What do you mean, “is there a reason for that”? Do you mean why is there a best approximation possible with small numbers, or why it’s that particular approximation, or why does pi have a continued fraction form, or why it’s that particular continued fraction, or why continued fractions produce good approximations to irrationals?
Bold statements, both of them. Probably too bold, in fact: I suspect that one can prove that the P-NP question is not Gödel-undecidable.
Funny enough, if you let go of the artificial difference between the horribly named real and complex they aren’t hard to deal with. You just have to quit using shortcuts that are only valid in a much simpler human construct.
sqrt(-1) or x^=-9 are just as valid as any other number. They are way more than a useful fiction.
i (or -i) are just what numbers “become” when rotated, it is the a human conventions that make them hard to deal with.
Complex numbers just require the work of Pythagoras, arithmetic, and a bit of algebra.
It does help to realize that all real numbers complex numbers, just the i is 0.
So the real 5 is just the complex 5 where i = 0
Electronic devices do reduce the burden of resorting to “FOIL” etc…
I did say “arguably” :). And calling it “the most famous and important unsolved problem” is ambiguous: was I making one claim, or two? It’s more defensible if you interpret it as a single claim.
My question may have been quite silly, but it wasn’t nearly as silly as any of the questions you’re trying to put into my mouth. 
pi, for example, enters into some remarkable approximations related to Heegner numbers. I wonder, for example, if those might be related to pi’s excellent rational approximation.
My statements may have been bold, but if “very bold” is a synonym of “probably wrong,” I think your “suspicion” is definitely the very boldest of all!
I consider both the original statements bold, as you do, but also consider your statement bold too! No one has yet proven the P-NP question is decidable in any particular standard formal system, nor proven it undecidable, nor proven it true, nor proven it false, nor proven it “provable if true”, nor proven it “disprovable if false”, nor any such thing. It’s all open.
(If we wish to distinguish between “Gödel-undecidable” and “undecidable” simpliciter, I’m not sure exactly how to formalize it, but no one’s made such a formal distinction and furthermore proven P vs. NP non-“Gödel-undecidable” either…)
I said that because the question is of such a form that, if P = NP, there must exist a proof of that fact. Though that doesn’t rule out the possibility that (as septimus asserts) P≠NP but that’s unprovable, it at least puts the question in a different category from the most-undecideable questions.
septimus, I didn’t mean to place silly questions into your mouth. I thought that I was simply exhaustively listing the things you could have meant by your question. But I guess I must have missed (at least) one, since you seem to be saying you weren’t asking any of those.
Thudlow Boink, do you mean that you were considering one single figure of merit, which combines importance and fame in some way, and then saying that the Riemann hypothesis scores highest on that hybrid figure of merit? That’s more defensible, though I suspect that the P-NP question might still be more famous, as well.
It’s not true that if P = NP, there must exist a proof of that fact. (As I said, P = NP has never been proven to be “provable if true”). It could be that there exists a program which does indeed solve NP-complete problems in polynomial time, but that there is no proof that this particular program always produces a correct answer in polynomial time.
Propositions with the “provable if true” property are what’s called Sigma_1. They look like “There exists a [finite object] such that it satisfies [some decidable property]”. For example, “The Goldbach conjecture” is false would be a Sigma_1 proposition: “There exists an even integer > 2 such that it is not a sum of two primes”. But P = NP is not of this form. It’s of the form Sigma_2: “There exists a [finite object] such that for all [some other finite object]s, they satisfy [some decidable relation]”. P = NP is the claim “There exists a computer program and polynomial such that for all instances of your favorite NP-complete problem, the program correctly solves the problem within the time prescribed by the polynomial”.
I haven’t read much about fractional approximations to pi. Not doubting you, but do you have any recommended reading?
ISTM the interesting point is not that 52163/16604 is better, but rather that 355/113 is so anomalously good. IOW in a measure of epsilon/digit count it’s a home run whereas its nearest neighbors are mediocre. Why? Are there other local spikes in epsilon efficiency farther out? Is there anything deeper than dumb luck at work here?
Here is a pdf article discussing the difficulty of proving P ≠ NP, though it hints that proofs might exist if one “thinks outside the box.”
Of course 52163/16604 is uninteresting — it was just a way to demonstrate the goodness of 355/113. Good approximations fall out of continued fractions (Google for details).
The continued fraction representation of pi is [3;7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,2,2,2,…] It is the early “292” that means there is an excellent early approximant.
