Appendix B Hints to Selected Exercises

(a) \(A \cap B = \{ 2 \}\text{;}\) (b) \(B \cap C = \{ 5 \}\text{.}\)

(a) \(A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}\text{;}\) (d) \(A \times D = \emptyset\text{.}\)

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, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)

\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)

\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)

(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.

(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. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)

(a) \(f(n) = n + 1\text{.}\)

(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.

\(f^{-1}(x) = (x+1)/(x-1)\text{.}\)

(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 \(A_2\) 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 \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)

(a) The relation fails to be symmetric. (b) The relation is not reflexive, since \(0\) is not equivalent to itself. (c) The relation is not transitive.

Let \(X = {\mathbb N} \cup \{ \sqrt{2}\, \}\) and define \(x \sim y\) if \(x + y \in {\mathbb N}\text{.}\)

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 \(n\text{.}\)

Follow the proof in Example 2.4.

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 2.3.16.

Use the Fundamental Theorem of Arithmetic.

Since \(\gcd(a,b) = 1\text{,}\) there exist integers \(r\) and \(s\) such that \(ar + bs = 1\text{.}\) Thus, \(acr + bcs = c\text{.}\)

Use the Principle of Well-Ordering and the division algorithm.

For (a), use Exercise 2.3.26. For (b), use the previous part with the pair of integers \(a/d\) and \(b/d\text{.}\) For (c), use part (b).

Every prime must be of the form \(2\text{,}\) \(3\text{,}\) \(6n + 1\text{,}\) or \(6n + 5\text{.}\) Suppose there are only finitely many primes of the form \(6k + 5\text{.}\)

(a) \(3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}\text{;}\) (c) \(18 + 26 \mathbb Z\text{;}\) (e) \(5 + 6 \mathbb Z\text{.}\)

(a) \(x \equiv 17 \pmod{55}\text{;}\) (c) \(x \equiv 214 \pmod{2772}\text{.}\)

(c) \({\mathbb Q}(\sqrt{2}\, )\) is a field; (f) \(R\) is not a ring.

True.

For (b), define the inverse like for complex numbers.

Compute \((a+b)^2\) and \((-1)^2\text{.}\)

Assume there is an isomorphism \(\phi: {\mathbb C} \rightarrow {\mathbb R}\) with \(\phi(i) = a\text{.}\)

False. Assume there is an isomorphism \(\phi: {\mathbb Q}(\sqrt{2}\, ) \rightarrow {\mathbb Q}(\sqrt{3}\, )\) such that \(\phi(\sqrt{2}\, ) = a\text{.}\)

(a) \(\phi(a) \phi(b) = \phi(ab) = \phi(ba) = \phi(b) \phi(a)\text{.}\)

Let \(a/b, c/d \in {\mathbb Z}_{(p)}\text{.}\) Then \(a/b + c/d = (ad + bc)/bd\) and \((a/b) \cdot (c/d) = (ac)/(bd)\) are both in \({\mathbb Z}_{(p)}\text{,}\) since \(\gcd(bd,p) = 1\text{.}\)

Suppose that \(x^2 = x\) and \(x \neq 0\text{.}\) Since \(R\) is an integral domain, \(x = 1\text{.}\) To find a nontrivial idempotent, look in \({\mathbb M}_2({\mathbb R})\text{.}\)

Use the Chinese Remainder Theorem.

(a) \(9x^2 + 2x + 5\text{;}\) (b) \(8x^4 + 7x^3 + 2x^2 + 7x\text{.}\)

(a) \(5 x^3 + 6 x^2 - 3 x + 4 = (5 x^2 + 2x + 1)(x -2) + 6\text{;}\) (c) \(4x^5 - x^3 + x^2 + 4 = (4x^2 + 4)(x^3 + 3) + 4x^2 + 2\text{.}\)

(a) No zeros in \({\mathbb Z}_{12}\text{;}\) (c) \(3\text{,}\) \(4\text{.}\)

Look at \((2x + 1)\text{.}\)

One factorization is \(x^2 + x + 8 = (x + 2)(x + 9)\text{.}\)

False.

Look at Euclid’s proof of the infinitude of prime numbers in Theorem 2.14.

(a) Reducible; (c) irreducible.

The integers \(\mathbb Z\) do not form a field.

Does the element \(x\) have a multiplicative inverse?

