[HN Gopher] A simple proof that pi is irrational [pdf] (1946)
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A simple proof that pi is irrational [pdf] (1946)
Author : pipopi
Score : 53 points
Date : 2024-08-07 05:46 UTC (4 days ago)
(HTM) web link (www.ams.org)
(TXT) w3m dump (www.ams.org)
| hughdbrown wrote:
| I don't know why this would qualify as simple.
|
| Compare it with the proof by contradiction that the square root
| of 2 is rational:
|
| $$\sqrt{2} = \frac{a}{b} 2 {b^2} = {a^2} {b^2} and {a^2} have an
| even number of factors of 2 Therefore 2 {b^2} has an odd number
| of factors of 2 and cannot equal {a^2}. $$
|
| I can explain this to a ten year old. The irrationality of pi
| proof has too many non-obvious steps and claims.
|
| [Hacker News can't render math?]
| lupire wrote:
| It's simpler than other proofs. Yes, it relies on real analysis
| (differential and integral calculus), including calculus of
| transcendental functions. But a high school calculus student
| could follow the proof, if they take some faith in
| infinitesimals.
| parpfish wrote:
| agreed. i think they were doing the math equivalent of code-
| golf to cram this proof onto a single page.
| red_trumpet wrote:
| As a mathematician, I find this proof reasonably written.
| Yes, not everything is spelled out, but the proof was
| published by the AMS for mathematician, and imo it can be
| expected that those know how to fill the gaps.
| klyrs wrote:
| It's simple for a proof that pi is irrational. It isn't simple
| as [?]2, but the number itself is quite a lot more complex --
| even defining pi to mathematical precision takes more work than
| the whole of your irrationality proof!
| prvc wrote:
| Simple to confirm, much harder to generate. What to strive for!
| lupire wrote:
| The key idea is a bit obscured.
|
| A rational number is a finite computation of integers. It is
| finite but in magnitude and in level of detail.
|
| Thus, if you can rewrite a computation as the same computation
| but with only a subset of the terms, or with smaller (absolute
| value) integer terms, in a repeatable way, then the computation
| must either collapse to 0, or be infinite in either number of
| terms or size of a term. That is, not rational.
|
| https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent
|
| Any finite set of rational numbers is equivalent to a set of
| integers.
| potbelly83 wrote:
| It's not immediately clear (though I'm sure it is true) from
| reading the article that F(\pi) + F(0) != 0 (assuming \pi =
| \frac{a}{b}). Given that F(x) is an alternating series, maybe
| F(\pi) = F(0) = 0. (again I'm sure this is not the case, but
| it's not discussed in the proof)
| ykonstant wrote:
| The article shows in (1) that F(\pi) + F(0) is the integral
| of f(x)sin(x) from 0 to pi; the integrand is non-negative
| there and thus so is the integral.
| red_trumpet wrote:
| Even stronger, the integrand is _positive_ , hence the
| integral is positive as well, so it cannot be zero.
| csense wrote:
| F(\pi) + F(0) is the integral of f(x) sin(x) from 0 to \pi.
| f(x) sin(x) is the product of four factors (not necessarily
| integers):
|
| f(x) sin(x) = (1/n!) (x^n) (a-bx)^n sin(x)
|
| Examine the signs of these terms on the interval 0 to \pi.
| All four factors are non-negative everywhere, so the integral
| has to be non-negative.
|
| Could the integral be zero? 1/n! is a nonzero constant. x^n =
| 0 only when x = 0. (a-bx)^n = 0 only when x = a/b = \pi. And
| sin(x) = 0 only when x = 0 or x = \pi.
|
| The four factors are all positive inside the interval, and
| nice continuous functions, so there's definitely a rectangle
| of positive area underneath their product (and the product
| can never be negative so you can never get a negative area to
| get the integral back down to 0).
| alimw wrote:
| Are you talking about this proof in particular? I see no
| connection.
| CaptainNegative wrote:
| I don't think that follows at all. Any series of rational
| numbers that converges to a rational (such as any appropriate
| geometric series) will still converge to a rational if you
| remove any finite or co-finite subset of terms.
| kazinator wrote:
| The problem is that the _sin_ and _cos_ functions the proof
| relies on are embroiled with _pi_. It 's not obvious whether or
| not that is a confounding issue or not.
| glowcoil wrote:
| It's not. Why would it be?
| moomin wrote:
| Sin and Cos are typically defined as "Here's some weird
| infinite sums" from which they deduce the behaviours. Indeed,
| pi is typically defined in terms of the first positive zero of
| the sin function.
| nawgz wrote:
| They had PDFs in 1946? Wow.
| fuzzer371 wrote:
| Yup. It was called a sheet of paper back then.
| cliffordc wrote:
| Sheet of what?
| rich_sasha wrote:
| Is the proof even valid? The claim is that the integral (1) is
| bounded between 0 and a positive number decaying to 0 as n tends
| to infinity, which sounds like a contradiction.
