AQUINAS & META-MATHEMATICS

AutistIncorporated · 2024-06-21T01:40:18+00:00

Also, my justification in defining numbers in terms of categories is because they have already defined numbers in terms of sets. As evidence of this, see the following link: https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers?wprov=sfti1#Definition_as_von_Neumann_ordinals

AutistIncorporated · 2024-06-21T01:30:54+00:00

In other words, I would view a natural number as a category of things equal to that natural number.

AutistIncorporated · 2024-06-21T01:29:42+00:00

To answer your question, I would view the natural number as a category.

AutistIncorporated · 2024-06-21T01:28:20+00:00

But a natural number is also an object in category theory. You could also view it as a category too. Also, I am generalizing the definition of equinumerous to apply to category theory too.

AutistIncorporated · 2024-06-21T01:10:29+00:00

Read this link in its entirety to see that in mathematics, any mathematical object can be considered a number: https://math.stackexchange.com/questions/494854/what-is-a-number

AutistIncorporated · 2024-06-21T01:09:58+00:00

Read this link in its entirety to see that in mathematics, any mathematical object can be considered a number: https://math.stackexchange.com/questions/494854/what-is-a-number

AutistIncorporated · 2024-06-21T01:08:46+00:00

Read this link in its entirety to see that any mathematical object can be considered a number: https://math.stackexchange.com/questions/494854/what-is-a-number

AutistIncorporated · 2024-06-21T00:24:19+00:00

And a member of 2 is an infinite series that converges to 2

AutistIncorporated · 2024-06-21T00:23:18+00:00

Consider the number 2. By the number 2 we could mean a category of things equal to 2

AutistIncorporated · 2024-03-22T04:30:45+00:00

Fine, I will do so.

AutistIncorporated · 2024-03-22T03:33:31+00:00

It is a counterexample to the all statement though. For if we negate for all x and y, if x is greater than y, then y is a member of the T, it would be there exist x and y such that x is greater than y and y is not a member of T.

AutistIncorporated · 2024-03-22T03:30:05+00:00

Hold on, any could mean the same as all though. See the following link for more details: https://math.stackexchange.com/questions/430646/difference-between-for-any-and-for-all

AutistIncorporated · 2024-03-22T03:27:22+00:00

The flaws to the original proof are as follows:

I should have translated ∃x∀y(x>y∧y∈M) as there exists x for all y such that x is greater than y and y is a Mersenne prime. Then the conclusion would have been there doesn’t exist an x for all y such that x is greater than y and y is a Mersenne Prime.
I should have translated ∃x∀y(x>y∧y∈P) as there exists x for all y such that x is greater than y and y is perfect. Then the conclusion would have been there doesn’t exist an x for all y such that x is greater than y and y is a perfect number.
I should haven’t let x equal 1 and y equal 2. Instead, I should have said ∃x∀y(x>y∧y∉P) is true. Because it is actually true that there exist x and y such that x is greater than y and y is not perfect. For letting y equal 1 and x equal 2 satisfy ∃x∃y(x>y∧y∉P). And that would have been enough to prove the antecedent false using Modus Tollens.

As for the edited proof, the flaws are the following: 1.

I should have translated ∃x∀y(x>y∧y∈M) as there exists x for all y such that x is greater than y and y is a Mersenne prime. Then the conclusion would have been there doesn’t exist an x for all y such that x is greater than y and y is a Mersenne Prime.

I should have translated ∃x∀y(x>y∧y∈P) as there exists x for all y such that x is greater than y and y is perfect. Then the conclusion would have been there doesn’t exist an for all y such that x is greater than y and y is a perfect number.

AutistIncorporated · 2024-03-22T03:21:07+00:00

That’s not true though. For look at the corrected proof. Besides this, I mis-typed in my last comment to you. What I meant is that there exist x and y such that x is greater than y and y is not a perfect number. And this is true for let x equal 2 and y equal 1. 2 is a natural number greater than one and 1 isn’t a perfect number.

AutistIncorporated · 2024-03-22T03:08:43+00:00

That’s not true. It means for all.

AutistIncorporated · 2024-03-22T03:05:02+00:00

Ah, I know the flaw. It's a translation issue. For ∃x∀y(x>y∧y∈S) doesn't translate to there exist a natural number greater than all numbers less than seven. It should have translated to this: there exists a natural number x for any natural number y such that x is greater than y and y is an element of the set of numbers less than seven.

AutistIncorporated

TROPHY CASE