2534 Solutions for Test 1A Fall 2009
Problem 1: (12pts) P is the statement, “You do all the problems” and A is the statement “you get an A”
a)
A P
A is Sufficient and P is necessary
b)
P A P A! "
P is sufficient and A is necessary
c)
P A A P "
d)
( )A P A P A P# " ! "
So we have that a, c, d are equivalent.
Problem 2: (8pt)
a) There is a particular name brand that all boys play with.
b) For every name brand there is a boy that plays with it.
Problem 3: (12pts) It is given that
[ ] ( ) B C A D #
is given to be false. Therefore the sufficient
condition
[ ]B C
is true and the necessary condition
( ) A D#
is false. Given that A and D have the
same truth value and the conjuction is false, A and D are both false. It also given that
B
is false. This
make B true and therefore C is true since the implication
B C
is true.
Problem 4: (14pt)
[ ( )] [( ) ( )]
[( ) ( )] By Absorption
[( ) ( )] Equivalent form of implication
[(
p q p p q p q
p p q p q
p p q p q
p p
# ! # # ! "
# # ! "
# # ! ! "
#
) ( )] By DeMorgan's Law
[ ( )] Distribution Law
[ ( )]
q p q
p p q q
p p T
# ! # "
# # ! "
# # "
Negation (Inverse) Law
( ) Identity
C
p p# "
Negation Law
Problem 5: (14pt) Theorem: If a and b are even integers, then 2 2
( 1) ( 1)a b a$ $ $ % is even
Proof: Since a is even then
2
a
= (a)(a) is also even since the product of two even integers is even.
Given that b is even the consecutive integer b+1 is odd since two consecutive integers have opposite
parity. We now have that 2
( 1)b$ is odd since the product of two odd integers is odd. We also note
that a-1 is also odd since it is consecutive to the even integer a. Therefore we have that the sum
2
( 1)b$ and (a-1) is even since the sum of two odd numbers is even. And this results added to
2
a
will
be even since the sum of even numbers is even.