Let \(\phi : R \rightarrow S\) be an isomorphism. Define \(\overline{\phi} : R[x] \rightarrow S[x]\) by \(\overline{\phi}(a_0 + a_1 x + \cdots + a_n x^n) = \phi(a_0) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.}\)

Note that \(z^{-1} = 1/(a + b\sqrt{3}\, i) = (a -b \sqrt{3}\, i)/(a^2 + 3b^2)\) is in \({\mathbb Z}[\sqrt{3}\, i]\) if and only if \(a^2 + 3 b^2 = 1\text{.}\) The only integer solutions to the equation are \(a = \pm 1, b = 0\text{.}\)

(a) \(5 = -i(1 + 2i)(2 + i)\text{;}\) (c) \(6 + 8i = -i(1 + i)^2(2 + i)^2\text{.}\)

Let \(z = a + bi\) and \(w = c + di \neq 0\) be in \({\mathbb Z}[i]\text{.}\) Prove that \(z/w \in {\mathbb Q}(i)\text{.}\)

Let \(a = ub\) with \(u\) a unit. Then \(\nu(b) \leq \nu(ub) \leq \nu(a)\text{.}\) Similarly, \(\nu(a) \leq \nu(b)\text{.}\)

Prove that if \(b\) is not a unit, then \(\nu(ab) > \nu(a)\) and then use induction on the valuation.

Show that 6 can be factored in two different ways.

(a) \(\{0 \}\text{,}\) \(\{0, 9 \}\text{,}\) \(\{0, 6, 12 \}\text{,}\) \(\{0, 3, 6, 9, 12, 15 \}\text{,}\) \(\{0, 2, 4, 6, 8, 10, 12, 14, 16 \}\text{;}\) (c) there are no nontrivial ideals.

If \(I \neq \{ 0 \}\text{,}\) show that \(1 \in I\text{.}\)

Let \(a \in R\) with \(a \neq 0\text{.}\) Then the principal ideal generated by \(a\) is \(R\text{.}\) Thus, there exists a \(b \in R\) such that \(ab =1\text{.}\)

(a) Not a group; (c) a group.

Pick two matrices. Almost any pair will work.

There is a nonabelian group containing six elements.

Look at the symmetry group of an equilateral triangle or a square.

The are five different groups of order 8.

Since \(abab = (ab)^2 = e = a^2 b^2 = aabb\text{,}\) we know that \(ba = ab\text{.}\)

\(H_1 = \{ \identity \}\text{,}\) \(H_2 = \{ \identity, \rho_1, \rho_2 \}\text{,}\) \(H_3 = \{ \identity, \mu_1 \}\text{,}\) \(H_4 = \{ \identity, \mu_2 \}\text{,}\) \(H_5 = \{ \identity, \mu_3 \}\text{,}\) \(S_3\text{.}\)

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, \((a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}\)

Look at \(\mathbb Z_6^+\text{.}\)

\(b a = a^4 b = a^3 a b = ab\)

(a) False; (c) false; (e) true.

(a) \(12\text{;}\) (c) infinite; (e) \(10\text{.}\)

(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) \(\{ 1, -1, i, -i \}\text{.}\)

(a) \(0\text{;}\) (b) \(1, -1\text{.}\)

\(1, 2, 3, 4, 6, 8, 12, 24\text{.}\)

(a) \(-3 + 3i\text{;}\) (c) \(43- 18i\text{;}\) (e) \(i\)

(a) \(\sqrt{3} + i\text{;}\) (c) \(-3\text{.}\)

(a) \(\sqrt{2} \cis( 7 \pi /4)\text{;}\) (c) \(2 \sqrt{2} \cis( \pi /4)\text{;}\) (e) \(3 \cis(3 \pi/2)\text{.}\)

(a) \((1 - i)/2\text{;}\) (c) \(16(i - \sqrt{3}\, )\text{;}\) (e) \(-1/4\text{.}\)

(a) \(292\text{;}\) (c) \(1523\text{.}\)

\(|\langle g \rangle \cap \langle h \rangle| = 1\text{.}\)

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 \(G\text{.}\)

(a) \((12453)\text{;}\) (c) \((13)(25)\text{.}\)

(a) \((135)(24)\text{;}\) (c) \((14)(23)\text{;}\) (e) \((1324)\text{;}\) (g) \((134)(25)\text{;}\) (n) \((17352)\text{.}\)

(a) \((16)(15)(13)(14)\text{;}\) (c) \((16)(14)(12)\text{.}\)

\((a_1, a_2, \ldots, a_n)^{-1} = (a_1, a_{n}, a_{n-1}, \ldots, a_2)\)

(a) \(\{ (13), (13)(24), (132), (134), (1324), (1342) \}\) is not a subgroup.

