Course 111: Algebra, October 2006
1. Given Z the set of integers, Q the set of rational numbers and R the
set of real numbers answer true or false to the following. If false give
an example of an element in the first set not in the second.
R ⊆ Q
Z ⊆ R
Q ⊆ Z
2. Prove de Morgan’s rules:
For a subset C of S let C 0 denote the complement of C in S.
For two subsets A and B of a set S
I (A ∪ B)0 = A0 ∩ B 0
II (A ∩ B)0 = A0 ∪ B 0
Proof (I)
First we show that (A ∪ B)0 ⊆ A0 ∩ B 0 .
Say x ∈ (A ∪ B)0 .
⇒
⇒
⇒
⇒
⇒
x∈
/ (A ∪ B)
x∈
/ Aand x ∈
/B
0
x ∈ A and x ∈ B 0
x ∈ A0 ∩ B 0 an element of 2 sets is an element of their intersection
(A ∪ B)0 ) ⊆ A0 ∩ B 0
Now we prove that A0 ∩ B 0 ⊆ (A ∪ B)0 .
As above we choose x ∈ A0 ∩ B 0 .
⇒ x ∈ A0 and x ∈ B 0
⇒ x∈
/ Aand x ∈
/B
If x not in A and x not in B then
⇒ x∈
/ (A ∪ B)
⇒ x ∈ (A ∪ B)0
Combining these results we have that (A ∪ B)0 ⊆ A0 ∩ B 0 and A0 ∩ B 0 ⊆
(A ∪ B)0 .
⇒ (A ∪ B)0 = A0 ∩ B 0 .
Proof II
As above. Begin by considering x ∈ (A ∩ B)0 then
⇒ x∈
/ (A ∩ B)00 = (A ∩ B) = (A0 ∪ B 0 )0 (using part 1)
⇒ x ∈ (A0 ∪ B 0 )
⇒ (A ∩ B)0 ⊆ (A0 ∪ B 0 ).
Now consider x ∈ (A0 ∪ B 0 ). Then
⇒ x∈
/ (A0 ∪ B 0 )0 = (A ∩ B) (using part 1)
⇒ x ∈ (A ∩ B)0
⇒ (A0 ∪ B 0 ) ⊆ (A ∩ B)0
And so (A ∩ B)0 = A0 ∪ B 0 .