CSCI 190 Exam I practice solutions
1)
a) If the streets are not wet, then it is not raining.
b)
All Mt. SAC students are not international students.
c) P: study hard, q: get A’s r: get rich.
p  (q  r )
~ q ~ r
__________
This is valid by Modus tollens
~P
2)
Prove or disprove each of the following:
a) For the domain of real numbers,
x( x  x 2 )
Solution: false. As a counter example, let x=0.5. Then
x  x2
b) For the domain of integers, xy ( xy  0)
Solution: this is true. Let x=0. Then 0(y)=0 for all y.
Solution:
Let x=1. Then for any positive number y, 1  y
3)
Prove that
3n  9 is odd if and only if n is even.
Solution: First show if
3n  9 is odd, then n is even.
n  2k 1 for some positive integer k. Then
3n  9  3(2k  1)  9  6k  3  9  2(3k  6) . Therefore 3n  9 is even.
Use contrapositive. Suppose n is odd. Thus
Next show if
n is even, then 3n  9 is odd.
n  2k for some positive integer k. Then
3n  9  3(2k )  9  6k  9  2(3k  4)  1 . Therefore 3n  9 is odd.
Suppose n is even. Thus Thus
4) Solution: Choose
x  A . Then since A  B  A , x  A  B , that is x  A and x  B . Therefore
x  B by addition.
5) Solution: Suppose there is the largest integer x. Then
x  1is an integer large than x. This
contradicts the statement that x is the largest integer. Therefore there is no largest
integer.
6)
(Four points each)
a) Show
3n 2  n is O(n 2 ) .
Use the fact that
b) Show n is not
3
n  n 2 . 3n 2  n  3n 2  n 2  4n 2 . Therefore 3n 2  n is O(n 2 ) .
O(n 2 )
Proof by contradiction: suppose n is
3
O(n 2 ) . Then n3  Cn 2 for some C for large n values. But
n3  Cn 2  n  C , which states all natural numbers are less than some constant C. But this is
impossible since there is no largest natural number.
7) (four points) Compute the Boolean Product
1 0 
0 0 
and B  
A  B for A  


0 1 
1 1 
0  0 0  0  0 0 
A B  


 0  1 0  1  1 1 
8) (Four points each)
 x  for x in [2, 2]
b) Show f : Z  Z  Z defined as f (m, n)  m  n is onto. Justify your answer.
a) Sketch
Proof: Let
a  Z (pick a typical element in the codomain. Then f (a, 0)  a  0  a
. Thus f is
onto.
9) (8 points) Find the close form (non-recursive form)of the recursive sequence
an  2nan1 ,
a0  5 .
an  2nan 1
 2n((2)( n  1)an  2 )  22 n(n  1)an  2
Solution:
 22 n(n  1)(2(n  2)an 3 )  23 n(n  1)(n  2)an 3
 2n n !a0
 2n n ! 5
10) (2 points each: true/false, short answers: No partial credit points will be given. You are
not required to show work.)
a) True/False: Let A, B be sets. If
Solution: False: For example,
A  B , then A  B
A  {1, 2}, B  {3, 4,5} A has fewer elements than B, but A is not a
subset of A
b) True/False:
x   x
for all real numbers x (including negative numbers)
Solution: True.
 x  is the greatest integer LESS THAN OR EQUAL TO x.
c) True/False: Determine if the following statement is true or false:
1  1  56 if and only if
30 16  25 :
Solution: This is a true statement:
p  q is true since p is false and q  p is true since q is
false.
d) Is the set of rational numbers countable or uncountable? ___Countable____________
e) Is the set of real numbers countable or uncountable? ____Uncountable.
f)
Rewrite the following sentence in “If P then Q” form. I pass CSCI 190 only if I study 10
hours a week.
________________________Solution: If I pass CSCI 190, then I study 10 hour a week. _______________
g) Find the power set of A   ,{} Solution: P( A)  { ,{},{{}} ,{} ______________
h) Find
i)
n
n
i 1
i 1
 Ai where Ai  {1,2,3i} Solution:
Compute
2
2
 (2i  j)
Ai  {1} ______
Solution: (2  1)  (2  2)  (4  1)  (4  2)  18
i 1 j 1
j)
Find the nth term of the sequence 0,3,8,15, 
Solution: an  n  1 _______________
2
f ([2, 2]) and f 1 ([0, 2]) : f ([2, 2])  all the y-values for x in f ([2, 2]) and [0, 2]  [0, 4] and
k)
f 1 ([0,2])  looking for all the x-values when y is in [0, 2]  [ 2, 2]
l)
Let A, B be
n n
matrices with C=AB. Find an expression for
cij (using

) given
n
A  [aij ], B  [bij ], C  [cij ] Solution: cij   aik bkj
k 1
11) Prove if a b and b c , then a c for nonzero integers a,b,c.
