Welcome to the last Mathologer video of the year. Almost everybody who watches

these videos knows that root 2, pi and e are irrational numbers, that these

numbers cannot be written as ratios of integers. Also a lot of you will be

familiar with proofs of these facts in the case of root 2 and e. On the other

hand, my guess is that very few of you will be able to say the same about

everybody's favorite number pi. In fact even many professional

mathematicians have not seen a proof that pi's irrational and they're really

just taking someone else's word for it in this respect. The main reason for this

sorry state of affairs is that all the known proofs for the irrationality of pi

are very technical and for the most part very unintuitive. My mission today is to

make one of these proofs accessible to you by maxing out what can be achieved

with the sort of animated algebra that I've been specializing in lately. The

proof I have in mind was published in 1761 by the Swiss polymath Johann

Lambert and that was actually the very first proof of the irrationality of Pi. Why

Lambert's prove and not any of the others? Well, unlike all the other proofs

it is quite easy to motivate Lambert's approach and the maths in it was really

calling out to me to be animated. I just couldn't resist. Anyway let's get

started. As a little warm-up let me show you what is probably the simplest

irrationality proof, a proof of the irrationality of log 3. If log 3 was

actually a fraction u/v that would be the same as saying that if I take 10

to the power of u/v then I get 3. But then raising both sides to the

v's power we get this equation here. But that is completely impossible because

the left side is odd and the right side is even. And so because our

assumption that log 3 is rational leads to this false

statement. And since nothing false can logically follow from anything true we

conclude that our assumption has to be false itself. So log 3 is not a

fraction. That means it is irrational. Q.E.D. Neat, huh? It's neat. Most of you like

puzzles and so here are some one-glance puzzles for you: which of these logs is

irrational? Leave your answers in the comments, I'll grade them later ... just

kidding! I really really, really hate grading! (Marty) I like grading. (Burkard) You are very strange, Marty is very

strange. So hopefully that was all fun and really easy but now on to the big

challenge, Lambert's proof that pi is irrational. Here's a three step summary

of the main ingredients. In step 1 Lambert proves a new miraculous formula

for the circle function tangent. So tan x is equal to x over 1 minus x squared

over 3 minus x squared over 5, and so on. All the odd numbers in there, ok.

In step 2 Lambert shows that if you plug in any rational number for x other than

0, you always get an irrational number. That's pretty amazing, isn't it? So tan of

1/2, tan of 3/4, all these numbers are irrational. But now the third and final

step is really, really easy. You all know from school that tan of pi/4 that's tan

or 45 degrees is equal to 1. But, of course, the number 1 is not irrational

and this implies that pi/4 cannot be written as a fraction. In

other words, pi/4 is irrational. But if pi/4 is irrational then pi itself

is 4 times an irrational number which is also an irrational number. So pi is

irrational. Now that's a really beautiful argument, isn't it? But, as usual, the devil

and the main part of the beauty of the proof is in the details. And, of course,

all the hardcore fans among you will ask for my head if I don't fill in those

details, right? (Burkard commenting on Marty giggling) He's just laughing! Well I'll do that in a moment, promise, but

first let me put Lambert's proof and this formula in its historical context

which will give you an idea of how Lambert may actually have discovered his

proof. Lambert was a contemporary of the mathematical superstar Leonhard Euler. In

fact, when Lambert came up with his proof Lambert and Euler were colleagues at the

Prussian Academy of Sciences in Berlin. At that time infinite nested fractions

like the one discovered by Lambert were are very much an in thing among

mathematicians. These continued fractions as they are usually called were made

popular by Euler who had succeeded in proving all sorts of amazing results

using them. For example, he showed that any number that can be written as an

especially simple infinite fraction in which all the numerators are equal to 1

and all the denominators are positive integers, any number like this is

irrational. For example this infinite fraction here

with the simple periodic pattern of denominators 12121212

must be irrational. In fact, this fraction

turns out to be root 3 which is not too difficult to show and which you can

attempt in the comments. Should they attempt it? (Marty) I thinks so. (Mathologer) I think so, too, it's a nice one.

So Euler's result leads to a very pretty proof that root 3 is irrational. Even better, all

Euler used the same irrationality of infinite fractions to give a first proof

that the number e is irrational and he did this by demonstrating that e can be

written as this amazing infinite fraction here with the numbers in the

denominators continuing in the infinite regular pattern 2 1 1 4 1 1 6 1 1 8 1 1

10 1 1, and so on. I already talked about all the stuff in another video so

check that one out too, that's an order! (Marty) Jawohl.

Okay, anyway, Lambert was undoubtedly aware of Euler's work when he came up

with a great idea to turn tan X into an infinite fraction. He starts out with

the basic fact that tan x is equal to sin x divided by cos x. Lambert then

uses another well-known fact that the functions sin x and cos x can be written

as these pretty infinite sums. Now I've already inflicted these sums

upon you many many times in other videos so here I just run with them, hope you

are okay with. (Marty) What if they are not okay with it? (Mathologer) Let's see what happens. Anyway this means that tan x

is equal to this beautiful monster here. Okay now we have to transform this

monster ratio into Lambert's infinite fraction for tan x. In Lambert's manuscript

this transformation is an absolute nightmare. So ready for the nightmare?

