MAT 115A, SSI16
Homework 1
This homework is due on Tuesday, June 28 in class. Your homework must be typed in LATEX.
1. (1.1.5) Prove that
3 is irrational using the Well-Ordering Principle.
2. (1.1.12) Show that for all x R, we have bxc + bx + 0.5c = b2xc.
MATH 115A Number Theory, Homework 3 Solutions
(1) Note that
(6k + 7)(1) + (3k + 4)(2) = 6k 7 + 6k + 8 = 1.
This shows the GCD of 6k+7 and 3k+4 must divide 1, which means (6k + 7, 3k + 4) = 1. By
definition, this means the fraction (6k + 7)/(3k + 4) is in
MATH 115A Number Theory, Homework 2 Solutions
(1) Consider this arguments inductive step in the case n + 1 = 2: the sets cfw_1 and cfw_2 dont have any
common members, so the argument doesnt work in this case.
(2) When n = 1, we just have a 2 2 chessboard.
MATH 115A Number Theory, HW5 Solutions
1. Suppose all of the powers in the prime power factorization of n are even. Then we can write n =
1 2n2
r
p2n
p2n
for some primes p1 , . . . , pr , and integers n1 , . . . , nr . But then
r
1 p2
1 2n2
r
p2n
= (pn1
MATH 115A Number Theory, Homework 1 Solutions
(1) (a) This is true. Let a/b be a rational number (so a and b are both integers, b 6= 0) and let x be an
irrational number. If a/b + x were a rational number, we could write it in the form
a/b + x = c/d
for s
PRACTICE PROBLEMS
1.
Part 1: Greatest Common Divisor, Fundamental Theorem of
Arithmetic
1. Prove or disprove: (a1 , ., ai ), (ai+1 , ., an ) = (a1 , ., an ).
2. Describe the set cfw_275n + 625mm, n Z as the set of all multiples of some xed
integer. Find t
STUFF TO REMEMBER 2
1.
On Polynomials over Z/pZ
Proposition 1. Suppose f (x) is an integer-coecient polynomial. x0 is a root of
multiplicity n if and only if f (x0 ) = f (x0 ) = . = f (n1) (x0 ) = 0, and f (n) (x0 ) 0.
Remark This is a result on regular p
LECTURE 22
1.
eg.
More Examples
Solve 3x 4x + 5 0(mod 5 72 ).
3
2
First, solve 3x3 4x+5 3x3 4x = x(3x2 4) 0(mod 5). Note that 3x2 4 0(mod
1
5) is equivalent to x2 3(mod 5) (3 = 2) and 3 is a quadratic non-residue mod 5.
So, x = 0 is the only root.
Let f (
LECTURE 14
1. Arithmetic Progression: a Summary
So far, we have proved various cases of the statement that there are innitely
many prime numbers of the form
aq + r,
for specic
a, r
such that
(a, r) = 1,
using
two techniques:
(1) If in
Z/aZ, i
is a unit su
STUFF TO REMEMBER
Proposition 1.
Proposition 2.
Proposition
3.
(a, b) = mincfw_ma + nbm, n Z, ma + nb > 0.
cfw_ma + nbm, n Z = (a, b). (n)
For any integer
n
is the set of all multiples of
n.)
that is not prime, it has a prime factor less or
n.
equal than
LECTURE 17
1. Warm-up
eg.
of
Classify all units of
12
Z/13Z
by their orders.
(13) = 12
and the positive factors
are 1,2,3,4,6,12.
Orders
Number of Units with this Order
1
2
3
4
6
12
You have to consider
List of them
(1) = 1
1
(2) = 1
1 = 12
(3) = 2
3, 9
(
LECTURE 13
1.
Application to Cryptography
As one important application of modular arithmetic, in this section we introduce
the basic idea behind cryptography.
1.1.
Caesar Cipher.
A most basic idea behind ciphering is that one can identify
Roman letters wi
LECTURE 9
Problem Session
1.
eg.
Since
eg.
In
q7 (9) = 2
and
3
2 = 1,
we get
Find a positive integer
Z/8Z, 3 3 = 1;
eg.
In
q7 (910000 ).
