University of Johannesburg
Quantitative Economics 4A
QTE8X01
Assignment 1
1.
(a) Check the Base Case The proposition is true for the base case n =
1 since
2n − 1 = 2(1) − 1 = 2 − 1 = 1 = 12 = n2
Inductive Hypothesis Assume the proposition holds for some k
∈N , so that
1 + 4 + 7 + ... + (2k − 1) = k2
Induction Step We now use the above result to show the
proposition holds for the next natural number n = k + 1:
1 + 4 + 7 + ... + [2(k + 1) − 1] = 1 + 4 + 7 + ... + [2k − 1] + [2(k + 1) − 1]
= k2 + [2(k + 1) − 1]
= k2 + [2k + 2 − 1]
= k2 + 2k + 1
= (k + 1)2
This is the desired result.
∴ For all n ≥ 1, 1 + 4 + 7 + ... + (2n − 1) = n2.
(b) Check the Base Case The base case n = 0 is true because
(1 + x)0 = 1 = 1 + 0x ≥ 1 + 0x
Inductive Hypothesis Assume the proposition holds for some k
∈N, so that
(1 + x)k ≥ 1 + kx
Induction Step We now use the above to show the proposition
true for the next natural number n = k + 1
(1 + x)k ≥ 1 + kx
(by induction)
We multiply both sides by (1 + x), leaving the inequality
unchanged since (1 + x) > 0
1
(by induction)
(1 + x)k(1 + x) ≥ (1 + kx)(1 + x)
(1 + x)k+1 ≥ 1 + kx + x + kx2
(1 + x)k+1 ≥ 1 + (k + 1)x + kx2
(1 + x)k+1 ≥ 1 + (k + 1)x
Where the last line holds because kx2 is non-negative.
∴ If 1 + x > 0 then (1 + x)n ≥ 1 + nx for all n ≥ 0.
2.
vowels = {a, e, i, o, u}
3.
Two sets are equal if they contain the same elements
{x : x2 − 3x + 2} = {x : (x − 2)(x − 1) = 0}
= {x : x − 2 = 0 or x − 1 = 0}
= {x : x = 2 or x = 1}
= {2,1}
= {1,2}
=E
4.
A and B are subsets of C.A is a proper subset of C.
5.
The complement of the set A is represented by the grey area
outsidethe circle but within the rectangle.
2
6.
The complement of a set A is the setAC = {x : x /∈ A}
7.
The complement of Q if R is the universal set is the set of all irrational
real numbers.
{x ∈R : x /∈Q}
8.
S = {{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}
9.
n(S) = 4 ∴ 2n(S) = 24 = 16
P(S) ={∅,{1},{2},{3},{4},
{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},
{1,2,3},{1,2,4},{1,3,4},{2,3,4}, {1,2,3,4}}
10. (a) A − B = {1, 5}
(b) A − C = {1, 3}
(c) B and C; C and D
(d) AC = {2, 4, 6}
(e) A ∪ B = {1, 3, 4, 5, 6} A ∩ B = {3}
11. A = {2,3,4,5}, B = {2}
B is a subset of A since every element in B is also an element of A.
A has four elements, B has only one.
This means that A contains B and A also has more elements than B.
B is a proper subset of A and therefore, cannot contain A.
12. A = {2,3,4,5}, C = {1,2,3,...,8,9}
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For all x, x ∈ A implies x ∈ C because every element in A is also in C.
Therefore, A is a subset of C.
If we find one element in C that is not in A, that will mean the two sets
are not equal, that A 6= C.
The element 1 ∈ C and 1 ∈/ A. Therefore, A 6= C.
Together the facts that , A is a subset of C4 and that A 6= C, show that A
is a proper subset of C.
13. (a) A set is null if it contains no elements. Since
A = {x : x2 = 9,2x = 4}
any x ∈ A must satisfy both conditions. We begin with:
2x = 4
x=2
So that 2 is the only x that satifies the second condition. If we plug
2 into x2 we get x2 = 4 6= 9. So it does not satisfy the first condition.
The conclusion is that no x matches both conditions and so A is
null.
