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Mathematics 411 Midterm
D.Collingwood-Autumn 2011
Name:
October 28, 2011
Instructions: This is a closed book exam, no notes or calculators allowed. Please turn off all cell phones and
pagers. If you run out of room, use the backs of pages. For this exam
Mathematics 411
Summer 2014
Practice Exam 1
Instructions: For this exam, clarity of exposition is as important as correctness of mathematics.
This practice exam is a bit longer than the actual exam, to give a larger sample of possible questions.
The actua
Mathematics 411
Summer 2014
Practice Exam 2
Instructions: For this exam, clarity of exposition is as important as correctness of mathematics.
The actual exam will be 6 questions, closed book, no notes or calculators allowed. There will
be room on the pape
Solutions
Math 411
HW 1
Ex. 2.5 Theorem 2.3 cannot be correct, so there must be an error in the proof. Study the
proof, determine where the error lies, and explain what is wrong.
explanation. The argument makes the implicit assumption that the two subsets
Solutions
Math 411
HW 0
Exercise 1.5: Analyze (6, 9, 20)-accessibility.
We partition the positive integers into 6 families, depending on what their remainders are
when divided by 6. For each list, we circle the rst (6, 9, 20)-accessible numbers.
Remainder
Math 411
@
HW6
(Answers to evennumbered problems are below. Answers to the odd-numbered
problems can be found at the back of the book. Please report any errors to
Sonali, sonaliuwatuniversity domain name.)
3.1.2: A general polynomial of degree 2 can be wr
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Math 411 Homework 2 Solutions
Summer 2012
June 26, 2012
Note that I will refer to the Arithmetic Properties of the Integers from page 10 as AP1
- AP6.
1. (1.2.2) For any integer a, prove that (1) a = a.
Proof
a =
=
=
=
=
=
=
=
a + 0 AP3
a + 0 a Propositio
Math 411 Homework 3 Solutions
Summer 2012
July 9, 2012
1. Let a, b, and c be three non-zero integers.
(i) We dene gcd(a, b, c) to be the largest integer which divides a, b, and c.
(ii) Theorem 1 for three integers:
Let a, b, and c be three integers, not a
Math 411 Homework 4 Solutions
Summer 2012
July 15, 2012
1. (2.5.4) See homework 3, 2.4.9.
2. (2.5.7) The units in Zm are those a Zm such that gcd(a, m) = 1. This was proved in
homework 3, 2.4.8.
3. (2.5.8) The units in Q[x], the ring of polynomials with r
Math 411 Homework 5 Solutions
Summer 2012
July 19, 2012
1. (3.4.4i) Determine if x4 + x + 1 is irreducible in Q[x].
Let f2 (x) = x4 + x + 1 in Z2 [x]. Then f2 (0) = 1 and f2 (x) = 1, so f2 (x) does
not have any roots in Z2 . Therefore f2 (x) does not fact
Math 412 Homework 2 Solutions
Summer 2012
August 5, 2012
1. Describe the containment lattice of all subgroups of D4 :
(recall the notation from the last homework)
The subgroups are:
I = cfw_I
H = cfw_I, H
V = cfw_I, V
D1 = cfw_I, D1
D2 = cfw_I, D2
R2
Math 412 Homework 3 Solutions
Summer 2012
August 12, 2012
1. (5.1.2) Let H = cfw_e, (13).
Left cosets:
eH
(123)H
(132)H
(23)H
(13)H
(12)H
=
=
=
=
=
=
H
cfw_(123), (23)
cfw_(132), (12)
cfw_(23), (123)
cfw_(13), e = H
cfw_(12), (132)
So the left cosets of H
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Mathematics 411 Practice Final
D.Collingwood-Summer 2012
Name:
July 18, 2012
Instructions: This is a closed book exam, no notes or graphing calculators allowed. Please
turn off all cell phones and pagers. If you run out of room, use the backs of pages.
L 33/.
i
ii
‘/
Letp(x) 2x4+x+1
This has no roots in Z2, since 10(0) 2 1 and p(1) = 1 in Z2. Thus, if this
polynomial is reducible in Zg, then it must factor into two quadratic factors.
Thus, we must have p(x) = f(w)g($) : (3:2 + cw: + 1)(ac2 + by: + 1). (