∞ is a free variable

[Magnus Anderson]

No you didn't. You asked for an integer. You didn't specify a representation. Why are you moving the goalposts?

Don't play dumb.

[Skepdick]

No you didn't. You asked for an integer. You didn't specify a representation. Why are you moving the goalposts?

Don't play dumb.

That's great advice! When will you begin practicing what you preach?

[Magnus Anderson]

That's great advice! When will you begin practicing what you preach?

When are you going to give us an integer that is neither even nor odd and that is expressed as a decimal numeral?

TREE(3) fails to meet both criteria.

[wtf]

Don't play dumb.

He's not playing

[Skepdick]

When are you going to give us an integer that is neither even nor odd and that is expressed as a decimal numeral?

TREE(3) fails to meet both criteria.

TREE(3) meets the original criterion.
It's an integer.
It's not odd (there isn't any proof of it being odd)
It's not even (there isn't any proof of it being even)

Therefore Tree(3) is NOT odd AND it's NOT even!

If something is NOT odd AND it's NOT even - then it's NEITHER odd NOR evem.


It fails to meet the criterion of being expressed as a decimal number, but I don't see how or why that's relevant. Are you implying that only integers expressed as a decimal numerals can be odd or even? That would be dumb. Are you being dumb?

[Skepdick]

Don't play dumb.

He's not playing

And yet you are

ALL integers are either odd or even. This instance of Excluded Middle can be proven inductively using the 2k/2k+1; or the modulo 2 definition.
Tree(3) is neither odd nor even because you can't figure out which pattern it follows.
This is not a contradiction because Excluded Middle is not assumed in general.

It's a trivial construction in sheaf theory to resolve this paradox and allow both the definite and indefinite parities to co-exist.
But you can't do this in a logic which can't represent indefinite states.

Queue some retort about how I am using terminology I don't understand.

[Skepdick]

Well yes, there are limits to what you can construct with any given axiom. My question is really whether there is any limit on what you can choose as an axiom.

Sure. Reverse Mathematics tackles that.

If you accept conclusion/theorem X work backwards to split appart the necessary from sufficient conditions.
Then you can work out which axioms are satisfactory and which aren't.

But if you are going the other way - you can choose literally any premise and go on an expedition about computing its consequences. Ad infinitum.

I gather you think maths is equally frivolous.

It's a useful tool. Insofar as it helps with conceptual clarity. If you know what you mean - you have a solid foundation to build from.

And it's all just tap-dancing around something being defined vs somethign being well-defined with ever-growing rigor or precision. It's just analytic philosphy on infinite difficulty.

I agree with Dijkstra: The purpose of abstraction is not to be vague, but to create a new semantic level in which one can be absolutely precise.

[Magnus Anderson]

TREE(3) meets the original criterion.
It's an integer.

No it does not. You don't know what the original criterion was. You misunderstood it. And you refuse to admit a mistake and learn.

So when are you going to give us an integer that is 1) represented as a decimal numeral, and 2) neither even nor odd? Why are you avoiding doing that?

It's not odd (there isn't any proof of it being odd)
It's not even (there isn't any proof of it being even)

Therefore Tree(3) is NOT odd AND it's NOT even!

Oh my, how clueless you are.

"Argument from ignorance" is not a valid argument. It's a logical fallacy.

Look it up.
https://en.wikipedia.org/wiki/Argument_from_ignorance

If you don't know that something is X, it does not follow that it is not X.

If you don't know whether any given number is even or odd, it absolutely does not follow that it is neither even nor odd.

Not to mention that there is a proof that EVERY integer is either even or odd.

And not to mention that there is a proof that TREE(3) is an even number.

It fails to meet the criterion of being expressed as a decimal number, but I don't see how or why that's relevant. Are you implying that only integers expressed as a decimal numerals can be odd or even? That would be dumb. Are you being dumb?

It was what I requested. And if there is an integer that is neither even nor odd, then why is it difficult for you to represent it as a decimal numeral?

We should keep things as simple as possible instead of unnecessarily talking about what a particular function with particular arguments evaluates to.

The point is to prevent distraction techniques which you're so accustomed to.

[Magnus Anderson]

Not to mention that there is a proof that EVERY integer is either even or odd.

And this proof is the only one that matters. And it follows from the definition of the terms.

An integer is said to be even if it can be expressed as "2 x n" where "n" is an integer.

Otherwise, it is odd.

Any given integer can either be expressed as "2 x n" or it cannot be. There is no other logical possibility.

Thus, every integer is either even or odd.

From this, we can conclude that TREE(3) is either even or odd. You not knowing which one it is is irrelevant and t's certainly not a proof that it is neither even nor odd.

