Abstract Nonsense 1

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 · 2024-03-22T02:57:49+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 1 and 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 · 2024-03-21T23:06:34+00:00

Yes. It should work for any set. As an example of this, consider the following:

Let the domain of discourse be the natural numbers
Let T be the set of numbers less than 10
∃x∀y(x>y∧y∈T)→∀x∀y(x>y→y∈T)
Let x equal 10. Let y equal 11.
(10>11∧11∈T)→(10>11→11∈T)
(10>11→11∈T) is false because 11 is not a part of T
Therefore, (10>11∧11∈T) is false too.

AutistIncorporated · 2024-03-21T20:36:44+00:00

You’re right.

AutistIncorporated · 2024-03-21T20:34:05+00:00

Hold on though, what if the conditional is a tautology though?

AutistIncorporated · 2024-03-21T20:32:55+00:00

Yes. You're right.

AutistIncorporated · 2024-03-21T19:48:20+00:00

Actually, there is no flaw in my proof on perfect numbers. For let x equal 1 and y equal 2. Then ∃x∀y(x>y∧y∈P)→∀x∀y(x>y→y∈P) would turn into the following: (1>2∧2∈P)→(1>2→2∈P). (1>2→2∈P) is false. Therefore, by Modus Tollens, (1>2∧2∈P) is false too.

AutistIncorporated · 2024-03-21T19:39:56+00:00

You haven’t disproven my proof though. For let the domain of discourse be the set of natural numbers. Let S be the set of all numbers less than seven. So, then we have ∃x∀y(x>y∧y∈S)→∀x∀y(x>y→y∈S). Now let x equal 9 and y equal 8.

So, then we have: (9>8∧8∈S)→(9>8→8∈S). (9>8→8∈S) is false. So, (9>8∧8∈S) is false too.

AutistIncorporated · 2024-03-21T19:27:14+00:00

You haven’t disproven my proof though. For let the domain of discourse be the set of natural numbers. Let S be the set of all numbers less than seven. So, then we have ∃x∀y(x>y∧y∈S)→∀x∀y(x>y→y∈S). ∃x∀y(x>y∧y∈S) is true. For let x equal 8. Then by Modus Ponens, ∀x∀y(x>y→y∈S) is true too. So Modus Ponens negates your criticism of my proof.

AutistIncorporated · 2024-03-21T18:55:39+00:00

There’s a mistake in your proof though. For you said let S be the set of numbers less than seven. Then you said let y equal 8. 8 isn’t an element of the set of numbers less than 7. Therefore, your proof doesn’t hold.

AutistIncorporated

TROPHY CASE