Meanwhile, to answer the actual question, mathematical proofs provide answers to specific, well-defined questions, for which no other process can provide an answer. An excellent example is the modern day rendering of Euclid's proof of the infinitude of the prime numbers.

We begin with some basic definitions. First:

Exactly divisible. Let two positive integers exist, a and b. Then a is exactly divisible by b, if a/b is itself a positive integer.

As a corollary, whenever a is exactly divisible by b, a can be written in the form k × b, where k is another positive integer. In algebra, we frequently use the juxtaposition shorthand kb for this, but this is merely a notational convention.

This leads to the corollary definition of:

Exact divisor. Given two positive integers a and b, b is an exact divisor of a, if a/b is itself a positive integer.

Next:

Prime number. A number p is a prime number, if and only if the sole exact divisors of p, are 1 and p itself. Any number a which has exact divisors

other than 1 and a itself, is a

composite number. so called because it can be composed by multiplying together those other exact divisors.

We now come to:

Theorem. The prime numbers form an infinite set.

Proof. We prove this result by contradiction. We first assert, by hypothesis, that the prime numbers constitute a finite set, and prove that this leads to a contradiction. This assertion consists of the statement that a set:

S = {p

1, p

2, p

3, ... , p

k}

containing k members, with each of the p

i in the set being the prime numbers from 2 upwards consecutively, comprises the totality of the prime numbers.

We now construct the following number:

P = p

1 × p

2 × p

3 × ... × p

kThis is a composite number (by definition), whose exact divisors are the prime numbers in the set S.

We now ask the elementary question, what are the exact divisors of (P+1)?

P+1 cannot have p

1 = 2 as a divisor, since P+1 is of the form 2K + 1, which is not exactly divisible by 2.

Likewise, P+1 cannot have p

2 = 3 as a divisor, since P+1 is of the form 3K + 1, which is not exactly divisible by 3.

Likewise, P+1 cannot have p

3 = 5 as a divisor, since P+1 is of the form 5K + 1, which is not exactly divisible by 5.

This continues all the way to ...

P+1 cannot have p

k as a divisor, since P+1 is of the form 3p

k + 1, which is not exactly divisible by p

k.

We are therefore left with two choices. Either:

[1] P+1 itself is prime, in which case we have found a prime number not contained in the original set S, contradicting our original assertion;

[2] P+1 is a composite number, and can therefore be written in the form (P+1) = rs. But the same logic above applies to r and s; neither of these can be divisible by any of the p

i in our set S above. If any one of r and s is prime, the result again follows. This procedure can be conducted recursively, until we find prime exact divisors of (P+1) that are not members of the set S.

The assertion that a finite set S as defined above is complete, is therefore refuted by contradiction, because for all such sets S, it is possible to construct a number, from the members of S, that yields a prime number that is not a member of S, and therefore it is

impossible to construct a

finite set of primes that is complete. Therefore, there exists an infinite number of prime numbers.

The proof is reliable, and rigorous. If anyone has a result from metaphysics matching this level of rigour and reliability, I'll be happy to learn of it. Until then, I think pure mathematics needs to take no notice of assertionists.