1. Prove DeMorgan's Laws. That is, let X be a set and let A and B be subsets of X. Show that (a) X – (AUB) = (X – A)n (X – B), and (b) X – (An B) = (X – A) U (X – B). a) X- CAUB) = (X - A)O(X- B) We have aE X- (A UB) So That is %3D ae X and a& A and agB. this only shows X – (AU В) сх — АПX - В. a¢ CAUB). QEX and 50, aE x-A and ae x-B. Hence ae Cx- A) N (x - B). Thus, X- LAU B) = \x-A)n(x-B) b) X- CANB) = (X - A)U (x- B) we have ae X- (ANBI so That is ae X and a& A or a¢B. a¢ CAn B). and this only shows X – (AN В) сх— АUX - B. 50, aE x-A a € x-B. Hence aG (x- A ) U(x-B). or Thus, X - LAN B) =(x-A)U(X-B)

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.2: Mappings
Problem 22E
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finish the argument is currently incomplete. ALso expalin why its incomplete 

1. Prove DeMorgan's Laws. That is, let X be a set and let A and B be subsets of X. Show that
(a) X – (AUB) = (X – A)n (X – B), and
(b) X – (AN B) = (X – A) U (X – B).
a) X- (AU B) = (X -A)N(x- B)
We have aE X- (A UB) So
That is
this only shows X – (AU
В) сх — Аnх- В.
and
ag (AU B).
QEX
and
a& A and a¢ B.
50,
aE x-A and ae x-B. Hence ac (x- A) N (x -B).
Thus, X - LAU B) = lX-A)n(x-B).
b) X- (AnB) = (X - A)U (x- B)
we have ae X- (ANB) So
That is
ae x and
a& A or a¢ B
a¢ CAn B).
and
this only shows X – (AN
В) сх - AUX — В.
So,
Thus, X - LAN B) = (x-A)U(x-B)
a E
X - A
a € X-B. Hence ae(x- A) U lx-B).
or
Transcribed Image Text:1. Prove DeMorgan's Laws. That is, let X be a set and let A and B be subsets of X. Show that (a) X – (AUB) = (X – A)n (X – B), and (b) X – (AN B) = (X – A) U (X – B). a) X- (AU B) = (X -A)N(x- B) We have aE X- (A UB) So That is this only shows X – (AU В) сх — Аnх- В. and ag (AU B). QEX and a& A and a¢ B. 50, aE x-A and ae x-B. Hence ac (x- A) N (x -B). Thus, X - LAU B) = lX-A)n(x-B). b) X- (AnB) = (X - A)U (x- B) we have ae X- (ANB) So That is ae x and a& A or a¢ B a¢ CAn B). and this only shows X – (AN В) сх - AUX — В. So, Thus, X - LAN B) = (x-A)U(x-B) a E X - A a € X-B. Hence ae(x- A) U lx-B). or
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