IS2 Problem Sheet II IS2 Problem Sheet II

Question 1
Let E = { 1,2, 3, 4, 5,6,7,8,9} be the universal set.
Let A = {1,2,3,5}, B = {2,4,7,8} and C = {3,4,5,8}.
Find:
(i) [`A],  (ii) A ÇC,   (iii) A ÈB,   (iv) (A ÈB) ÇB,   (v) [`((A ÇB))],   (vi) A ÈÆ,   (vii) A ÈE.

Question 2
27 professional photographers were asked whether they work in black and white or in colour. 25 said they work in colour and 13 said they use both. Use the Inclusion-Exclusion Principle to find how many work in black and white only.

Question 3
In a group of 29 people, there are 14 dog owners and 9 cat owners. If 4 people own both a cat and a dog, how many people own neither?

Question 4
Let E = { 1,2,3,4,5,6,7,8,9}.
(i) Find the representation of Æ, {1,3,9}, {2,4,5}, {9}, {2,4,5,7} as bit strings.
(ii) Write down the sets represented by the following bit strings  001110101, 010110001, 110101101, 101101011, 111100001
(iii) Find the sets got by bitwise AND-ing the following together

001110101 
AND
 010110001,
110101101 
AND
 101101011,
111100001 
AND
 001001010,
(iv) Find the sets got by bitwise OR-ing the following together
001110101 
OR
 010110001
110101101 
OR
 101101011
111100001 
OR
 001001010
(iv) Find the sets got by bitwise NOT-ing the following
NOT
 010110001
NOT
 101101011
NOT
 001001010

Question 5
Represent the following relations as (i) directed graphs and (ii) relation matrices.
(a) R = { (1,2), (2,1), (3,3), (1,1), (2,2)} on X = {1,2,3}
(b) R = { (1,2), (2,3), (3,4), (4,1)} on X = {1,2,3,4}
(c) R  on {1,2,3,4} defined by (x,y) Î R if x2 ³ y
(d) R = { (a,b), (b,a), (b,d), (a,c), (c,c), (c,d)} on X = {a,b,c,d}

Question 6
Give the set on which the following relations are defined. Write the relations as sets of ordered pairs.

Figure

Question 7
Determine if the following relations are equivalence relations on {1,2,3,4,5}, and if they are, determine the associated equivalence classes.
(i) R = { (1,1), (2,2), (3,3), (4,4), (5,5), (1,3), (3,1)}
(ii) R = { (1,1), (2,2), (3,3), (4,4), (5,5), (1,5), (5,1), (3,5), (5,3), (1,3),(3,1)}
(iii) R = { (1,1), (2,2), (3,3), (4,4)}
(iv) R = { (1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (4,3) (4,4), (5,5)}
(v) R = { (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1) (3,2), (3,3), (4,4), (5,5) }

Question 8
Determine if the following relations are reflexive, irreflexive, symmetric, antisymmetric or transitive. If possible classify them as equivalence relations or partial orders.
(i) R = {(x,y) | x = y} defined on the set of positive integers.
(ii)R = { (1,1), (2,2), (3,3), (4,4), (1,2), (2,3), (2,1), (3,2) } defined on {1,2,3,4}
(iii)R = { (1,1), (2,1), (2,2) } defined on {1,2,3,4}
(iv) R = {(x,y) | 3 divides x+y } on the set {1,2,3,4,5}

Question 9
Determine if the following relations from X = {1,2,3,4} to Y = {a,b,c,d} are functions. If they are functions are they one-to-one or onto?
(i) R = { (1,a), (2,a), (3,c), (4,b)}
(ii) R = { (1,c), (2,a), (3,b), (4,c), (2,d)}
(iii) R = { (1,c), (2,d), (3,a), (4,b)}
(iv) R = { (1,d), (2,d), (4,a)}
(v) R = { (1,b), (2,b), (3,b),(4,b)}

Question 10
Given function g from X = {1,2,3} to Y = {a,b,c,d} defined by

g : = { (1,b), (2,c), (3,a)}
and function f from Y to Z = {w,x,y,z} defined by
f : = { (a,x), (b,x), (c,z), (d,w)}
Write f°g as a set of ordered pairs.

