∞ is a free variable

[Skepdick]

I'd like to understand your reasoning more clearly.

Suppose I make the following definition:

An integer is even if it's divisible by 2.

Here "divisible by 2" has its obvious meaning.

This is precisely where we drift apart. It's not obvious to me what is so "obvious" to you and why.

Do you remember me asking you in the past whether the integer 3 is the same number as the real number 3?
If you want a chuckle - the phrase "the same" has its obvious meaning.

Without loss of generality: is integer X the same number as real number X? Yes - I know their domains differ but 5 is 5 is 5 (or so you said)

In the mathematical expression f(x) = 5, 5 is a constant. Its value does not "range" over anything.

If we accept that x:N ≡ x:R is a theorem then by implication there exists no function F such that F(x:N) !≡ F(x:R) is inhabited.
So what is this function F (divisible by 2) inhabiting it?

What is this function "divisible by 2" which is true for 3:R but false for 3:N ?

It seems to me the important philosophical question here is "What does it mean for a number to be NON-divisible by 2"?

The integer is 0 mod 2. Or it has remainder 0 upon integer-division by 2.

That's not "integer division". As our friend Magnus Dorkinson was so kind to inform us (using various reputable sources) integer division means "floor division". Integer division discards the remainder. The function you are talking about isn't doing any discarding.

Surely you see a difference between a Mathematical operator with type signature N -> N -> N and another Mathematical operator with type signature N -> N -> N x N ?

They are not the same (where "the same" has its obvious meaning).

I hope we don't have to play word games about the meaning of divisible by 2.

How do you do any Mathematics without word games??!?

Here we are... playing a word game about two things being "the same" a.k.a equivalent.

Suppose I make the additional definition:

An integer is odd if either

a) It's not divisible by 2

Yeah. Fine. If you want to prove the negation of "divisible by 2" all you have to do is prove that the co-domain of "divisible by 2" is uninhabited.

b) Its divisible-by-2 status is in some kind of intederminate state.

Which kind of indeterminate state? There are many kinds of indeterminacy.

https://en.wikipedia.org/wiki/Kind_(type_theory)

I am going to address this further below but...

Is it indeterminate that the number is divisible by 2; or is it indeterminate that the number is NOT divisible by 2?

Without excluded middle it's not impossible that both indeterminacies hold.

EDIT: Such as the indeterminacy of TREE(3).

I have two questions:

1) What's wrong with that definition? A number is odd if it's not even; namely, if it's either definitely not divisible by 2, or else it's in some kind of intuitionistic twilight state.

The exact same thing that would be wrong with the definition "a number is even if it's not odd" - it's circular. You are defining the complement by exploiting an unproven involution.

The english word "Either" is just another way to express Excluded Middle a.k.a the axiom of choice.

Either(A,B) -> {Left Projection, Right Projection}

So I have an "Even machine." I enter a number. If it's divisible by 2, the machine outputs "Even." If it's definitely not divisible by 2, or if the machine can't figure out if it's divisible by 2, it outputs "Odd."

What's wrong with that?

Nothing's right or wrong with it. It does what it does.

It's not possible to determine which Mathematical operator your machine implements until further empirical testing is performed on it. It's a black box and I am going to fuzz the hell out of that thing to eliminate some options.

Suppose I enter "2" an the machine says "Odd"... OK... the machine can't figure out if 2 is divisible by 2.
Suppose I enter 3" and the machine says "Even"...OK.... the machine knows how to divide 3 by 2.
Suppose I enter Tree(3) and the machine says "Odd". Was it actually odd or is the machine signaling it failed to figure out if it's divisible by 2?

Do observe that you are trapped in Excluded Middle again. You are pre-supposing Odd or Even as all-exhaustive categories.
Which is precisely why you had to make a choice about the inndeterminate state. The default choice.

Why do you return Odd and not Even if you can't figure out if it's divisible by 2 or not divisible by 2?

2) I can't believe Brouwer and all the other clever early intuitionists couldn't define an even number. Convince me that this is the correct interpetation, and that you really think I can't define what an even number is without the law of the excluded middle.