ETA: Contrast this with the continued fraction representation of the Golden Ratio: [1;1,1,1,1,1,1,1,1,1,1,1,1…]. The Golden Ratio has no good small approximant.
Okay, point taken, and thank you. 
BTW, why do we need to know the sqrt of -1?
Yes, the beauty of mathematics is not only the way it serves us in the practical world but also as a source of continual analysis and philosophical discourse. What more can you ask? No wonder it has been described as beautiful.
I see your point.
Basically anytime a problem involves a sine wave, the imaginary number is used.
You wouldn’t have cell phones, reliable power in your home, or the internet without that sqrt of -1
Or rotation, the ability to turn 90 degrees and such…
I read your proof, and I don’t quite get it. Or rather, I get most of the steps in a hand-waving manner, but some are fairly obscure and the others I can’t quite fully justify to myself. This is just a reflection of my own limits and not your description, of course.
That said, Niven’s proof on the Wikipedia page was understandable to the point that I can describe it from memory. I’ll repeat it in my own words since I did have to think through a few things that confused me a bit initially.
This is a proof by contradiction, where we start by assuming pi=a/b (a and b being positive integers). We examine this function:
f(x) = x[sup]n[/sup](a - bx)[sup]n[/sup] / n!
Technically this should be f[sub]n/sub, since n is a parameter, but I’ll leave it off and we’ll just keep in mind that we can change n to whatever we want.
First, note that f(0) = f(pi). If you substitute x => (a/b - x) and do a little algebra, you get the same equation (i.e., f(x) = f(pi - x)). So whatever we conclude about f(0) will apply to f(pi) later.
Let’s examine the derivatives: we can see that the lowest order term (of x) is x[sup]n[/sup], which means that f(0) is equal to 0 for the first n-1 derivatives. Note that as we’re doing this, we’re progressively canceling out the n! term at the bottom. At the nth derivative, there’s finally a constant term (not multiplied by a power of x), but nothing left in the denominator. After that, no fractions are introduced in further derivatives. At the (2n)th derivative, there’s only a constant term, and after that nothing.
The big implication here is that f(0), f’(0), etc. are all integers. We don’t know (or care) what their values are, just that there are no fractions in there. Furthermore, f(pi) (and its derivatives) is an integer since it’s equal to f(0).
The next step involves this formula:
F(x) = f(x) - f’'(x) + f[sup]4/sup - … -1[sup]n[/sup]f[sup]2n/sup
It’s pretty straightforward to see that this holds:
F’’ + F = f
Since the terms alternate in sign, F’’ mostly cancels the terms in F, with the exception of f(x) and the final term. But that final term is the 2n+2 derivative, which we know to be 0. So it’s really just f.
The next step is to show this is true:
(F’sin - Fcos)’ = f*sin
A little application of the product rule (and linearity of derivatives) get you this:
F’'sin + Fsin = f*sin
And then:
(F’’ + F)sin = f*sin
So we see that the formula works. Going back, we can also turn this (va fundamental theorem of calculus):
(F’sin - Fcos)’ = f*sin
Into this:
Int[0…pi] f*sin = [0…pi] (F’sin - Fcos)
Looking at the right hand side, we see that we can ignore the F’sin term because sin is 0 at 0 and pi. From before, F(0) and F(pi) are integers, and cos(0)=1 and cos(pi)=-1. From all of this, we can conclude that:
Int[0…pi] f*sin = [0…pi] (F’sin - Fcos) = an integer
But let’s try to compute an upper bound for:
Int[0…pi] f*sin
We’ll so this by looking at the upper bounds of the different components. sin(x) of course has a maximum of 1.0 in that range. f is a little trickier, but let’s do a little transformation:
x[sup]n[/sup](a - bx)[sup]n[/sup] / n! = (ax - bx[sup]2[/sup])[sup]n[/sup] / n!
That bx[sup]2[/sup] term is negative… so it can only make the value smaller. We’ll ignore it. Therefore, one upper bound is:
(ax)[sup]n[/sup] / n!
Put together, this is an upper bound for the integral:
pi * (ax)[sup]n[/sup] / n!
But look: that n! grows much faster than (ax)[sup]n[/sup]. No matter what a is, and even with the maximum value of pi for x, n! can be bigger yet. So we can always choose an n where our integral is less than 1.
However, we already decided that the integral must evaluate to an integer. If it can’t even be as large as 1, it must be zero.
Well, it can’t be that either: f(x) and sin(x) are all positive for 0<x<pi. Therefore the integral is positive. Small, maybe, but not zero.
