Suppose that
\begin{align*} A & = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},\\ B & = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},\\ C & = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of 5}\}. \end{align*}Describe each of the following sets.
\(A \cap B\)
\(B \cap C\)
\(A \cup B\)
\(A \cap (B \cup C)\)
(a) \(A \cap B = \{ 2 \}\text{;}\) (b) \(B \cap C = \{ 5 \}\text{.}\)
If \(A = \{ a, b, c \}\text{,}\) \(B = \{ 1, 2, 3 \}\text{,}\) \(C = \{ x \}\text{,}\) and \(D = \emptyset\text{,}\) list all of the elements in each of the following sets.
\(A \times B\)
\(B \times A\)
\(A \times B \times C\)
\(A \times D\)
(a) \(A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}\text{;}\) (d) \(A \times D = \emptyset\text{.}\)
Prove \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
If \(x \in A \cup (B \cap C)\text{,}\) then either \(x \in A\) or \(x \in B \cap C\text{.}\) Thus, \(x \in A \cup B\) and \(A \cup C\text{.}\) Hence, \(x \in (A \cup B) \cap (A \cup C)\text{.}\) Therefore, \(A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)\text{.}\) Conversely, if \(x \in (A \cup B) \cap (A \cup C)\text{,}\) then \(x \in A \cup B\) and \(A \cup C\text{.}\) Thus, \(x \in A\) or \(x\) is in both \(B\) and \(C\text{.}\) So \(x \in A \cup (B \cap C)\) and therefore \((A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)\text{.}\) Hence, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
Prove \(A \cup B = (A \cap B) \cup (A \setminus B) \cup (B \setminus A)\text{.}\)
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
Prove \(A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)\text{.}\)
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
Which of the following relations \(f: {\mathbb Q} \rightarrow {\mathbb Q}\) define a mapping? In each case, supply a reason why \(f\) is or is not a mapping.
\(\displaystyle f(p/q) = \frac{p+ 1}{p - 2}\)
\(\displaystyle f(p/q) = \frac{3p}{3q}\)
\(\displaystyle f(p/q) = \frac{p+q}{q^2}\)
\(\displaystyle f(p/q) = \frac{3 p^2}{7 q^2} - \frac{p}{q}\)
(a) Not a map since \(f(2/3)\) is undefined; (b) this is a map; (c) not a map, since \(f(1/2) = 3/4\) but \(f(2/4)=3/8\text{;}\) (d) this is a map.
Determine which of the following functions are one-to-one and which are onto. If the function is not onto, determine its range.
\(f: {\mathbb R} \rightarrow {\mathbb R}\) defined by \(f(x) = e^x\)
\(f: {\mathbb Z} \rightarrow {\mathbb Z}\) defined by \(f(n) = n^2 + 3\)
\(f: {\mathbb R} \rightarrow {\mathbb R}\) defined by \(f(x) = \sin x\)
\(f: {\mathbb Z} \rightarrow {\mathbb Z}\) defined by \(f(x) = x^2\)
(a) \(f\) is one-to-one but not onto. \(f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}\text{.}\) (c) \(f\) is neither one-to-one nor onto. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)
Define a function \(f: {\mathbb N} \rightarrow {\mathbb N}\) that is one-to-one but not onto.
Define a function \(f: {\mathbb N} \rightarrow {\mathbb N}\) that is onto but not one-to-one.
(a) \(f(n) = n + 1\text{.}\)
Let \(f : A \rightarrow B\) and \(g : B \rightarrow C\) be maps.
If \(f\) and \(g\) are both one-to-one functions, show that \(g \circ f\) is one-to-one.
If \(g \circ f\) is onto, show that \(g\) is onto.
If \(g \circ f\) is one-to-one, show that \(f\) is one-to-one.
If \(g \circ f\) is one-to-one and \(f\) is onto, show that \(g\) is one-to-one.
If \(g \circ f\) is onto and \(g\) is one-to-one, show that \(f\) is onto.
