COMP2012 Discrete Mathematics
Lab 3(week 4) Exercises
Model Answers
Q1)
Proof:
For an arbitrary 𝑥, by definition, 𝑥 ∈ (𝐴 ∪ 𝐵)′ ↔ 𝑥 ∉ (𝐴 ∪ 𝐵)
↔ 𝑥 ∉ 𝐴 and 𝑥 ∉ 𝐵
↔ 𝑥 ∈ 𝐴′ and 𝑥 ∈ 𝐵′
↔ 𝑥 ∈ 𝐴′ ∩ 𝐵′
Q2)
i)
[3 4
13
]
∙
[
2
8
6
9 7
7 4
4 0
15
6]
3
= [3 × 13 + 4 × 8 + 2 × 6 3 × 9 + 4 × 7 + 2 × 4 3 × 7 + 4 × 4 + 2 × 0 3 × 15 + 4 × 6 + 2 × 3]
= [39 + 32 + 12
= [83 63
ii)
37
27 + 28 + 8 21 + 16 + 0 45 + 24 + 6]
75]
Calculate (𝐴 + 𝐵) ∙ 𝐶 and 𝐴 ∙ 𝐶 + 𝐵 ∙ 𝐶, where 𝐴 = [
𝐶=[
2 1
3 1
], 𝐵 = [
] and
3 2
0 1
1 4
]
3 2
(𝐴 + 𝐵) ∙ 𝐶 = ([2 1] + [3
3 2
0
2 1 1
𝐴∙𝐶+𝐵∙𝐶 =[
]∙[
3 2 3
11 24
=[
]
12 18
1
1
]) ∙ [
1
3
4
3
]+[
2
0
4
5 2 1
]=[
]∙[
2
3 3 3
1 1 4
5
]∙[
]=[
1 3 2
9
4
11 24
]=[
]
2
12 18
6 14
10
]+[
]
3 2
16
iii)
Let 𝐴 = [𝑎𝑖𝑗 ], 𝐵 = [𝑏𝑖𝑗 ] and 𝐶 = [𝑐𝑖𝑗 ] be real-valued matrices of dimension 𝑚 × 𝑛,
𝑚 × 𝑛 and 𝑛 × 𝑝. Prove the distributive property for matrix addition over
multiplication:
(𝐴 + 𝐵) ∙ 𝐶 = 𝐴 ∙ 𝐶 + 𝐵 ∙ 𝐶
Proof:
Since the dimensions of 𝐴, 𝐵 and 𝐶 are 𝑚 × 𝑛, 𝑚 × 𝑛 and 𝑛 × 𝑝 respectively, both
(𝐴 + 𝐵) ∙ 𝐶 and 𝐴 ∙ 𝐶 + 𝐵 ∙ 𝐶 are of the dimension 𝑚 × 𝑝.
Let 𝐷 = [𝑑𝑖𝑗 ] = (𝐴 + 𝐵) ∙ 𝐶 and 𝐸 = [𝑒𝑖𝑗 ] = 𝐴 ∙ 𝐶 + 𝐵 ∙ 𝐶 be two matrices with
dimension 𝑚 × 𝑝.
For the (𝑖, 𝑗) element in 𝐷 = [𝑑𝑖𝑗 ] = (𝐴 + 𝐵) ∙ 𝐶,
𝑑𝑖𝑗 = ∑ ( 𝑎𝑖𝑘 + 𝑏𝑖𝑘 ) ∙ 𝑐𝑘𝑗 = ∑ 𝑎𝑖𝑘 ∙ 𝑐𝑘𝑗 + ∑ 𝑏𝑖𝑘 ∙ 𝑐𝑘𝑗 = 𝑒𝑖𝑗
𝑘=1,…,𝑛
𝑘=1,…,𝑛
𝑘=1,…,𝑛
Each element in 𝐷 and 𝐸 is identical, and dim(𝐷) = dim(𝐸). So 𝐷 = 𝐸, and
(𝐴 + 𝐵) ∙ 𝐶 = 𝐴 ∙ 𝐶 + 𝐵 ∙ 𝐶
Q3)
Proof:
Let 𝑥, 𝑦, 𝑧 ∈ ℕ
If (𝑥, 𝑦) ∈ 𝑅𝐷𝐼𝑉 ∧ (y, z) ∈ 𝑅𝐷𝐼𝑉 , by definition of 𝑅𝐷𝐼𝑉 , there exists integers 𝑘 and 𝑚 such that
𝑥 = 𝑘𝑦 and 𝑦 = 𝑚𝑧
So 𝑥 = 𝑘𝑦 = 𝑘(𝑚𝑧) = (𝑘 ∙ 𝑚)𝑧, where 𝑘 ∙ 𝑚 is also an integer, i.e. 𝑥 is divisible by 𝑧.
Thus (x, z) ∈ 𝑅𝐷𝐼𝑉 , and 𝑅𝐷𝐼𝑉 is transitive.
Q4) Is the relation ∈ between sets:
i)
Reflexive?
ii)
Symmetric?
iii)
Transitive?
If so, prove the corresponding property. If not, give a counterexample.
i)
No. ∅ ∉ ∅.
The empty set does not contain any element, and it cannot be an element of itself
either. (Note that ∅ ∉ ∅ but ∅ ⊆ ∅.) So ∈ is not reflexive.
ii)
No. ∅ ∈ {∅} but {∅} ∉ ∅.
{∅} denotes the set of empty set, i.e. a set which has empty set as its only element. So
there is ∅ ∈ {∅}.
The empty set does not have any element, so the set of empty set cannot be an
element of the empty set, i.e. {∅} ∉ ∅. So ∈ is not symmetric.
iii)
No. ∅ ∈ {∅} and {∅} ∈ {{∅}}, but ∅ ∉ {{∅}}.
∅ is the only element in {∅} , and {∅} is the only element in {{∅}}. Hence there is
∅ ∈ {∅} and {∅} ∈ {{∅}}.
∅ is not an element of {{∅}}. So ∈ is not transitive.