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

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?

2. Originally Posted by 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?

3. Originally Posted by A Toy Windmill
Originally Posted by 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?

4. Originally Posted by 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.

5. 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.

6. Originally Posted by A Toy Windmill
Originally Posted by 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.

7. Originally Posted by 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.
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).

8. 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?

9. Originally Posted by Angra Mainyu
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.

10. Originally Posted by A Toy Windmill
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

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