## View Poll Results: Is the deduction from an inconsistent set of premises in the argument in the OP, valid?

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• Yes, it is valid.

2 66.67%
• No, it is not valid.

1 33.33%

0 0%
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# Thread: The square root of two (an argument with inconsistent premises)

1. Originally Posted by Speakpigeon Then it seems to me your reading of the proof has to be formalised as follows:
(A → (B ∧ (B → ¬B)) ⊢ ¬A
Thank you either to confirm or propose your own formalisation.
EB
A is the only assumption that occurs in the proof, so it's the only thing to appear on the left of a turnstile. A subproof shows

A ⊢ B ∧ ¬B

We can discharge the A, leaving just

⊢ ¬A  Reply With Quote

2. Originally Posted by A Toy Windmill
I think your third remark is an odd way of phrasing what's going on, and I think it's still odd if you change "premise" to "statement." The contradiction you have is between having n1 and m1 not both even and then having n1 and m1 both even. It's a contradiction between two steps in the proof. That's the contradiction you point to, and the one that I think most people who understand the proof would point to. But at the end, you claim the contradiction is between your opening premise and all unstated mathematical facts.
The contradiction you have is between having n1 and m1 not both even and then having n1 and m1 both even, yes. It's a contradiction between two statements I derived at different parts of the proof. However, I derived them from the assumption that there were n,m such that

and other mathematical statements that are not explicitly stated in the proof, but are accepted by the target audience (here, I expect the audience to already know that the square root of two is not a rational number, but if I were to use this proof to convince people who don't know it, I would also expect them to accept those facts).
So, the way I see it, it still sounds natural to make my third remark. But if that terminology is unusual, I would have no problem replacing "premises" for "statements", and it's still natural the way I see it. Originally Posted by A Toy Windmill

There isn't a unique way to talk about the logic symbolically, but we're not butchering things much by giving it the broad structure:

1) Suppose the square root of 2 is the quotient of integers
2) Then it can be put into a simplest form.
3) Then the simplest form is not the simplest form.
4) Therefore the square root of 2 is not the quotient of integers (contradiction from 1-3)

Here, it's clear that the contradiction is between statements 2 and 3, and not between 1 and other unstated statements.
Yes, that is the contradiction I derived in the proof, but what I am saying is that I derived that contradiction from the aforementioned assumption and some known mathematical facts. The assumption alone would not have sufficed. Whether those facts and the assumption are called "premises" in the argument I used to derived the contradiction, or the assumption called "premise", and the other known facts are statements that, together with the stated premise, are used to derive the contradiction, but are not called "premises", seems to be a terminological matter, but does not affect what is going on.

At the end of the day, I used an inconsistent set of statements (the assumption in question, and some other, known stuff, some explicit, some not) to derive a contradiction. In the title, I said "an argument with inconsistent premises", because I was inclined to call "premises" the statements I use to derive a conclusion, even if some of them are not made explicit. But if that terminology is unusual, I would have no problem saying "an argument in which a contradiction is derived from inconsistent statements".

So, it seems to me our disagreement is the terminology (though if you say it's unusual, I would be willing to change it), and what we find to be the natural way of talking about what is going on, though not about what follows from what.  Reply With Quote

3. Originally Posted by Angra Mainyu So, it seems to me our disagreement is the terminology (though if you say it's unusual, I would be willing to change it), and what we find to be the natural way of talking about what is going on, though not about what follows from what.
One way to settle this sort of dispute is to look at what happens if you take it down to the level of formalized mathematics, such as with first-order Peano Arithmetic. There, ambient facts will be axioms and theorems of the theory, which have a very different behaviour in the mechanics of proof than assumptions made for the purpose of being discharged. In such settings, it is inarguable that there is exactly one assumption that appears in your proof, and it is discharged to produce a theorem with the assumption negated.

However, one way to save the idea that assumptions, axioms and theorems have the exact same status is declare your proof to be of the pure logical theorem

A1 → A2 → ... An → C

where A1...An include whatever axioms or theorems you want to select together with the assumption that a rational squares to 2. But then, there is no definite reason to say that the theorem proves the irrationality of the square root of 2: it's just showing that a particular set of statements is contradictory and so has no interpretation, in natural numbers, or anything else.

