I don't want to write Lambert's biography here which you may find elsewhere. Instead, I want to spend some time with the irrational numbers, especially with the proof that the number\[

{\Huge \pi \approx 3.14159\dots}

\] is irrational, a proof that Lambert presented in 1761, i.e. 252 years ago. What people could have achieved in the 18th century is pretty amazing. I wonder how many people know the proof that \(\pi\) is irrational.

Rational numbers are numbers of the form \(p/q\) for \(p,q\in\ZZ\). Almost all real numbers are irrational, i.e. not rational, which makes the terminology surprising. Shouldn't rational reasoning be able to deal with all numbers or almost all numbers? If it's so, why are almost all numbers irrational?

Well, an answer we could give to this question is that the word "rational" has actually nothing to do with the rational reasoning. Instead, it boils down to the word "ratio" because that's how we call numbers of the form \(p/q\).

We have apparently gotten rid of the linguistic paradox but we shouldn't have. It's a fact that ancient mathematicians actually did believe that all numbers should be accessible to rational reasoning – and they thought that the condition was equivalent to their being rational in the modern sense.

The ancient Greek mathematicians apparently used the universal word "logos" for a ratio – a word that can mean "a ground", "a plea", "an opinion", "an expectation", "word", "speech", "account", "reason", i.e. pretty much anything as long as it sounds fundamental enough. This type of "logos" got translated as "ratio" to early Latin which may mean that some extra noise was added. Medieval writers actually improved the terminological mess by coining a new word, "proportio", for a ratio (and proportionality for identities relating ratios) but modern English and other languages have largely abandoned these medieval improvements.

Pythagoras himself was the most famous defender of the myth I have already referred to. All numbers should be rational. However, Pythagoras had a student called Hippasus who presented a geometric proof that \(\sqrt{2}\) was irrational. His proof was probably geometric but an algebraic proof is easier. If we could write \(\sqrt{2}=p/q\), then\[

2 = \frac{p^2}{q^2}, \quad p^2 = 2q^2.

\] But if you decompose the integers \(p^2\) into products of powers of primes, the exponent above the prime number \(2\) will be even for numbers \(p^2,q^2\) because those numbers are squares but it will be odd for \(2q^2\) because of the extra factor of two. So the numbers \(p^2\) and \(2q^2\) just can't be the same.

Pythagoras' belief was shattered. He couldn't find an error in Hippasus' proof of an irrational number but at the end, he found a way to undo the effects of Hippasus' devastating insight: he threw Hippasus overboard and drowned him. ;-) Leepeters Shmoits resuscitated a similar method to "undo" the results of string theory in recent years.

**The case of \(\pi\)**

Lambert's proof that \(\pi\) was irrational as well is harder, of course. Incidentally, in Czech, we often use the term Ludolphine number for \(\pi\) because Ludolph calculated 20 (in 1596) and later 35 digits. It's a bit surprising that it doesn't seem to be used outside the broader German-speaking realm.

Lambert was using Brook Taylor's expansions that had been around for half a century or so. His proof of irrationality has two steps. First, he showed that \(\tan(x)\) may be written using a clever unlimited continued fraction. Second, he showed that such continued fractions have to be irrational numbers.

Assuming that you know the Taylor expansions for sines and cosines, you may write\[

\Large

\tan x = \frac{\sin x}{\cos x} = \frac{x}{ \frac{\sum_{n=0}^\infty \frac{(-x^2)^n}{(2n)!}}{\sum_{n=0}^\infty \frac{(-x^2)^n}{(2n+1)!} } }

\] They're the usual expansions for sines and cosines. I just picked \(x\) out of the sine. And yes, the coefficient is \((-x^2)\). You see that the denominator is a ratio of two similar sums. The sum in the denominator of the denominator is a bit larger so the large denominator may be written as \(1+R_1\) for some negative \(R_1\). We will have\[

\tan x = \frac{x}{1+R_1}.

\] If you compute what \(R_1\) is, you get\[

R_1 = -x^2 \frac{ \sum_{n=1}^\infty \frac{(-1)^{n-1}(2n)x^{2n-2}}{(2n+1)!} }{ \sum_{n=1}^\infty \frac{(-1)^{n}x^{2n}}{(2n+1)!} } = \frac{-x^2}{ \frac{ \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n}}{(2n+1)!} }{ \sum_{n=0}^\infty \frac{(-1)^n (2n+2)x^{2n}}{(2n+3)!} } }

\] Again, the large denominator is a ratio of two similar sums. If you ignore all the terms in the sums except for the leading terms, the ratio is \(3\). The exact ratio is again a bit smaller and you may write it as \(3+R_2\) for a negative small correction \(R_2\), and so on. This process may be continued indefinitely and you may derive the following continued fraction for the tangent:\[

\tan(x) = \cfrac{x}{1 - \cfrac{x^2}{3 - \cfrac{x^2}{5 - \cfrac{x^2}{7 - {}\ddots}}}}.

\] Don't forget that \(x\) is in the natural units for angles, the radians. This is a pretty cute formula. You may verify it numerically. The odd numbers appearing in the formula may remind you of the expansion for the arctangent.

That was hard. The second part of the proof is hard, too. But it's funny. Lambert proved that such continued fractions can't be rational by contradiction. If they were rational, we could construct an infinite sequence of positive numbers \(p_i/q_i\) such that \(|p_i/q_i|\lt 1\) and the positive numerators \(p_i\) form a decreasing sequence. That's clearly impossible because a positive number can't decrease infinitely many times while staying positive.