(b) Assume B is nonempty, then there exists some x such that x 6= x.
This can be rewritten as x = x + d where d is some number different
to zero.
x=x+d
x − x = d x(1
− 1) = d x(0) =
d
0=d
Which is absurd since we set d 6= 0, and so we must reject the
original assumption as false and conclude that B is null.
(c) Assume C is nonempty, then there exists some x such that
4
x+8=8x=8
−8x=0
Therefore, x = 0 is an element of C, so C is not null.
14. (A ∪ B) = {(x,y) ∈R2 : [x − y = 0] or [x + y = 0]}
We prove that two sets are equal by showing that each is a subset of the
other.
We first show that (A ∪ B) ⊆ C, that A ∪ B is a subset of C, by
demonstrating that (x,y) ∈ (A ∪ B) implies (x,y) ∈ C.
Let (x,y) ∈ (A ∪ B), then
x−y=0
or
x+y=0
This can be rewritten as
(x − y)(x + y) = 0 x2
− xy + xy − y2 = 0 x2 − y2
=0
Therefore, (x,y) ∈ C, and we conclude that (A ∪ B) ⊆ C.
These steps can be reversed to show that the converse is also true. Let
(x,y) ∈ C, then
x2 − y2 = 0
(x − y)(x + y) = 0
Which can be rewritten as
x−y=0
or
x+y=0
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So that (x,y) ∈ (A ∪ B), and C ⊆ (A ∪ B).
These two results, (A∪B) ⊆ C and C ⊆ (A∪B), together imply that (A ∪
B) = C.
15. A counterexample is used to prove that S is not closed under this new
addition.
Choose x = 1 and y = −2. Then x ∈ S and y ∈ S.
Substituting, we get
x y=x+y+1
= (1) + (−2) + 1
=0
Such that
because the 0 ∈/ S.
This counterexample proves that S is not closed under addition.
16. (a) True, F1 is a σ-algebra
(b) True, F2 is a σ-algebra
(c) False, F3 is not a σ-algebra since it is not closed under countable
unions
17. σ(J) = {∅,{5},{6},{5,6},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,6},{1,2,3,4,5,6}}
18. We rephrase the proposition by stating the contrapositive: If b is even,
then b2 is even.
Assume b is even, then b = 2n for some n ∈Z. Therefore,
b2 =4n2 b2
=2(2n2)
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The number 2n2 is an integer, since it is just the product of integers. So
b2 can be written as
b2 = 2m
where m = 2n2 ∈Z. Therefore, b2 is even. This proves the contrapositive.
We can then conclude that if b2 is odd, then b is odd.
√
19. As a preliminary step we prove that 6 is an irrational number.
√
Assume 6 rational, then
where q 6= 0, and p,q ∈Z and are coprime.
Then
6q2 = p2
2(3q2) = p2
∴ p2 is even, so p is even.
We can then write,
p = 2m
where m ∈Z.
Therefore,
∴ q = 0 or q /∈Z. Either case leads to a contradiction since we chose q
6= 0 and q ∈Z. √
Therefore 6 is irrational.
√
7
√
To prove the main result, assume
3+
2 is rational, then
where x,y ∈Z are coprime, and y 6= 0.
Therefore,
The RHS is a product and quotient of integer terms since√ p,q ∈Z, so
the RHS is rational. Therefore,√ 6 is rational. But we proved in the
preliminary result that 6 is irrational. This amounts to a contradic√ √ tion, and is enough to conclude that
3+
20. Assume x ∈Z is odd so that we can write
x = 2n + 1
where n ∈Z.
Then
x2 = (2n + 1)2
= 4n2 + 4n + 1
= 2(2n2 + 2n) + 1
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2 is therefore irrational.
Let m = 2n2 + 2n, then m is a product of integers so m ∈Z.
Therefore,
x2 = 2m + 1
This proves that x2 is odd.
∴ The square of an odd integer is odd.
21. A probability mapping in a sample space S is a function P with domain
D such that D is a σ-algebra and, i) P(A) ≥ 0 for all sets A ∈ D ii) P(S) = 1
iii) If A1,A2,A3,...,∈ D are pairwise disjoint, then
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