(a) Let \(x, y \in A\text{.}\) Then \(g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))\text{.}\) Thus, \(f(x) = f(y)\) and \(x = y\text{,}\) so \(g \circ f\) is one-to-one. (b) Let \(c \in C\text{,}\) then \(c = (g \circ f)(x) = g(f(x))\) for some \(x \in A\text{.}\) Since \(f(x) \in B\text{,}\) \(g\) is onto.
Define a function on the real numbers by
\begin{equation*} f(x) = \frac{x + 1}{x - 1}. \end{equation*}What are the domain and range of \(f\text{?}\) What is the inverse of \(f\text{?}\) Compute \(f \circ f^{-1}\) and \(f^{-1} \circ f\text{.}\)
\(f^{-1}(x) = (x+1)/(x-1)\text{.}\)
Let \(f: X \rightarrow Y\) be a map with \(A_1, A_2 \subset X\) and \(B_1, B_2 \subset Y\text{.}\)
Prove \(f( A_1 \cup A_2 ) = f( A_1) \cup f( A_2 )\text{.}\)
Prove \(f( A_1 \cap A_2 ) \subset f( A_1) \cap f( A_2 )\text{.}\) Give an example in which equality fails.
Prove \(f^{-1}( B_1 \cup B_2 ) = f^{-1}( B_1) \cup f^{-1}(B_2 )\text{,}\) where
\begin{equation*} f^{-1}(B) = \{ x \in X : f(x) \in B \}. \end{equation*}Prove \(f^{-1}( B_1 \cap B_2 ) = f^{-1}( B_1) \cap f^{-1}( B_2 )\text{.}\)
Prove \(f^{-1}( Y \setminus B_1 ) = X \setminus f^{-1}( B_1)\text{.}\)
(a) Let \(y \in f(A_1 \cup A_2)\text{.}\) Then there exists an \(x \in A_1 \cup A_2\) such that \(f(x) = y\text{.}\) Hence, \(y \in f(A_1)\) or \(f(A_2) \text{.}\) Therefore, \(y \in f(A_1) \cup f(A_2)\text{.}\) Consequently, \(f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\text{.}\) Conversely, if \(y \in f(A_1) \cup f(A_2)\text{,}\) then \(y \in f(A_1)\) or \(f(A_2)\text{.}\) Hence, there exists an \(x \in A_1\) or there exists an \(x \in A_2\) such that \(f(x) = y\text{.}\) Thus, there exists an \(x \in A_1 \cup A_2\) such that \(f(x) = y\text{.}\) Therefore, \(f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\text{,}\) and \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
Determine whether or not the following relations are equivalence relations on the given set. If the relation is an equivalence relation, describe the partition given by it. If the relation is not an equivalence relation, state why it fails to be one.
\(x \sim y\) in \({\mathbb R}\) if \(x \geq y\)
\(m \sim n\) in \({\mathbb Z}\) if \(mn > 0\)
\(x \sim y\) in \({\mathbb R}\) if \(|x - y| \leq 4\)
\(m \sim n\) in \({\mathbb Z}\) if \(m \equiv n \pmod{6}\)
(a) NThe relation fails to be symmetric. (b) The relation is not reflexive, since 0 is not equivalent to itself. (c) The relation is not transitive.
