Maths & Order

[Speakpigeon]

What would you say is the necessary condition for:

(a < b) → (c < d)​

EB

a is less than b. If it's an equation you can give a and b any value you like.

You have to assume (a < b) is true so you can't give a and b any value you like. Once you choose one it becomes a limitation on the other.
EB

- - - Updated - - -

Isn't the relationship between c and d the only important one?

That's progress.
EB

 

[DrZoidberg]

a is less than b. If it's an equation you can give a and b any value you like.

You have to assume (a < b) is true so you can't give a and b any value you like. Once you choose one it becomes a limitation on the other.

Sure, but we don't know the values for c and d either. All we know is that they don't have the same values of a or b.

I still don't see any conflict. This is an if/else statement. If (argument) then (argument). Since the first part is true, then the second part is also true.

 

[Speakpigeon]

You have to assume (a < b) is true so you can't give a and b any value you like. Once you choose one it becomes a limitation on the other.

Sure, but we don't know the values for c and d either. All we know is that they don't have the same values of a or b.

We in fact don't know that. All we can do is assume, but we can assume a bit more than that, namely that ((a < b) → (c < d)) is true.

I still don't see any conflict. This is an if/else statement. If (argument) then (argument). Since the first part is true, then the second part is also true.

So you think there's no necessary condition to ((a < b) → (c < d)) being true except itself?
EB

 

[DrZoidberg]

Sure, but we don't know the values for c and d either. All we know is that they don't have the same values of a or b.

We in fact don't know that. All we can do is assume, but we can assume a bit more than that, namely that ((a < b) → (c < d)) is true.

Sure we know it. If the first statement is true then the second statement is true. This is maths. We can know things for certain in maths.

I still don't see any conflict. This is an if/else statement. If (argument) then (argument). Since the first part is true, then the second part is also true.

So you think there's no necessary condition to ((a < b) → (c < d)) being true except itself?
EB

It's maths. Either it follows or it doesn't. Since this follows therefore it's true. And there's no logical self contradiction. We don't know what any of the numbers mean. But we don't need to know that.

 

[Malintent]

That's a start. The 'must' is. Inasmuch as it conveys the idea of necessity.

"f(a) = c" too is an idea.

or you are expressing a false dichotomy (like I originally said).

I have to say I don't see the relevance of a false dichotomy in here.
EB

If Apples are smaller than Boulders, then what can this tell us about the relationship between Cars and Dishwashers?

Well, nothing, unless there is some relationship between Apples and Cars or Boulders and Dishwashers.

I guess you are right in that it is not a false dichotomy... it is simply 'illogical'... the conclusion does not follow from the premise.

 

[Speakpigeon]

Ok.

And thanks for your time and for the answers.
EB

 

[Malintent]

You're welcome.

I'm afraid I can't help but leave this thread with the feeling that someone was looking for a means to logically show something as true that fails to follow from its premises... which is why no one could find "the answer" (there isn't a way to logically prove that which is illogical)

 

[Kharakov]

What is the necessary condition for x%30 ≡ 0?? That x is a multiple of 30. It's the same thing as saying x%30 = 0 though, so if you're going for something deeper, you'll have to name what you want.

I want necessary conditions, nothing more profound than that. It's a maths problem, not some kind of philosophical puzzle. And if you can't find any, it's also a legitimate answer.

So what exactly do you want?

The only absolutely necessary condition for your first if a<b then c<d statement is trivial as well- c<d.

Ok...

Oops! No!!! For gawdsake, ((a < b) → (c < d)) could be true and (c < d) still false. You write that down one hundred times before you can come back, you hear?!

{ c<d || a>b } * 100 Either of these conditions, if true, will render the statement true. The point is you only need one of these 2 conditions to be true for your statement to always be true, unless you meant something like....

if and only if (a<b) then (c<d)?

 

[Kharakov]

I still don't see any conflict. This is an if/else statement. If (argument) then (argument). Since the first part is true, then the second part is also true.

So you think there's no necessary condition to ((a < b) → (c < d)) being true except itself?
EB

It's maths. Either it follows or it doesn't. Since this follows therefore it's true. And there's no logical self contradiction. We don't know what any of the numbers mean. But we don't need to know that.

If it's true, it's true. What a letdown. Can't you work some form of the liar's paradox into it somewhere?

 

[DrZoidberg]

I still don't see any conflict. This is an if/else statement. If (argument) then (argument). Since the first part is true, then the second part is also true.

So you think there's no necessary condition to ((a < b) → (c < d)) being true except itself?
EB

It's maths. Either it follows or it doesn't. Since this follows therefore it's true. And there's no logical self contradiction. We don't know what any of the numbers mean. But we don't need to know that.

If it's true, it's true. What a letdown. Can't you work some form of the liar's paradox into it somewhere?

The letters are symbols. They can represent anything. Unless stated, we have to infer the relative values from the equation.

I understand the basic idea in the OP was that the letters represent the hierarchy of the alphabet. But he didn't state that. If the equation stands by itself then the correct inference is that the letters do not represent their position in the alphabet.

 

[Malintent]

I still don't see any conflict. This is an if/else statement. If (argument) then (argument). Since the first part is true, then the second part is also true.

So you think there's no necessary condition to ((a < b) → (c < d)) being true except itself?
EB

It's maths. Either it follows or it doesn't. Since this follows therefore it's true. And there's no logical self contradiction. We don't know what any of the numbers mean. But we don't need to know that.

If it's true, it's true. What a letdown. Can't you work some form of the liar's paradox into it somewhere?

The letters are symbols. They can represent anything. Unless stated, we have to infer the relative values from the equation.

