Welcome to another Mathologer video. Mathematical seatbelts on? Airbags working?

Crash helmets comfy? Okay, let's go. In recent videos we proved that e and pi

are irrational, that they cannot be written as ratios of integers.

We now want to go much much further. We'll prove that e and pi are transcendental numbers.

Now I've already done a couple of videos on transcendental numbers.

In one of them I showed you an ingenious but also very accessible proof by Georg

Cantor that the vast majority of real numbers are in fact transcendental numbers.

If you throw a (mathematical) dart at the real number line you're almost certain to hit

a transcendental number. On the other hand, transcendence proofs for individual numbers

like e and pi are notoriously weird and unfamiliar to even most

professional mathematicians. The proofs are usually of "the what the hell" variety,

coming with very little explanation of how on Earth anybody could ever come up

with them. So, our aim for this video was to take the best transcendence proofs

for e and pi, to rework them from scratch, to complement them and to animate them

into Mathologerised proofs that are as visual, simple, motivated and accessible

as possible. Well you'll be the judge. Having said that, this video is one of

our most ambitious and it's definitely not suitable for everybody. You should be

okay if you're a longtime follower of Mathologer and/or if you know a little

bit of calculus. But for those still questioning whether

0.999.. .= 1, it may be better to postpone watching this video.

Okay, great, you've chosen to follow me on the path to transcendence. The YouTube

doors are locked and there's no turning back. To mark the

stages of this mathematical quest, we'll break up our journey into seven levels

of enlightenment. The spiritual guide announcing the levels will be my friend

the Buddhabrot fractal who I already introduced to you in the Mandelbrot

video. Okay here we go, level one. Level one is a warm-up.

To motivate our transcendence proofs, let's have another close look at the simplest

and most beautiful proof that e is irrational. Yep,

proving e is irrational is the warm up! Pretty crazy, right? To begin we have to

say exactly what e is and for this proof we use the stunning infinite sum here.

Remember, the exclamation marks mean factorial. For example, 3! means

1 times 2 times 3 which equals 6, then 4! is that times 4 so 24, and 5

factorial equals 120, and so on, with these numbers growing very quickly. One

simple property of factorials that we'll use over and over is that every

factorial is a multiple of all the smaller factorials. So, for example, 5!

is a multiple of the smaller 3! because 5! is equal

to 3! times 4 times 5. Now back to our infinite sum with its rapidly

shrinking terms. We're going to prove that e is irrational. So we start by

assuming the opposite that e is a fraction a/b, with a and b positive

integers. We then have to show that this leads to a contradiction, to some obvious

nonsense which then implies that our assumption that e is a fraction was wrong.

Ok, so assuming that e is a fraction that means a/b is equal to our infinite

sum, right? We now use the denominator b to decide a cut-off point, to think of

the infinite sum as the first bit up to b, plus the infinite tail.

Okay, now there are two key observations. First, because the yellow denominator b!

is a multiple of all the previous factorials, we can write this

finite sum up there as a single fraction with denominator b!. The second

observation is that the infinite tail is a very tiny number. Well let's see how

tiny it is. To start with, all these terms have a common factor b! in the

denominator. So let's take out that common factor here.

There we go. Out there, out another time and out a third time. Pull all of them in front and..., well the remaining bit

in the brackets is messy and pretty scary.

However, this mess is obviously positive, right, and it's actually easy to show

that it's also less than a geometric series with sum 1/b. For some extra

enlightenment credit for you, fill in geometric details in the comments. And so

in total, from our assumption that e is equal to the fraction a/b we

conclude that a/b can be written like this. But that's simply not possible.

Now to clearly see the contradiction we'll just multiply both sides by b!

and fire up the auto algebra. So autoalgebra is firing up. There you can see

all sorts of stuff canceling out, ... integer left over there, ... there's something else

cancelling out and there's another integer there. But then the equation

reads that an integer equals an integer plus a number between 0 and 1 which is

of course impossible. And there that's our contradiction.