|
| But the integral (1) is itself a function of n, since both f and
| F are. Not a fixed number. So the inequality is eminently
| satisfiable.
|
| In other words, if we call (1) I, they show
|
| 0 < I < p^n a^n / n!
|
| which is bad. But I depends on n so this really should read
|
| 0 < I_n < p^n a^n / n!
|
| which is fine... as far as the proof shows.
|
| But perhaps I'm missing something...
| ducttapecrown wrote:
| Think of a collection of integrals, one for each n, rather than
| a function depending on multiple variables. The contradiction
| is that the number F(pi) + F(0) is both a positive integer and
| arbitrarily small.
|
| There is no contradiction until n is large enough that F(pi) +
| F(0) is bounded below 1, hence the statement at the start:
| we'll pick a large enough value of n later.
| rich_sasha wrote:
| But F(pi) + F(0) depends on n, rather than being a number,
| just this isn't explicitly spelled out. So all the proof
| shows is that the integral (1) converges to 0 rather quickly,
| _as a function of n_.
|
| The claim of the proof is that I=F(0)+F(p) is simultaneously
| positive and arbitrarily small, which is not possible for a
| fixed number, but totally possible if you see I as a sequence
| indexed by n.
| alimw wrote:
| > The claim of the proof is that I=F(0)+F(p) is
| simultaneously positive and arbitrarily small, which is not
| possible for a fixed number, but totally possible if you
| see I as a sequence indexed by n.
|
| This is not possible because I_n must be an integer for
| every n.
| rich_sasha wrote:
| Ah yes, that's the key bit. Thanks!
| Someone wrote:
| Wikipedia has this proof with a few more intermediate steps:
| https://en.wikipedia.org/wiki/Proof_that_p_is_irrational#Niv....
|
| That may be a bit easier to follow.
| Y_Y wrote:
| I did a project on proofs of the irrationality of pi during my
| undergrad. Strangely I didn't find anything that I would have
| considered "from the book"[0]. Even this "simple" proof is not
| really elementary and gets a lot longer if you have to spell it
| out.
|
| In trying to find a proof with some explanatory value, like
| saying _why_ pi is irrational the best I came across was
| Lindemann-Weierstrass[1].
|
| [0] https://en.wikipedia.org/wiki/Proofs_from_THE_BOOK
|
| [1]
| https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_...
| hsolatges wrote:
| Oh man, the following proof of p's transcendence is beautiful.
| Thanks.
| earlgray wrote:
| That sounds like a cool project. Do you know of any
| transcendental number that does admit an intuitive proof of
| irrationality (or, more generally, of not being algebraic to
| any specific degree)?
|
| My gut feeling is that this shouldn't typically be the case,
| but I'm tired and struggling to convey why without descending
| into downright criminal levels of vagueness. I'd be more
| hopeful about an algebraic number of degree >1 having a 'nice'
| reason for specifically being irrational.
| brcmthrowaway wrote:
| How was this typeset so well in 1946?
| lovecg wrote:
| Enjoy some history of math typesetting
| http://www.practicallyefficient.com/2017/10/13/from-boiling-...
|
| Famously, Knuth lamented about the state of math typesetting at
| the time
| https://www.ams.org/journals/bull/1979-01-02/S0273-0979-1979...
| and in some ways TEX/Metafont was inspired by older examples
| from the early 1900s
| moomin wrote:
| Okay, here's where I find it gets weird.
|
| Being rational means you're a solution to a linear equation.
| sqrt(2) is irrational.
|
| Being algebraic means you you're a solution to an algebraic
| equation. Pi is transcendental, which means it's not only
| irrational but non-algebraic.
|
| Being computable means you can write a terminating program to
| compute the nth digit of the number. Pi is computable, I have no
| idea what isn't computable.
|
| But... just as naturals are countable, so are rationals, so are
| algebraic numbers, so are computable numbers. Which in turn means
| that the non-computable numbers that I can't even describe are
| the VAST majority of numbers on the real number line.
| geertj wrote:
| > I have no idea what isn't computable.
|
| The busy beaver function S(n) is not computable, at least, not
| for every n. If it were, it could be used to determine if an
| arbitrary Turing machine halts, solving the halting problem.
|
| To get from this non-computable function to a non-computable
| number, you can construct the finite number R =
| 0.S(1)S(2)S(3)... that concatenates all busy beaver numbers.
| Its first few digits are R = 0.1621 ... (thanks to ChatGPT for
| this step).
| lisper wrote:
| A.K.A.Chatin's constant:
|
| https://en.wikipedia.org/wiki/Chaitin%27s_constant
| hsolatges wrote:
| That is remarkable. Thanks.
| dullcrisp wrote:
| Or put more plainly, most infinite strings of digits don't
| represent anything meaningful.
| cpp_frog wrote:
| For anyone interested who could not follow the reasoning in the
| paper, it is explained extensively in this video [0] by Michael
| Penn.
|
| [0] https://www.youtube.com/watch?v=dFKbVTHK4tU
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