Math 3311 Abstract Algebra I
Review Problems for Quiz 4
Section 13:
In Exercises 1 through 15, determine whether the given map is a homomorphism. [Hint: The straightforward way to proceed is to check whether (ab) = (a) (b) for all a and b in the domain of
5
1.1. THE IMPORTANT STUFF
f
c
a
q
b
Figure 1.4: Right triangle with sides a, b and c.
1.1.4
Math: Trigonometry
You will also need some simple trigonometry. This wont amount to much more than relating
the sides of a right triangle, that is, a triangle wit
6
CHAPTER 1. MATHEMATICAL CONCEPTS
B
C
B
A
A
Figure 1.5: Vectors A and B are added to give the vector C = A + B.
y
A
Ay
Ax
x
Figure 1.6: Vector A is split up into components.
Vectors are represented by arrows which show their magnitude and direction. The
10
CHAPTER 1. MATHEMATICAL CONCEPTS
Likewise change 35 min to seconds:
35 min = (35 min)
60 s
1 min
= 2100 s
The total is
1 h + 35 min = 3600 s + 2100s = 5700 s
(b) Change one day to seconds; use the unit factors:
1 day = (1 day)
24 h
1 day
60 min
1h
60 s
8
CHAPTER 1. MATHEMATICAL CONCEPTS
y
C
B
By
A
Ay
Ax
Bx
x
Figure 1.8: Vectors A and B add to give the vector C. The x components of A and B add to give the x
component of C: Ax + Bx = Cx . Likewise for the y components.
Summing up, many problems involving
9
1.2. WORKED EXAMPLES
(b) Using the fact that a milligram is a thousandth of a gram: 1 mg = 103 g, and our
answer from (a), we nd
m = 5 103 g = (5 103 g)
1 mg
103 g
= 5 mg
(c) Using the fact that a microgram is 106 (one millionth) of a gram: 1 g = 106 g
11
1.2. WORKED EXAMPLES
6.20
3.50
q
x
Figure 1.9: Right triangle for example 5.
7.10
y
36o
x
Figure 1.10: Right triangle for example 6.
We can use the Pythagorean theorem to nd x. Pythagoras tells us:
x2 + (3.50)2 = (6.20)2
Solving for x gives
x2 = (6.20)
12
CHAPTER 1. MATHEMATICAL CONCEPTS
y
o
q
o
q
45
45
2.60
1.50
2.60
1.50
(a)
(b)
Figure 1.11: Right triangle for example 7.
Likewise, we can write a relation involving the adjacent side and the cosine of the
angle,
x
cos 36 =
=
x = (7.10) cos 36 = 5.74
7.1
13
1.2. WORKED EXAMPLES
192 m
2.0 o
x
Figure 1.12: Gateway Arch is viewed from car.
The situation is diagrammed in Figure 1.12. (Of course the ground is not exactly at and
your eyeballs are not quite at ground level but these details dont make much dieren
14
CHAPTER 1. MATHEMATICAL CONCEPTS
30o
y
q
2.0 m
q
1.0 m
(b)
(a)
Figure 1.13: Isocelestriangle shaped Christmas tree
y
290 N
52
o
x
Figure 1.14: Force vector for Example 10.
1.2.3
Vectors and Vector Addition
10. A force vector points at an angle of 52 ab
15
1.2. WORKED EXAMPLES
y (N)
R
0.5 m
2.1 m
20 o
5.0 m
x (E)
Figure 1.15: Displacements of the golf ball in Example 11.
(b) We also have
tan(52 ) =
Then solve for Fx :
Fx =
Fy
(290 N)
=
Fx
Fx
(290 N)
= 227 N
(tan 52 )
11. A golfer, putting on a green, req
4
CHAPTER 1. MATHEMATICAL CONCEPTS
R
R
D
(b)
(a)
Figure 1.2: (a) Circle; C = D = 2R; A = R2 . (b) Sphere; A = 4R2; V = 4 R3. Youve seen these
3
formulae before. Oh, yes you have.
A
R
h
(a)
h
(b)
Figure 1.3: (a) Circular cylinder of radius R and height h.
3
1.1. THE IMPORTANT STUFF
y
z
x
x
(a)
y
(b)
Figure 1.1: (a) Rectangle with sides x and y. Area is A = xy. I hope you knew that. (b) Rectangular box
with sides x, y and z. Volume is V = xyz. I hope you knew that too.
If we have to convert 3.68 104 s to mi
Math 3311 Abstract Algebra I
Solutions to Quiz 4 review problems
Section 14:
In Exercises 1 through 8, nd the order of the given factor group.
