VKS comments on Rationality Quotes September 2012 - Less Wrong

7 03 September 2012 05:18AM

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Comment author: 04 September 2012 11:51:02PM *  23 points [-]

After I spoke at the 2005 "Mathematics and Narrative" conference in Mykonos, a suggestion was made that proofs by contradiction are the mathematician's version of irony. I'm not sure I agree with that: when we give a proof by contradiction, we make it very clear that we are discussing a counterfactual, so our words are intended to be taken at face value. But perhaps this is not necessary. Consider the following passage.

There are those who would believe that every polynomial equation with integer coefficients has a rational solution, a view that leads to some intriguing new ideas. For example, take the equation x² - 2 = 0. Let p/q be a rational solution. Then (p/q)² - 2 = 0, from which it follows that p² = 2q². The highest power of 2 that divides p² is obviously an even power, since if 2^k is the highest power of 2 that divides p, then 2^2k is the highest power of 2 that divides p². Similarly, the highest power of 2 that divides 2q² is an odd power, since it is greater by 1 than the highest power that divides q². Since p² and 2q² are equal, there must exist a positive integer that is both even and odd. Integers with this remarkable property are quite unlike the integers we are familiar with: as such, they are surely worthy of further study.

I find that it conveys the irrationality of √2 rather forcefully. But could mathematicians afford to use this literary device? How would a reader be able to tell the difference in intent between what I have just written and the following superficially similar passage?

There are those who would believe that every polynomial equation has a solution, a view that leads to some intriguing new ideas. For example, take the equation x² + 1 = 0. Let i be a solution of this equation. Then i² + 1 = 0, from which it follows that i² = -1. We know that i cannot be positive, since then i² would be positive. Similarly, i cannot be negative, since i² would again be positive (because the product of two negative numbers is always positive). And i cannot be 0, since 0² = 0. It follows that we have found a number that is not positive, not negative, and not zero. Numbers with this remarkable property are quite unlike the numbers we are familiar with: as such, they are surely worthy of further study.

• Timothy Gowers, Vividness in Mathematics and Narrative, in Circles Disturbed: The Interplay of Mathematics and Narrative
Comment author: 06 September 2012 10:01:40PM *  2 points [-]

The two examples are not contradictory, but analogous to one another. The correct conclusion in both is the same, and both are equally serious or ironic.

1. Suppose x² -2=0 has a solution that is rational. That leads to a contradiction. So any solution must be irrational.

2. Suppose x² +1=0 has a solution that is a number. That leads to a contradiction. So any solution must not be a number. Now what is a "number" in this context? From the text, something that is either positive, negative, or zero; i.e. something with a total ordering. And indeed we know (ETA: this is wrong, see below) that such solutions, the complex numbers, have no total ordering.

I see no relevant difference between the two cases.

Comment author: 07 September 2012 07:54:57AM 4 points [-]

The only relevant difference that I can see is that, in the first paragraph, the solutions are explicitly limited to the rational numbers; in the second case, the solutions are not explicitly limited to the reals.

Comment author: 06 September 2012 10:25:29PM *  4 points [-]

You can work the language a little to make them analogous, but that's not the point Gowers is making. Consider this instead:

"There are those who would believe that all equations have solutions, a view that leads to some intriguing new ideas. Consider the equation x + 1 = x. Inspecting the equation, we see that its solution must be a number which is equal to its successor. Numbers with this remarkable property are quite unlike the numbers we are familiar with. As such, they are surely worthy of further study."

I imagine Gowers's point to be that sometimes a contradiction does point to a way in which you can revise your assumptions to gain access to "intriguing new ideas", but sometimes it just indicates that your assumptions are wrong.

Comment author: 06 September 2012 10:49:11PM *  8 points [-]

"There are those who would believe that all equations have solutions, a view that leads to some intriguing new ideas. Consider the equation x + 1 = x. Inspecting the equation, we see that its solution must be a number which is equal to its successor. Numbers with this remarkable property are quite unlike the numbers we are familiar with. As such, they are surely worthy of further study."

Yes, yes they are.

Comment author: 06 September 2012 10:53:51PM *  1 point [-]

Consider the equation x + 1 = x.

(Edited again: this example is wrong, and thanks to Kindly for pointing out why. CronoDAS gives a much better answer.)

