Search
- Filter Results
- Location
- Classification
- Include attachments
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)This book is intended for a one semester introduction to abstract algebra. We assume that students have some familiarity with basic set theory linear algebra and calculus. But very little of this natu...This book is intended for a one semester introduction to abstract algebra. We assume that students have some familiarity with basic set theory linear algebra and calculus. But very little of this nature will be needed To a great extent the course is self contained except for the requirement of a certain amount of mathematical maturity. And hopefully the student's level of mathematical maturity will increase as the course progresses.
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/11%3A_The_QuaternionsThis means that if we define for \(a \in \mathbb{R}\) and \((x,y,z,w) \in \mathbb{R}^4\) the scalar by vector product \[a(x,y,z,w)=(ax,ay,az,aw),\] the quaternion \(q=(x,y,z,w)\) may be written unique...This means that if we define for \(a \in \mathbb{R}\) and \((x,y,z,w) \in \mathbb{R}^4\) the scalar by vector product \[a(x,y,z,w)=(ax,ay,az,aw),\] the quaternion \(q=(x,y,z,w)\) may be written uniquely in the form \[q=x1+yi+zj+wk.\] Now if we abbreviate \(x=x1\), the quaternion takes the form \[q= x+yi+zj+wk.\] Addition now becomes \[(x+yi+zj+wk) + (a+bi+cj+dk) = (x+a) +(y+b)i+(z+c)j+(w+d)k.\] Products of the basis elements \(1,i,j,k\) are defined as follows: \[1q=q1=q \mbox{ for all } q \in \…
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/02%3A_Appendices/2.02%3A_FunctionsHere we collect a few basic facts about functions. Note that the words function, map, mapping and transformation may be used interchangeably. Here we just use the term function. We leave the proofs of...Here we collect a few basic facts about functions. Note that the words function, map, mapping and transformation may be used interchangeably. Here we just use the term function. We leave the proofs of all the results in this appendix to the interested reader.
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/02%3A_Appendices/2.01%3A_Some_Rules_of_Logic\[\begin{align} P \Longleftrightarrow Q \label{A3} \end{align}\] as an abbreviation for the two statements \[P \Longrightarrow Q \quad \mbox{ and } \quad Q \Longrightarrow P\] So, for example, if you ...\[\begin{align} P \Longleftrightarrow Q \label{A3} \end{align}\] as an abbreviation for the two statements \[P \Longrightarrow Q \quad \mbox{ and } \quad Q \Longrightarrow P\] So, for example, if you need to prove \(P \Longleftrightarrow Q\) you really have two things to prove: both \(P \Longrightarrow Q\) and \(Q \Longrightarrow P\).
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/02%3A_Appendices
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.03%3A_The_Symmetric_GroupsLet \(\sigma\) and \(\tau\) be defined as follows: \[\sigma= \left ( \begin{array} {ccc} 1&2&3\\ 2&1&3 \end{array} \right ), \quad \quad \tau = \left ( \begin{array} {ccc} 1&2&3\\ 2&3&1 \end{array} \r...Let \(\sigma\) and \(\tau\) be defined as follows: \[\sigma= \left ( \begin{array} {ccc} 1&2&3\\ 2&1&3 \end{array} \right ), \quad \quad \tau = \left ( \begin{array} {ccc} 1&2&3\\ 2&3&1 \end{array} \right )\] It follows that \[\begin{array} {c c c c c c c} \sigma\tau(1) &=&\sigma(\tau(1))&=&\sigma(2)&=1 \\ \sigma\tau(2) &=&\sigma(\tau(2))&=&\sigma(3)&=3\\ \sigma\tau(3) &=&\sigma(\tau(3))&=&\sigma(1)&=2 \end{array}\] Thus we have \[\sigma\tau = \left ( \begin{array} {ccc} 1&2&3\\ 1&3&2 \end{arra…
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.09%3A_Introduction_to_Ring_TheoryProblem 9.11 Assume there is a positive element \(\sqrt{2} \in \mathbb{R}\) such that \[(\sqrt{2})^2 =2.\] Define the following subset of \(\mathbb{R}\): \[\mathbb{Q}(\sqrt{2}) = \{ a+b\sqrt{2} \ | \ ...Problem 9.11 Assume there is a positive element \(\sqrt{2} \in \mathbb{R}\) such that \[(\sqrt{2})^2 =2.\] Define the following subset of \(\mathbb{R}\): \[\mathbb{Q}(\sqrt{2}) = \{ a+b\sqrt{2} \ | \ a, b \in \mathbb{Q}\}.\] Prove that \(\mathbb{Q}(\sqrt{2})\) is a subfield of \(\mathbb{R}\). (The tricky part is showing that all non-zero elements are units.)
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.06%3A_Direct_Products_of_GroupsIf \(G_1, G_2, \dots, G_n\) is a list of \(n\) groups we make the Cartesian product \(G_1\times G_2 \times \dots \times G_n\) into a group by defining the binary operation \[(a_1,a_2, \dots, a_n) \cdo...If \(G_1, G_2, \dots, G_n\) is a list of \(n\) groups we make the Cartesian product \(G_1\times G_2 \times \dots \times G_n\) into a group by defining the binary operation \[(a_1,a_2, \dots, a_n) \cdot (b_1, b_2, \dots, b_n) = (a_1 \cdot b_1, a_2 \cdot b_2, \dots, a_n \cdot b_n).\] Here for each \(i \in \{ 1, 2, \dots, n \}\) the product \(a_i \cdot b_i\) is the product of \(a_i\) and \(b_i\) in the group \(G_i\).
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.02%3A_Introduction_to_Groups\((M_2(K),+)\) where \(K\) is any one of \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{Z}_n\) is a group whose identity is the zero matrix \[\nonumber \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{arra...\((M_2(K),+)\) where \(K\) is any one of \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{Z}_n\) is a group whose identity is the zero matrix \[\nonumber \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right ]\] and the inverse of the matrix \[A=\left [ \nonumber \begin{array} {c c} a & b \\ c & d \end{array} \right]\] is the matrix \[\nonumber -A = \left [ \begin{array} {c c} -a & -b \\ -c & -d \end{array} \right].\]
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.01%3A_Binary_OperationsRecall that the set \(\mathcal{P}(X)\) is called the power set of \(X\); and, if \(A\) and \(B\) are sets, then \(A \cup B\) is called the union of \(A\) and \(B\) and \(A \cap B\) is called the inter...Recall that the set \(\mathcal{P}(X)\) is called the power set of \(X\); and, if \(A\) and \(B\) are sets, then \(A \cup B\) is called the union of \(A\) and \(B\) and \(A \cap B\) is called the intersection of \(A\) and \(B\).
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.07%3A_Isomorphism_of_GroupsOn the other hand by Theorem 4.2 since \(o(a) = n\) we have \[\langle a \rangle = \{ a^0, a^1, \dots , a^{n-1} \}.\] So the mapping \(\varphi:\mathbb{Z}_n \to \langle a \rangle\) defined by the rule \...On the other hand by Theorem 4.2 since \(o(a) = n\) we have \[\langle a \rangle = \{ a^0, a^1, \dots , a^{n-1} \}.\] So the mapping \(\varphi:\mathbb{Z}_n \to \langle a \rangle\) defined by the rule \(\varphi(i) = a^i\) for \(i = 0, 1, 2, \dots , n-1\), is clearly one-to-one and onto.