What a fantastic proof, don't you think? Now, the key to this proof and what's really

really important for our next level of enlightenment is the fact that for all

those infinitely many values of b we can write e like this, where the bit in red

is a very small number. Okay, another way of expressing this is

to say that the fractions on the left approximate e spectacularly well,

considering the size of their denominators. And to give all this a

paradoxical flavor, we can also say that in some sense we were able to prove that

e is irrational because e can be approximated by fractions better than

fractions can be approximated by fractions. How weird is that? Starting to

feel enlightened? Well, for the next step we begin by reshuffling our

irrational "not equality", that guy there. Okay, here we go, really easy stuff.

Okay, now just look at this right. So what does it say? Well e being irrational is

the same as e not being the solution of a non-trivial linear equation with integer

coefficients, that one there. And what comes after linear equations? Quadratics!

So our next challenge is to prove that e is not the solution of a non-trivial

quadratic equation with integer coefficients.To get a sense of what this

means notice that root 2 is irrational but it

IS a solution of a simple quadratic equation, that one there. So, in a

sense, we're proving that e is even more irrational than root 2. How can we show

that this quadratic equations is impossible? Well, following our strategy at level 1 we

begin by assuming that it IS possible.

Now let's reshuffle the equation a little. Autopilot again.

Cancel stuff, all right, as before we can think of e as really really close to a

fraction. Remember the integer m indicates the beginning of the tiny tail.

Also remember that we get to choose m in the larger m the smaller the small

numerator. Now, as a lot of you will know 1/e can be written as an infinite

plus/minus counterpart to our infinite sum for e, that one there. And an

argument similar to the one we used on e shows that 1/e can also be

written in the form fraction plus tiny bit. Now, substituting these expressions

for e and 1/e into the equation we set the stage for another contradiction.

We then let the auto-algebra go to work. Again let's just do this. Okay so here we're

multiplying through with m! gives this, and just sorting out things ...

an integer over there, and another integer over there. Well this is

definitely looking good since, as in our first a proof, it's very easy to make the

size of the small stuff less than 1 by cranking up the m. But this time we have

a little problem and you can probably see what it is right? It seems unlikely

but what if all the small stuff sums to 0? Well then there'd be no contradiction.

Now, in fact, we can ensure the sum of the small stuff to be nonzero. It's a bit

fiddly but most of you who've made it this far will be able to handle it. And

that's our second challenge for you which, as usual, you can respond to in the

comments. And for those wanting to save their mental energy, irrational number

expert Marty has written (Marty protesting in the background)... irrational numbers expert Marty has written up a

nice detailed proof which you can find linked in the video description.

Say hello Marty. Marty: I am not an expert. Burkard: He's an expert. Ok so not only is e irrational, it is a quadratic

irrational. And what comes after quadratics? Everything. Saying e is

transcendental means that e is not the solution of any non-trivial polynomial

equation with integer coefficients. This was first proved in 1873 by the great

French mathematician Charles Hermite. Whoa that hair:) So the first grand goal for this

video is to give a proof of Hermite's theorem. For this third level of enlightenment

we'll just summarize our plan of attack. Just to simplify the details we'll focus

on cubic equations. So to set up the familiar proof by contradiction

let's assume that e is a solution of a cubic equation with

integer coefficients. We want to show that this is impossible. Well actually it

is possible: just make all the coefficients zero, right? But of

course, that's cheating. To foil the cheaters we'll demand the constant term a

is not zero. Can we demand that? Yes we can, though it may take a moment to

realize why and we'll leave that as a detail for you to ponder. Now we'll get

on with proving that e cannot solve this cubic. As you follow along, you

should be able to see that what we're doing will also work for nonzero

polynomials of any degree. So, though it may look like cheating, we're not really

oversimplifying by restricting ourselves so the cubic case. I've linked to another

article by transcendence specialist Marty (Marty protesting in the background) that gives the full gory details.

Transcendence specialist, yes :) Now how are are we going to prove that this equations

is impossible? Motivated by our first irrationality proof, we'll have a close

look at e and its powers up to e cubed and we'll show that they are

simultaneously very close to fractions. What's the simultaneousness? Well the

point is that all the expressions have the same denominator big A. Then

substituting these expressions into our cubic equation gives this. Multiplying

through by big A and firing up the auto-algebra will zap the denominators as

usual. Okay here we go.

