Question

- is this proof acceptably ?
- If n is a natural number >= 3, prove that allways will be one natural naber ,,a" and a natural number ,,b" such that this statement n=a+b+1 is true

- so than :
assume there exists a natural number k such that k >= 3 and there exists a pair of natural numbers, a_k and b_k, such that (a_k + b_k + 1) = k.
Let a_(k+1) = a_k.
So, a_(k+1) is a natural number.
Let b_(k+1) = (b_k) + 1.
So, b_(k+1) is a natural number.
(a_(k+1) + b_(k+1) + 1) = [a_k + ((b_k) + 1) + 1] = [(a_k + b_k + 1) + 1] = (k + 1).
And (k + 1) is a natural number >

Answer 1

And (k + 1) is a natural number >= 3.
So, (k + 1) is a natural number >= 3 such that there exists a pair of natural numbers, a_(k+1) and b_(k+1), such that (a_(k+1) + b_(k+1) + 1) = (k + 1)
If there exists a natural number k such that k >= 3 and there exists a pair of natural numbers, a_k and b_k, such that (a_k + b_k + 1) = k,
then, for every natural number n >= k, there exists a pair of natural numbers, a_n and b_n, such that a_n + b_n + 1 = n.
Let a_3 = 1, which is a natural number.
Let b_3 = 1, which is a natural number.
(a_3 + b_3 + 1) = 3, which is a natural number >= 3.
THUS, for every natural number n >= 3, there exists a pair of natural numbers, a_n and b_n, such that a_n + b_n + 1 = n.