\((12345)(678)\text{.}\)

Calculate \((123)(12)\) and \((12)(123)\text{.}\)

Consider the cases \((ab)(bc)\) and \((ab)(cd)\text{.}\)

For (a), show that \(\sigma \tau \sigma^{-1 }(\sigma(a_i)) = \sigma(a_{i + 1})\text{.}\)

The order of \(g\) and the order \(h\) must both divide the order of \(G\text{.}\)

The possible orders must divide \(60\text{.}\)

This is true for every proper nontrivial subgroup.

False.

(a) \(\langle 8 \rangle\text{,}\) \(1 + \langle 8 \rangle\text{,}\) \(2 + \langle 8 \rangle\text{,}\) \(3 + \langle 8 \rangle\text{,}\) \(4 + \langle 8 \rangle\text{,}\) \(5 + \langle 8 \rangle\text{,}\) \(6 + \langle 8 \rangle\text{,}\) and \(7 + \langle 8 \rangle\text{;}\) (c) \(3 {\mathbb Z}\text{,}\) \(1 + 3 {\mathbb Z}\text{,}\) and \(2 + 3 {\mathbb Z}\text{.}\)

\(4^{\phi(15)} \equiv 4^8 \equiv 1 \pmod{15}\text{.}\)

Let \(g_1 \in gH\text{.}\) Show that \(g_1 \in Hg\) and thus \(gH \subset Hg\text{.}\)

Show that \(g(H \cap K) = gH \cap gK\text{.}\)

If \(\gcd(m,n) = 1\text{,}\) then \(\phi(mn) = \phi(m)\phi(n)\) (Exercise 2.3.25 in Chapter 2).

LAORYHAPDWK

Hint: V = E, E = X (also used for spaces and punctuation), K = R.

\(26! - 1\)

(a) \(2791\text{;}\) (c) \(112135 25032 442\text{.}\)

(a) \(31\) (c) \(14\text{.}\)

(a) \(n = 11 \cdot 41\text{;}\) (c) \(n = 8779 \cdot 4327\text{.}\)

If \(E\) is defined by \(y^2 = x^3 + ax + b\text{,}\) how many zeros does the polynomial\(x^3 + ax + b\) have?

Example 13.1: \(0\text{,}\) \({\mathbb R}^2 \setminus \{ 0 \}\text{.}\) Example 13.2: \(X = \{ 1, 2, 3, 4 \}\text{.}\)

(a) \(X_{(1)} = \{1, 2, 3 \}\text{,}\) \(X_{(12)} = \{3 \}\text{,}\) \(X_{(13)} = \{ 2 \}\text{,}\) \(X_{(23)} = \{1 \}\text{,}\) \(X_{(123)} = X_{(132)} = \emptyset\text{.}\) \(G_1 = \{ (1), (23) \}\text{,}\) \(G_2 = \{(1), (13) \}\text{,}\) \(G_3 = \{ (1), (12)\}\text{.}\)

(a) \({\mathcal O}_1 = {\mathcal O}_2 = {\mathcal O}_3 = \{ 1, 2, 3\}\text{.}\)

\((3^4 + 3^1 + 3^2 + 3^1 + 3^2 + 3^2 + 3^3 + 3^3)/8 = 21\text{.}\)

The group of rigid motions of the cube can be described by the allowable permutations of the six faces and is isomorphic to \(S_4\text{.}\) There are the identity cycle, 6 permutations with the structure \((abcd)\) that correspond to the quarter turns, 3 permutations with the structure \((ab)(cd)\) that correspond to the half turns, 6 permutations with the structure \((ab)(cd)(ef)\) that correspond to rotating the cube about the centers of opposite edges, and 8 permutations with the structure \((abc)(def)\) that correspond to rotating the cube about opposite vertices.

\((1 \cdot 2^6 + 3 \cdot 2^4 + 4 \cdot 2^3 + 2 \cdot 2^2 + 2 \cdot 2^1)/12 = 13\text{.}\)

\((1 \cdot 2^8 + 3 \cdot 2^6 + 2 \cdot 2^4)/6 = 80\text{.}\)

Use the fact that \(x \in g C(a) g^{-1}\) if and only if \(g^{-1}x g \in C(a)\text{.}\)