The square root of two (an argument with inconsistent premises)

[Angra Mainyu]

The following is an argument meant to prove that the square root of two is not a rational number - in other words, it is not a quotient of integers.

Suppose there are integers m, n (both different from zero, obviously) such that

\( \begin{equation} \sqrt{2}=\frac{n}{m}. \end{equation} \)

If n and m are both even numbers, then we simplify them, obtaining n₁, m₁ such that
\( \begin{equation} \sqrt{2}=\frac{n_1}{m_1}, \end{equation} \)
and at least one of the numbers n₁ or m₁ or is odd (note that if either n is odd or m is odd, we can just let n₁ :=n and m₁ or:=m. Hence, in any case, we get that the equation above holds with either n₁ or odd, or m₁ or odd).
It follows from the previous equation that
\( \begin{equation} 2m_1^2=n_1^2. \end{equation} \)
This implies that n₁² is even. Hence, n₁ is even. Therefore, there is an integer n₂ such that
\( \begin{equation} n_1=2n_2. \end{equation} \)
It follows from the previous equations that
\( \begin{equation} 2m_1^2=(2n_2)^2=2^2n_2^2, \end{equation} \)
which entails that
\( \begin{equation} m_1^2=2n_2^2. \end{equation} \)
Therefore, m₁² is even, and so m₁ is even. Now we have obtained that both n₁ and m₁ are even, which is a contradiction because we had simplified above so that at least one of the two is odd.


In short, our argument was as follows: First, we assumed there were integers m, n such that

\( \begin{equation} \sqrt{2}=\frac{n}{m}. \end{equation} \)

From that assumption and known facts about numbers, we derived a contradiction. We conclude that there are no integers n, m for which that equation holds (in other words, the square root of two is not a rational number).

In the proof above, we made a deduction with an inconsistent set of premises, namely the premise that the square root of two is a quotient of integers, and known facts about numbers. From that, we derived a contradiction. Is that deduction valid?

 

[A Toy Windmill]

From that assumption and known facts about numbers, we derived a contradiction. We conclude that there are no integers n, m for which that equation holds (in other words, the square root of two is not a rational number).

In the proof above, we made a deduction with an inconsistent set of premises, namely the premise that the square root of two is a quotient of integers, and known facts about numbers. From that, we derived a contradiction. Is that deduction valid?

My question is why you concluded that there are no integers n, m for which the equation holds. Why not conclude that one of your known facts about numbers is wrong?

 

[Angra Mainyu]

From that assumption and known facts about numbers, we derived a contradiction. We conclude that there are no integers n, m for which that equation holds (in other words, the square root of two is not a rational number).

In the proof above, we made a deduction with an inconsistent set of premises, namely the premise that the square root of two is a quotient of integers, and known facts about numbers. From that, we derived a contradiction. Is that deduction valid?

My question is why you concluded that there are no integers n, m for which the equation holds. Why not conclude that one of your known facts about numbers is wrong?

Well, because they're known, so they can't be wrong.

But alright , since the validity of the argument with contradictory premises does not depend on that, so let's say that either the square root of two is not a quotient of two integers, or else one of our basic beliefs about integers is mistaken, and the point I'm trying to make holds.

For the fun of it, what could be the false belief?

Consider the following:

S1: 1 is not even.
S2. For every integer n, either n is even, or n+1 is even.
S3. If n² is even, then n is even (n is any integer, etc.)

They all seem obvious, though I can argue for them if you want. For example, let's say S1 and S2 are true, but S3 is false.

So, let's say n² is even, but n is not even. Now, n+1 is even because n is not even. Thus, there is n₂ such that

\( n+1=2n_2 \)
Hence,

\( n^2+2n+1=(n+1)^2=(2n_2)^4=4n_2^2. \)

It follows that
\( 1=4n_2^2-2n-n^2. \)
Now, n² is even, so there is n₃ such that

\( n^2=2n_3, \)
which, together with the above equation, yields that

It follows that
\( 1=4n_2^2-2n-2n_3=2(2n_2^2-n-n_3), \)
implying that 1 is even, a contradiction.

So, S3 is true if S1 and S2 are true (yes, I used contradictory premises, but then, that was part of the original OP argument anyway; those who reject that will not be persuaded). Is there any other suspect, or do you think S1 or S2 might be so?

 

[A Toy Windmill]

So, S3 is true if S1 and S2 are true (yes, I used contradictory premises, but then, that was part of the original OP argument anyway; those who reject that will not be persuaded). Is there any other suspect, or do you think S1 or S2 might be so?

Maybe I suspect distributivity of multiplication over addition.

