Why can't we prove psychology mathematically?

content
“Direct evidence
»Indirect or contradiction proof
“Example of a contradiction proof

Direct evidence

The straightforwardness of direct proof makes it an intuitive approach to proof. At the beginning there are axioms, mathematical propositions that have already been proven and the requirements of the proposition to be proven. Then logical inferences are drawn until the proposition of the sentence is shown. The inferences are often of the form: We know that the implication "If \ (A \) then \ (B \)" is true. The statement \ (A \) is one of our premises. So \ (B \) is also true, and we can use \ (B \) to prove further statements.

A small example: We show that \ (25 \) is not a prime number. (If you don't know what a prime number is, you can read it here: [Link]) We know that \ (5 \ cdot 5 = 25 \). Therefore \ (5 \) is a divisor of \ (25 \), and it is \ (1 \ neq 5 \ neq 25 \). So \ (25 \) cannot be a prime number.

Indirect or contradiction proof

The contradiction proof (or indirect proof) works fundamentally different. Here it is initially assumed that the statement to be proven is false. A contradiction is then brought about through logical inferences, e.g. one shows that one of the premises (which are assumed to be true) is false. Since that cannot be the case, the assumption must be false and the statement to be proven must be true.

Here, too, a small example: We show again that \ (25 \) is not a prime number. Assume that \ (25 \) is a prime number after all. Then \ (1 \) and \ (25 \) must be the only divisors of \ (25 \). But it is \ (5 \ cdot 5 = 25 \) is, so \ (5 \) is a divisor of \ (25 \). Contradiction!

Example of a contradiction proof

As a longer example we would like to show Euclid's theorem. This says: There are infinitely many prime numbers.

We assume: There are only finitely many prime numbers. Now let's try to create a contradiction.

If there are only finitely many prime numbers \ (2, 3, 5, 7, \ dots \), then we can find them in the set \ (P: = \ {p_1, \ dots, p_k \} \) with a \ ( k \ in \ mathbb {N} \). \ (P \) is the set of all prime numbers that exist!

Now we consider the product of all prime numbers and add 1, i.e. \ (n: = p_1 \ cdot \ dots \ cdot p_k +1 \). According to the fundamental theorem of arithmetic, this number has a prime factorization.

For every prime number \ (p_i \ in P \), however, \ (p_i \) is not a divisor of \ (n \), because when \ (n \) is divided by \ (p_i \), the remainder remains 1 . If \ (q \) is one of the factors from the prime factorization of \ (n \), then \ (q \ notin P \), but \ (q \) is a prime number. This contradicts the fact that \ (P \) is the set of all prime numbers.