mathrm{\;{cm}}\right\rbrack ,{4.9}\left\lbrack \mathrm{\;{cm}}\right\rbrack ,{5.0}\left\lbrack \mat... | Again, the best estimate should be given by the sample mean of these 5 samples, \[ \widehat{\mu } = \frac{{x}_{1} + {x}_{2} + \cdots + {x}_{N}}{N} \] \[ = \frac{{5.1}\left\lbrack \mathrm{\;{cm}}\right\rbrack + {4.9}\left\lbrack \mathrm{\;{cm}}\right\rbrack + {4.7}\left\lbrack \mathrm{\;{cm}}\right\rbrack + {4.9}\left\l... | Yes |
A study (Murphy and Abbey, Cancer in Families, 1959) addressed the question of whether cancer runs in families. The investigator identified 200 women with breast cancer and another 200 women without breast cancer and asked them whether their mothers had had breast cancer. Of the 400 women in the two groups combined, 10... | The proper way, assuming total initial ignorance, is to use the Beta distribution:\n\n\[ P\left( {{\theta }_{\text{cancer }} \mid \text{ data }}\right) = \operatorname{Beta}\left( {h = 7, N = {10}}\right) \]\n\nwhich has a median of \( {\widehat{\theta }}_{\text{cancer }} = {0.68} \), but a \( {95}\% \) credible interv... | Yes |
Consider \( E = A\left( {B \cup {C}^{c}}\right) \cup {A}^{c}{\left( B \cup {C}^{c}\right) }^{c} \) and \( F = {A}^{c}{B}^{c} \cup {AC} \) of the example above, and suppose the respective minterm probabilities are \[ {p}_{0} = {0.21},{p}_{1} = {0.06},{p}_{2} = {0.29},{p}_{3} = {0.11},{p}_{4} = {0.09},{p}_{5} = {0.03},{p... | Use of a minterm map shows \( E = M\left( {1,4,6,7}\right) \) and \( F = M\left( {0,1,5,7}\right) \) . so that \[ P\left( E\right) = {p}_{1} + {p}_{4} + {p}_{6} + {p}_{7} = p\left( {1,4,6,7}\right) = {0.36}\text{ and }P\left( F\right) = p\left( {0,1,5,7}\right) = {0.37} \] | Yes |
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