There’s no integer between 0 and 1, so we have our contradiction. Pi can’t be expressed as a/b and is irrational.
Well, if you like, I’d be happy to clarify the details you found obscure or were unable to justify. I’d be curious to hear what they were, at any rate.
For what it’s worth, the proof I gave actually has the same fundamental idea as the Niven proof (or as all the other proofs given on that Wikipedia page; perhaps even as all known proofs of this irrationality), but is framed in the way which made things clearest and most generalizable to me in understanding the underlying phenomenon. [Of course, different people have different preferences for how to understand things]
There’s a good chance that simply formulating the questions clearly will enable me to not need them :).
The inductive step I’m sure I can figure out if I sit down with some paper, but so far I just did a few derivatives by hand; enough to make the argument plausible.
I’m pretty fuzzy on this part “If y were rational with lowest-terms denominator d, then the smallest nonzero value integer polynomials of degree O(N) in y could produce is 1/dO(N)”. I get the overall idea (since as you said it’s the same basic idea as the Niven proof) but somehow it’s not quite gelling. Though as I said, even formulating the question right will probably lead me to the answer.
I’m actually a little surprised there isn’t a more trivial proof, at close to the level of the sqrt(2) one. It seems like the trig functions, with their endless zeroes at multiples of pi, could more directly show that pi can’t be a ratio. Something along the lines of showing that a periodic function with rational zeroes can’t have the properties of sin/cos. I dunno.
I’ll try pre-emptively adding some more details for everyone which may not be obvious:
The first third of the proof:
The gist:
Let f(x) = cos(sqrt(x)), and, as is conventional, let f’ be its derivative, f’’ its second derivative, etc., using fsup[/sup] to indicate its N-th derivative.
Note that fsup[/sup] = P(y)f + Q(y)f’, where y = 1/(4x) and P and Q are integer coefficient polynomials of degree growing at rate O(N).
Short further explanation:
This follows inductively from f’’ having this form, which is just the fact that cos’’ = -cos, translated to this reparametrization.
Longer further explanation:
Specifically, just applying mechanical differentiation, we find that f’ = -1/2 * sinc(sqrt(x)). And then, re-differentiating, we find that f’’ = (sinc(sqrt(x)) -cos(sqrt(x)))/(4x) = (-2f’ - f)/(4x).
If we give 1/(4x) the name y(x), then we have the convenient relationship f’’ = (-2f’ - f) * y, and also conveniently that y’ = -1/(4x[sup]2[/sup]) = -4y[sup]2[/sup].
What’s so convenient about this? Well, the specific details don’t matter very much, but the fact that f’’ = A(y) * f + B(y) * f’ while y’ = C(y), where A, B, and C are all integer coefficient polynomials… That is very convenient.
Why is that?
Well, suppose you knew that fsup[/sup] = P(y)f + Q(y)f’, and now you wanted to calculate f[sup](N + 1)[/sup] by differentiating yet again. Mechanically working it out using the standard differentiation rules and the above forms for f’’ and y’, you get P’(y)C(y)f + P(y)f’ + Q’(y)C(y)f’ + Q(y)[A(y)f + B(y)f’]. Collecting the terms on f and on f’, you get [P’(y)C(y) + Q(y)A(y)]f + [P(y) + Q’(y)C(y) + Q(y)B(y)]f’.
So each of f and f’ is again multiplied by a polynomial of y, and… here’s the kicker… the degrees on the polynomials involved can only have risen by so much. Whatever the degree of the polynomials involved before (P and Q), the degrees of the polynomials involved now has risen by at most max(deg© - 1, deg(A), deg(B)). This is some fixed finite quantity.
So we’ve established that fsup[/sup] is some polynomial of y times f + some polynomial of y times f’, for all N, and, furthermore… since each differentiation causes the degrees to only rise by at most the same amount, we have that the degree of the polynomials involved in fsup[/sup] grows at rate at most O(N) as N keeps going up.
[In our particular case, tracing the particulars of the above reasoning, we find that the polynomials involved in fsup[/sup] specifically have degree at most N - 1 for N ≥ 2. But the specific constants of the O(N) aren’t actually important for us. Indeed, what I’ve written out above still contains more gory details worked out than is actually necessary. You don’t have to actually calculate the derivative of P(y)f + Q(y)f’ in any detail. You can tell ahead of time that it’s going to work out with the properties we want, just by considering that derivatives of polynomials are polynomials, etc.]
Oh, right. Sounds deep.