Answers to homework #14, due May 2
for grader
p. 215 #2. K , since every element has order 2.
p. 215 #3. (Z4 , +) since it is cyclic.
p. 215 #5. (Z4 , +) since it is generated by a single element with order 4.
OOB1. Since o(i) = 4, < i >= cfw_1, i, 1, i i
Answers to homework due April 28
for grader
OOB1. (a) is a group: an addition table with a few comments will show this.
(b) is not closed under the operation (e.g., 32 = 0) and not every element has an inverse (e.g.,
3 has no inverse).
(c) is not closed (
QUIZ 1
Question 1: Prove that for all n 4, n! 2n .
Proof. For n = 4, 4! = 24, and 24 = 16. Hence 24 16.
Assume that for n = k 4, k ! 2k .
When n = k + 1,
(k + 1)! = k ! (k + 1) 2k (k + 1) 2k 2 2k+1 .
Hence by mathematics induction (1), the statement is tr
HOMEWORK 1
Question 1: Prove that 13 + 23 + + n3 = [n(n + 1)/2]2 for all n 1.
Question 2: Prove that
1 + 2 + 22 + + 2n1 = 2n 1
for every n 1.
Question 3: Prove that for any real number x and for all numbers n > 1,
xn 1 = (x 1)(xn1 + xn2 + + xnr + + x + 1)
Answers to hw 10
for grader
OOB 1. stw + rtw + rsw + rst = 1. rstw = 5.
1
r
+1+1+
s
t
1
w
=
stw+rtw+rsw+rst
rstw
=
1
5
= 1.
5
OOB 2. a a + b = 2 = b, a2 + ab ab = 1 = a2 . So a = 1, b = 2. As a check, note
that ab = 2, and indeed 12 (2) = 2.
OOB 3. (a) #G
for grader
p. 230 #16. The elements are cfw_0, 1, 2, , + 1, + 2, 2, 2 + 1, 2 + 2. Since G is commutative, all subgroups are normal. Subgroups other than the trivial group and G itself are:
cfw_0, 1, 2, cfw_0, , , cfw_0, + 1, 2 + 2, cfw_0, + 2, 2 + 1. Ther
Answers to Math 558 S13 Midterm
1. (a) z 4 = 81(cos 3 + i sin 3 ) = 81
(b) o(w) = 8.
2. (a) No. 10, 24 are not relatively prime.
(b) Yes. 7, 24 are relatively prime.
(c) In Z24 , 5 5 = 1, so 51 = 5.
(d) By (c), o(5) = 2.
3. P is irreducible i Q, R, if P =
Math 558 Spring 2013
Answers to nal review questions
1. (a) Since
30
5
is the smallest multiple of
3
5
which is also a multiple of 2 , o(z ) = 10.
Since 10 = 2 5, o(z 2 ) = 5.
10
(b) |z | = 32 + a2 = 10. So |z 10 | = 10 = 105 = 100, 000.
2. (a) No. P (0)
This is not a sample exam. For example, it is longer than the exam. And the format is
dierent from the exam. But if you can answer these questions you can be condent that you really
know the material. You should start working on these problems after class
Math 558 Spring 2013
Answers to midterm review
y
1. x2 = cos + i sin , xy = cos 76 + i sin 76 , x = cos + i sin
6
6
2. |x| =
13, |x5 | =
5
13 .|xy | = 13.| x | = 1.
y
3. (a) If z is a root of unity, o(z ) is the least m so z m = 1.
(b)For k, n Z, gcd(k,
Math 558
Answers to homework #11
for grader
p. 90 #2. In Z11 if a = 0 then a10 = 1. So 02000 = 0 and if a = 0 then a2000 = (a10 )200 = 1.
p. 90 #4. We work in Z7 . 0 + 0 + 5 = 0, so 0 is not a solution. If a = 0 then a6 = 1. So
7777
a6
= (a6 )66.6 = 1. He
Answers to homework due Tuesday March 12
for grader
OOB 1. Since the degree is 3, you only have to check that there is no zero. The polynomials
to look at are: P1 = x3 + x2 + 1, P2 = x3 + x2 + 2, P3 = x3 + 2x2 + 1, P4 = x3 + 2x2 + 2. 0 is
not a zero of an