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Math 266, Exam #1, Summer 2010
Name _______________________________________________
Instructions: Show all work. On proofs, clearly explain your reasoning. Unexplained leaps of logic, even
if correct, will be treated as if it is false. On take home quizzes, all work must be your own; you may not
work together.
1. Construct a truth table for the proposition ( p  q)  ( p  q) . Use the table below. You may
or may not need the whole thing. (5 points)
q
( p  q)
( p  q)
( p  q)  ( p  q)
T
T
F
T
T
T
F
T
F
F
F
T
T
F
F
F
F
F
F
T
p
2. Construct a truth table for the proposition ( p  q)  (p  r ) (5 points)
p
q
r
p
( p  q)
(p  r )
( p  q)  (p  r )
T
T
T
F
T
T
T
T
T
F
F
T
T
T
T
F
T
F
F
T
F
T
F
F
F
F
T
F
F
T
T
T
T
T
T
F
T
F
T
T
F
F
F
F
T
T
T
T
T
F
F
F
T
T
F
F
3. Show that ( p  r )  ( q  r ) and ( p  q)  r are not logically equivalent. (8 points)
p
q
r
( p  r)
(q  r)
( p  r)  (q  r)
( p  q)
( p  q)  r
T
T
T
T
T
T
T
T
T
T
F
F
F
F
T
F
T
F
T
T
T
T
T
T
T
F
F
F
T
T
T
F
F
T
T
T
T
T
T
T
F
T
F
T
F
T
T
F
F
F
T
T
T
T
F
T
F
F
F
T
T
T
F
T
4. Translate in two ways each of these statements into logical expressions using predicates,
quantifiers, and logical connectives. Let the domain consist of all people. (2 points each)
a. Everyone has a cellular phone.
Answers will vary, examples include: xH ( x, cellular phone)
xC ( x )
b. Somebody has seen a foreign movie.
Answers will vary, examples include: xF ( x )
xS ( x, foreign movie)
c. Everyone is your friend and is perfect.
Answers will vary, examples include: xF ( x )  xP( x )
x ( F ( x )  P( x ))
d. All students can solve quadratic equations.
Answers will vary, examples include: x ( S ( x )  Q ( x ))
x(Q ( x )  S ( x ))
e. Not everybody is your friend, or someone is not perfect.
Answers will vary, examples include: (xF ( x ))  xP( x )
(xF ( x ))  (xP( x ))
Many more can be found with various applications of DeMorgan’s Laws
5. Express each of these statements using mathematical and logical operators, predicates, and
quantifiers, where the domain consists of all integers. (3 points)
a. The sum of two negative integers is negative.
xy ( x, y  Z  )  ( x  y )  Z  ) 
b. The difference of two positive integers is not necessarily positive.
xy ( x, y  Z  )  ( x  y  0) 
c. The sum of the square of two integers in greater than or equal to the square of their sum.
xy  x 2  y 2  ( x  y )2 
d. The absolute value of the product of two integers is the product of their absolute values.
xy | xy || x |  | y |
6. What is wrong with this argument? Let S(x,y) be “x is shorter than y”. Given the premise
xS ( x, Max ) , it follows that S ( Max, Max ) . Then by existential generalization, it follow that
xS ( x, x ) , so that someone is shorter than himself. (5 points)
The error is in jumping from xS ( x, Max ) to S ( Max, Max ) . This is an incorrect application of
existential instantiation since this can only be done for a general c, not a specific one.
7. Identify the error or errors in this argument that supposedly shows that if x ( P( x )  Q ( x )) is
true then xP( x )  xQ ( x ) is true. (There may be more than one error.) (6 points)
1.
2.
3.
4.
5.
6.
7.
x ( P( x )  Q ( x ))
P(c)  Q (c)
P(c)
xP( x )
Q (c)
xQ ( x )
x ( P( x )  xQ ( x ))
Premise
Universal instantiation, from 1
Simplification, from 2
Universal generalization, from 3
Simplification, from 2
Universal generalization, from 5
Conjunction, from 4 & 6
All three of these lines are incorrect. They would be correct if the operator was a conjunction, but not a
disjunction.
8. Prove that if n is a perfect square, then n+2 is not a perfect square. (10 points)
Let n = m2. Then n +2 = m2 + 2. But there are no perfect squares with a difference of two between
them. Consecutive squares have one even number and one odd number, which means that the
difference between them must be odd.
9. Prove that m2 = n2 if and only if m = n or m = -n. (10 points)
Suppose m=n, then m2 = n2, or if m = -n, then m2 = (-n)2 = n2.
Conversely, suppose that m2 = n2. Then m2 – n2 = 0. This has factors of (m-n)(m+n)=0. Therefore either
m = n or m = -n, and by the fundamental theorem of algebra, there are only two possible solutions to a
second degree equation.
10. Prove that there are no solutions in integers x and y to the equations 2x2 + 5y2 = 14. (10 points)
This proof can be done exhaustively, or if we solve for y, we have y  
14  2 x 2
. This will only be a
5
2
real number if 7  x  0 . The only possible integer solutions for a real value of y are x=0,1,2. Testing
14
12
6
y


,

,

these in the equation, we have possible y values of
5
5
5 . None of which are
themselves integers.
11. Explain why A  B  C and ( A  B )  C are not the same. (6 points)
Elements of A  B  C are (a,b,c) which is a triple, but elements of ( A  B )  C are a pair of the form
((a,b),c), where a  A, b  B, c  C
12. Translate each of these quantifications into English and determine its truth value. (3 points
each)
a.
x  R( x 3  1)
There exists at least one x in R so that x3 = -1; True
b.
x  Z ( x 2  2)
There exists an x in Z so that x2 = 2; False
c.
x  Z ( x  1  Z )
For every x in Z, the expression x-1 is also in Z; True.
d. x  Z ( x 2  Z )
For every x in Z, the expression x2 is also in Z; True.
13. Prove the complement laws by showing that: (6 points each)
a.
A A  U
Let x  A  A . Then by definition of the union, x is in A or x is in A , but by definition of A , any x that
is not in A is in A  U  A . Therefore, x is in U.
Conversely, let x  U . Then by x is either in A or in A since A  U  A . Therefore, x is in A  A .
Therefore, A  A  U .
b.
x  A A
First,   A  A since the empty set is a subset of every set.
Conversely, suppose x  A  A , then x must be in both A and in A , but by definition of the
complement, any element in A is not in A . This is a contradiction, thus there are no x  A  A , so
x  A A.