Download Math 2534 Test 1B Fall 2008 Name

Math 2534 Test 1B Fall 2008
solutions
Problem 1: (12pts)
Given the statement: October means that leaves will change to orange and maroon.
a) Put into implication form.
b) Give the contrapositive and express in an English sentence.
c) Give the contradiction and express in an English sentence.
a) Let O : it is October and L: the leaves will change O  L
b) Contrapositive is L  O
The leaves have not turned, so it is not October.
c) Contradiction is [O  L]  O  L
It is October and the leaves have not changed.
Problem 2: (16pts)
Put the following statements into symbolic logic and then determine which
statements are equivalent. Explain your reasoning.
Solution:
a)
b)
c)
d)
If you pass this course then you will graduate.
You will not graduate if you do not pass this course.
Only if you pass this course will you graduate.
Pass this course or do not graduate.
Solution: Let P : You pass the course, and G: You graduate
a) P  Q
b) P  G  G  P
c) G only if P  G  P
d) P  G  P  G  G  P
b), c) and d) are equivalent statements.
Problem 3:(12pts)
Use any of the following theorems that are needed to prove that if b is an odd integer,
then b2  a(a  1)  b is even.
[Use these theorems as given and do not reprove them].
a) The product of an even integer and any other integer is even
b) The sum of an even and odd integer is odd.
c) Consecutive integers have opposite parity
d) The product of two odd integers is odd.
e) The sum of two even integers is even
f) The sum of two odd integers is even
g) A prime number greater than 2 is odd
Solution: Proof: Given that b is an odd integer then b2 = (b)(b) is odd since the
product of two odd integers is odd. You will also note that a(a+1) is the product of
two consecutive integers. Consecutive integers always have opposite parity, so one
must be even. This gives us that the product a(a+1) is even, since the product of an
even integer and any other integer is always even. If we add b2 and a(a+1) we get a
value that is odd since the sum of an odd integer and an even integer is always odd.
If we now add b to this sum we will get a value that is even since the sum of two odd
integers is even.
Problem 4: (8pts)
Is the following argument valid. Explain why or why not.
All CS majors take a Discrete Math course
Lucy takes a Discrete Math Course
Therefore Lucy is a CS major.
Solution: Let CS: all CS majors and M: take a Discrete math course and L is Lucy
Then we have the following argument
CS  M
M ( L)
 CS ( L)
This argument is invalid since you have the necessary condition only (Lucy took
Discrete math) and this will not guarantee the sufficient condition (being a CS
major) this is the Converse error.
Problem 5 (12pts)
Use algebra of logic to prove the following equivalence.
[(P  Q)  ( P  Q)]  ( P  Q)  P
Solution:
TH :[(P  Q)  ( P  Q)]  ( P  Q)  P
Proof:
[(P  Q)  ( P  Q)]  ( P  Q) 
[(P  Q)  ( P  Q)]  ( P  Q)  Equivalent form
[(P  Q)  (P  Q)]  ( P  Q)  De Morgan's Law
[(P  (Q  Q)]  ( P  Q) 
Distribution Law
[(P  (T )]  ( P  Q ) 
Negation or Inverse
[(P]  ( P  Q) 
Identity
P  ( P  Q) 
Double Negative
P
Absorption Law
Problem 6: (12pts)
Prove the following by contradiction:
If r is rational and not zero and w is irrational then prove that the product
rw is irrational.
Solution:
Theorem: If r is rational and not zero and w is irrational then prove that the
product rw is irrational.
Proof:
Proof by contradiction: Assume r is rational and not zero and w is irrational
and the product rw is irrational.
a
Since r is rational, then by definition r  , a and b are integers and b  0 . We
b
c
have assumed that rw is also rational rw  , c and d are integers and d  0
d
Now consider
c
rw 
d
a
c
w
b
d
cb m
w

da n
Where m = cb and n = da are integers and w is rational by definition of rational.
This contradicts that w is given to be irrational.
Therefore the product rw is also irrational.
Problem 7: (6pts)
Write the following expressions in natural sounding English and indicate if
you think the statement is true or false. Let C be the set of all nations and let F
be the set of all national flags. Let P(x, y) = x is classified by y when
x C and y F
a) (x, x C )(y, y F ) P( x, y )
b) (y, y F )(x, x C ) P( x, y )
Solutions: a) Every country has it’s own national flag
b) There is only one national flag that represents all countries.
Problem 8: (12 pts) Use the contrapositive to prove that:
If n does not divide (a – b) then n does not divide a or does not divide b. Use
definitions only and no previous theorems.
Solution: Theorem: If n does not divide (a – b) then n does not divide a or does not
divide b.
Proof by contrapositive method: Restate theorem to read as follows: If n divides a
and n divides b, then n divides (a – b).
Since n divides a and n divides b we have, by the definition of divisible, that
nd = a and nk = b where d and k are integers.
Now consider a – b = nd – nk = n(d – k) = nw where w = d – k is an integer.
Therefore by definition of divisible, we have that n divides a – b. Since I have
proven that the contrapositive is true, we know that the equivalent original
statement is also true.
Problem 9: (12pts)
Below is a set of premises with a conclusion. Verify that the conclusion is correct. You
must explain your reasoning by referring to conjunction, disjunction, implication and
how these operations determine the truth values of components of each premise. Be clear
an write in complete sentences.
a ) p  ( r  s )
b) t  s
c ) u  p
d ) w
e) u  w
f ) Therefore t
The conclusion is valid. Since w is true, w must be true. Since u  w is a true
statement and w is false, then by disjunction u must be true. Notice that u is the sufficient
condition of the statement u  p . The true sufficient condition, u, guarantees the
necessary condition p to be true. If p  (r  s) and p is the sufficient condition
and guarantees the necessary condition (r  s ) is also true. Since this is a conjunction
and true, both r and s true. Since s is true, s must be false. Given that t  p and p
is the necessary condition and false, in order for the implication to be true, t must be false
and t is true.