Pf: Since a b ,
b  ak for some integer k. Similarly, since b c , c  bs for some integer s. But
b  ak and c  bs  c  (ak ) s  a(ks) . Since ks is an integer, a c .
12) Show
f (n)  n2  3n  1 is (n 2 )
Solution:
f (n)  n2  3n  1  n2  3n2  n2  5n2 . Thus f (n)  n2  3n  1 is O(n2 )
On the other hand,
f (n)  n2  3n  1  n2 . Thus f (n)  n2  3n  1 is (n2 ) . Therefore,
f (n)  n2  3n  1 is (n 2 )
13)
  {1,2,3} . True. The empty set is a subset of any set.
b) (1 point) True/False: }  {1,2,3} False. {1, 2,3} does not contain the element 
c) (2 points) Find the power set of {a , b} : Solution: { ,{a},{b},{ab}}
a) (1 point) True/False:
d) (2 points) For A  {1,2} and B  {3,7} , find A  B .
Solution {(1,3), (1, 7), (2,3), (2, 7)}
14) Find a non-recursive expression for an  2an 1  1, a0  1
Solution:
an  2an 1  1
 2(2an  2  1)  1  22 an  2  2  1
 22 (2an 3  1)  2  1  23 an 3  2 2  2  1
 2n a0  2n 1  2n  2
1
2n  1
2 1
n 1
 2 1
 2n 
15) Determine if f : Z  Z  Z defined by f (m)  6m  2n is onto.
Solution: No, it is not onto. It is impossible to find an integer m with f (m, n)  3 since
6m  2n  2(3m  1) is always even.
16) Prove that ~ ( p  q ) is logically equivalent to p  (~ q )
Solution:
P
q
T
T
F
F
T
F
T
F
pq
T
F
T
T
~ ( p  q)
~q
F
T
F
F
F
T
F
T
p  (~ q )
F
T
F
F
~ ( p  q) is logically equivalent to p  (~ q ) since their entries in the truth table are identical.
17)
Prove that if a b , then a b c for integers a,b,c.
Pf: Since a b ,
b  ak for some integer k. Multiply both sides by c. b  ak  bc  a(kc) . Since
kc is an integer, a b c .
18) Prove or disprove the following:
xy ( y  x) , where the domain is the set of all natural numbers 1, 2, 3, ….
Solution: True. Let x=1. Then for any natural number y,
19) Construct a truth table to determine if
P
T
T
F
F
Solution:
q
T
F
T
F
( p  q)
F
T
T
F
p  ( p  q)
T
T
T
F
They are not equivalent.
20) Show that if
A  B then A  B  B
p  ( p  q)  p
y  1.
A  B  B . Let x  A  B . We need to show x  B . x  A  B , so x  A or
x  B . Case1) If x  A , then since A  B , x  B Case 2) x  B . Then we are done. In
both cases x  B
c) Next show B  A  B . Pick x  B . We need to show x  A  B . But if x  B then a fortiori
x  A or x  B . Therefore x  A  B .
Pf: a) First show
Since
A  B  B and B  A  B , A  B  B .
21) Suppose x is a nonzero real number. Prove that if
1
is an irrational number, then x is also
x
an irrational number.
Pf: Use contradiction. Suppose x is a rational number. Then x 
Then
a
for nonzero integers a, b.
b
1 b
1
 with a nonzero a. Contradiction since
is an irrational number. Thus x is an
x
x a
irrational number
22) Prove or disprove:
a) For the domain of real number xy
( x  y  0)
Solution: False: This says you can find x such that x plus any number is 0. For x = 0,
then let y=1. Then x+y is not 0. For x not 0, let y=x. Then x+y is 2x, which is not 0.
b) For the domain of real numbers,
xy ( x  y  0)
Solution: true. This says any real number has an additive inverse. For any x, set y=-x.
Then x+y=0.
23) Prove that there is no positive integer satisfying
x 3  4 x  20
Pf : Use exhaustive proof: You only need t try x=1 and x=2 since
33 exceeds 20.