Curiously enough, the Peano axioms don't seem to say that S(n)!=n. Lo, a finite model of Peano:

X = {0, 1} Where: 0+0=0; 0+1=1+0=1+1=1 And the usual equality operation.

In this model, x+1=1 has a solution, namely x=1. Not a very interesting model, but it serves to illustrate my point below.

sometimes a contradiction does point to a way in which you can revise your assumptions to gain access to "intriguing new ideas", but sometimes it just indicates that your assumptions are wrong.

Contradiction in conclusions always indicates a contradiction in assumptions. And you can always use different assumptions to get different, and perhaps non contradictory, conclusions. The usefulness and interest of this varies, of course. But proof by contradiction remains valid even if it gives you an idea about other interesting assumptions you could explore.

And that's why I feel it's confusing and counterproductive to use ironic language in one example, and serious proof by contradiction in another, completely analogous example, to indicate that in one case you just said "meh, a contradiction, I was wrong" while in the other you invented a cool new theory with new assumptions. The essence of math is formal language and it doesn't mix well with irony, the best of which is the kind that not all readers notice.

Comment author: 07 September 2012 05:05:09AM 3 points [-]

But that's the entire point of the quote! That mathematicians cannot afford the use of irony!

Comment author: 07 September 2012 08:54:39AM *  0 points [-]

Yes. My goal wasn't to argue with the quote but to improve its argument. The quote said:

But could mathematicians afford to use this literary device? How would a reader be able to tell the difference in intent between what I have just written and the following superficially similar passage?

And I said, it's not just superficially similar, it's exactly the same and there's no relevant difference between the two that would guide us to use irony in one case and not in the other (or as readers, to perceive irony in one case and serious proof by contradiction in the other).

Comment author: 06 September 2012 11:24:02PM *  2 points [-]

Your model violates the property that if S(m) = S(n), then m=n, because S(1) = S(0) yet 1 != 0. You might try to patch this by changing the model so it only has 0 as an element, but there is a further axiom that says that 0 is not the successor of any number.

Together, the two axioms used above can be used to show that the natural numbers 0, S(0), S(S(0)), etc. are all distinct. The axiom of induction can be used to show that these are all the natural numbers, so that we can't have some extra "floating" integer x such that S(x) = x.

Comment author: 07 September 2012 08:47:01AM 0 points [-]

Right. Thanks.

Comment author: 14 September 2012 07:30:35AM *  1 point [-]

There are lots of total orderings on the complex numbers. For example:

a + bi [>] c + di iff a >= c or (a = c and b >= d).

In fact, if you believe the axiom of choice there are "nice total orders" for any set at all.

Comment author: 14 September 2012 01:36:49PM 2 points [-]

Importantly, however, the complex numbers have no total ordering that respects addition and multiplication. In other words, there's no large set of "positive complex numbers" closed under both operations.

This is also the reason why the math in this XKCD strip doesn't actually work.

Comment author: 14 September 2012 11:05:16PM 0 points [-]

You can still find divisors for Gaussian integers. If x, y, and xy are all Gaussian integers, which will be trivially fulfilled for any x when y=1, then x, y both divide xy.

You can then generalize the \sigma function by summing over all the divisors of z and dividing by |z|.

The resulting number \sigma(z) lies in C (or maybe Q + iQ), not just Q, but it's perfectly well defined.

Comment author: 15 September 2012 01:07:06AM 2 points [-]

If you sum over all the divisors of z, the result is perfectly well defined; however, it's 0. Whenever x divides z, so does -x.

Over the integers, this is solved by summing over all positive divisors. However, there's no canonical choice of what divisors to consider positive in the case of Gaussian integers, and making various arbitrary choices (like summing over all divisors in the upper half-plane) leads to unsatisfying results.

Comment author: 16 September 2012 03:24:48AM 0 points [-]

That's like saying the standard choice of branch cut for the complex logarithm is arbitrary.

And?

When you complexify, things get messier. My point is that making a generalization is possible (though it's probably best to sum over integers with 0 \leq arg(z) < \pi, as you pointed out), which is the only claim I'm interested in disputing. Whether it's nice to look at is irrelevant to whether it's functional enough to be punnable.

Comment author: 16 September 2012 04:21:00AM 0 points [-]

You're right -- the generalization works.