This is all nitpicking, for which I apologize. I'm just throwing it out there in case it turns out to be relevant in your arguments with Speakpigeon.  Reply With Quote

4. Originally Posted by A Toy Windmill
This is all nitpicking, for which I apologize. I'm just throwing it out there in case it turns out to be relevant in your arguments with Speakpigeon.
No problem, at least this is a more interesting discussion than Speakpigeon's stuff - though if you want to know, it turns out this whole thread was not needed, since Speakpigeon replied in the other thread that the premises are an inconsistent set of statements (whether they're called "premises" is not relevant I think, but Speakpigeon called them that even as I have accepted your point about terminology), and accepted that sort of argumentation, so I just pointed out that the squid argument is precisely one with an inconsistent set of premises, none of which is contradictory on its own, so it's valid by Speakpigeon's own claims (some of Speakpigeon's claims anyway; others imply otherwise). Originally Posted by A Toy Windmill
One way to settle this sort of dispute is to look at what happens if you take it down to the level of formalized mathematics, such as with first-order Peano Arithmetic. There, ambient facts will be axioms and theorems of the theory, which have a very different behaviour in the mechanics of proof than assumptions made for the purpose of being discharged. In such settings, it is inarguable that there is exactly one assumption that appears in your proof, and it is discharged to produce a theorem with the assumption negated.
That does not settle the dispute, it seems , because when I look at it that way, I see that in order to derive a contradiction, I need the extra assumption plus either axioms or theorems of a theory that are not logically valid first order formulas. So, I would say that the assumption in question + the axioms of the theory form an inconsistent set of statements, from which a contradiction is derived. Originally Posted by A Toy Windmill

However, one way to save the idea that assumptions, axioms and theorems have the exact same status is declare your proof to be of the pure logical theorem

A1 → A2 → ... An → C

where A1...An include whatever axioms or theorems you want to select together with the assumption that a rational squares to 2. But then, there is no definite reason to say that the theorem proves the irrationality of the square root of 2: it's just showing that a particular set of statements is contradictory and so has no interpretation, in natural numbers, or anything else.
I think there might be, though we need to step outside the formalism and consider questions such as "Why do we accept the axioms?"

So, in this particular case, I would ask: why do we accept some axioms of Peano Arithmetic, in a first-order formulation? Those aren't classically logically valid first-order formulas, after all. For each one of them, there is a consistent first-order model that has the negation of one of those formulas as an axiom. So, why accept them as true? And true of what?

I would say that the natural numbers are not defined by the first-order model, but rather, the first-order formulation is an attempt to model the natural numbers, which are defined intuitively and ostensively: that's one object, that's two, that's three, etc. In that manner, we (i.e., humans) talk about a hypothetical abstract scenario (which we call "natural numbers"), and about addition, etc. Humans did that for thousands of years before any first-order formalization was attempted, or even first-order formalizations were devised. There are basic facts about the set of natural numbers that are immediately apprehensible by a human who grasps the meaning of the words (defined ostensively as before) and intuitive logic. Some of those facts are captured by the Peano axioms, and then there are first-order formulations of them (with adequate interpretations of the formulas).

So, in a way, intuitive logic and linguistic competence give us the Peano Axioms. I grant that some find the use of intuition suspect, but I would say it's at least to a considerable extent a matter of terminology, but in any case, in the end we always inevitably rely on that at some basic level (and it is proper to do so).

That would give us a reason to accept the axioms. From that, we get the theorems. And from those axioms and theorems + the assumption in question, we derive a contradiction. So, the assumption is false. It won't give us that there is a square root of two (that might require another discussion in the meta-theory and philosophy of math!), though, but I would say the OP argument proves that the square of any fraction of natural numbers (easily extended to any fraction of integers if those are known, though we need a different argumentation to explain the integers if someone asks!) does not equal two.  Reply With Quote

5. Originally Posted by A Toy Windmill  Originally Posted by Speakpigeon Then it seems to me your reading of the proof has to be formalised as follows:
(A → (B ∧ (B → ¬B)) ⊢ ¬A
Thank you either to confirm or propose your own formalisation.
EB
A is the only assumption that occurs in the proof, so it's the only thing to appear on the left of a turnstile. A subproof shows

A ⊢ B ∧ ¬B

We can discharge the A, leaving just

⊢ ¬A
A - Suppose the square root of 2 is the quotient of integers
B - Then it can be put into a simplest form.
C - Then the simplest form is not the simplest form.
D - Therefore the square root of 2 is not the quotient of integers (contradiction from 1-3)
I take it my C here is your ¬B.

