Complete Solutions I(b) 1
Complete Solutions to Exercise I(b)
1. (a) We can draw the given sets A   x 
B  x 
0  x  10 and
 10  x  10 on the number line as follows:
0  x  10
-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
1
-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
2
1
3
4
5
Set A
6
7
8
9 10 11
2 3 4 5 6 7 8 9 10 11
-10  x  10
Set B
Clearly the set A is a subset of B, that is A  B .
0  x  10 and
(b) Again drawing the given sets A   x 
B  x 
0  x  10 we have
0
1
2
3
4
5
6
7
8
9
10
The set A is the set of all the integers (whole numbers) between 0 to 10, that is
A  0, 1, 2, , 10
whilst the set B is all the real numbers between 0 to 10, that is all the line in the above
diagram. Again the set A is a subset of the set B, that is A  B .
(c) This is similar to the sets in part (b) but we have switched sets A and B. Therefore
we have B  A .
 10  x  10 mean?
(d) What does the set notation A   x 
It means that x is a natural number between 10 to 10. What numbers does this set
include?
Remember natural numbers are whole numbers greater than or equal to 1, therefore
1  x  10 represent?
A  1, 2, 3, , 10 . What does B   x 
It is the same as the set A because we have integers between 1 to 10, that is
B  1, 2, 3, , 10
Hence we have A  B and B  A which means we have A  B .
(e) Both the given sets are the same but one of them includes the end points:
Set A
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
Set B
6
7
8
9
10
Clearly B  A because the set B does not include the end points 0 and 10.
Complete Solutions I(b) 2
2. Before we evaluate which of the given sets are subsets we need to write down the
elements of each given set.
, A  2, 3, 5


B  x
x 2  4  0  2, 2
C  x 
x is prime and less than 10  2, 3, 5, 7
D  x 
0  x  10  0, 1, 2, 3,
, 10
Now we check for subsets.
(a) For the given sets  and A we have   A because the empty set  is a subset
of every set.
(b) Is A subset of A?
Yes we have A  A . [Every set is a subset of itself].
(c) We need to examine the given sets A  2, 3, 5 and C  2, 3, 5, 7 . Clearly
all the elements of the set A which are 2, 3 and 5 are in the set C therefore A is a
subset of C, that is A  C .
(d) Similarly we have C  2, 3, 5, 7 and D  0, 1, 2, 3, , 10 . Again all the
elements of the set C which are 2, 3, 5 and 7 are in the set D therefore C is a subset of
D, that is C  D .
(e) Is the set B subset of the set C?
We have B  2, 2 and C  2, 3, 5, 7 and the element 2 in the set B is not in
the set C therefore B  C . What does this notation mean?
The set B is not a subset of the set C.
3. We need to write out the elements of each of the given sets.
A  x 
0  x  5  1, 2, 3, 4
B  x 
x is an even number  2, 4, 6, 8,
C  x 
x is a multiple of 2  2, 4, 6, 8,
D  x 
x  x  
E  x
x3  1, 23 , 33 , 43 ,
F  x 
0  x  2  1


  1, 8,


27, 64,

Note that the set D is the empty set because there is no real number x such that x  x .
(a) Since A  1, 2, 3, 4 and B  2, 4, 6, 8,  therefore A  B because the
elements 1, 3, 5, … are not in the set B.
(b) Similarly we have A  1, 2, 3, 4 and C  2, 4, 6, 8,  therefore A  C .
(c) We have B  2, 4, 6, 8,  and C  2, 4, 6, 8,  which means we have
BC.
(d) B and C are the same sets as in part (c), that is
B  2, 4, 6, 8,  and C  2, 4, 6, 8,
therefore C  B . In fact B  C .

Complete Solutions I(b) 3
(e) Since D   and A  1, 2, 3, 4 therefore the set A cannot be a subset of the
empty set  which means we have A  D
(f) Because D   and the empty set is a subset of every set therefore D  A .
(g) We have the sets E  1, 8, 27, 64,  and F  1 . Thus the set
E  1, 8, 27, 64,
 cannot be a subset of the set F  1 . We have E  F .
(h) As part (g) we have E  1, 8, 27, 64,  and F  1 and since the element 1
is in the set E  1, 8, 27, 64,  therefore F  E .
4. What is the power set of a, b, c ?
It is all the subsets of a, b, c . These are
, a , b, c, a, b, a, c, b, c, a,
b, c
5. What is the power set of a, b, c, d  ?
Again it is all the subsets of a, b, c, d  :
,

a,

a,
 a,

a , b , c , d  ,
b , a, c , a, d  , b, c , b, d  , c, d  ,
b, c , a, b, d  , a, c, d  , b, c, d  ,
b, c , d 







6. What does the term cardinality mean?
Cardinality is the number of elements in the set and is denoted by A .
(a) Since  denotes the empty set which means it has no elements therefore
 0
(b) We are given the set A  a, b, c therefore A  3 because the set has 3
elements.
(c) What is the size of P  A where A  a, b, c ?
Remember P  A represents the power set and in the main text we have that if A  n
then P  A   2n . In this case we have A  3 therefore P  A   23  8 . The
cardinality of the power set is 8.
(d) We are given A  x 
3x 2  x  0 and we need to find the elements of this


set A. Solving the given quadratic
3x 2  x  0
x  3x  1  0
 Factorising 
1
3
Since x  therefore members of this set A can only be integers (whole numbers)
which means only 0 is a member. Thus A  0 . What is the cardinality of this set?
x  0, x 
Since the set is a singleton (only one element) therefore the cardinality A  1 .
Complete Solutions I(b) 4
(e) What are the elements of the given set A   x 
x  x  1 ?
Since for any real number x we have x  x 1 therefore there are no x values which
satisfy the equation x  x 1 . Hence the set A is empty, that is A   and so A  0 .
7. We are given the sets A  1, 2, 3, 4, 5 and B   x 
x is prime  5 .
What are the elements of the set B?
B  x 
x is prime  5  2, 3, 5
How can we show A  B ?
Remember by definition (I.7) we have A  B if and only if every element of set A is
also in the set B. In this case we have 1  A but 1 B therefore A  B .
How do we show B  A ?
Again by using the definition (I.7) we show that every element of the set B is also in
the set A. Since B  2, 3, 5 and all 3 elements 2, 3 and 5 are in the set
A  1, 2, 3, 4, 5 therefore B  A .
3 

