Download Math 2534 Test 1A solution Spring 2010

Math 2534 Test 1A solution Spring 2010
Problem 1: Put each statement into implication symbolic logic. Define your variables and then
determine if any of the following are equivalent statements.
a) Birds head south when the weather gets cold.
b) Only if the weather is cold, do birds head south.
c) Birds do not head south or the weather is getting cold.
d) The weather is getting cold if the birds head south.
Solution: (12pts)
Let B be the statement that Birds head south and C be the statement the weather gets cold.
a) C  B
b) B  C
c) B  C  B  C
d) B  C
The sufficient and necessary conditions are the same for b, c, d.
Problem 2: Is the following argument valid? Put statements into symbolic logic and then
justify your reasoning.
If John is cheating then he is not ethical. To be ethical implies being honest.
Therefore if John is honest then John is not cheating.
Solution: (10pts) Let C be the statement that John is cheating, E be the statement that John is
ethical and H be the statement that John is honest.
Statement 1: C  E
Statement 2: E  H
Conclusion:  H  C
Notice that H is the necessary condition in statement 2 and will give no conclusive information
about the sufficient condition. This is the converse error.
Problem 3: (14pts) Prove the following using the method of contrapositive. Do not use any
previous theorems. Use definitions only.
Theorem: For all real numbers r, if r2 is irrational then r is irrational.
Proof by contradposition: If r is rational then r2 is rational.
Since we have assumed that r is rational, then using the definition of rational we may express
a
r  where a, b are non zero integers. Now consider
b
a a a2 m
 2 
where m = a 2  Z and n = b 2  Z therefore r2 is rational by definition
bb b
n
of rational. Since the contrapositive is true the equivalent original statement is also true.
r2  r *r 
Problem 4: (10pts)
Explain why every prime number, except 2, may be represented in the form 6q + 1,
6q + 3 or 6q + 5 for some integer q.
Solution: If you consider the integers Zmod6, you can partition Z into six distinct groups:
6q,6q  1,6q  2,6q  3,6q  4,6q  5 , Clearly the groups 6q, 6q+2, and 6q+4 will contain even
integers only. The odd integers will be in the groups 6q+1, 6q+3 and 6q+5. If n is prime other
that 2, we have proved that n must be odd and will therefore be in one the groups 6q+1, 6q+3
and 6q+5.
Problem 5: (14pts) Prove the following using the method of contradiction.
Theorem: For any integer n, 4n + 3 is not divisible by 4. (Use definition of divisible)
Proof by contradiction: Assume that for some integer n, 4n+3 is divisible by 4.
If 4 divides 4n + 3 and then there exist and INTEGER q such that 4q = 4n + 3. Solving for q
we get q = n + ¾. But this is not an integer and we have a contradiction to our definition of
divisible. Therefore the original statement is true and 4 does not divide 4n + 3.
Problem 6: (14pts) Put the following into symbolic logic with multiple quantifiers. Define
variables and domain.
a) For each real number there is another real number that will produce the sum of zero.
Consider arbitrary real numbers x and y with the Predicate P(x,y) is x + y = 0.
x  R, y  R P( x, y)
b) There is a real number that acts as an additive identity for every other real number.
Consider arbitrary real numbers x and y with the Predicate P(x,y) is x + y = y.
x  R, y  R P( x, y)
Problem 7: (12pts) Use Algebra Logic to prove the following equivalence. Justify each step.
Theorem:
[ p  (q  p)]  (q p) 
( q  p)
Proof:
[ p  (q  p )]  (q 
[
p) 
p  (q  p )]  (q 
[ p  (q  p )]  (q 
[ p]  (q
( p
p)  q 
( p)  q 
( p
q) 
( q  p) 
 ( p
p) 
Given
p) 
p) 
Equivalent form of impliction
Double negative
Absorption
Commutativity and Associativity
Idempotent
DeMorgan's Law
Commutativity
q )  [ p  (q  p )]  (q
p)
Problem 8: (14pts) Prove the following using definitions and/or theorems. You must restate
clearly any theorems used.
Theorem: If a is even, b + c is odd and a – b is also even then c2 is odd.
Proof: Since a is even then by definition of even we have that a = 2p for some integer p. It is
also given that b + c is odd and by definition of odd b + c = 2k +1 for some integer k. We also
have that a – b is even and a – b = 2h for an integer h. Now consider the following algebra.
a – b = 2p – b = 2h and b = 2p – 2h = 2m where m = p-h is an integer. Therefore by definition of
even we have that b is even. This in turn gives us that b + c = 2m + c = 2k + 1 and c = 2k + 1 -2m
= 2 (k-m) + 1 = 2f + 1 where f = k-m is an integer. So by definition for odd we have that c is odd.
Now consider c2 = c*c which is also odd since the product of two odd integers is always odd.