epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 145
Therefore
(^1 n ^2 n... n ^„ n An+x)'^ {{Ax n yi2 n... n A„) n An+it
Using the fact that we are now deahng with two sets (namely the set in
parentheses and ^„+i), we have
(^1 n ^2 n... n ^„ n An+x) = [{Ax n ^2 n... n ^„) n An+x^
= (^inA2n...n^yuA;+i.
The inductive hypothesis and the associative property of intersection
yield
(^1 n ^2 n... n ^ n An+x)' = [{Ax n ^2 n... n A„) n A^+i]'
= {AxnA2r\,..f\An)'yJA!^^^
^(4u4u...u4)u^;^i
= 4u4u...u4u4^i.
Therefore, by the principle of mathematical induction, the equahty is
true for all n > 2.
9. a. The smallest value of n we can use is 0. If a set has 0 elements it is
the empty set. The only subset of the empty set is itself. Therefore, a
set with 0 elements has 1 subset, and 1 = 2^. So the statement is
true for n = 0.
b. Assume that if A = {xx,X2,.. .,Xn}, then A has 2" subsets,
c. Let B= {bx.bi,... ,bn,bn^x] be a set with n+1 elements. Does B
have 2"+^ subsets? We can write B = {fci,&2,• •. ,fcn} U {&„+i}. The
set {bx,b2,. ..,bn} has n elements; therefore, it has 2" subsets,
Bx, B2, B3,..., J52«. None of these subsets includes bn+x- Every one of
these subsets is a subset of B as well, and it uses only the first n
elements of B. Thus, we can construct more sets of B using bn+x-
Thus, we have the following subsets of B: Bx, B2, B3,..., B2n,
Bx U {fc„+i}, B2 U {fc„+i}, 53 U {b„+i},.. .,^2" U {fe„+i}. Therefore, B
has 2" + 2" = 2"+^ subsets. Thus, by the principle of mathematical
induction, the statement is true for all n>0.
EQUALITY OF NUMBERS
1. Part 1. Let us assume that {x - yf -\-{x- y)^ = 0. Will this imply that
x = yl Using the distributive property, we can rewrite this equahty as
{x — y)^[{x — yf' + 1] = 0. The product of several factors is equal to
zero if and only if at least one of the factors equals zero. Therefore,
either {x -yf = 0 or [{x - yf -{-!] = 0. The first equahty implies