I understand the basic idea in the OP was that the letters represent the hierarchy of the alphabet. But he didn't state that. If the equation stands by itself then the correct inference is that the letters do not represent their position in the alphabet.

That's silly. It's like showing that the following equation for determining the area of a circle is incorrect:

Pi*R^2

Show

Pies aren't SQUARE!!!! Pies are ROUND!!!!!

 

[Kharakov]

The letters are symbols. They can represent anything. Unless stated, we have to infer the relative values from the equation.

I understand the basic idea in the OP was that the letters represent the hierarchy of the alphabet. But he didn't state that. If the equation stands by itself then the correct inference is that the letters do not represent their position in the alphabet.

Yeah, I'm not understanding the point of the if then format?

EB?

 

[Speakpigeon]

The letters are symbols. They can represent anything. Unless stated, we have to infer the relative values from the equation.

I understand the basic idea in the OP was that the letters represent the hierarchy of the alphabet. But he didn't state that. If the equation stands by itself then the correct inference is that the letters do not represent their position in the alphabet.

Yeah, I'm not understanding the point of the if then format?

EB?

What difference does it make what a, b, c and d are taken to represent?
EB

 

[Kharakov]

What if you say that things with greater value are placed in lower positions in the list?

[] = value comparison
{} = list order comparison

[a > b] → {a < b}

 

[Speakpigeon]

What if you say that things with greater value are placed in lower positions in the list?

[] = value comparison
{} = list order comparison

[a > b] → {a < b}

First of all, there's always something left that won't be specified. I guess it's because language needs actual human beings to get a life. Second, I agree we can specify whatever we fancy (why would we limit ourselves?). Third, really, if I had meant something unusual I would have specified it (I'm not a psychopath). Fourth, your idea here is definitely unusual (we can tell because you need special bracketing to express it) and I would definitely have specified that sort of thing.

So, no, just take '<' to mean the usual thing. And then a, b, c and d could be taken to mean some numerical values or indeed anything at all as long as it is susceptible of being ordered, i.e. "a < b" has to make sense, i.e. it has to be understood as either true or false. If it doesn't, there's not even the shadow of a problem to start with.
EB

 

[DrZoidberg]

What if you say that things with greater value are placed in lower positions in the list?

[] = value comparison
{} = list order comparison

[a > b] → {a < b}

First of all, there's always something left that won't be specified.

Not in maths. Nor logic. That's why garbage in, garbage out, applies.

I guess it's because language needs actual human beings to get a life. Second, I agree we can specify whatever we fancy (why would we limit ourselves?). Third, really, if I had meant something unusual I would have specified it (I'm not a psychopath). Fourth, your idea here is definitely unusual (we can tell because you need special bracketing to express it) and I would definitely have specified that sort of thing.

So, no, just take '<' to mean the usual thing. And then a, b, c and d could be taken to mean some numerical values or indeed anything at all as long as it is susceptible of being ordered, i.e. "a < b" has to make sense, i.e. it has to be understood as either true or false. If it doesn't, there's not even the shadow of a problem to start with.
EB

Problems don't need to be hard.

 

[Speakpigeon]

First of all, there's always something left that won't be specified.

Not in maths. Nor logic. That's why garbage in, garbage out, applies.

Quis custodiet ipsos custodes?

You need a metalanguage (some ordinary language will usually do) to talk about any logical language you want to consider and you would need a meta-metalanguage to talk about the metalanguage. This stops somewhere because it would become tedious. But no language is self-explaining so you will always need a human being (or something like one) to make sense of linguistic expressions. Something will be left unspecified. That's also apparent in the axiomatic presentation of logic you'll find in all logic textbooks.
EB

 

[DrZoidberg]

Not in maths. Nor logic. That's why garbage in, garbage out, applies.

Quis custodiet ipsos custodes?

You need a metalanguage (some ordinary language will usually do) to talk about any logical language you want to consider and you would need a meta-metalanguage to talk about the metalanguage. This stops somewhere because it would become tedious. But no language is self-explaining so you will always need a human being (or something like one) to make sense of linguistic expressions. Something will be left unspecified. That's also apparent in the axiomatic presentation of logic you'll find in all logic textbooks.
EB

That's what's so nice about maths. What you see is what you get. meta-metalanguage only applies to natural language. It does apply to logic. But as Whitehead so apptly put it, philosophy is asking questions like children and answering them like lawyers. The lawyer bit is the crucial here. Logic just looks at what is in the sentence. Any context or clauses not in the logical phrasing doesn't enter into the deduction.

The OP was clearly phrased as maths or logic, since it employed those symbols (metalanguage) which allows us to draw conclusions.

 

[Speakpigeon]

Quis custodiet ipsos custodes?

You need a metalanguage (some ordinary language will usually do) to talk about any logical language you want to consider and you would need a meta-metalanguage to talk about the metalanguage. This stops somewhere because it would become tedious. But no language is self-explaining so you will always need a human being (or something like one) to make sense of linguistic expressions. Something will be left unspecified. That's also apparent in the axiomatic presentation of logic you'll find in all logic textbooks.
EB

That's what's so nice about maths. What you see is what you get. meta-metalanguage only applies to natural language. It does apply to logic. But as Whitehead so apptly put it, philosophy is asking questions like children and answering them like lawyers. The lawyer bit is the crucial here. Logic just looks at what is in the sentence. Any context or clauses not in the logical phrasing doesn't enter into the deduction.

The OP was clearly phrased as maths or logic, since it employed those symbols (metalanguage) which allows us to draw conclusions.

Sorry, I think I'm lost as to what your point might be.

You're leaving too many things unspecified I think.
EB