13  4(1)  20, 23  4(2)  20 . Therefore there is no positive integer satisfying x 3  4 x  20
24) (2 points each) Do NOT show work.
a) True/False   {1,2}
Solution: True. The empty set is a subset of every set.
b) Write the negation of
Solution:
x y p( x, y)
c) Find an integer
Solution: a=11
xy p( x, y) . Make sure the negation appears next to p(x,y).
6  a  13 such that a  4(mod 7) ______________
d) [5,9]
e) Negate the following sentence: Nancy is tall and beautiful.
___________Solution: Nancy is not tall or Nancy is not beautiful.
f)
Pf:
1.2   1.2   =_______________________
1.2   1.2    1.2  (2)    0.8  0
m)
1
f 1 ({0,1, 2})  looking for all the x-values when y is in 0,1, 2 : f ( x)  0  3n  1  0  n  ,
3
2
not an integer. f ( x)  1  3n  1  1  n  0 , f ( x )  2  3n  1  2  n  , not an integer.
3
1
f ({0,1,2})  {0}
g) P= there is gas in the car, q= I will not drink beer, r= I will go shopping.
pq
qr
~r
No conclusions can be drawn.
_____
X
25)
a) (7 points) Find a formula (in closed form) for the recursive sequence
an  an 1  n, a0  1
Solution:
an  an 1  n,
 an  2  (n  1)  n
 an 3  (n  2)  (n  1)  n
 a0  1  2 
n
n(n  1)
2
2
n n2

2
 1
b) (3 points) Find the nth term of the sequence 2, 6,18,54,
 an  2(3n 1 )
xy( x 2  y 2  4) , where the domain is the set of all real numbers.
2
Solution: This statement is false. Let x=3. Then 9  y  4 has no solution.
26) Prove or disprove:
27) Prove or disprove:
f : R  R defined by f ( x)  x3  1 is a bijection.
Solution: This is true. A) 1-1: Suppose
c) Onto: Pick y in the codomain.
f ( x)  f ( y ) . Then x3  1  y 3  1  x  y
x3  1  y  x  3 y  1 : i.e. f ( 3 y 1)  y . Thus f is onto.
28)
a) For a function
f : R  R defined by f ( x)  x 2  1 , find a) f 1[1,5] and b)
f [1,5]
Solution:
a)
f 1[1,5]  {x : f ( x) [1,5]} : that is , all the x-values whose corresponding y-values falls in
the range [1,5]. Using common sense,
b)
f 1[1,5]  [2, 2]
f [1,5]  { y : f ( x) for 1  x  5} : that is , all the y-values whose corresponding x-values
falls in the range [1,5]. Using common sense,
29) Prove that
2 is irrational.
Pf: Proof by contradiction. Suppose
number,
f [1,5]  [2, 26]
2
2 is rational.
Then by the definition of a rational
a
for integers a and b. We may assume the gcd of a and b is 1.
b
a
 a  2b . Next square both sides to get a 2  2b 2 . Thus a 2 even. So a is even. Write
b
a as a  2k for some k. a 2  2b 2 now becomes (2k )2  2b2  2k 2  b2  b is also even. But
2
this is a contradiction since the gcd of a and b is 1. Therefore
30)
If
m  n is odd, then n and m are both odd.
2 is irrational.
Pf: Proof by contrapositive: Show if m is odd and n is odd, then mn is odd.
Pf: Since m and n are odd, m  2k  1, n  2l  1, k .l  Z Then mn  (2k  1)(2l  1)  2(2 ml  m  l )  1 is
odd.
31)
a) Show n!  O(n n )
Pf: n!  n(n  1)(n  2)
(1)  n(n)(n)
n  nn . Thus n!  O(n n ) with C=1.
b) Show 2  O(n!)
n
Pf: 2n  2(2)(2)
C) Show
(2)  2(1(2)(3)
n)  2n! . Therefore 2n  O(n!) with C=2
3n  O(2n ) . Proof by contradiction. Suppose 3n  O(2n ) . Then there is a constant C
n
3
such that 3  C 2 for large n. Dividing both side by 2 , 3  C 2     C . But this is
2
impossible since an exponential function with base greater than 1 increases to  as n
n
n
grows. Thus 3  O(2 ) .
n
n
n
n
n
2
32) Prove that if a b , then a b for integers a,b,
2
Pf: since a b , b  ak for some integer k. Square both side b  ak  b  a k . Since k
2
2
integer, a b .
2
2
2
2
is an