Finding the statement of n = (2 – y), for all positive natural numbers of ℕ.
(∀n ϵ ℕ) (∃y ϵ ℕ) (n = 2 – y).
(∀n ϵ ℕ).
n/2 + y/2 = 1
n/2 = 1 – y/2
n = 2(1 – y/2)
n = 2–2 y/2
n = 2–2y/2
n = 2 – y
Finding the k, element of ℕ, where n = 2k, for all even numbers.
(∀n ϵ ℕ) (∃k ϵ ℕ) (n = 2k) for some number (k ϵ ℕ).
For all even numbers.
n = 2k = 2 – y
Where;
k = (2 – y)/2
= 1– y/2
For all even numbers.
Finding the k, element of ℕ, where n = 2k + 1, for all prime numbers.
(∀n ϵ ℕ) (∃k ϵ ℕ) written as n = 2k + 1, for all prime numbers (k ϵ ℕ).
For all prime numbers.
n = 2k + 1 = 2 – y
Where;
2k + 1 = 2 – y
2k = 2 – y – 1
2k = 1 – y
k = 1/2 – y/2
For all prime numbers.
Adding two prime numbers together will result in a (y ϵ k) value for the resulting even number.
2k + 1 + 2k + 1 = 0
2(1/2 – y/2) + 1 + 2(1/2 – y/2) + 1 = 0
(1–2/2y) + 1 + (1–2/2y) + 1 = 0
– y + 2 + 1 – y + 1 = 0
–2y + 4 = 0
–2y = – 4
y = – 4/– 2
y = 2
Substituting this (y ϵ k) from the addition of two primes as the (y ϵ k) for the resulting even number gives an (n ϵ k).
n = 2k
= 2(1– y/2)
= 2(1–2/2)
= 2 (0)
= 0
Substituting the (n ϵ k) of the resulting even number from the addition of two primes into the (n ϵ ℕ) gives a (y ϵ ℕ).
n = 2 – y
0 = 2–0
0–2 = – y
– 2 = – y
2 = y
When (n ϵ k) equals zero, (y ϵ ℕ) equals two.
Demonstrating the proof for the first even number two where the (n ϵ ℕ) = 2 and produces a (y ϵ ℕ).
n = 2 – y
n – 2 = – y
2–2 = – y
0 = – y
Substituting this (y ϵ ℕ) into the (n ϵ ℕ) equation n = 2 – y, proves the conjecture.
n = 2 – y
= 2 – – y