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 homework, and subsequently these solutions, is to give y
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 them. (a) 15x+1O=4inZ 20
(5 6/ 15-jq M J t
:2Ck 4r ,ie
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: For each positive integer n, the only possible remainder
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 ,evitagen si c fI .evorp ot erom gnihton si ereht os )1.1 mero
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. Suppose a b (mod n). Then, by definition, we have a - b =
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 divisor > 1. Then 2= a2 b2 2b2 = a2 . (1)
Since 2b2 is ev
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 solutions of each congruence: (b) 3x 1 (mod 7) (d) 6x 10 (mod 1
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 algorithm to factor primes that doesn’t use quantum
co
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 . The basis for induction P1 is the statement that 12 =
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 give us more examples of rings to think about. Denition. A
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 step. Now suppose that the equation is true for n = k ,
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) + (2k + 3) + (2k + 5) + + (4k 1) = 3k 2 . Well prove it h
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 = 19 Solution: (a) 302 = 19 15 + 17, q = 15, 0 r = 17 <
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 following relation on the set R of real numbers is an eq
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 = 15 · 19 + 17, so q = 15 and r = 17
(b) −302 = −16 ·