Find the error in the following argument by providing a counterexample. “The reflexive property is redundant in the axioms for an equivalence relation. If \(x \sim y\text{,}\) then \(y \sim x\) by the symmetric property. Using the transitive property, we can deduce that \(x \sim x\text{.}\)”
Let \(X = {\mathbb N} \cup \{ \sqrt{2}\, \}\) and define \(x \sim y\) if \(x + y \in {\mathbb N}\text{.}\)
Prove that
\begin{equation*} 1^2 + 2^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \end{equation*}for \(n \in {\mathbb N}\text{.}\)
The base case, \(S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2\) is true. Assume that \(S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6\) is true. Then
\begin{align*} 1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 & = [k(k + 1)(2k + 1)]/6 + (k + 1)^2\\ & = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6, \end{align*}and so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)
Prove that \(n! \gt 2^n\) for \(n \geq 4\text{.}\)
The base case, \(S(4): 4! = 24 \gt 16 =2^4\) is true. Assume \(S(k): k! \gt 2^k\) is true. Then \((k + 1)! = k! (k + 1) \gt 2^k \cdot 2 = 2^{k + 1}\text{,}\) so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)
Prove the Leibniz rule for \(f^{(n)} (x)\text{,}\) where \(f^{(n)}\) is the \(n\)th derivative of \(f\text{;}\) that is, show that
\begin{equation*} (fg)^{(n)}(x) = \sum_{k = 0}^{n} \binom{n}{k} f^{(k)}(x) g^{(n - k)}(x). \end{equation*}Follow the proof in Example Example 2.1.4.
If \(x\) is a nonnegative real number, then show that \((1 + x)^n - 1 \geq nx\) for \(n = 0, 1, 2, \ldots\text{.}\)
The base case, \(S(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x\) is true. Assume \(S(k): (1 + x)^k -1 \geq kx\) is true. Then
\begin{align*} (1 + x)^{k + 1} - 1 & = (1 + x)(1 + x)^k -1\\ & = (1 + x)^k + x(1 + x)^k - 1\\ & \geq kx + x(1 + x)^k\\ & \geq kx + x\\ & = (k + 1)x, \end{align*}so \(S(k + 1)\) is true. Therefore, \(S(n)\) is true for all positive integers \(n\text{.}\)
The Fibonacci numbers are
\begin{equation*} 1, 1, 2, 3, 5, 8, 13, 21, \ldots. \end{equation*}We can define them inductively by \(f_1 = 1\text{,}\) \(f_2 = 1\text{,}\) and \(f_{n + 2} = f_{n + 1} + f_n\) for \(n \in {\mathbb N}\text{.}\)
Prove that \(f_n \lt 2^n\text{.}\)
Prove that \(f_{n + 1} f_{n - 1} = f^2_n + (-1)^n\text{,}\) \(n \geq 2\text{.}\)
Prove that \(f_n = [(1 + \sqrt{5}\, )^n - (1 - \sqrt{5}\, )^n]/ 2^n \sqrt{5}\text{.}\)
Show that \(\lim_{n \rightarrow \infty} f_n / f_{n + 1} = (\sqrt{5} - 1)/2\text{.}\)
Prove that \(f_n\) and \(f_{n + 1}\) are relatively prime.
For (a) and (b) use mathematical induction. (c) Show that \(f_1 = 1\text{,}\) \(f_2 = 1\text{,}\) and \(f_{n + 2} = f_{n + 1} + f_n\text{.}\) (d) Use part (c). (e) Use part (b) and Exercise Exercise 2.3.16.
Let \(x, y \in {\mathbb N}\) be relatively prime. If \(xy\) is a perfect square, prove that \(x\) and \(y\) must both be perfect squares.
Use the Fundamental Theorem of Arithmetic.
Define the least common multiple of two nonzero integers \(a\) and \(b\text{,}\) denoted by \(\lcm(a,b)\text{,}\) to be the nonnegative integer \(m\) such that both \(a\) and \(b\) divide \(m\text{,}\) and if \(a\) and \(b\) divide any other integer \(n\text{,}\) then \(m\) also divides \(n\text{.}\) Prove that any two integers \(a\) and \(b\) have a unique least common multiple.