Mainly what I don't like about it is that \sigma(z) no longer has the nice properties it had over the integers: for example, it's no longer multiplicative. This doesn't stop Gaussian integers from being friendly, though, and the rest is a matter of aesthetics.

Comment author: 14 September 2012 10:53:57AM 0 points [-]

The well-ordering principle doesn't really have any effect on canonical orderings, like that induced by the traditional less-than relation on the real numbers.

This doesn't affect the truth of your claim, but I do think that DanArmak's point was quite separate from the language he chose. He might instead have worded it as having no real solution, so that any solution must be not-real.

Comment author: 14 September 2012 09:30:57AM 0 points [-]

Gah. You're quite right. I should refrain from making rash mathematical statements. Thank you.

Comment author: 11 September 2012 02:37:45AM 1 point [-]

The first one shows that assuming that there's a rational solution leads to contradiction, then drops the subject. The second one shows that assuming that there's a real solution leads to a contradiction, then suggests to investigate the non-reals. How are you supposed to tell which drops the subject and which suggests investigation?

Comment author: 07 September 2012 04:40:08PM 1 point [-]

I see no relevant difference between the two cases.

Isn't that the entire point? I see this as a mathematical version of the modus tollens/ponens point made elsewhere in this page.

Comment author: 07 September 2012 06:05:00PM 1 point [-]

The quote says,

could mathematicians afford to use this literary device? How would a reader be able to tell the difference in intent between what I have just written and the following superficially similar passage?

This seems to me to mean: the two cases are different; the first is appropriately handled by serious proof-by-contradiction, while the second is appropriately handled by irony. But readers may not be able to tell the difference, because the two texts are similar and irony is hard to identify reliably. So mathematicians should not use irony.

Whereas I would say: the two cases are the same, and irony or seriousness are equally appropriate to both. If readers could reliably identify irony, they would correctly deduce that the author treated the two cases differently, which is in fact a wrong approach. So readers are better served by treating both texts as serious.

I'm not saying mathematicians should / can effectively use irony; I'm saying the example is flawed so that it doesn't demonstrate the problems with irony.

Comment author: 07 September 2012 09:55:23PM 3 points [-]

The difference is that mathematicians apply modus tollens and reject sqrt2 being rational, but apply modus ponens and accept the existence of i; why? Because apparently the resultant extensions of theories justify this choice - and this is the irony, the reason one's beliefs are in discordance with one's words/proof and the reader is expected to appreciate this discrepancy.

But what one regards as a useful enough extension to justify a modus tollens move is something other may not appreciate or will differ from field to field, and this is a barrier to understanding.

Comment author: 07 September 2012 10:57:56PM 2 points [-]

I hadn't considered that irony. I was thinking about the explicit irony of the text itself in its proof of sqrt(2) being irrational. The reader is expected to know the punchline, that sqrt(2) is irrational but that irrational numbers are important and useful. So the text that (ironically) appears to dismiss the concept of irrational numbers is in fact wrong in its dismissal, and that is a meta-irony.

...I feel confused by the meta levels of irony. Which strengthens my belief that mathematical proofs should not be ironical if undertaken seriously.

Comment author: 07 September 2012 11:03:36PM 1 point [-]

Yes, I feel similarly about this modus stuff; it seems simple and trivial, but the applications become increasingly subtle and challenging, especially when people aren't being explicit about the exact reasoning.

Comment author: 07 September 2012 10:51:53PM 0 points [-]

If mathematicians behaved simply as you describe, then those resultant extension theories would never have been developed, because everyone would have applied modus tollens regarding in a not-yet-proven-useful case. (Disclaimer: I know nothing about the actual historical reasons for the first explorations of complex numbers.)

Therefore, it's best for mathematicians to always keep the M-T and M-P cases in mind when using a proof by contradiction. Of course, a lot of time the contradiction arises due to theorems already proven from axioms, and what happens if any one of the axioms in a theory is removed is usually well explored.

Comment author: 07 September 2012 10:48:26PM 2 points [-]

You're drawing the parallel differently from the quote's author. The second example requires assuming the existence of complex numbers to resolve the contradiction. The first example requires assuming, not the existence of irrational numbers (we already know about those, or we wouldn't be asking the question!), but the existence of integers which are both even and odd. As far as I know, there are no completely satisfactory ways of resolving the latter situation.