[Skepdick]

No it does not. You don't know what the original criterion was. You misunderstood it. And you refuse to admit a mistake and learn.

And now we are at the gaslighting phase...

You asked for an integer that is neither odd nor even.
I gave you Tree(3).
1. It's an integer.
2. It's not even (which does NOT imply it's odd)
3. It's not odd. (which does NOT imply it's even)

So when are you going to give us an integer that is 1) represented as a decimal numeral, and 2) neither even nor odd? Why are you avoiding doing that?

Because that's not what you asked for. Why are you moving the goal posts?

When are you going to give me the set of ALL integers represented as decimal numerals?
All you gave me were some dots.

Oh my, how clueless you are.

"Argument from ignorance" is not a valid argument. It's a logical fallacy.

I am not making an argument FROM ignorance. I am making a factual argument ABOUT ignorance.
YOUR Ignorance.

YOU don't know that Tree(3) is Odd.
YOU don't kow that Tree(3) is Even.

Therefore the epistemic status of Tree(3) in YOUR OWN HEAD is "NEITHER ODD NOR EVEN"

If you don't know that something is X, it does not follow that it is not X.

Which is why nobody is making such claims. not(X) and (not X) are different claims in Constructive Mathematics.

If you don't know whether any given number is even or odd, it absolutely does not follow that it is neither even nor odd.

It trivially follows in a Constructive setting.

You have neither a proof of it being odd nor a proof of it being even.

Neither its oddness nor its evenness can be asserted.

Not to mention that there is a proof that EVERY integer is either even or odd.

A general property of ALL integers is not a particular property of any given integer.

The number 1 is not "either odd or even". It's just odd.
The number 2 is not "either odd or even". It's just even.

And not to mention that there is a proof that TREE(3) is an even number.

Wonderful! An existence claim!

Please construct the proof.

It was what I requested.

No, it's not. Why are you lying? Maybe it's what you INTENDED to request, but it's most definitely not what you requested.

It's on record and everything.

And if there is an integer that is neither even nor odd, then why is it difficult for you to represent it as a decimal numeral?

Because there's not enough universe to hold its representation?

The point is to prevent distraction techniques which you're so accustomed to.

UNcharirtable as always. Like the wanker that you are.

[Skepdick]

Not to mention that there is a proof that EVERY integer is either even or odd.

And this proof is the only one that matters. And it follows from the definition of the terms.

An integer is said to be even if it can be expressed as "2 x n" where "n" is an integer.

Otherwise, it is odd.

Any given integer can either be expressed as "2 x n" or it cannot be. There is no other logical possibility.

Thus, every integer is either even or odd.

Everything in the universe is ether an elephant or it's not an elephant!

This is circular reasoning. You are just re-wording the Law of Excluded Middle.

From this, we can conclude that TREE(3) is either even or odd.

You can also conclude that TREE(3) is either an elephant or it isn't.

You not knowing which one it is is irrelevant and t's certainly not a proof that it is neither even nor odd.

You have proven neither the oddness of TREE(3) nor the evenness of TREE(3).

You have provided no continous map (a proof!) from TREE(3) -> {Odd, Even}

[Magnus Anderson]

And now we are at the gaslighting phase...

No gaslighting whatsoever, merely your desire to explot the slightest inexactness in other people's claims.

Because that's not what you asked for. Why are you moving the goal posts?

That's exactly what I asked for.

When are you going to give me the set of ALL integers represented as decimal numerals?
All you gave me were some dots.

Why shuld I do that?

YOU don't know that Tree(3) is Odd.
YOU don't kow that Tree(3) is Even.

Therefore the epistemic status of Tree(3) in YOUR OWN HEAD is "NEITHER ODD NOR EVEN"

The epistemic status of TREE(3) in YOUR head is "I don't know whether it's even or odd".

"I don't know whether it's even or odd" is not the same as "It's neither even nor odd".

It trivially follows in a Constructive setting.

Why should anyone care what follows in your imaginary, made up, fantasy version of logic?

In real logic, it does not follow. It's an argument from ignorance, a logical fallacy, a sophistical argument.

You have neither a proof of it being odd nor a proof of it being even.

Neither its oddness nor its evenness can be asserted.

I actually do. But let me just add that just because neither its oddness nor its evenness can be asserted by you, it does not follow that it's neither even nor odd.

You actually need a proper proof, not merely an argument from ignorance, that TREE(3) is neither even nor odd.

You haven't provided one so far.

A general property of ALL integers is not a particular property of any given integer.

It very much is.

The number 1 is not "either odd or even". It's just odd.
The number 2 is not "either odd or even". It's just even.

They are both.

Wonderful! An existence claim!

Please construct the proof.

I don't have to.

Because there's not enough universe to hold its representation?

