A simple proof that pi is irrational [pdf] (1946)
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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?]
You can say it's obvious today to a high school calculus student, but the elapsed time between understanding of the two proofs is about two millennia.
* λ: [reasonably] well [written] - the lambda, written in stern red Biro, in your double spaced manuscript, indicates you missed out a word. Reasonably is an adjective that requires a comparative (reasonably ... what).
* [imo]: sp - imo is an abbreviation so why not use some capital letters? You are a mathematician: you deploy many alphabets routinely so why piss around with the Roman one? I doubt you muddle your h and H.
* Final clause: expected meaning can be derived but why make it hard?
A grammarian would know that. Not to mention their nephew in the fifth grade.
Also, "grammatitian" is an irregular word used by people who don't know about "grammarian" and "grammatist", likely because they don't read books and can't be bothered to use a dictionary.
Honestly, this approach also sounds like it has the potential to be insufferable, though I've seen it done skillfully. But I guess it must have been better than the prevailing norm for Marcus Arelius to cherish the wisdom. Maybe to Marcus it was more about the humility therein than about playing the role of a grammar tutor.
If your conversation partner is made to feel like a human on equal footing with you, then you can just mention the mistake candidly without any condescension implied or perceived. Maybe don't try it in settings where you would risk accidentally embarrassing someone.
That may be a bit easier to follow.
He notes that p(x) is a polynomial of degree 2n, and then applies that general formula for multiple derivatives of a product that he proves at the start of the video. That involves a bunch of fiddling with binomial coefficients. He does this so he can write down an exact equation for the polynomial and its derivatives as a sum of powers of x. Around 20% of the video is spend on proving that general derivative formula and later applying it.
That's overkill because we do not need to know the exact coefficient of each power of x.
Simpler is to note that since p(x) is a polynomial of degree 2n and the whole thing is divided by n! it is a sum of terms of the form Ci x^i / n! for i = 0...2n. Also note that because it has a factor of x^n, Ci = 0 for i < n.
Consider then some arbitrary term of the polynomial, Ci x^i / n!, i >= n.
When x = 0, that term is 0 because of x^i.
Differentiating lowers the exponent of x by 1, so the first i-1 derivates all still have a factor of x in them, so are all 0 at x = 0.
The i'th derivative finally gets rid of the x. In particular, the i'th derivative of x^i is i!, so the i'th derivative of the term is Ci i! / n!. Remember that i >= n, so i!/n! is an integer and so is Ci i! / n!.
Further derivatives of the term are all 0 because they are derivatives of a constant.
In summary, the polynomial is 0 at x = 0, because all terms have a factor of x in them. As you repeatedly differentiate the polynomial each individual term of the derivative is 0 at x = 0 until you've differentiated enough times to get rid of all its factors of x. At that point the term is a constant integer. Further derivatives of the term are derivatives of a constant which is 0.
To show that the polynomial and its derivatives are integers at pi note that f(x) = f(pi-x), and so f'(x) = - f'(pi-x), f''(x) = f''(pi-x), ..., f^(m)(x) = (-1)^m f^(m)(pi-x). The polynomial and all its derivatives at pi are either the same as they are at 0 or the negative of what they are at 0, and since they are all integers at 0 they are also all integers at pi.
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_...
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.
Assume e is algebraic of degree m, with A_m e^m + A_(m-1) e^(m-1) + ... + A_0 = 0 where the A_i's are integers, A_0 != 0.
1. Define a polynomial f(x) = x^(p-1) (x-1)^p (x-2)^p ... (x-m)^p / (p-1)! where p is an odd prime to be specified later.
2. Then define F(x) as f(x) + f'(x) + f^(2)(x) + ... + f^(mp+p-1)(x).
3. Show that for 0 < x < m we have |f(x)| < m^(mp+p-1)/(p-1)!.
4. Show that the derivative of e^(-x) F(x) = -e^(-x) f(x).
5. Show that A_j integral( e^(-x) f(x), 0, j) = A_j F(0) - A_j e^(-j) F(j).
6. Multiple that by e^j and then sum over j = 0, 1, ..., m giving sum( A_j e^j integral(e^(-x)f(x), 0, j), j = 0 -> m) = - sum( sum( A_j f^(i)(j), i= 0 -> mp+p-1), j= 0 -> m).
7. Show that f^(i)(j) is an integer for all i, j in that sum, and that each is divisible by p except in the case j = 0, i = p-1.
8. Show that f^(p-1)(0) is not divisible by p if we choose p > m.
9. Show that if we chose p > |A_0| the double sum on the right in #6 consists of multiples of p except for for the term -A_0 f^(p-1)(0), and so must be a non-zero integer.
10. Use the inequality from #3 in the left side in #6 to conclude that |left side| < 1 if p is sufficiently large. That contradicts #9.
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.
The busy beaver function Σ(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.Σ(1)Σ(2)Σ(3)... that concatenates all busy beaver numbers. Its first few digits are R = 0.1621 ... (thanks to ChatGPT for this step).
