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48 Pages

### v43-f10productsfinal

Course: MATH 301, Spring 2011
School: NYU
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Word Count: 22896

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c Notes: F.P. Greenleaf 2003 - 2010 v43-f10products.tex, version 12/1/10 Algebra I: Chapter 6. The structure of groups. 6.1 Direct products of groups. We begin with a basic product construction. 6.1.1 Definition (External Direct Product). Given groups A1 , . . . , An we define their external direct product to be the Cartesian product set G = A1 . . .An equipped with component-by-component multiplication of...

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NYU - MATH - 301
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NYU - MATH - 301
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NJIT - CHEM 126 - 126
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University of Toronto - MAT - 102
1.13. Let x A. Then x = 2k - 1 for some k Z. Let k1 = k - 1. We then have 2k1 + 1 B, but also 2k1 + 1 = 2(k - 1) + 1 = 2k - 1 = x, so x B. Hence A B. Conversely, let x B so that x = 2m + 1 for some m Z and let m1 = 2m - 1. Then 2m1 - 1 A and x = 2m + 1 =
University of Toronto - MAT - 102
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University of Toronto - MAT - 102
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University of Toronto - MAT - 102
2.3. In symbols, the sentence is &quot;(a R)(x R)P (a, x) Q(x).&quot; This is false since P (0, 1) is true, but Q(1) is false. One way to change the quantifiers to make a true statement is &quot;(a R - cfw_0)(x R)P (a, x) Q(x).&quot; Another way is &quot;(a R)(x R)P (a, x) Q(x).&quot;
University of Toronto - MAT - 102
2.5. (a) We claim x = (y - b)/m is a solution. We have m y-b + b = (y - b) + b = y, mso this is indeed a solution to y=mx+b as claimed. To show uniqueness, suppose x1 and x2 are solutions. Then mx1 + b = y = mx2 + b mx1 = mx2 x1 = x2 . In other words, al
University of Toronto - MAT - 102
3.1. Let P (n) be &quot;n &lt; 100&quot;. 3.2. Let Q(n) be P (n). By induction, Q(n) is true for all n N, so P (n) is false for all n N. 3.6. This is false. Let P (n) be &quot;n &gt; 1.&quot; 3.11. We use induction on n. If n = 1, then a set with n elements has precisely 2 = 21 su
University of Toronto - MAT - 102
3.33. The largest length for a subinterval as described in the exercise is n - 1 (consider [1, n] itself) and the smallest is 0 (obtained by the intervals of the form [i, i] , 1 i n). Fixing i N, 0 i n - 1, we wish to count the number of subintervals of l
University of Toronto - MAT - 102
4.5. There is such a bijection if and only if A contains at least two elements. If A contains only one element or is empty, then the only function from A to A is the identity function. If A does contain at least two elements, then let x, y A, x = y. Defin
University of Toronto - MAT - 102
4.12 (a) False. One counterexample is f (x) = e-x . (b) False. Take f (x) = 0. (Any constant function will work.) (c) False. Consider x if 0 = x = 1, 1 if x = 0, f (x) = 0 if x = 1. (d) True. If |f (x)| M for every x R, then M + 1 is not in the image of f
University of Toronto - MAT - 102
4.26 Suppose f (x) = f (y). Then |f (x) - f (y)| = 0, so c |x - y| 0. Since c &gt; 0 and &gt; 0, this implies that |x - y| = 0 and hence x = y. Thus f is injective. 4.32 We have f (-x) = -(-x) = x and g(y -1 ) = (y -1 )-1 = y for every x F and y F - cfw_0, so f
University of Toronto - MAT - 102
4.33 (a) Let f : A B, g : B C be injections. Suppose (g f )(x) = (g f )(y). This means g(f (x) = g(f (y). Since g is injective, this implies that f (x) = f (y). But f is also injective, so x = y. It follows that g f is injective. (b) Let f : A B, g : B C
University of Toronto - MAT - 102
Homework 12 - MATH 310 0101 - Summer I 2010 - Due by Thursday, June 241. Prove that if A is countable, then Ak is countable for every k N. (Hint: Use induction. You will want to use the fact that Ak+1 = Ak A ; you must prove this in your solution.) 2. Pr
University of Toronto - MAT - 102
