Download Equivalent form of implication

2534 Solutions for Test 1B Fall 2009
Problem 1: (8pt)
a) For every name brand there is a boy that plays with it.
b) There is a particular name brand that all boys play with.
Problem 2: (14pt)
[ p ∧ (q ∨ p )] ∧ [( p ∧ q)∨  (  p → q)] ≡
p ∧ [( p ∧ q )∨  ( p → q )] ≡
p ∧ [( p ∧ q )∨  ( p ∨ q)] ≡
p ∧ [( p ∧ q ) ∨ ( p ∧  q )] ≡
p ∧ [ p ∧ (q ∨  q )] ≡
By Absorption
Equivalent form of implication
By DeMorgan's Law
Distribution Law
p ∧ [ p ∧ (T )] ≡
p ∧ ( p) ≡
Negation (Inverse) Law
Identity
C
Negation Law
Problem 3: (12pts) It is given that [ B → C ] → ( A ∧ D ) is given to be true but the sufficient condition
B → C is false. Therefore the necessary condition ( A ∧ D) is undetermined since it could be true or
false. If B → C is false, then the sufficient condition B must be true and the necessary condition C is
false.
Problem 4: (12pts) P is the statement, “You do all the problems” and A is the statement “you get an A”
a)  P ∨ A ≡ P → A P is sufficient and A is necessary
b)  P → A ≡ A → P
c) A → P A is Sufficient and P is necessary
d)  ( A∧  P ) ≡  A ∨ P ≡ A → P
So we have that b, c, d are equivalent.
Problem 5: (14pt) Theorem: If a does not divide mb, then a does not divide b for integers a, b and m.
Proof: We will use method of contrapositive. If a divides b, then a divides mb.
Since a divides b, then by definition of divisible b = aq for some integer q. Now multiply both sides by m
to get mb = maq = a(mq) = ak where k = mq is an integer. Therefore a divides mb.
Since the contrapositive form is true the equivalent form and the original statement is also true.
Problem 6: (14pt) Theorem: All even integers greater than 2 are not prime.
Proof: Using method of contradiction, we will assume there is an integer that is even and is prime. But
this would mean that this integer is divisible by 2. By definition of prime, a prime number can only be
divided by itself or one. If a prime number is also even then it can also be divided by 2 which contradicts
the definition of prime. Therefore no even integer greater than two is prime.
Note: Using a counter example such as the number 4 is not a counter example to the statement of
contradiction.
Problem 7: (14pt) Theorem: If a and b are even integers, then a 2 + (b + 1) 2 + (a − 1) is even
Proof: Since a is even then a 2 = (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 (b + 1) 2 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
(b + 1) 2 and (a-1) is even since the sum of two odd numbers is even. And this results added to a 2 will
be even since the sum of even numbers is even.
Problem 8: (12pt) Theorem: The product of any two consecutive integers can be represented by the
forms 4k and 4k+2
Proof: By the QRT, the set of integers can be represented in one of the forms 4k, 4k+1, 4k+2, and 4k+3
Since the product of two consecutive integers will always be even then the only possible representation
for an even number mod 4 is 4k or 4k+2.