Download Boolean Algebra Solution to Homework 11 Spring 2014

Boolean Algebra Solution to Homework 11
Spring 2014
Problem 1: Consider the set A of all divisors of 24, A = { 1, 2, 3, 4, 6, 8, 12, 24}, with the operations
defined as follows: a♥b = LCM(a, b), a♦b = GCD(a,b) and the complement( or negation) is defined to be
ac = 24/a.
Intuitively do the following examples indicate that A is a Boolean Algebra or not.
SOLUTIONS:
1) Is the following property true for 3♥(4♦8) = (3♥4)♦(3♥8) ?
3♥(4♦8) = (3♥4)♦(3♥8)
3♥(4) = (12)♦(24)
12 = 12 so the distributive property holds for this example.
2) Find the identities for each of the operations. Justify each choice with an example.
The identity for LCM is 1 and the identity for the GCD is 24.
Example the LCM (6, 1 ) = 6 and GCD (6, 24) = 6
3) Evaluate 6♥24 = ? and 3♦1 = ?
LCM(6,24) = 24 and GCD(3, 1) = 1 which illustrates the Dominance Law
4) Evaluate 6♥6c = ? and 3♦3c = ?
LCM(6, 6c) = LCM(6, 4) = 12, GCD(3,3c) = GCD(3, 8) = 1
This is not a Boolean Algebra, although the results is correct for GCD(3, 3c) it is not the correct
results for LCM(6, 6c) which should have been 24.
Problem 2: Given elements a, b in the Boolean algebra B with operations  and  where m is the
identity for  and p is the identity for , Justify each step of the proof below . The inverse of any
element a is a/.
Theorem: For a, b in B, (a  b)/  (b  b) = m
Proof:
(a  b)/  (b  b) =
__ given
/
(a  b)  b =
__Idempotent Law
/
/
(a  b )  b =
_ DeMorgan’s Law
/
/
a  (b  b) =
__Associative Law
/
a m =
__Inverse (Complement) Law
m
__Dominance Law
Problem 3: : Given elements a, b in the Boolean algebra B with operations  and  where k is the
identity for  and h is the identity for . Let b be the complement of b. Justify each step of the proof
below.
Theorem: For a, b in B, [b  (b
a)]
[b  (b
(b
a)  a
a)]
(b
a) 
Given
[b  (b  a )]
(b
a) 
DeMorgan's Law
[(b  b)  a)]
[(b)  a)]
[b  a )]
(b
(b
(b
a) 
a) 
Associative Law
Idempotent Law
(b  a ) 
DeMorgan's Law
b)  a 
Distributive Law
( h)  a 
Inverse Law
a
Identity Law
[b  (b
a)]
(b
a)  a