Well hopefully the animations, which only took me 10 million hours to create, JUST

FOR YOU, will make it all a lot easier to follow. In any case, now is a really,

really good time to fasten your mathematical seat belts, promise.

First, let's write all those powers of x as products like this, very pretty isn't

it? Pull out a common factor in the numerators. Okay, now we're dealing with a

product of x and a fraction. At this point I have to remind you of a little

fraction trick that we'll be using over and over to build the infinite fraction.

The product over there is the same as the quotient of x and the reciprocal of

the fraction. Just think about it for a second.

You're ok with that, great! Now, the quotient at the bottom we can also write

like this big nested fraction here. All good. Of course this is simple junior

school algebra but we'll be using this trick over and over with the green and

yellow expressions both monster sums so I thought I better make absolutely sure

that we're on the same page. Now let's unleash the reciprocal trick for the first time.

Just watch it. Here we've got our product and abracadabra this real magic as far

as I'm concerned. So we're off to a good start,

great. Now if this 1 here was not there, we could move the common factor

x squared out in front of the fraction at the bottom and repeat the reciprocal

trick, right? But since the 1 is there we'll first

get rid of it using another fraction trick I call it the add-and-subtract-the

same-stuff-from-the-numerator trick. (Marty) Catchy title :) Okay so let's just launch into it.

For the moment don't worry too much about following every detail, okay. We add and

subtract the green denominator to and from the yellow numerator. Whoa, looks

scary, but really isn't. (Marty) Yes, it is. (Mathologer) Isn't! Just think about it, we've just added and subtracted

the same thing to and from the numerator that does not change the value of the

numerator. Now swap the top two sums and do some housekeeping. First in the pink

box pull the minus into the brackets. Okay, now 1-1 in the pink is

zero, so we're about to get rid of the blue 1, as planned.

Also, for later, note that we get rid of the blue 1 using the other 1 here

which is the constant of the sum in the denominator, the green sum. Anyway we

just got rid of the 1 we wanted to get rid of. It's not clear how this will help

but just wait for it. Now some quick routine housekeeping on autopilot.

Okay that's minus two thirds put it together, okay. Do it again and again

again. At this point we spot a pattern and can confidently say, etc. Now the blue

boxes contain the same stuff and so the red fraction is equal to blue / blue

plus pink / blue but blue / blue is 1 and so we get 1 plus pink / blue. Now

again some easy housekeeping in the numerator. Pull out

the minus, there you go, pull out the common factor there, cancel

the common factors that I've highlighted here and just, generally, clean up a bit.

Very nice, now we are ready to unleash the reciprocal trick for the second time.

just watch it - magic again. Magic, magic magic, magic. Yes, now rinse and repeat

well perhaps there was a little too much for you to handle the next rinse on your

own so lets us go through the whole thing one more time real quick as before

we'll get rid of the constant in the numerator of the red fraction using the

constant in the denominator now the blue is equal to three times the pink and so

this time we add and subtract three times the denominator to and from the

numerator. Okay, now just follow our nose again, there we go. Okay and this is the

point where the numbers at the front get cancelled again. Alright now just some

housekeeping again and we're back to this picture here. Our red fraction at

the bottom is now three times blue / blue + pink / blue but three times blue

/ blue is three and so we get three plus that. Clean up again really quick. This

time I promise quick. quick, quick, quick, good. And we are ready for the reciprocal

trick again. So let's watch it again: magic. magic :)

And what's next? Well we need to kill the 1/3 with the 1 / 3 times 5 and since

five times the pink is equal to the blue after the next reciprocal trick we get

the five out in front, and so on, pretty neat, huh?

Phew, that was a hundred slides since I told you to fasten your seat belts :) Anyway all these

manipulations are definitely quite complicated but in the end also very

natural and something that a lot of you would have ended up

doing if I had challenged you with the input output of this manipulation.

Especially those of you who know how to divide power series should have

experienced an eerie feeling of deja vu while you were watching me. Maybe

one of you can explain the connection in the comments. Anyway all this suggests

to Lambert and to us that somehow this equality should hold. Well we're not

going to get anywhere with making sense of this formula unless I finally tell

you how mathematicians actually assign a value to the infinite fraction. To

motivate our rule for doing this let's chop off the fraction at the first

denominator. Okay, so we end up with a finite expression which evaluates to the

function x. Let's graph both tan x and x, here we go. As you can see, the blue

function x approximates the orange tan x very well close to 0. Let's chop the

infinite fraction at the second denominator and plot the new function.

Amazing right? The new partial fraction approximates tan

spectacularly well. Let's keep going: third partial fraction, fourth partial

fraction, fifths, and so on. Overall these partial fractions converge to tan x and it

is in this sense that tan x really appears to be equal to the infinite fraction.