Let \(S = \{s \in {\mathbb N} : a \mid s\text{,}\) \(b \mid s \}\text{.}\) Then \(S \neq \emptyset\text{,}\) since \(|ab| \in S\text{.}\) By the Principle of Well-Ordering, \(S\) contains a least element \(m\text{.}\) To show uniqueness, suppose that \(a \mid n\) and \(b \mid n\) for some \(n \in {\mathbb N}\text{.}\) By the division algorithm, there exist unique integers \(q\) and \(r\) such that \(n = mq + r\text{,}\) where \(0 \leq r \lt m\text{.}\) Since \(a\) and \(b\) divide both \(m\text{,}\) and \(n\text{,}\) it must be the case that \(a\) and \(b\) both divide \(r\text{.}\) Thus, \(r = 0\) by the minimality of \(m\text{.}\) Therefore, \(m \mid n\text{.}\)
Let \(a, b, c \in {\mathbb Z}\text{.}\) Prove that if \(\gcd(a,b) = 1\) and \(a \mid bc\text{,}\) then \(a \mid c\text{.}\)
Since \(\gcd(a,b) = 1\text{,}\) there exist integers \(r\) and \(s\) such that \(ar + bs = 1\text{.}\) Thus, \(acr + bcs = c\text{.}\) Since \(a\) divides both \(bc\) and itself, \(a\) must divide \(c\text{.}\)
Prove that there are an infinite number of primes of the form \(6n + 5\text{.}\)
Every prime must be of the form 2, 3, \(6n + 1\text{,}\) or \(6n + 5\text{.}\) Suppose there are only finitely many primes of the form \(6k + 5\text{.}\)
Find all \(x \in {\mathbb Z}\) satisfying each of the following equations.
\(3x \equiv 2 \pmod{7}\)
\(5x + 1 \equiv 13 \pmod{23}\)
\(5x + 1 \equiv 13 \pmod{26}\)
\(9x \equiv 3 \pmod{5}\)
\(5x \equiv 1 \pmod{6}\)
\(3x \equiv 1 \pmod{6}\)
(a) \(3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}\text{;}\) (c) \(18 + 26 \mathbb Z\text{;}\) (e) \(5 + 6 \mathbb Z\text{.}\)
Which of the following multiplication tables defined on the set \(G = \{ a, b, c, d \}\) form a group? Support your answer in each case.
(a) Not a group; (c) a group.
Give a multiplication table for the group \(U(12)\text{.}\)
Give an example of two elements \(A\) and \(B\) in \(GL_2({\mathbb R})\) with \(AB \neq BA\text{.}\)
Pick two matrices. Almost any pair will work.
Prove or disprove that every group containing six elements is abelian.
There is a nonabelian group containing six elements.
Give a specific example of some group \(G\) and elements \(g, h \in G\) where \((gh)^n \neq g^nh^n\text{.}\)
Look at the symmetry group of an equilateral triangle or a square.
Give an example of three different groups with eight elements. Why are the groups different?
The are five different groups of order 8.
Show that there are \(n!\) permutations of a set containing \(n\) items.
Let
\begin{equation*} \sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ a_1 & a_2 & \cdots & a_n \end{pmatrix} \end{equation*}be in \(S_n\text{.}\) All of the \(a_i\)s must be distinct. There are \(n\) ways to choose \(a_1\text{,}\) \(n-1\) ways to choose \(a_2\text{,}\) \(\ldots\text{,}\) 2 ways to choose \(a_{n - 1}\text{,}\) and only one way to choose \(a_n\text{.}\) Therefore, we can form \(\sigma\) in \(n(n - 1) \cdots 2 \cdot 1 = n!\) ways.
Let \(a\) and \(b\) be elements in a group \(G\text{.}\) Prove that \(ab^na^{-1} = (aba^{-1})^n\) for \(n \in \mathbb Z\text{.}\)
Show that if \(a^2 = e\) for all elements \(a\) in a group \(G\text{,}\) then \(G\) must be abelian.
Since \(abab = (ab)^2 = e = a^2 b^2 = aabb\text{,}\) we know that \(ba = ab\text{.}\)
Find all the subgroups of the symmetry group of an equilateral triangle.