1 First we outline the proof that Ak+1 = Ak A . We define a map (a1 , a2 , . . . , ak+1 ) (a1 , a2 , . . . , ak ), ak+1 ). This map is easily shown to be a bijection from Ak+1 to Ak A. Now we prove the main result by induction on k. For the base case, we
University of Toronto - MAT - 102
Homework 13 - MATH 310 0101 - Summer I 2010 - Due by Thursday, June 25For each of the following relations, determine whether it is reflexive, symmetric, and/or transitive. In each case either prove it satisfies the property or display a counterexample. U
University of Toronto - MAT - 102
1. Reflexive, symmetric, and transitive. Equivalence relation. 2. Reflexive and symmetric, but not transitive. Not an equivalence relation. 3. Transitive, but neither reflexive nor symmetric. Not an equivalence relation. 4. Symmetric, but neither reflexiv
University of Toronto - MAT - 102
Homework 14 - MATH 310 0101 - Summer I 2010 - Due by Wednesday, June 301. Exercise 7.6 2. Exercise 7.8 3. Let A and B be partial orders on A and B respectively. Define a relation on A B by (a, b) (a , b ) if and only if either a A a and a = a or a = a an
University of Toronto - MAT - 102
Homework 15 - MATH 310 0101 - Summer I 2010 - Due by Wednesday, June 301. Let S R. Prove that if S has a minimum element , then S has an infimum with inf S = . 2. Prove that inf(a, b) = a. 3. Exercise 13.8 4. Exercise 13.19 5. Exercise 13.20 6. Exercise
University of Toronto - MAT - 102
1. By the definition of a minimum, x for every x S, so is a lower bound for S. If is an another lower bound for S, then, in particular, since S. Hence must be the greatest lower bound of S. In other words, = inf S. 2. a is a lower bound for (a, b) simply
University of Toronto - MAT - 102
13.12 This is false. Let S = cfw_0, 1 and set xn = 1, yn = 0. 13.23 Since sup A a and sup B b for every a A and b B, sup A+sup B a+b for every a+b C. Thus sup A+sup B is an upper bound for C. By Proposition 13.15, there are sequences an in A and bn in B s
University of Toronto - MAT - 102
Homework 17 - MATH 310 0101 - Summer I 2010 - Due by Thursday, July 81. Exercise 14.1 2. Exercise 14.2 3. Exercise 14.29 4. Define f : R R by f (x) = -1 if x &lt; 0, 1 if x 0.Show that f is not continuous at 0. Do this in two ways: (a) Use the &quot; - &quot; defini
University of Toronto - MAT - 102
1. n is an unbounded sequence with no bounded, and hence no convergent, subsequence. an defined by an = n if n is odd 0 if n is evenis an unbounded sequence that has 0 as a convergent subsequence. 2. (a) Let an = (b) Let an =n i=1 n i=11/n. n.(c) Let
University of Toronto - MAT - 102
Homework 18 - MATH 310 0101 - Summer I 2010 - Due by Thursday, July 81. Show that the discrete metric is a metric. 2. Show that (a, b) is open in R. 3. Show that the union of open sets is open. 4. Show that a finite subset of a metric space is closed. 5.
University of Toronto - MAT - 102
1. We simply verify the properties of a metric. Positive definite: This is just a trivial observation from the definition. Symmetric: Obviously d(x, x) = d(x, x). If x = y, then d(x, y) = 1 = d(y, x). Triangle Inequality: If x = z, then d(x, z) = 0 = 0+0
University of Toronto - MAT - 102
MAT102S - Introduction to Mathematical Proofs - UTM - Spring 2010 Solutions to Selected Problems from Problem Set DFor the first two questions (about fields) see solutions to Quiz #2. 2.2. Take a = 0 and b = 1. The statement claims that there are integer
University of Toronto - MAT - 102
University of Toronto - MAT - 102
University of Toronto - MAT - 102
MAT102S - Introduction to Mathematical Proofs - UTM Problem Set A - Spring 2010Here are the Exercises assigned: 1.2. Fill in the blanks. The equation x2 +bx+c = 0 has exactly one solution when and it has no solutions when . 1.5. (-) Consider the Celsius
University of Toronto - MAT - 102
University of Toronto - MAT - 102
University of Toronto - MAT - 102
University of Toronto - MAT - 102
University of Toronto - MAT - 102
MAT102S - Introduction to Mathematical Proofs - Spring 2010 - UTM Problem Set E - TO BE SUBMITTED TO YOUR TADue:Monday, February 8, in tutorials. This assignment must be submitted to your TA at the beginning of the tutorial on the above date. Marking s