Question 11
If function f: R ® R is defined by f(x) = x2+3x and function g:R ® R is defined by g(x) = 7x+2, evaluate the following
(i) f °g
(ii) g °f
(iii) f °f
(iv) f °g °f
(v) g °f °g

Question 12
Determine if the following functions have inverses, and if they do, find the inverse:
(i) y = 0.5x+1
(ii) y = x2     x ³ 0
(iii) y = x3 +1
(iv) y = x4     x ³ 0
(v) y = 1/2x -7/2

Question 13
Evaluate the first six terms for the following recursively defined sequences:
(i) t(1) = 7,     t(n) = t(n-1) +3   (n > 1)
(ii) t(1) = 2,     t(n) = 1/2t(n-1) +3   (n > 1)
(iii) t(1) = 0,     t(n) = 3t(n-1) +n   (n > 1)

Question 14
Evaluate t(3), t(4), t(5) and t(6) for the following:
(i) t(1) = 2     t(2) = 4     t(n) = 2t(n-1) +t(n-2)    (n > 2)
(ii) t(1) = 3     t(2) = 1     t(n) = 1/2[t(n-1) +t(n-2)]    (n > 2)
(iii) t(1) = 7     t(2) = 11     t(n) = 3t(n-1) +2t(n-2)    (n > 2)

Question 15
For each of the sequences in Question 13 write an iterative algorithm to output the first n terms of the sequence.

Question 16
Prove the following statements are true for all natural numbers n using induction.
(i) 1(2) +2(3) + 3(4) + ... +n(n+1) = [(n(n+1)(n+2))/ 3]
(ii) 1(1!) +2(2!) + 3(3!) + ... +n(n!) = (n+1)! -1
(iii) 12 -22 + 32 - ... +(-1)n+1n2 = (-1)n+1 [(n(n+1))/ 2]
(iv) 13 + 23 + 33 + ... +n3 = ([(n(n+1))/ 2])2

Question 17
Let t(n) be defined recursively as follows:
t(1) = 4        t(n) = 2t(n-1) +3
Prove by induction that t(n) = 7·2n-1 -3

Question 18
Let t(n) be defined recursively as follows:
t(1) = 7        t(n) = 3t(n-1) +11
Prove by induction that t(n) = 3n-1(12.5) - 5.5

Question 19
Find a recursive expression for the output of each of the following algorithms. Hence prove by induction that the algorithm produces the output claimed.
(i) Algorithm to output a table of values of [(i(4i2 +18i +11))/ 3] i = 1,2,3,... n

1. input n 
2. value <- 0 
3. for i=1 to n do 
	3.1 j <- (2i+1)(2i+1)
	3.2 value <- value + j
	3.3 output i, value

(ii) Algorithm to output a table of values of [(i(i-1))/ 2], i = 1,2,3,...,n:

1. input n 
2. value <- 0 
3. for i=1 to n do 
	3.1. j <- i-1
	3.2 value <- value +j
	3.3 output i, value

Answers

Question 1
(i) [`A] = {4,6,7,8,9}
(ii) A ÇC = {3,5}
(iii) A ÈB = {1,2,3,4,5,7,8}
(iv) (A ÈB) ÇB = {2,4,7,8}
(v) [`((A ÇB))] = {1,3,4,5,6,7,8,9}
(vi) A ÈÆ = A
(vii) A ÈE = E.

Question 2
Two use black and white only.

Question 3
Ten people do not own either a dog or a cat.

