Assigment 3 MATH1061
Jakob Kerin-Bird
May 2023
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Question 1
(6 marks) Consider the sequence {an }n1 defined recursively by
a1 = 2 and ak = 3ak1 + 2 for each integer k2.
a)
Calculate the values of a2 , a3 , a4 and a5 .
For a2 , substitute k = 2 into the formula:
a2 = 3a1 + 2
It is known that a1 = 2, as such we can substitute 2 into the formula.
a2 = 3 · 2 + 2 = 6 + 2 = 8
Thus, a2 is 8.
For a3 , substitute k = 3 into the formula:
a3 = 3a2 + 2
It is known that a2 = 8, as such we can substitute 8 into the formula.
a3 = 3 · 8 + 2 = 24 + 2 = 26
Thus, a3 is 26.
For a4 , substitute k = 4 into the formula:
a4 = 3a3 + 2
It is known that a3 = 26, as such we can substitute 26 into the formula.
a4 = 3 · 26 + 2 = 78 + 2 = 80
Thus, a4 is 80.
For a5 , substitute k = 5 into the formula:
a5 = 3a4 + 2
It is known that a4 = 80, as such we can substitute 80 into the formula.
a5 = 3 · 80 + 2 = 240 + 2 = 242
Thus, a5 is 242.
Therefore, it is now known that:
a1 = 2 a2 = 8 a3 = 26 a4 = 80 a5 = 242
b)
Guess an explicit formula for this sequence and prove that your guess is correct.
The following sequence can be observed in the samples a1 through to a5 :
a1 = 2 = 31 − 1
a2 = 8 = 32 − 1
a3 = 26 = 33 − 1
a4 = 80 = 34 − 1
a5 = 242 = 35 − 1
Thus, the guessed explicit formula takes the form:
an = 3n − 1
2
Now, using induction, a proof for the formula can be developed:
Base case n = 1:
a1 = 31 − 1
a1 = 3 − 1
a1 = 2
Thus, the base case holds.
Inductive Step:
Assume that the formula holds for n = k:
ak = 3k − 1
We need to prove that the formula also holds for n = k + 1:
ak+1 = 3k+1 − 1
We can modify the original recursive formula for ak+1 :
ak = 3ak−1 + 2
ak+1 = 3ak + 2
Now, the proposed explicit formula can be substituted into the formula
for ak :
ak+1 = 3(3k − 1) + 2
ak+1 = 3k+1 − 3 + 2
ak+1 = 3k+1 − 1
Therefore, the explicit formula an = 3n − 1 holds true for all n ≥ 1.
Question 2
(4 marks) Prove the following statement using set identities. For all sets A, B
and C, that are subsets of a universal set U ,
A − (B ∪ C) = (A − B) − C.
Show your work by naming the identities you use.
First, let’s focus on simplifying the right-hand side of the equation:
(A − B) − C
(A − B) − C = (A ∩ B c ) − C
Using the Set Difference Law.
(A ∩ B c ) ∩ C c
Using the Set Difference Law again.
A ∩ (B c ∩ C c )
Using De Morgan’s Law.
Now, we will simplify the left side of the equation:
A − (B ∪ C)
A ∩ (B ∪ C)c
Using the Set Difference Law.
A ∩ (B c ∩ C c )
Using De Morgan’s Law.
Therefore, it is shown that:
A − (B ∪ C) = (A − B) − C
A ∩ (B c ∩ C c ) = A ∩ (B c ∩ C c )
Therefore, the statement has been proven using set identities.
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Question 3
(4 marks) Define sets S, T and E as follows:
S = {n ∈ Z|n ≡ 0 (mod 4)}
T = {n ∈ Z|n ≡ 2 (mod 4)}
E = {n ∈ Z|n ≡ 0 (mod 2)}
Prove that S × T ⊂ E × E.
Define Cartesian Products:
S × T:
The Cartesian product is (s, t), where s ∈ S and t ∈ T .
E × E:
The Cartesian product is (e1 , e2 ), where e1 , e2 ∈ E.
Now, it can be seen that ∀s ∈ S, ∃e ∈ E : s = e as if a number, s, is
divisible by 4 it must also be divisible by 2 and as such even and a part of
the set E.
Likewise, for ∀t ∈ T, ∃s ∈ S : t = s ± 2, as we know all even numbers, s,
whenever we have an offset, remainder, of 2, that new number t will also be
even and as such a part of the set E.
Thus, we have shown that for any arbitrary element in S × T , there is a
corresponding element in E × E. Therefore, S × T ⊂ E × E.
Question 4
(4 marks) Define a function f : P ({0, 1}) × P ({1, 2})ß{0, 1, 2} by f ((A, B)) =
|A ∩ B|.
a)
Is the function f injective? Justify your answer.
Proof by Counterexample:
Firstly, lets define the following:
A1 = 0, B1 = 1 and A2 = 1, B2 = 2
Then, we have:
f ((A1 , B1 )) = |A1 ∩ B1 | = |∅| = 0
f ((0, 1)) = |0 ∩ 1| = |∅| = 0
And we have:
f ((A2 , B2 )) = |A2 ∩ B2 | = |∅| = 0
f ((1, 2)) = |1 ∩ 2| = |∅| = 0
Thus, we have f ((A1 , B1 )) = f ((A2 , B2 )), but (A1 , B1 ) ̸= (A2 , B2 ).
Therefore, f is not injective.
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b)
Is the function f surjective? Justify your answer.
Direct Proof:
Firstly, let’s consider each element, y, of the codomain {0, 1, 2}:
y = 0:
A = 0 and B = 2
Then, A ∩ B = ∅, so |A ∩ B| = 0 and f ((A, B)) = 0
y = 1:
A = 1 and B = 1
Then, A ∩ B = 1, so |A ∩ B| = 1 and f ((A, B)) = 1
y = 2:
A = 0, 1 and B = 1, 2
Then, A ∩ B = 1, so |A ∩ B| = 1 and f ((A, B)) = 1
In these cases, we cannot find an element (A, B) in the domain for
which f ((A, B)) = 2.
Therefore, the function f is not surjective.
Question 5
(6 marks) Let A and B be the following intervals of real numbers:
A = [0, 2] and B = (2, 6).
Use the Schröder-Bernstein Theorem to prove that |A| = |B|.
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