However, your A ⊢ B ∧ ¬B is still missing the real conclusion, D: Therefore the square root of 2 is not the quotient of integers.

In effect, you read A ⊢ (B ∧ ¬B) where I read A → (B ∧ ¬B), and you leave my own ⊢ ¬A as a kind of afterthought even though it is clearly the real conclusion.

I know you are merely following the script of the standard proof in mathematical logic in this case, but yours seems a rather inadequate formalisation of the original mathematical reasoning.

Your formalisation is also at odds with your own use of the key word "Therefore" in clause D.
EB  Reply With Quote

6. I voted "not valid".

This is in fact rather obvious. Still, I won't explain given that I'm no longer interested in convincing you. I am interested however in other people's votes and possible explanations.

The problem is rather simple.

A implies B and C implies D.

Does therefore A and C implies B and D?

EB  Reply With Quote

7. I don't think you understand the turnstile relation.  Reply With Quote

8. Originally Posted by Speakpigeon I voted "not valid".
It's a standard proof. So, you wanted to know about the impact of having what you believe to be the "wrong" logic on mathematics? Well, as I pointed out, it's pervasive. This proof is not weird, or suspect, or anything like that. This is (pretty much, not every detail; I don't remember every detail, plus I simplified it a little because the only prime I consider is 2) a proof that was given in a basic algebra course I took (when talking about the applications of prime factorization, iirc). That was before I took a course in mathematical logic. I understood the proof. All of the other students understood the proof. Very probably (almost certainly), none of them had taken a course in mathematical logic, either (that's generally for considerably more advanced students). We all reckoned the proof was correct. That is our intuitive sense of logic.

If you want evidence that this is a standard proof, then look it up: look for proofs of the irrationality of the square root of two, and you'll see that, other than details, they're essentially of the same sort. I don't think you will find any that is valid in your logic. For example: just google "square root 2 irrational" without quotation marks to see the arguments.  Reply With Quote

9. 2^1/2 = n/m
N cannot be less than or equal m otherwise you get a number less than 1 or equal to 1.
N and m must both be positive or negative.
For n > m n cannot be an integr number of m.

And so on. The problem is not properly bounded.

2^1/2 = n/m
N^2 = 2m^2
n = sqrt(2m^2)
n/m = 2 ^1/2

The manipulation says nothing about whether it is solvable. That is not a proof. You transpositions of m and n are not valid, the original form remains.

The question is can any arbitrary irrational number be expressed as the ratio of two integers. I think there was a math thread on this.

This implies that n12 is even. Hence, n1 is even. Therefore, there is an integer n2 such that 2m1^2 = n1^2
No it does not. You have to show that if this were a mathematical proof.  Reply With Quote

10. Originally Posted by steve_bank 2^1/2 = n/m
N cannot be less than or equal m otherwise you get a number less than 1 or equal to 1.
N and m must both be positive or negative.
For n > m n cannot be an integr number of m.

And so on. The problem is not properly bounded.

2^1/2 = n/m
N^2 = 2m^2
n = sqrt(2m^2)
n/m = 2 ^1/2

The manipulation says nothing about whether it is solvable. That is not a proof. You transpositions of m and n are not valid, the original form remains.

The question is can any arbitrary irrational number be expressed as the ratio of two integers. I think there was a math thread on this.

This implies that n12 is even. Hence, n1 is even. Therefore, there is an integer n2 such that 2m1^2 = n1^2
No it does not. You have to show that if this were a mathematical proof.
I regret you do not understand it, but it is a standard proof. It's the sort of proof you'll get in a basic algebra course, when talking about prime factorization. I do not understand why you are not able to follow it (I simplified it, so that only division by 2 is considered, leaving aside other primes), but I suggest you just google "square root two not rational" (without the quotation marks) or something like that, and you will find proofs pretty much like this one, except for details.  Reply With Quote

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