Question 4
(i)

Æ
=
000000000
{1,3,9}
=
101000001
{2,4,5}
=
010110000
{9}
=
000000001
{2,4,5,7}
=
010110100
(ii)
001110101
=
{3,4,5,7,9}
010110001
=
{2,4,5,9}
110101101
=
{1,2,4,6,7,9}
101101011
=
{1,3,4,6,8,9}
111100001
=
{1,2,3,4,9}
(iii)
001110101 AND 010110001
=
000110001 = {4,5,9}
110101101 AND 101101011
=
100101001 = {1,4,6,9}
111100001 AND 001001010
=
001000000 = {3}
(iv)
001110101 
OR
 010110001 = 011110101 = {2,3,4,5,7,9}
110101101 
OR
 101101011 = 111101111 = {1,2,3,4,6,7,8,9}
111100001 
OR
 001001010 = 111101011 = {1,2,3,4,6,8,9}
(iv)
NOT
 010110001 = 101001110 = {1,3,6,7,8}
NOT
 101101011 = 010010100 = {2,5,7}
NOT
 001001010 = 110110101 = {1,2,4,5,7,9}

Question 5
(a) R = { (1,2), (2,1), (3,3), (1,1), (2,2)} on X = {1,2,3}

1
2
3
1
T
T
F
2
T
T
F
3
F
F
T
(b) R = { (1,2), (2,3), (3,4), (4,1)} on X = {1,2,3,4}
1
2
3
4
1
F
T
F
F
2
F
F
T
F
3
F
F
F
T
4
T
F
F
F
(c) R  on {1,2,3,4} defined by (x,y) Î R if x2 ³ y
1
2
3
4
1
T
F
F
F
2
T
T
T
T
3
T
T
T
T
4
T
T
T
T
(d) R = { (a,b), (b,a), (b,d), (a,c), (c,c), (c,d)} on X = {a,b,c,d}
a
b
c
d
a
F
T
T
F
b
T
F
F
T
c
F
F
T
T
d
F
F
F
F

Question 6
(i) R = { (1,1), (2,2), (3,3), (4,4), (5,5), (3,5), (5,4), (4,3)} on {1,2,3,4,5}.
(ii) R = { (b,c), (c,b), (d,d)} on {a,b,c,d}
(iii) R = { (1,1), (1,2), (1,3), (1,4), (2,2) (2,3), (3,3), (3,4), (4,4) } on {1,2,3,4}.

Question 7
(i) Equivalence relation: [1] = [3] = {1,3}, [2] = {2}, [4] = {4}, [5] = {5}
(ii) Equivalence relation: [1] = [3] = [5] = {1,3, 5}, [2] = {2}, [4] = {4}
(iii) Not an equivalence relation since (5,5) \not Î R.
(iv) Equivalence relation: [1] = [2] = {1,2}, [3] = [4] = {3,4}, [5] = {5}
(v) Equivalence relation: [1] = [2] = [3] = {1,2,3}, [4] = {4}, [5] = {5}

Question 8
(i) This is reflexive, symmetric, antisymmetric and transitive. It is not irreflexive.
This means that it is an equivalence relation and a partial order.
(ii)This is reflexive, symmetric and transitive. It is not irreflexive and antisymmetric. It is an equivalence relation
(iii) This is antisymmetric and transitive. It is not reflexive, irreflexive, symmetric.
(iv) This is symmetric. It is not reflexive, irreflexive, transitive or antisymmetric.

Question 9
(i) It is a function, it is neither one-to-one or onto.
(ii) It is not a function since (2,a) and (2,d) are in R.
(iii) It is a function. It is one-to-one and onto.
(iv) It is not a function since 3 is not mapped onto the range.
(v) It is a function. It is neither one-to-one or onto.

Question 10
f°g = { (1,x), (2,z), (3,x) }

Question 11
If function f: R ® R is defined by f(x) = x2+3x and function g:R ® R is defined by g(x) = 7x+2, evaluate the following
(i)

f °g
=
f(7x+2)
=
(7x+2)2 +3(7x+2)
=
49x2 + 28x + 4 +21x+6
=
49x2 +49x +10
(ii)
g °f
=
g(x2+3x)
=
7(x2+3x) + 2
=
7x2 + 21x +2
(iii)
f °f
=
f(x2 +3x)
=
(x2 +3x)2 + 3(x2+3x)
=
x4 + 6x3 +9x2 + 3x2 + 9x
=
x4 +6x3 + 12x2 + 9x
(iv)
f °g °f
=
f(g(x2+3x))
=
f (7x2 + 21x +2)
=
(7x2 + 21x +2)2 + 3 (7x2 + 21x + 2)
=
49x4 + 147x3 + 14x2 +147 x3 + 441x2 +42x +14x2 +42x = 4 + 21x2 +63x +6
=
49x4 +294x3 +490x2 +147x +10

