Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

30.1 VENN DIAGRAMS


gets beyond three it becomes impossible to draw a simple two-dimensional Venn


diagram, but this does not change the results.


The 2nregions will break down inton+ 1 types, with the numbers of each type

as follows§


no events, nC 0 =1;
one event but no intersections, nC 1 =n;
two-fold intersections, nC 2 =^12 n(n−1);
three-fold intersections, nC 3 =3!^1 n(n−1)(n−2);
..
.
ann-fold intersection, nCn=1.

That this makes a total of 2ncan be checked by considering the binomial


expansion


2 n=(1+1)n=1+n+^12 n(n−1) +···+1.

Using Venn diagrams, it is straightforward to show that the operations∩and

∪obey the following algebraic laws:


commutativity, A∩B=B∩A, A∪B=B∪A;
associativity, (A∩B)∩C=A∩(B∩C), (A∪B)∪C=A∪(B∪C);
distributivity, A∩(B∪C)=(A∩B)∪(A∩C),
A∪(B∩C)=(A∪B)∩(A∪C);
idempotency, A∩A=A, A∪A=A.

Show that(i)A∪(A∩B)=A∩(A∪B)=A,(ii) (A−B)∪(A∩B)=A.

(i) Using the distributivity and idempotency laws above, we see that


A∪(A∩B)=(A∪A)∩(A∪B)=A∩(A∪B).

By sketching a Venn diagram it is immediately clear that both expressions are equal to
A. Nevertheless, we here proceed in a more formal manner in order to deduce this result
algebraically. Let us begin by writing


X=A∪(A∩B)=A∩(A∪B), (30.4)

from which we want to deduce a simpler expression for the eventX. Using the first equality
in (30.4) and the algebraic laws for∩and∪,wemaywrite


A∩X=A∩[A∪(A∩B)]
=(A∩A)∪[A∩(A∩B)]
=A∪(A∩B)=X.

§The symbolsnCi,fori=0, 1 , 2 ,...,n, are a convenient notation for combinations; they and their
properties are discussed in chapter 1.
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