However here's a little bit of bad news. Although Lambert's calculation starts

with a valid formula for tan x and ends up with this infinite fraction and despite

what our graphing suggests, we cannot be absolutely sure that the equal sign

between tan x and the infinite fraction is really justified. What's the problem? Well

let's just quickly switch back to the point where we said "and so on". Here you

see at any stage of the calculation there is one of these infinite monster

fractions at the bottom. When we say, etc. we simply discard these monsters and

effectively hope for the best. Well so what's really still needed at this point

is a bullet proof proof that we really do have equality between

tan x and the infinite fraction in the sense that I just described. This part of

the proof involves looking very carefully at all those chopped off

partial fractions. Lambert actually does that. The calculations are again quite

laborious but in the end pretty much as straightforward as the ones that spit

out the infinite fraction in the first place. I'll skip those calculations here

but link to an article that has the full details and if you ask me really

nicely I can also put a supplementary video on Mathologer 2 where I nut out

some more details. Anyway for our purposes I hereby declare Lambert's tan

formula under control. Okay, hands up if you're still here!

Okay, fantastic, you've all earned the Mathologer algebra

merit badge but there's a little bit more work to be done and another real

highlight to look forward to: an infinite descent into mathematical hell. Okay, so

let's plug in a nonzero fraction u/v for x and let's see how Lambert

concludes that the tangent of this fraction is irrational. That then

finishes the proof remember the infinite fractions like the

one for root 3 or for e I that I showed you earlier were of a special simple type,

all the numerators were 1 and all the denominators began with positive

integers and pluses and being of this special type Euler had shown that these

infinite fractions always represent irrational numbers. Lambert's infinite

fractions are obviously not so simple so Euler's result does not automatically

apply to prove the irrationality of Lambert's fractions. This may seem like

just the technicality but in fact there are infinite fractions that amount to

rational numbers. For example, as a puzzle for you, you can try to show that the

infinite fraction down there equals 1. As a harder puzzle try to come up with

infinitely many infinite fractions that are equal to 1.

Anyway to get somewhere with the irrationality proof, let's see how close

we can massage Lambert's infinite fraction to Euler's special form. Getting

rid of the fractions in the numerators is easy. We'll get rid of them one at a

time. This animation of the transformation should be fairly

self-explanatory.

Okay now let's show why, as long as both u and v are positive integers the

infinite fraction is always equal to an irrational number. That will then finish

Lambert's proof. Now, apart from the very first numerator u, all the other

numerators are equal to u squared, all right. On the other hand, the denominators

get larger and larger. This means that from some point on the denominators are

all going to be a lot larger than the numerators. Now there's a general theorem

which says that any infinite fraction of this form is irrational.

However, waving that magic wand now to finish off would be a bit unsatisfying

don't you think. Wouldn't be good, right? So let's see if we can convince

ourselves of this directly. Just so that we can keep things within a small window

let's suppose that 5 times v is at least 2 larger than u squared. So, from

this point on the denominators are all at least 2 larger than the numerators.

Okay, alright now it's a bit fiddly but with this extra plus 2 wiggle room in

the denominators it's now not too hard to show that just like u squared over

five v is a positive number less than 1, this whole infinite fraction here which

I've highlighted in blue is also equal to a positive number that is smaller

than 1. And the same is true for this infinite fraction here and for the

following ones. What we'll show is that this positive and less than one property

of all these nested infinite fractions implies that the first blue infinite

fraction is an irrational number which then implies that the number we are

really interested in namely this one here is irrational - Why? Because

three times an integer minus an irrational number is irrational, a positive integer

divided by an irrational number is irrational, and so on, Okay, so why is the

blue infinite fraction irrational. Well let's assume that it was actually equal

to the fraction B over A. Remember the infinite fraction is special in that it

is positive and less than 1. But that means that B is less than A, right? Let's

note this down somewhere. Now let's consider the size of the next infinite

fraction. To do this we just solve it. Okay here we go, solve, solve, solve, magic magic,

magic and all the bits in the numerator are integers and this means that the

whole numerator is an integer, let's call it C. But the yellow number was also

special in that it is positive and less than one and that means that C is less

than B. Let's also note that down and now we repeat and solve for the next

infinite fraction which will generate a positive integer D less than C, there we

go. And now repeating this infinitely often gives us an infinite sequence of

ever-decreasing positive integers and now for the

punchline: There is no such thing as an infinite sequence of ever-decreasing

positive integers. Why? Well suppose A was 1,000. Then we could count down for a

while but not forever. Eventually we'd have to go below zero

and the same is true no matter how large an A we start with. We can only descend

finitely many steps, each ending on positive integers, until we hit zero. And

so, because our assumption that the blue infinite fraction is rational leads to

this impossible infinite descent of positive integers, we conclude that our

assumption has to be false itself that is the blue stuff cannot be a fraction,

and so must be irrational. What an absolutely beautiful argument to

finish of things, don't you think? Lovely. Anyway, this was the last piece of

our puzzle so I hope you enjoyed my take on Lambert's beautiful proof. As usual

let me know in the comments what did and did not work for you. If you made it all

the way to here you earned yourself a special Mathologer seal of approval.

Merry Christmas, frÃ¶hliche Weihnachten and I'll see you in

the new year :)