2. Z4
Z12 = (h2i
h2i)
Solution: We observe that
jh2ij =
jh2ij =
4
4
= = 2 in Z4 ; and
gcd (2; 4)
2
12
12
=
= 6 i
Math 3311 Abstract Algebra I
Solutions to Review Problems for Quiz 1
Section 1
Exercise 1-9: Compute the given arithmetic expression and give the answer in the form a + bi for
a; b 2 R
3. i23
Solution: We observe that
i2 =
1; i3 = i2 i =
i; i4 = i2 i2 = (
Math 3311 Abstract Algebra I
Review Problems for Quiz 3
Section 9
Exercise 7-9: In exercises 7 through 9, compute the indicated product of cycles that are permutations
of f1; 2; 3; 4; 5; 6; 7; 8g.
7. (1; 4; 5)(7; 8)(2; 5; 7)
Solution: Let
= (2; 5; 7);
= (
Math 3311 Abstract Algebra I
Quiz 5 Review
Section 18:
In the following exercises, decide whether the indicated operations dened on the given set, give a
ring structure. If a ring is not formed, tell why? If a ring is formed, state whether it is commutati
Math 3311 Abstract Algebra
Solutions to Review Problems for Quiz 2
Section 3:
Exercise 2-10: Determine whether the given map
the second. If it is not an isomorphism, why not?
2. hZ; +i with hZ; +i ; where
Solution: The map
(n) =
is an isomorphism of the r
College Algebra by Beecher, Penna, Bittinger textbook.
1314 Project
Domain and Range of a Square Root Function ~ Student Handout
This activity is to be completed in conjunction with studying section 1.2.
Students will work in groups of two or three. Each
Calculus Review
Texas A&M University
Dept. of Statistics
In This Review
Differentiation
Differentiation formulas
Examples
Trig. functions
Integration
Indefinite integrals
Definite integrals
Trig. Functions
Also In This Review
Calculus in statisti
Chapter 1
Mathematical Concepts
1.1
1.1.1
The Important Stu
Measurement and Units in Physics
Physics is concerned with the relations between measured quantities in the natural world. We
make measurements (length, time, etc) in terms of various standards f
2
CHAPTER 1. MATHEMATICAL CONCEPTS
1.1.2
The Metric System; Converting Units
To make the SI system more convenient we can associate prexes with the basic units to
represent powers of 10. The most commonly used prexes are given here:
Factor
1012
109
106
10
7
1.1. THE IMPORTANT STUFF
A
y
y
x
x
A
(a)
(b)
Figure 1.7: Vectors can have negative components when theyre in the other quadrants.
clockwise from the +x axis, then the component of this vector that runs along x has length
Ax , where the relation between
16
CHAPTER 1. MATHEMATICAL CONCEPTS
+y
B
A
20.0
o
60.0
o
+x
C
Figure 1.16: Vectors for Example 12.
and the direction of the net displacement, as measured in the usual way (North of East)
is given by , where
(1.22)
Ry
=
= 0.175
tan =
Rx
(6.97)
so that
= t
17
1.2. WORKED EXAMPLES
y
R
x
Figure 1.17: Vector R lies in quadrant II.
and
Cx = 0
Cy = 4.00 m
The resultant (sum) of all three vectors (which we call R) then has components
Rx
Ry
=
=
Ax + Bx + Cx = 4.698 m + 2.500 m + 0 m = 2.198 m
Ay + By + Cy = +1.710
30
CHAPTER 2. MOTION IN ONE DIMENSION
v0 = +30.0 m/s
v0 = -30.0 m/s
Figure 2.8: Two pellet guns shoot pellets; pellet from Gun A goes up then down. Pellet from B goes
straight down.
Hoo! This one sounds complicated. And they didnt even tell us how high th
31
2.2. WORKED EXAMPLES
We get:
(60 m )
s
= 6.1 s
t=
m
(9.80 s2 )
Summing up, it takes 6.1 s for pellet A to go up and back down to the original height;
this is the amount of time it spends in the air longer than the time B is in the air. So pellet
A hits
Chapter 3
Motion in Two Dimensions
3.1
3.1.1
The Important Stu
Motion in Two Dimensions, Coordinates and Displacement
We will now deal with more general motion, motion which does not take place only along a
straight line.
An example of this is shown in Fi
35
3.1. THE IMPORTANT STUFF
is called the (instantaneous) speed of the particle. Speed is always a positive number and
like velocity it has units of m .
s
The instantaneous x and y accelerations are dened by:
ax =
vx
for small t
t
ay =
vy
for small t
t
(3