You can define what an odd and even number is without the law of excluded middle.
But you aren't doing that. You are defining it USING excluded middle.

You are presupposing an involution between Odd and Even..

The way we'd do it constructively is we'd define odd and even as separate properties.
Once you've proven that the two categories are all exhaustive within the domain then the involution trivially follows.

In your mode of reasoning not Even -> Odd is an axiom.
In my mode of reasoning it's a theorem.

Surely you understand the difference between axioms and theorems?

After all, dividing an integer by 2 can be completed by a Turing machine in a finite number of steps.

Sure, but how many steps would a Turing machine take to NOT divide an integer by 2?

So we can always have a Yes/No machine for the question, "Is n even?"

I don't think so. How many steps would the Turing machine take to NOT divide n by 2?

This is the problem with classical reasoning. It fails to grasp the difference between dedicable and semi-decidable problems.

There's a turing machine that can decide whether the number n is divisible by 2.
There is a turing machine that can decide whether a number n is NOT divisible by 2.

These are two different turing machines.

So I just don't understand what an indeterminate answer could be in this context.

The intermediate answer is neither of the two turing machines has halted.

When "divisibleBy2" OR "NotDivisibleBy2" halts you'll know what kind of number n is.

It's a race... who will win?

Wait and see.

Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true.--Bertrand Russel

I think in the end you could just parrot a tautology: Even numbers are those for which division by 2 is defined, and odd numbers are those for which division by 2 is undefined.

All the Reals are even...
Half the integers aren't.

[Skepdick]

The set of all integers is { ..., -3, -2, -1, 0, +1, +2, +3, ... }.

The set of all even integers is { ..., -4, -2, 0, +2, +4, ... }.

The set of all odd integers is { ..., -3, -1, +1, +3, ... }.

Hey Mr Abstract thinker. Is the cardinality of these sets odd or even?

Can he name a single integer that is neither even nor odd?

Sure thing, cupcake.

Using the axiom of choice pick a random element from the set of all integers and I name it A.
e.g let A:= RandomChoice({ ..., -3, -2, -1, 0, +1, +2, +3, ... })

Is A odd or even?

His argument so far has been that if you define the term even number as an integer that leaves no remainder when you divide it by 2 using integer ( = Euclidean ) division that it follows that every integer is an even number.

That's not my argument. That's your strawman of my argument.

It's pure sophistry employing equivocation in an effort to construct a seemingly good argument that is in actuality terribly flawed.

I am not equivocating. You are. Literally.

You using the name "division" to refer to meaningfuly different Mathematical operators.

The flaw consists in him not really understanding what a remainder is, mistakenly thinking that if a remainder is not included within the result of the division, that it's not there, that it's 0.

Your flaw consists in you not really understanding what "understanding" is.

And you continue to equivocate between division WITH remainder and division WITHOUT remainder.

At the same time, he has ignored that there are alternative intensional definitions of the term even number, at least two that don't rely on the concept of remainder.

Q.E.D equivocation.

Different (intensional) definitions of "evenness" are different definitions of "evenness"

For example, an even number can be defined as an integer that when divided by 2 using real number division results in a number with no fractional part. Alternatively, you can define it in terms of modulo operation.

But we should not worry, I am sure he can construct equally if not more idiotic arguments against these other definitions all with the aim to make it look like he has refuted extremely basic widely accepted beliefs.

Why don't you just tell me how you define "odd" and "even" so you can determine if the cardinality of { ..., -3, -2, -1, 0, +1, +2, +3, ... } is odd or even?

[Magnus Anderson]

Can you tell me if the cardinality of these sets is odd or even?

Can you tell me why anyone should bother answering your random questions?

Can he name a single integer that is neither even nor odd?

Sure thing

So when are you going to do it?

Asking me whether a randomly chosen integer is odd or even isn't it.

But it's good that you remind us that asking dumb, irrelevant, questions is one of your favorite tactics.

That's not my argument. That's your strawman of my argument.