8. We have A  1, 3 , B  1, 3, 3, 1 and C  1, 3, ,  . Remember from
1 

section A that a set such as a, b is the same as a, b, a, b . Here we have
B  1, 3, 3, 1  1, 3
3 

C  1, 3, ,   1, 3, 3, 1  1, 3
1 

Thus we have A  B  C  1, 3 .
9. We need to prove  B \ A  B . How?
Let x be an element of the set B \ A and then show that this element x is also in the set
B.
Proof. Let x be an arbitrary element of the set B \ A . What does this mean?
By definition (I.5) we have x  B \ A means that x  B but x  A . Since x  B
therefore  B \ A  B which is our required result.
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10. We need to prove two results  A  B   A and  A  B   B . We prove
 A  B   A and leave  A  B   B
for the student. How do we prove  A  B   A ?
As for question 9 we let an arbitrary element be in the set A  B and then show it is
in the set A.
Proof. Let x  A  B be an arbitrary element. The notation x  A  B means that
x  A and x  B . Since x  A therefore  A  B   A which is our required result.
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Complete Solutions I(b) 5
11. We are required to prove that  B \ A  Ac . How?
Let x be an arbitrary element of B \ A and then show that x is also an element of Ac .
Proof. Let x be an arbitrary member of B \ A . What does this mean?
By definition (I.5) we have x  B but x  A . Since x  A therefore x  Ac , this
follows by the definition (I.4) Ac   x x  A . We have x  Ac which means that
 B \ A  Ac .
12. The proof of  A \ B   B is very similar to the proof of question 11,
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c
 B \ A  Ac .
Proof. Let x  A \ B where x is an arbitrary element. Since x  A \ B therefore x  B
which means that x  B c . Thus we have  A \ B   Bc which is our required result.
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13. How do we prove A   A  B   A ?
To show equality of sets we need to prove A   A  B   A and A  A   A  B  .
How do we show each of these results?
Let x be an arbitrary element of A   A  B  and then show x is a member of the set
A. Next let y be an arbitrary element of the set A and prove that y is also in
A   A  B .
Proof. Let x be an arbitrary member of the set A   A  B  . This means that x  A or
x  A  B . In either case x  A therefore A   A  B   A .
Let y be a member of the set A. Since y  A therefore y  A   A  B  which means
that A  A   A  B  .
Thus combining these 2 together, A   A  B   A and A  A   A  B  , we have
our result
A   A  B  A
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14. How do we prove A   A  B   A ?
Similar to question 13. Let x be an arbitrary element of the Left Hand Side set
A   A  B  and show that it is in the Right Hand Side set A. Then let y be in the
Right Hand Side set A and show that it is in the Left Hand Side set A   A  B  .
Proof. Let x  A   A  B  be an arbitrary element. What does this mean?
Means that x is a member of the set A and the set A  B . Since x  A therefore we
have A   A  B   A .
Let y be an arbitrary member of the set A. This means that the member y is in both
sets, that is y  A and y  A  B . Therefore y  A   A  B  . Thus we have
A  A   A  B .
Complete Solutions I(b) 6
Since we have A   A  B   A and A  A   A  B  therefore by definition (I.11)
we have our required result A   A  B   A .
15. We need to prove  B \ A    A  B  .
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c
Proof.
Let x  B \ Ac be an arbitrary member. What does this x  B \ Ac mean?
By definition (I.5) x  A \ B means that x  A but x  B so in this case x  B \ Ac
means that x  B but x  Ac . Since x  Ac therefore x  A . We have x  A and
x  B therefore x  A  B . We have established  B \ Ac    A  B 
Let y be an arbitrary member of the set A  B which means that y  A and y  B .
Since y  A therefore y  Ac . We have y  B and y  Ac which means that
y  B \ Ac . Thus  A  B    B \ Ac  .
We have  B \ Ac    A  B  and  A  B    B \ Ac  which means that
 B \ A    A  B .
c
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16. We need to prove A   B  C    A  B    A  C  .
Proof.
Let x be an arbitrary member of the set A   B  C  . Thus x  A or x  B  C . If
x  A then x   A  B    A  C  . If x  B  C then x  B and x  C . This means
that x   A  B    A  C  . In either case the arbitrary element x is in the set
 A  B    A  C  which means that we have
A   B  C    A  B   A  C  .
Going the other way, let y be an arbitrary member of the Right Hand Side set
 A  B    A  C  . This means that y is a member of the set A  B and A  C . If
y  A then clearly y  A   B  C  . If y  A then y  B and y  C because
y   A  B    A  C  . Since y  B and y  C therefore y  B  C which implies
that y  A   B  C  . Thus we have  A  B    A  C   A   B  C  .
Since we have established that A   B  C    A  B    A  C  and
 A  B    A  C   A   B  C  therefore we have our result
A   B  C    A  B   A  C 
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