\(H_1 = \{ id \}\text{,}\) \(H_2 = \{ id, \rho_1, \rho_2 \}\text{,}\) \(H_3 = \{ id, \mu_1 \}\text{,}\) \(H_4 = \{ id, \mu_2 \}\text{,}\) \(H_5 = \{ id, \mu_3 \}\text{,}\) \(S_3\text{.}\)
Prove that
\begin{equation*} G = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \text{ and } a \text{ and } b \text{ are not both zero} \} \end{equation*}is a subgroup of \({\mathbb R}^{\ast}\) under the group operation of multiplication.
The identity of \(G\) is \(1 = 1 + 0 \sqrt{2}\text{.}\) Since \((a + b \sqrt{2}\, )(c + d \sqrt{2}\, ) = (ac + 2bd) + (ad + bc)\sqrt{2}\text{,}\) \(G\) is closed under multiplication. Finally, \((a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}\)
Prove or disprove: If \(H\) and \(K\) are subgroups of a group \(G\text{,}\) then \(H \cup K\) is a subgroup of \(G\text{.}\)
Look at \(S_3\text{.}\)
Let \(a\) and \(b\) be elements of a group \(G\text{.}\) If \(a^4b = ba\) and \(a^3 = e\text{,}\) prove that \(ab = ba\text{.}\)
Since \(a^4b = ba\text{,}\) it must be the case that \(b = a^6 b = a^2 b a\text{,}\) and we can conclude that \(ab = a^3 b a = ba\text{.}\)
Prove or disprove each of the following statements.
All of the generators of \({\mathbb Z}_{60}\) are prime.
\(U(8)\) is cyclic.
\({\mathbb Q}\) is cyclic.
If every proper subgroup of a group \(G\) is cyclic, then \(G\) is a cyclic group.
A group with a finite number of subgroups is finite.
(a) False; (c) false; (e) true.
Find the order of each of the following elements.
\(5 \in {\mathbb Z}_{12}\)
\(\sqrt{3} \in {\mathbb R}\)
\(\sqrt{3} \in {\mathbb R}^\ast\)
\(-i \in {\mathbb C}^\ast\)
72 in \({\mathbb Z}_{240}\)
312 in \({\mathbb Z}_{471}\)
(a) 12; (c) infinite; (e) 10.
List all of the elements in each of the following subgroups.
The subgroup of \({\mathbb Z}\) generated by 7
The subgroup of \({\mathbb Z}_{24}\) generated by 15
All subgroups of \({\mathbb Z}_{12}\)
All subgroups of \({\mathbb Z}_{60}\)
All subgroups of \({\mathbb Z}_{13}\)
All subgroups of \({\mathbb Z}_{48}\)
The subgroup generated by 3 in \(U(20)\)
The subgroup generated by 5 in \(U(18)\)
The subgroup of \({\mathbb R}^\ast\) generated by 7
The subgroup of \({\mathbb C}^\ast\) generated by \(i\) where \(i^2 = -1\)
The subgroup of \({\mathbb C}^\ast\) generated by \(2i\)
The subgroup of \({\mathbb C}^\ast\) generated by \((1 + i) / \sqrt{2}\)
The subgroup of \({\mathbb C}^\ast\) generated by \((1 + \sqrt{3}\, i) / 2\)
(a) \(7 {\mathbb Z} = \{ \ldots, -7, 0, 7, 14, \ldots \}\text{;}\) (b) \(\{ 0, 3, 6, 9, 12, 15, 18, 21 \}\text{;}\) (c) \(\{ 0 \}\text{,}\) \(\{ 0, 6 \}\text{,}\) \(\{ 0, 4, 8 \}\text{,}\) \(\{ 0, 3, 6, 9 \}\text{,}\) \(\{ 0, 2, 4, 6, 8, 10 \}\text{;}\) (g) \(\{ 1, 3, 7, 9 \}\text{;}\) (j) \(\{ 1, -1, i, -i \}\text{.}\)
Find the subgroups of \(GL_2( {\mathbb R })\) generated by each of the following matrices.
