MODEL ANSWERS TO THE SECOND HOMEWORK
1. Label the vertices of the square A, B, C, D, where we start at
the top left hand corner and we go around the square clockwise. In
particular A is opposite to C and B is opposite to D.
There are three obvious types o
Midterm 2
Zonglin Jiang
December 2, 2014
1.
Suppose Z12 Z9 is cyclic, i.e. its equal to < ([x]12 , [y]9 ) > for some x, y Z. Hence
|Z12 Z9 | = | < ([x]12 , [y]9 ) > | = o([x]12 , [y]9 )
However, since 12|36x and 9|36y
36([x]12 , [y]9 ) = ([36x]12 , [36y]9
Midterm 1
Zonglin Jiang
December 2, 2014
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?
1. Suppose 3 is rational, that is 3 m for some m, n P Z. Taking square on both
n
2
sides, we have 3 m2 , 3n2 m2 . Applying the function 3 , we have
n
3 p3n2 q 3 pm2 q
3 p3q ` 3 pnq ` 3 pnq 3 pmq ` 3 pmq
1 ` 23
Math 100A, Fall 2012, Quiz, 10/22/12
Instructions. Answer all questions.
1. (10 points) Let p be a prime number of the form 4n + 3 for n an integer. Show
there is no integer x such that
x2 1 mod p.
(Hint: Use Fermats little theorem which says xp x mod p.)
Math 100A, Fall 2012, Midterm, 11/9/12
Instructions. Answer all questions.
1. (a) (5 points) Show that the remainder of 4121 when divided by 17 is 4.
(b) (10 points) Solve the following system of congruences:
x 3 mod 5, x 4 mod 7.
Solutions. (a) 4121 = (4
MATH 100A SYLLABUS
AUTUMN 2016
Lectures
MWF 1:00-1:50, CENTR 113
Instructor
James Mc Kernan, APM 6260, phone (858)-534-6347
Office Hours
M 2:00-4:00PM
or by appointment, if you cannot make these times.
Teaching Assistants Iacopo Brivio [email protected]; D
MODEL ANSWERS TO THE FOURTH HOMEWORK
2. Chapter 3, Section 1: 1 (a)
1 2 3 4 5 6
.
4 5 2 1 3 6
(b)
1 2 3 4 5
.
3 1 2 4 5
(c)
1 2 3 4 5
.
1 4 3 2 5
5. It suffices to find the cycle type and take the lowest common multiples of the individual lengths of a cyc
MODEL ANSWERS TO THE FIFTH HOMEWORK
1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z,
(ab) = [ab]
= [a][b]
= (a)(b).
This map is clearly surjective but not injective. Indeed the kernel is
easily seen to be nZ.
(b) No. Suppose that G is not abelian and
MODEL ANSWERS TO THE SIXTH HOMEWORK
1. Chapter 3, Section 6: 1 There are two cosets. The first coset is
[1] = N , the second is the coset containing 1, which is the set of all
negative real numbers.
[1] [1] = [1], [1] [1] = [1] [1] = [1] and [1] [1] = [1]
100A: Main Results
Zonglin Jiang
November 20, 2014
1
Elementary Number Theory
Most of this section are already tested in the rst midterm. However, its possible to be tested again
according to the syllabus.
Theorem 1 (Euclidean algorithm) m Z, n Z \ cfw_0,