. a KlsiNmrKnT- x \@ DashboardiCz x \M VourGoogleF' x/ p Caunungnne x \@ Coulse: MATHE x "\(, Loang. x \5 richard brualdi x "\g Downloads x . ><
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SECTION: 5.3
Q#7
First draw (B-C)
Now draw A-(B-C)
Now ven diagram for (A-B) is
Now for (A-B)-C
From figures
(b)
Q#11
(b)
(b
(e)
SECTION# 5.4
Q#4
Q#07
(a) AxB=cfw_1xcfw_2=cfw_(1,2)
BXA=cfw_2Xcfw_1=cfw_(2,1)
S0 AXB BXA
(b) AX(BXC)=cfw_1X(cfw_2,cfw_3)
=(1,(

Chapter VIII
Ordered Sets, Ordinals and Transfinite Methods
1. Introduction
In this chapter, we will look at certain kinds of ordered sets. If a set \ is ordered in a reasonable way,
then there is a natural way to define an order topology on \ . Most inte

Catalan Numbers
Tom Davis
tomrdavis@earthlink.net
http:/www.geometer.org/mathcircles
February 19, 2016
We begin with a set of problems that will be shown to be completely equivalent. The solution to
each problem is the same sequence of numbers called the

CURRENT HISTORY
September 2016
The absence of effective regional institutions means that the geopolitical competition between the United States
and China is likely to play out without the constraining influence such organizations can provide . . .
Can the

ICS 241: Discrete Mathematics II (Spring 2015)
9.6 Partial Orderings
A relation R on a set S is called a partial ordering or partial order if it is reflexive, antisymmetric,
and transitive.
Poset
A set S together with a partial ordering R is called a part

Math 378 Spring 2011
Assignment 5
Solutions
Brualdi 7.1.
Solution.
Conjecture (a). f1
f3
f2n
Let n 1, f1 1 f2 .
f2n1
Proof. [By Induction] Base case:
f2N 1 f2N for all N n. Note that
f2n
f2n
f2n
f2n
f2n1
f2n1
f2n1
f2n1
Now, assume that f1
f3
f2n2
f2p

Name
MATH 385
Quiz 2 (20 points)
(1) Prove that a partially ordered set is a total order if and only if it is a linear order.
1
2
(2) Let R be a relation from R R to R R defined as
(a, b)R(c, d) if and only if either a < c or both (a = c and b d).
Here th

Exercise#4.1
Q#3(b)
Let P (k) be the statement 1+5+9+ (4k-3) = k (2k-1)
Basis Step: Show true for n=1.
LHS: 1
RHS: n(2n-1) = 1*(2*1-1) = 1*1 = 1
Since RHS = LHS, weve proved the basis step.
Inductive Step: Show that P(k) P(k+1)
Suppose P(k): 1+5+9+(4k-3)