However, if you include the number of digits in each number, that is
R = 0.|Σ(1)|Σ(1) |Σ(2)|Σ(2) |Σ(3)|Σ(3)... = 0.11 14 16 213
In your version of R, you can unambiguously get Σ(n) from R, therefore, the halting problem proves that R itself is not computable. Nice!
Actually, my version has an issue, which is that (while you can read the Σ(n) unambiguously after reading |Σ(n)|) you can't unambiguously get the |Σ(n)| without knowing their length.
The most efficient way, I believe, is to just include ||Σ(n)|| as well, and so on recursively down until you have only one digit. But then you'll also need to keep track of the length of the sequence, so it's a mess.
Better just write everything in unary. So (1, 4, 6, 13) gets encoded as
R = 10 11110 1111110 11111111111110 ...
Despite its obvious flaw of not having a rigorous proof but it is quite intuitive and easy to understand, and no need for definition of trigonometry functions, just need to have the concept of infinitesimal.
Edit: I just found out that pi can be expressed as infinite series of continued fractions, hence proving its irrationality. Found it interesting because it is almost exactly same as I thought. I think of that "proof" when I was in high school.
It's basically the same proof except with the following changes.
1. Instead of f(x) = x^n (a - bx)^n / n!, where pi = a/b, it defines f(x) = x^n (1-x)^n / n!. It notes that for 0 < x < 1 we have 0 < f(x) < 1/n!.
2. It uses a lemma that was proven a few pages earlier to show that the j'th derivative of f at 0, f^(j)(0), is an integer for all j. Here's the lemma and proof [1].
3. It notes that f(x) = f(1-x), and so f^(j)(1) is also an integer for every j.
4. As in the pi irrationality proof, it defines F(x) as an alternating series of derivatives of f(x), but the F(x) is a little different:
F(x) = b^n { pi^(2n) f(x) - pi^(2n-2) f^(2)(x) + pi^(2n-4) f^(4)(x) - ... + (-1)^n f^(2n)(x)}
where pi^2 = a/b.
5. It takes the derivative of F'(x) sin(pi x) - pi F(x) cos(pi x), giving pi^2 a^n f(x) sin(pi x).
6. Integrate that divided by pi from 0 to 1 to give pi a^n integral(f(x) sin(pi x), 0, 1) = F(1) + F(0).
7. Note that F(1) + F(0) is an integer.
8. From 0 < f(x) < 1/n! (see item #1 above), we have that 0 < pi a^n integral(f(x) sin(pi x), 0, 1) < pi a^n/n! < 1 for sufficiently large n, contradicting item #7.
[1] Lemma: if h(x) = x^n g(x) / n! where g(x) is a polynomial with integer coefficients, then h^(j)(0) is an integer for j = 0, 1, 2, ..., and with the possible exception of the case j = n that integer is divisible by n+1. No exception is needed in the case j = n if g(0) = 0.
Proof: h^(j)(0) = Cj j! / n! where Cj is the coefficient of x^j in x^n g(x) so Cj is an integer by hypothesis. (That's all that is needed in the pi^2 irrationality proof, so you can stop here if that's all you are interested in).
For j < n we have Cj = 0 so h^(j)(0) = 0 which is divisible by n + 1.
For j > n we have a factor of n + 1 in j! / n!, so h^(j)(0) will also have a factor of n + 1.
For j = n, we have h^(j)(0) = Cn. If g(0) = 0, then Cn = 0, which is divisible by n + 1.
If one wants to get it from there I'd recommend either the 6.2 MB PDF of the 6th printing, or the 8.0 MB PDF that doesn't state the printing in the description, but is the only 8.0 MB PDF.
Both of those are searchable PDFs, and pretty good scan quality. The 8.0 MB one has the best scan quality, although the pages are a little yellowed which might be a bit annoying.
The 8.0 MB one is actually just a copy of the Internet Archive's scan, which is available for borrowing directly from the IA [1].
There's a 3.9 MB PDF but the scan quality is noticeably worse than the others and it is not searchable [2].
I've got a searchable PDF I made myself from my physical copy of the book using a CZUR ET24 Pro. The IA scan text is little better, but not by much, and their file is a little smaller (mine is 9.1 MB), but overall I like my scan more. I've been quite impressed with the CZUR scanner.
[1] https://archive.org/details/irrationalnumber00nive/page/n3/m...
[2] By "not searchable" I mean it doesn't appear to have text information included to allow for fast search. Some PDF readers can still search it but I think they may do their own OCR on it to do so, and that takes a while. The ones I've said are searchable can search the whole document almost instantly.
The infinite sums give us ways to calculate their approximations, and show some of their properties as functions.
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.
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).
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 < π^n a^n / n!
which is bad. But I depends on n so this really should read
0 < I_n < π^n a^n / n!
which is fine... as far as the proof shows.
But perhaps I'm missing something...
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.
The claim of the proof is that I=F(0)+F(π) 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.
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