(v)

g °f °g
=
g(f(7x+2))
=
g( 49x2 +49x +10)
=
7(49x2 +49x +10) +2
=
343x2 + 343 x + 70 +2
=
343x2 + 343 x + 72

Question 12
(i) f-1 (x) = 2(x-1)
(ii) f-1 (x) = +Öx
(iii) f-1 (x) = 3Ö{x-1}
(iv) f-1 (x) = +4Ö{x}
(v) f-1 (x) = 2x +7

Question 13
(i)

t(1)
=
7
t(2)
=
7+3 = 10
t(3)
=
10+3 = 13
t(4)
=
13+3 = 16
t(5)
=
16+3 = 19
t(6)
=
19+3 = 22

(ii)

t(1)
=
2
t(2)
=
1
2
·2 + 3 = 4
t(3)
=
1
2
·4 + 3 = 5
t(4)
=
1
2
·5 + 3 = 5 1
2
t(5)
=
1
2
·5 1
2
+ 3 = 5 3
4
t(6)
=
1
2
·5 3
4
+ 3 = 5 7
8

(iii)

t(1)
=
0
t(2)
=
3(0) + 2 = 2
t(3)
=
3(2) + 3 = 9
t(4)
=
3(9) + 4 = 31
t(5)
=
3(31) + 5 = 98
t(6)
=
3(98) + 6 = 300

Question 14
(i)

t(3)
=
2(4) + 2 = 10
t(4)
=
2(10) + 4 = 24
t(5)
=
2(24) + 10 = 58
t(6)
=
2(58) + 24 = 82

(ii)

t(3)
=
1
2
[ 3+1] = 2
t(4)
=
1
2
[ 2+1 ] = 1 1
2
t(5)
=
1
2
[ 1 1
2
+ 2 ] = 1 3
4
t(6)
=
1
2
[ 1 3
4
+ 1 1
2
] = 1 5
8

(iii)

t(3)
=
3(11) + 2( 7) = 47
t(4)
=
3(47) + 2(11) = 163
t(5)
=
3(163) + 2(47) = 583
t(6)
=
3(583) + 2(163) = 2075

Question 15
(i)

1. input n
2. value <- t(1)=7
3  output value
4. for i = 1 to n do 
	4.1 value <- value + 3
	4.2 output value

(ii)

1. input n
2. value <- 2
3  output value
4. for i = 1 to n do 
	4.1 value <-  0.5(value) + 3
	4.2 output value

(iii)

1. input n
2. value <- 2
3  output value
4. for i = 1 to n do 
	4.1 value <-  3(value) + n
	4.2 output value