Oh, so only extremely large numbers are neither even nor odd? How convenient! Well, in that case, you should choose a random integer that is neither even nor odd and tell us its last decimal digit. You don't have to give us all of its digits. Just the last one. If it happens to be 0, 2, 4, 6 or 8, the integer is even. Otherwise, it is odd. I'm looking forward to this magical last decimal digit.

[Magnus Anderson]

Everything in the universe is ether an elephant or it's not an elephant!

Correct. And your point is?

This is circular reasoning.

So you also don't know what circular reasoning is.

You can also conclude that TREE(3) is either an elephant or it isn't.

Correct. And your point is?

You have proven neither the oddness of TREE(3) nor the evenness of TREE(3).

You have provided no continous map (a proof!) from TREE(3) -> {Odd, Even}

I have proven that TREE(3) is either even or odd.

You yourself have done nothing other than employ argument from ignorance in an effort to prove that TREE(3) is neither even nor odd.

[Skepdick]

No gaslighting whatsoever, merely your desire to explot the slightest inexactness in other people's claims.

You are complaining about excatness when discussing Mathematics ?!?!

That's exactly what I asked for.

I guess you have no idea what exactness is then, huh?

Why shuld I do that?

I don't know. Why should you do that? Why should I do anything?

YOU don't know that Tree(3) is Odd.
YOU don't kow that Tree(3) is Even.

Therefore the epistemic status of Tree(3) in YOUR OWN HEAD is "NEITHER ODD NOR EVEN"

The epistemic status of TREE(3) in YOUR head is "I don't know whether it's even or odd".

That's precisely exactly what the epistemic claim "it's neither odd nor even" means.

In our heads (the only place where Mathematics exists) - it's NEITHER odd NOR even.

"I don't know whether it's even or odd" is not the same as "It's neither even nor odd".

They are epistemically identical statements. Unless you are a silly Platonist who thinks numbers exist somewhere out there.

Why should anyone care what follows in your imaginary, made up, fantasy version of logic?

Because my imaginary, made up, fantasy version of logic is stricter, more expressive and more precise than your imaginary made up version of logic.

In real logic, it does not follow. It's an argument from ignorance, a logical fallacy, a sophistical argument.

Your imaginari version of logic is "REAL"?

Take your meds.

I actually do. But let me just add that just because neither its oddness nor its evenness can be asserted by you, it does not follow that it's neither even nor odd.

That's precisely what follows.

1. I can't say that it's Odd - because I have no reason to believe it
2. I can't say that it's Even - because I have no reason to believe it.

I have no reason to believe it's either one!

You actually need a proper proof, not merely an argument from ignorance, that TREE(3) is neither even nor odd.

I gave you a proper proof.

1. There is NO proof for it being odd.
2. There is NO proof for it being even.
3. There is NO reason to believe it's either odd OR even because NEITHER one is proven true.

The instance of LEM with respect to TREE(3)'s parity has NOT been proven.

The negation of EITHER is NEITHER.

You haven't provided one so far.

Yes I have.

It very much is.

You think 1 is odd OR even; and not just odd ?!?
You think 2 is odd OR even; and not just even ?!?

What's wrong with you?

They are both.

No they are not. odd OR even presents a choice.

ODD-only or EVEN-only presents NO choice.

I don't have to.

Is this the Mathematics sub-forum or the Clown subforum?

Oh, so only extremely large numbers are neither even nor odd? How convenient!

It's decidedly inconvenient. Since you want to know the answer.

Well, in that case, you should choose a random integer that is neither even nor odd and tell us its last decimal digit.

Sure thing! (with dripping irony) I'll give you EXACTLY the answer you've given me

It's either 0, or 1, or 2; or 3; or 4; or 5; or 6; or 7; or 8; or 9; or 0.

You don't have to give us all of its digits. Just the last one. If it happens to be 0, 2, 4, 6 or 8, the integer is even. Otherwise, it is odd. I'm looking forward to this magical last decimal digit.

Yeah... you don't have to give us ALL the options on Parity.

I am looking to this magical Odd or Even answer.

[Magnus Anderson]

"Odd" means the same thing as "not even". And "even" means the same thing as "not odd". It's by definition. So when you say, "It's neither even nor odd" you're saying "It's odd and even". Obvious nonsense.

"I don't know whether X is even or odd" is a claim about one's knowledge. "X is neither even nor odd" is a claim about X. Two very different claims. Certainly, the latter does not follow from the former.

Since 1 is odd, it is also either even or odd.
Since 2 is even, it is also either even or odd.
And so on.

You have too many issues with language, so you should stay away from philosophy as far as possible. With your cardinal lack of adequate intellectual capacity, you can only demolish it.