complete assignment

Assignment
Question 1
Two sets are identical if they have the same elements
In our case, we need a model of sets in which the extensionality axiom does not hold
i.e. a≠ b
It follows that:
U= [{a, b}, {a}, {a}. ϵ]
Where a ϵ {a}, a ϵ {a, b} and because a≠ b; then {a} ≠ {a, b} but for all x ϵ U, we have
X ϵ {a} ↔ x, ϵ {a, b}
Using the above relation, we thus have sets b & c and sets a & d as identical.
Question 2
Part 1
To prove that [{a}, {a, b}] = [{c}, {c, d}] for a= c and b= d
We first define axiom extensionality
Ɐx Ɐy (x =y ↔ Ɐz (z ϵ x↔ z ϵ y)
Assuming all x=a and y=b
Then;
A ϵ {a} and a ϵ {a, b}
Now that a=b, we have {a}= {a, b}
But for every x ϵ U
We get x ϵ {a} ↔ x ϵ {a, b}
Applying the initial principle for axioms thus, we get that
[{a}, {a, b}] = [{c}, {c, d}]
Part 2
Set of ordered pairs
Since *n is a multiple m* on the set of natural numbers {0, 1, 2,…}
We have
{0}, {0, 1}, {0, 1, 2}, {0, 1, 2, 3,…, n}
Sets of ordered pairs for every *m sets up to n* pairs
The relation is reflexive since every base set say {0} is duplicated in the subsequent set
Proof of reflexive property
A set of natural numbers is reflexive if the set of positive integers (x, y) ϵ R i.e. x + 2y= 1 is not satisfied by any (x, y) hence the reflexive property holds.
Part 3
Show by induction (11n – 6) is divisible by 5
We set the base case n= 0
We then have 110 – 60 = 1- 1= 0
Since every non- zero number divides 0 we conclude that it is true
We thus assume 7|11k – 6k and 11k – 6k set to a number say m
11k – 6k = 7m
11k +1 – 6k + 1 = 11 × 11k – 6 × 6k
= (7 + 6) × 11k – 6 × 6k
= 7 × 11k + 6 × (11k – 6k)
= 7 × (11k – 4 × m)
Proved by induction
Question 3
Inductive definition
We show that n= 1 is true and again n= k is true
We assume that n= k – 1
For 1- tuples, we define (x1)= (y1) for x1 – y1
Thus for n + 1 tuples, we have
(x1,…, xn, xn + 1)= (y1,…, yn, yn + 1) to mean
(x1,…, xn)= (y1,…, yn)
and
xn + 1 = yn + 1