Find
n
such that
so one can take
Find a positive integer
m
3333
q7 (910000 ) = 1
n
3 =1
in
2 = 2.
Z/8Z.
n = 2.
such t
LECTURE 11
1.
Warm-up
eg. Show that there are innitely many prime numbers of the form
Proof.
6n + 5.
1, 5 are the only units in Z/6Z. So an odd prime number greater
N
6n + 1 or of the form 6n + 5. Since 1 5 and 2, 3 do
6n + 5, we conclude that an integer
MATH 115A Number Theory, Quiz 3
1. Show that every positive integer can be written as the product of a square (possibly 1) and a square-free
integer. A square-free integer is an integer that is not divisible by any perfect squares other than 1.
Solution.
MATH 115A Number Theory, Homework 3 Solutions
(1) (a) We know there exist integers m, n and x, y such that
ma + nb = 1 = xa + yc.
Then, multiplying the two equations, we see
1 = (ma + nb)(xa + yc) = maxa + mayc + nbxa + nbyc = (max + myc + nbx)a + (ny)(bc
MAT 115A
HW3
Joshua Sumpter
HW 3 Solutions
1
Book Problems
3.5.12) Let n be a positive integer. Show that the power of a prime p occurring in the prime factorization of n! is
given by [n/p] + [n/p2 ] + [n/p3 ] + . . .
Proof. We consider the product form o
MAT 115A
HW1
Joshua Sumpter
HW 1 Solutions
1b) Given that fn An , show that lim
n!1
fn+1
= .
fn
fn+1
An
= 1 and that lim
= 1.
n+1
n!1 A
n!1 fn
Proof. By our conjecture, we know that lim
It follows that
fn+1
An
1 = lim
lim
n!1 An+1
n!1 fn
An
fn+1
= lim
n!1
MAT 115A
HW2
Joshua Sumpter
HW 2 Solutions
1
book problems
2.3.18) Use identity (2.2) with n = 4, and then with n = 2, to multiply (10010011)2 and (11001001)2 .
Solution:
We have the following identity: ab = (22n + 2n )A1 B1 + 2n (A1 A0 )(B0 B1 ) + (2n +
Dept. of Computer Science, University of California, Davis
ECS132
Instructor: Rob Gysel
7/14/17
Homework #3: Due Monday, 7/17/17 by
11:59pm
Written homeworks are to be turned in on Gradescope by the due date.
Due to the odd summer schedule, the R program
MATH 115A Number Theory, HW8 Solutions
1. The units digit of 7999,999 is given by the least positive residue of 7999999 mod 10. Note that (7, 10) = 1,
(10) = 4, and 999999 = 249999 4 + 3, so by Eulers theorem
7999999 = (74 )24999 73 1 73
973
mod 10
mod 10
LECTURE 2
1.
eg.
Divisibility
Every non-zero integer divides 0. The only positive divisor of 1 is 1. If
positive divisor of another positive number
b,
then
a
b a 0.
is a
Here are some simple facts:
Lemma 1.
3 integers. If
(2)Let
Suppose
(1)Let a, b, c be
LECTURE 6
1.
eg.
Problem Session
Pell's equation re-visited.
2
x
ny 2 =
1, where n > 0 is not a perfect square, we are studying units in the ring (Z[ n], +).
As we've discussed earlier, when we solve for solutions to the equation
Denition 1.
Let
R
be a r
LECTURE 1
1.
Introduction to this course
Structures are the weapons of the
mathematician.
Nicolas Bourbaki
1.1.
The Study of Structures.
As many of you might know, number theory is the
study of integers. This is a good starting point, but what do we mean
LECTURE 7
1. Motivation for Problem 10, HW2
Prime numbers are the central objects of study in number theory. However, it
is hard to write down a simple formula that produces lots of prime numbers. Euler
produces the following quadratic formula:
f (x) = x2
LECTURE 18
1. Warm-up
eg.
Solve
x2 + 1 0(mod
100).
First of all, take the prime factorization of 100:
100 = 22 52 .
This is equivalent to solving the system
x2 + 1 0(mod 22 )
2
x + 1 0(mod 52 ).
The rst equation does not have a solution since Z/4Z does n