UCLA Mathematics 110A: selected solutions from homework #1
David Wihr Taylor July 2, 2010
Introduction
When reading these solutions always keep in mind the common techniques being used. The point of h
1. Solve the following congruences. If there are no solutions, say so, and give some justification. (You do not need to give a complete proof.) If there are multiple solutions, be sure to find all of
Solutions to Assignment 1
Question 1. [Exercises 1.1, # 6] Use the division algorithm to prove that every odd integer is either of the form 4k + 1 or of the form 4k + 3 for some integer k. Solution: F
1
. | c| < r 0 d n a r + qc- = a yltnelaviuqe ,ro , | c| < r 0 dna r + q|c| = a taht hcus r dna q sregetni euqinu tsixe ereht :mhtiroglA noisiviD eht ylppa nac ew dna evitisop si |c| = c- neht ,evitag
Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2)
Problems from 2.1 2.1.1. Prove that a b (mod n) if and only if a and b leave the same remainder when divided by n. Proof. Su
THREE PROBLEMS
PROBLEM 1: Prove that 2 is irrational. a Proof: Assume to the contrary that 2 is rational, that is 2 = , where a and b are integers b and b = 0. Moreover, let a and b have no common div
Introduction to Ring Theory. Math 228
Unless otherwise stated, homework problems are taken from Hungerford, Abstract Algebra, Second Edition.
Homework 4 - due February 9
2.1.11 b, d: Find all solution
Math 110A Homework #2
David Wihr Taylor
Summer 2006
1. Problem 1.3.1: Express each number as the product of primes: (a)5040, (b) −2345, (c) 45670,
(d) 2042040.
Answer: To my knowledge there is no fast
HW 1 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson 1.1: Prove 12 + 22 + . . . + n2 = 1 n(n + 1)(2n + 1) for all natural 6 numbers n. The proof is by induction. Call the nth proposition Pn
Chapter 2, Congruence in Z and modular arithmetic. This leads us to an understanding of the kernels and images of functions between rings (ideals, quotient rings, ring homomorphisms). It will also giv
Math 453 Problem Solutions, Fall 2006 HW# 1 1.3 #12. Show that
n j =1
j j ! = (n + 1)! 1 for every positive integer n. When n = 1, both sides of the equation are
k j =1
1, and this suces for the base
Homework 2 Solutions
1. (1.10) Prove (2n + 1) + (2n + 3) + (2n + 5) + + (4n 1) = 3n2 for all n N, n 1. Solution: Base step: When n = 1, we get 3 = 3, which is true. Inductive step: Suppose (2k + 1) +
Math 310, Section 2, Fall 2009
Remarks on HW Section 1.1
From Section 1.1: 1a,b,c, 6, 8
1 Find the quotient and remainder when a is divided by b: (a) a = 302, b = 19 (b) a = -302, b = 19 (c) a = 0, b
Introduction to Ring Theory. Math 228
Unless otherwise stated, homework problems are taken from Hungerford, Abstract Algebra, Second Edition.
Homework 3 - due February 2, 2010
D.3(a): Prove that the f
Math 110A Homework #1
David Wihr Taylor
Summer 2006
1. Problem 1.1.1: Find the quotient and remainder when a is divided by b:
(a) a = 302, b = 19
(b) a = −302, b = 19
(c) a = 0, b = 19
Answer:
(a) 302