Question 16
(i) True for n = 1 since

(1)(2) = 1(2)(3)
3
Assume true for n = k i.e.
1(2) + 2(3) + ... + k(k+1) = k(k+1)(k+2)
3
Need to show this assumption implies it is true for n = k+1:
1(2) + 2(3) + ... + k(k+1) + (k+1)(k+2)
=
k(k+1)(k+2)
3
+ (k+1)(k+2)
=
k(k+1)(k+2) + 3(k+1)(k+2)
3
=
(k+3)[(k+1)(k+2)]
3
=
(k+1)(k+2)(k+3)
3
As this is the original formula with n = k+1 this means that the expression is true for n = k+1 if it is true for n = k. Hence, by induction, it is true for all natural numbers n.
(ii) True for n = 1 since
(1)(1!) = (1+1)! -1
Assume true for n = k i.e.
1(1!) + 2(2!) + ... + k(k!) = (k+1)! -1
Need to show this assumption implies it is true for n = k+1:
1(1!)+ 2(2!) + ... + k(k!) + (k+1)((k+1)!)
=
(k+1)! -1 + (k+1)((k+1)!)
=
((k+1) +1)(k+1)! -1
=
(k+2)(k+1)! -1
=
(k+2)! -1
As this is the original formula with n = k+1 this means that the expression is true for n = k+1 if it is true for n = k. Hence, by induction, it is true for all natural numbers n.
(iii) True for n = 1 since
12 = (-1)1+1 1(1+1)
2
[ = (-1)2 1(2)
2
= 1]
Assume true for n = k i.e.
12 - 22 + 33 - ... (-1)k+1kk = (-1)k+1 k(k+1)
2
Need to show this assumption implies it is true for n = k+1:
12 - 22 + 32 - ... + (-1)k+1k2 + (-1)k+2(k+1)2
=
(-1)k+1 k(k+1)
2
+ (-1)k+2 (k+1)r2
=
(-1)k+1 [ k(k+1)
2
- (k+1)2]
=
(-1)k+1 (k+1) [ k
2
- (k+1)]
=
(-1)k+1 (k+1) [ k -2k -2
2
]
=
(-1)k+1 (k+1) [ -k -2
2
]
=
(-1)k+2 (k+1) [ k +2
2
]
=
(-1)k+2 (k+1)(k +2)
2
]
As this is the original formula with n = k+1 this means that the expression is true for n = k+1 if it is true for n = k. Hence, by induction, it is true for all natural numbers n.
(iv) True for n = 1 since
13 = æ
ç
è
1(1+1)
2
ö
÷
ø
2

 
Assume true for n = k i.e.
13 + 22 + 32 + ... + k3 = æ
ç
è
k(k+1)
2
ö
÷
ø
2

 
Need to show this assumption implies it is true for n = k+1:
13 + 22 + 32 + ... + k3 +(k+1)3
=
æ
ç
è
k(k+1)
2
ö
÷
ø
2

 
+(k+1)3
=
(k+1)2 k2)
4
+(k+1)3
=
(k+1)2 æ
ç
è
k2
4
+(k+1) ö
÷
ø
=
(k+1)2 æ
ç
è
k2 + 4k+4
4
ö
÷
ø
=
(k+1)2 æ
ç
è
(k+2)2
4
ö
÷
ø
=
æ
ç
è
(k+1)(k+2)
2
ö
÷
ø
2

 
As this is the original formula with n = k+1 this means that the expression is true for n = k+1 if it is true for n = k. Hence, by induction, it is true for all natural numbers n.

Question 17
True for n = 1 since

t(1) = 7·20 -3
Assume true for n = k i.e.
t(k) = 7·2k-1 -3
Need to show this assumption implies it is true for n = k+1:
t(k+1)
=
2t(k) + 3
=
2(7·2k-1 -3) +3
=
7·2k -2(3) +3
=
7·2k -2(3) +3
=
7·2k -3
As this is the original formula with n = k+1 this means that the expression is ture for n = k+1 if it is true for n = k. Hence, by induction, it is true for all natural numbers n.

Question 18
True for n = 1 since

t(1) = 31-1(12.5) - 5.5 = 12.5 -5.5 = 0
Assume true for n = k i.e.
t(k) = 3k-1(12.5) - 5.5
Need to show this assumption implies it is true for n = k+1:
t(k+1)
=
3t(k) +11
=
3(3k-1(12.5) - 5.5) + 11
=
3k(12.5) - 3(5.5) + 11
=
3k(12.5) - 16.5 + 11
=
3k(12.5) - 5.5
As this is the original formula with n = k+1 this means that the expression is ture for n = k+1 if it is true for n = k. Hence, by induction, it is true for all natural numbers n.

Question 19
(i) t(1) = 9, t(k) = t(k-1) + (2k+1)2 . Need to use induction to show

t(k) = k(4k2 +18k +11)
3
(ii) t(1) = 0, t(k) = t(k-1) + k -1 . Need to use induction to show
t(k) = k(k-1)
2


File translated from TEX by TTH, version 2.10.
On 19 Apr 1999, 14:49.