Download Math 2534 Test 1B Solutions

Math 2534 Test 1B Solutions
Problem 1: Using one or more of the following previously proved theorems below,
prove the following Theorem:
List of previously proved theorems:
Refer to needed theorems only. Do not try to reprove any of these theorems.
Two consecutive integers have opposite parity
The product of two even (odd) integers is even (odd).
The sum of any two odd integers or any two even integers is even.
Any prime number greater than 2 is an odd integer.
The sum of any two integers is an integer.
The sum of an even and odd integer is odd.
Solutions: (12pts):
If a > 2 is a prime number and (b + 1) is an even number, then a 4 + b is even.
Proof: If a is a prime number greater than 2, then a is odd. If a is odd then a2 = (a)(a) is
also odd since the product of two odd numbers is always odd. We also conclude
that a4 = a2a2 is also odd since the product of two odd numbers is odd. We are given that
b+1 is even. Then b will be odd since two consecutive integers has opposite parity.
Therefore a4 + b will be even, since the sum of two odd numbers will be even.
Problem 2: Are the following arguments valid? Justify your conclusion.
1) An apple a day keeps the doctor away. Since Fred is sick, he did not eat an
apple everyday.
2) All discrete math students can tell a valid argument from an invalid one. All
thoughtful people can tell a valid argument from an invalid one. Therefore all discrete
math students are thoughtful.
Solutions: (14pts)
1) Let A be the statement, He eats an apple a day
Let H be the statement He is healthy
A→ H
¬H ( F )
∴¬A( F )
This argument is valid since it is the contrapositive form
2) Let D be Discrete math students
Let T be Thoughtful people
Let V be valid arguments
D →V
T →V
∴D →T
This argument is invalid since the necessary condition can not guarantee the sufficient
condition. This is converse error.
Problem 3:
Put the following statement into symbolic logic with multiple quantifiers.
Each student at VT takes at least one math course.
Solution: (8pts)
Let S be the set of all students at VT
Let M be the set of all math courses
P(x,y): x take y
∀x ∈ S , ∃y ∈ M P( x, y ) = ∀x, ( x ∈ S ) → [∃y y ∈ M ∧ P( x, y )]
Problem 4: Max wishes to prove the following conjecture using the method of proof by
contradiction. The statement must be rewritten to fit the method. Rewrite the statement
for Max. (Do not actually do the proof)
Theorem: Given integers a and b, if (ab)2 is even then a is even or b is even.
Solution: (8pts)
Statement of the contradiction: Given integers a and b, assume (ab)2 is even and a is
odd and b is odd.
Problem 5
If the operation of "nor" is defined to be p ↓ q ≡ ∼ (p ∨ q). Then use the algebra of
logic to prove the following theorem and justify each step.
Theorem: (p ↓ q) ↓ ( p ↓ q) ≡ p ∨ q.
Solution: ( 12pts)
Theorem: (p ↓ q) ↓ ( p ↓ q) ≡ p ∨ q.
Proof:
(p ↓ q) ↓ ( p ↓ q) =
¬[¬( p ∨ q ) ∨ ¬( p ∨ q )] ≡
Definition of "nor"
( p ∨ q) ∧ ( p ∨ q) ≡
De Morgan's Law
p∨q
Idempotent
Therefore (p ↓ q) ↓ ( p ↓ q) ≡ p ∨ q.
Problem 6: Use method by contrapositive to prove the following theorem.
Theorem: For all natural numbers n, If 7n 2 + 5 is even, then n is odd.
Solution: (12pts)
Theorem: For all natural numbers n, If 7n 2 + 5 is even, then n is odd.
Proof by contrapositive: For all natural numbers n, If If n is even, then 7n 2 + 5 is odd.
Assuming that n is even then by definition of even we have that n = 2k for some integer
n. We now have 7n2 + 5 = 7(2k)2 + 5 = 28k2 + 5 = 28 k2 + 4 + 1 = 2(14 k2 +2) + 1 =
2p + 1 where p = 14 k2 +2 is an integer. Therefore by definition of odd we have that
7n2 + 5 is odd. Since the contrapositive is true, the original equivalent statement is also
true.
Problem 7: Use direct method to prove the following theorem:
Theorem: If ∀a, b, c ∈ Z , If ab c, then a c
Solution: (12pt)
Theorem: If ∀a, b, c ∈ Z , If ab c, then a c
Proof: Since ab c , by definition of divisible we have that (ab)q = c where q is some
integer. By commutative law we have a(bq) = c to get ap = c where c = bq is an integer.
So by definition of divisible we now have that a c .
Problem 8 Use direct method to prove the following theorem.
Theorem: Every integer is a rational number.
Solution (12pt)
Theorem: Every integer is a rational number.
a
. This is clearly a ratio of integers
1
which fulfills the definition of a rational number which states that all rational numbers
b
can be expressed as a ratio of integers , c ≠ 0 . Therefore all integers are rational.
c
Proof: Given any integer a, it can be expressed as
Problem 9
Are the following two statements equivalent? Justify your answer using symbolic logic
along with some English sentences.
a) I will not dance or I will not sing.
b) I will sing only if I do not dance.
Solution: (10pts)
Let D be the statement, I will dance.
Let S be the statement, I will sing.
a) ¬D ∨ ¬S ≡ D → ¬S by equivalent form of implication
b)
If a > 2 is a prime number and (b + 1) is an even number, then a 4 + b is even.
S → ¬D ≡ D → ¬S by contrapositive
Therefore both statements are equivalent.