Assume that a similar – perhaps more general – continued fraction is\[

x = \frac{p}{q} = \frac{a_1}{b_1+p_1}.

\] The form \(p/q\) follows from the assumption of rationality while \(a_1,b_1\) are the initial integers defining the continued fraction. From that, we may calculate the remainder in the denominator\[

p_1 = \frac{qa_1-pb_1}{p} = \frac{r}{p}.

\] Because we're assuming that \(x\) as well as all the intermediate results for the continued fractions are smaller than one, an assumption I forgot to mention which is right for the continued fraction computing \(\tan x\) for \(x=1\), we may conclude that \(|r|\lt |p|\). Further iterations allow us to construct rational numbers whose numerators are decreasing – an increasing sequence.

Allow me the following idiosyncratic but, in my opinion, illuminating explanation of the philosophy of the proof. As you compute the unlimited continued fraction more accurately, i.e. going deeper into the hierarchical iterations, you are actually getting "increasingly irrational" rational numbers i.e. numbers \(p/q\) with increasing values of \(p\). But if the final limit of the unlimited continued fraction, the exact result, were a rational number, you could prove that the hierarchy goes in the opposite way: adding additional iterations should make the number "increasingly rational". But it's not possible to switch to an "increasingly rational" approximation of a number between zero and one infinitely many times.

This presentation of mine is far from complete but you may find the complete story at various places of the web or try to complete it using your own mental tools.

Later, other people proved that \(\pi^2\) was irrational, too, and so on. Somewhat different proofs of the irrationality of \(\pi\) were found by other mathematicians.

I like to think that our present maths is much more sophisticated than the maths of the mid 18th century. It's sort of humbling and refreshing to see a proof constructed in 1761 that is hard to even follow – and there can't be a doubt that it had to be harder to construct it for the first time.

Between 1882 and 1885, people also proved that \(\pi\) and \(e\) were transcendental. Not only they can't be written as \(p/q\); they are not solutions to any algebraic (polynomial) equation with rational coefficients. It's fascinating to see how developed mathematics was at the end of the 19th century – at roughly the same time when political crackpots such as Karl Marx were still inventing communism and similar junk.

Very cute article and demo. Of course pi is (by definition) also a ratio - just not a ratio of integers.

ReplyDeleteYou have again written and published some to me previously not encountered - including some eyebrow-raising and shockingly pointed - information! Am not only but especially referring to the psychopathically insane dreadfulness of Pythagoras' "indiscretion". %-O

ReplyDeleteThanks, Peter. This story that I heard decades ago - see last paragraphs of

ReplyDeletehttps://en.wikipedia.org/wiki/Hippasus

could very well be just a legend. However, the message that dogmatism sucks and that using the same words for pet beliefs as for "rational reasoning" can't make the pet beliefs right, is surely right and eternally valid.

I would say that the sum in the denominator of the denominator is a bit smaller - no larger - in order to large denominator could be written as 1+R1. Right, or am I out?

ReplyDeleteDon't you really want to compute the continued fraction for arctan(x) and then use the relation

ReplyDeleteArctan(1) = pi/4

perform the proof?

Dear Lubos

ReplyDeleteThis

It's fascinating to see how developed mathematics was at the end of the 19th century – at roughly the same time when political crackpots such as Karl Marx were still inventing communism and similar junkmade my day :)

As for irrational numbers, there is a much more general (and not so hard) result.

Consider f(x) = 1/([x]-x) where [x] is integer part of x.

For example we have f(Pi) = 1/3-3.14159... etc.

By composing f you create a continued fraction of increasing depth and you note that the leading term is by definition always an integer.

You prove easily that for any x=p/q there exists a finite unique n such as x= [x] + f^n(x) (the proof uses the fact that you rightly noticed that a strictly decreasing suite of integers can't get below 0).

.

Now a natural extension of this result is to consider x irrational.

Then you prove that the suit f^n(x) converges to x what is slightly harder but establishes actually an isomorphism between irrational numbers and infinite continued fractions. The proof goes by constructing a strictly increasing suit Pn(x) such as Pn(x)x

Finally using the isomorphism and a formula with Pi inside (works f.ex for Arctan too) where n is obviously infinite thus defining an infinite continuous fraction you trivially conclude that Pi is irrational.

Same method applies of course to prove that any number that can be immediately expressed as an infinite continued fraction is irrational

So the proof is not so hard but I find it impressive that Lambert got there without knowing what is an isomorphism.

Of course the proof considers that irrational numbers exist but that is something that we knew about Sqrt(2) for 2 thousands years.

.

Btw there is an amusing corollary.

The continued fraction f^n(x) for given n is the closest rational approximation of the irrational number x with the smallest denominator and explains why somebody (Pythagoras ?) took for Pi 22/7 because there is no p/q with denominator <=7 that approaches Pi better.

Indeed [Pi] + f²(Pi) = 22/7.

This is also a general result - if you compute [e] + f^3(e) you'll obtain a rational number best approaching e with a denominator <= f^3(e)

Etc.

Ah yes I forgot, there is also another known amusing result.

ReplyDeleteThe continued infinite fraction 1/(1+1/(1+1/ ...) is (1+sqrt(5))/2 which is the golden mean, solution of x= 1 + 1/x.

Again an irrational number which is expressed by the simplest infinite continued fraction containing only 1s.

This leads to the question what is the irrational represented by an infinite continued fraction containing only Ps (1/(P+1/(P+1/ ...).

Etc.