\(\displaystyle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\)
\(\displaystyle \begin{pmatrix} 0 & 1/3 \\ 3 & 0 \end{pmatrix}\)
\(\displaystyle \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}\)
\(\displaystyle \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\)
\(\displaystyle \begin{pmatrix} 1 & -1 \\ -1 & 0 \end{pmatrix}\)
\(\displaystyle \begin{pmatrix} \sqrt{3}/ 2 & 1/2 \\ -1/2 & \sqrt{3}/2 \end{pmatrix}\)
(a)
\begin{equation*} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. \end{equation*}(c)
\begin{equation*} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ -1 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}. \end{equation*}Find all elements of finite order in each of the following groups. Here the “\(\ast\)” indicates the set with zero removed.
\({\mathbb Z}\)
\({\mathbb Q}^\ast\)
\({\mathbb R}^\ast\)
(a) \(0, 1, -1\text{;}\) (b) \(1, -1\)
If \(a^{24} =e\) in a group \(G\text{,}\) what are the possible orders of \(a\text{?}\)
1, 2, 3, 4, 6, 8, 12, 24.
Evaluate each of the following.
\((3-2i)+ (5i-6)\)
\((4-5i)-\overline{(4i -4)}\)
\((5-4i)(7+2i)\)
\((9-i) \overline{(9-i)}\)
\(i^{45}\)
\((1+i)+\overline{(1+i)}\)
(a) \(-3 + 3i\text{;}\) (c) \(43- 18i\text{;}\) (e) \(i\)
Convert the following complex numbers to the form \(a + bi\text{.}\)
\(2 \cis(\pi / 6 )\)
\(5 \cis(9\pi/4)\)
\(3 \cis(\pi)\)
\(\cis(7\pi/4) /2\)
(a) \(\sqrt{3} + i\text{;}\) (c) \(-3\text{.}\)
Change the following complex numbers to polar representation.
\(1-i\)
\(-5\)
\(2+2i\)
\(\sqrt{3} + i\)
\(-3i\)
\(2i + 2 \sqrt{3}\)
(a) \(\sqrt{2} \cis( 7 \pi /4)\text{;}\) (c) \(2 \sqrt{2} \cis( \pi /4)\text{;}\) (e) \(3 \cis(3 \pi/2)\text{.}\)
Calculate each of the following expressions.
\((1+i)^{-1}\)
\((1 - i)^{6}\)
\((\sqrt{3} + i)^{5}\)
\((-i)^{10}\)
\(((1-i)/2)^{4}\)
\((-\sqrt{2} - \sqrt{2}\, i)^{12}\)
\((-2 + 2i)^{-5}\)
(a) \((1 - i)/2\text{;}\) (c) \(16(i - \sqrt{3}\, )\text{;}\) (e) \(-1/4\text{.}\)
Calculate each of the following.
\(292^{3171} \pmod{ 582}\)
\(2557^{ 341} \pmod{ 5681}\)
\(2071^{ 9521} \pmod{ 4724}\)
\(971^{ 321} \pmod{ 765}\)
(a) 292; (c) 1523.
If \(g\) and \(h\) have orders 15 and 16 respectively in a group \(G\text{,}\) what is the order of \(\langle g \rangle \cap \langle h \rangle \text{?}\)
\(|\langle g \rangle \cap \langle h \rangle| = 1\text{.}\)
Let \(G\) be an abelian group. Show that the elements of finite order in \(G\) form a subgroup. This subgroup is called the torsion subgroup of \(G\text{.}\)
The identity element in any group has finite order. Let \(g, h \in G\) have orders \(m\) and \(n\text{,}\) respectively. Since \((g^{-1})^m = e\) and \((gh)^{mn} = e\text{,}\) the elements of finite order in \(G\) form a subgroup of \(G\text{.}\)
Prove that if \(G\) has no proper nontrivial subgroups, then \(G\) is a cyclic group.
If \(g\) is an element distinct from the identity in \(G\text{,}\) \(g\) must generate \(G\text{;}\) otherwise, \(\langle g \rangle\) is a nontrivial proper subgroup of \(G\text{.}\)