Oh really?

I am not equivocating. You are. Literally.

You are calling multiple different Mathematical operations using the same name "division".

You need to learn what the term "equivocation" means.

And you continue to equivocate between division WITH remainder and division WITHOUT remainder.

And you continue to be Mr. Completely Irrelevant Trying to be Extra Smart but Failing Miserably Guy.

[Magnus Anderson]

So where is this integer that is neither even nor odd?

[Skepdick]

Can you tell me why anyone should bother answering your random questions?

<random question. unanswered>

Can he name a single integer that is neither even nor odd?

Yes. The random integer A.

Asking me whether a randomly chosen integer is odd or even isn't it.

Contradiction. By the axiom of choice "A" is a valid name for a random integer.

And if you like the random integer a lot you can even call it something nice. Like Barney.

You need to learn what the term "equivocation" means.

Yes you do.

Using the name "division" to reffer to DIFFERENT Mathematical operators is an instance of equivocation.

And you continue to be Mr. Completely Irrelevant Trying to be Extra Smart but Failing Miserably Guy.

So irrelevant your ego can''t shut up

[Skepdick]

So where is this integer that is neither even nor odd?

The randomly chosen integer A.

https://en.wikipedia.org/wiki/Axiom_of_choice

It's either odd or even.

Which one?

[Magnus Anderson]

Yes. The random integer A.

"A" is not an integer. Try again.

Using the name "division" to reffer to DIFFERENT Mathematical operators is an instance of equivocation.

Learn what equivocation is then come back.

[Skepdick]

Yes. The random integer A.

"A" is not an integer. Try again.

Contradiction. A is selected from the set of integers. It's impossible for it to NOT be an integer.

If you reject the axiom of choice say so.

Learn what equivocation is then come back.

Instruction not understood. I already know what equivocation is so I am staying here.

[Magnus Anderson]

So where is this integer that is neither even nor odd?

The randomly chosen integer A.

https://en.wikipedia.org/wiki/Axiom_of_choice

It's either odd or even.

Which one?

Give us a number, dummy. Stop playing games.

[Skepdick]

Give us a number, dummy. Stop playing games.

I gave you a number dummy.

It's a randomly chosen one. From the set of all integers.

It's a fucking number! What else could it be?

[Magnus Anderson]

These are the integers, imbecile.

{ ..., -3, -2, -1, 0, +1, +2, +3, ... }

There is no "A" in there. No letters whatsoever. Only numbers.

Tell us which one is neither even nor odd.

[Skepdick]

These are the integers, imbecile.

{ ..., -3, -2, -1, 0, +1, +2, +3, ... }

There is no "A" in there. No letters whatsoever. Only numbers.

Tell us which one is neither even nor odd.

Ok fine! How about A := TREE(3) ?

[Magnus Anderson]

These are the integers, imbecile.

{ ..., -3, -2, -1, 0, +1, +2, +3, ... }

There is no "A" in there. No letters whatsoever. Only numbers.

Tell us which one is neither even nor odd.

Ok fine! How about A := TREE(3) ?

Are you really that dumb or are you intentionally pretending that you don't understand what you're asked to do?

[Skepdick]

Are you really that dumb or are you intentionally pretending that you don't understand what you're asked to do?

Are you realy that dumb or are you intentionally pretending that you don't understand I've done exactly what you asked me to do?

You asked for an integer. I gave you one. TREE(3)

https://en.wikipedia.org/wiki/Kruskal%2 ... E_function

You can neither prove it's odd nor prove it's even.

[wtf]

Yeah. Fine. If you want to prove the negation of "divisible by 2" all you have to do is prove that the co-domain of "divisible by 2" is uninhabited.

I'll let you have the last word. You''ve picked up some terminology that you don't understand and you are making the claim that, for example, Tree(3) has no determinate parity. That's wrong. There's a Turing machine that executes the division by 2 algorithm and halts on Even or Odd after finitely many steps. It makes no difference that the number of steps is too large to be practical.

The rest of your post was childish sophistry. I'm out.