 

[A Toy Windmill]

My complaint here is that I don't think the proof in the OP is based on contradictory premises, because I don't regard mathematical facts as premises in mathematical arguments. The only premise in your original argument is that the square root of 2 is rational, and it is clear that you make this a premise in order to show that it is false.

This is how premises work in nearly all treatments of formal logic for mathematics. They are only in force within the proof. At some point in the proof, they are discharged. This is not the case for the ambient mathematical facts, being axioms, or previously proven theorems. Those are never discharged.

So the way I would phrase your proof in the OP is to say that it derives a contradiction from a single false premise, not from multiple contradictory premises. With the premise discharged, you have shown that "the rationality of the square root of 2 implies a falsehood", which is to say that the square root of 2 is not rational.

This explains the asymmetry. It explains why you conclude that the square root of 2 is irrational, and you don't conclude that multiplication doesn't distribute over addition. The first was a discharged premise. The second is not a premise at all.

 

[Angra Mainyu]

So, S3 is true if S1 and S2 are true (yes, I used contradictory premises, but then, that was part of the original OP argument anyway; those who reject that will not be persuaded). Is there any other suspect, or do you think S1 or S2 might be so?

Maybe I suspect distributivity of multiplication over addition.

Fair enough. I suspect that not many readers will suspect that, but if some do, I guess they might not be persuaded that no rational number squared equals two. But that will not change the facts about the validity of the argument.

 

[Angra Mainyu]

My complaint here is that I don't think the proof in the OP is based on contradictory premises, because I don't regard mathematical facts as premises in mathematical arguments. The only premise in your original argument is that the square root of 2 is rational, and it is clear that you make this a premise in order to show that it is false.

This is how premises work in nearly all treatments of formal logic for mathematics. They are only in force within the proof. At some point in the proof, they are discharged. This is not the case for the ambient mathematical facts, being axioms, or previously proven theorems. Those are never discharged.

So the way I would phrase your proof in the OP is to say that it derives a contradiction from a single false premise, not from multiple contradictory premises. With the premise discharged, you have shown that "the rationality of the square root of 2 implies a falsehood", which is to say that the square root of 2 is not rational.

This explains the asymmetry. It explains why you conclude that the square root of 2 is irrational, and you don't conclude that multiplication doesn't distribute over addition. The first was a discharged premise. The second is not a premise at all.

But I'm deriving the contradiction from the premise in question and other statements. I can't do it with that false premise alone. I would be able to do that if the premise were of the form, say (P and ¬P), but it's not of that sort (though in that case, we would still have an inconsistent set of statements). So, okay, if they are not called "premises", we can call them "statements", but it remains the case that I derived a contradiction from an inconsistent set of statements - one that I made explicit (the premise), and others that were background knowledge and were used as needed or convenient. What we call them is not essential, I think, to the point I was trying to make, which is that I was reasoning from an inconsistent set of statements in order to get a contradiction and in that manner, proved one of those statements false (because we knew the others were true).

 

[Angra Mainyu]

A Toy Windmill,

I just want to clarify I have no problem conceding the point, but I want to ask whether I'm reading your complaint right, because it seems to me it's a matter of terminology. I mean, we do not seem to disagree about what follows from what. I'm using a premise, and some other statements that are not called "premises" in order to derive a contradiction. The former is the statement I intend to show is false, and the latter are statements I expect the audience to accept, and which I'm not questioning in the context of the original argument - though I might defend them if some of the readers challenge them, depending on how long it would take, whether I think they will be persuaded, etc.

Am I getting this right, or is there a further objection you're raising?

 

[A Toy Windmill]

at least one of the numbers n₁ or m₁ or is odd

Now we have obtained that both n₁ and m₁ are even, which is a contradiction because we had simplified above so that at least one of the two is odd.

In the proof above, we made a deduction with an inconsistent set of premises, namely the premise that the square root of two is a quotient of integers, and known facts about numbers.

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.

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.

 

[Speakpigeon]

In the proof above, we made a deduction with an inconsistent set of premises, namely the premise that the square root of two is a quotient of integers, and known facts about numbers.

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.

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.

I'd like to make sure I understand how you read the proof.

If I relabel your propositions 1 to 4 as follows:

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)

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 Toy Windmill]

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

 

[Angra Mainyu]

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


\( \sqrt[2]=\frac{n}{m} \)
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.

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.

 

[A Toy Windmill]

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.

 

[Angra Mainyu]

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).

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.

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.

 

[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

 

[Speakpigeon]

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

 

[A Toy Windmill]

I don't think you understand the turnstile relation.

 

[Angra Mainyu]

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.

 

[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.

 

[Angra Mainyu]

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.