(Music)

The number in the black box will definitely be an integer. What about the gray box? The numbers there may be

positive or negative but we can definitely arrange for its size to be

less than 1. And, except for one minor but very very annoying hiccup, that will

be our contradiction. And I'm sure many of you have already guessed what that

hiccup will be. How are we going to capture e and its powers? Obviously our

first attempt would be to keep going with our infinite sum identity, right?

Unfortunately, there is no obvious way to make this work. But there are many other

expressions for e that could potentially help us with our master plan. These

identities will definitely look familiar to Mathologer regulars and the last

infinite fraction is actually a natural place to hunt for a transcendence proof.

However, our plan will employ a different super famous piece of integral magic.

Many of you will be familiar with this amazing identity under the title of the

gamma function. I usually set proving this identity as an integration by parts

exercise in a first year calculus course. Anyway, let's just take this

formula as a given and have a close look at what it tells us. So there is e and

our friend n! Overall what this identity says is that n! is the

area under the graph of the function x^n times e^(-x)

all the way from x=0 to infinity. Here's what this graph

looks like in a special case of n=2. So the orange area is equal

to 2! which of course also happens to be 2. Here's the curve for

n=3. So the area here is 3! which is equal to 6. Then n=4

, etc. Each curve consists of a single wave which peters out to 0. The

integrals relate e to integers but how exactly does that help? Let's

first make things more complicated, wait for it, by slotting in a

polynomial p(x) with integer coefficients. What a mess.

But after expanding out we'll just have a sum of basic gamma integrals. So our

polynomial integral will be an integer. In fact, since all the x exponents will

be at least n the integral will be n! times integer. And now we're

getting there. Remember, we are after those coefficients big A, B, C and D. In a

moment we'll make a very clever choice of the polynomial p(x) and then our

common denominator a will be 1/n! times the integral. From what

we've noted above it will definitely be an integer, right? Now for the numerators.

Let's take a look at e cubed. Okay, so we want to write e cubed

as a fraction with denominator A. Hmm well a no-brainer way of making this

happen is to just write e cubed times A divided by A that's e cubed. Big help hmm?

Surprisingly, it is? We just have to figure out how to write the numerator e cubed

times A as almost an integer, an integer plus a tiny bit. Let's have a close look

at this numerator. Let's plug in our A ok. That e cubed bit in front slides right into

the integral, like that. But this integral can now be split into two bits, the part

from x=0 to 3 and the part beyond 3 to infinity. There we go. And now,

as long as we choose p(x) carefully that first bit will be small and the

second bit will be an integer. Neat, hmm? So why should the first bit be small?

Well if n is huge then we have that super huge n! in the

denominator and sure x^n in the integral will also be semi-super

huge but the integral is over a finite chunk from 0 to 3, of course there's some

calculations to do to show this all works. But we've seen

the factorial-beats-everything game a number of times in previous videos. Well

we've actually cheated a little bit. That polynomial p(x) will also contain x^n

bits, but the factorial will beat them as well, in exactly the same way.

Okay, great. Now, what about the second bit? We need to convince ourselves that this

is an integer. Here we can perform a standard calculus trick. Note that the

integral starts at x=3. This means that if we shift everything to the

left 3 units, then we can think of this as a new integral going from 0 to

infinity and this shifting amounts to replacing x by x+3. Here we go. Ok

that's looking good since we know integrating polynomials times e^(-x)

results in integers, and uh-oh we've lost our factor x^n,

the bit that guarantees us the n! The x^n has

been replaced by (x+3)^n which just won't do the job. Bummer!

What do we do? Well, I know what to do, ok. Now our polynomial comes to the

rescue. Up to this stage p(x) could have been anything but we can

choose it to be whatever we like. In particular, if we make sure that p(x)

has a factor (x-3)^n, then the shifting by 3 will get us a

very special p(x+3). Ok, here we go. There is a factor of x^n.

This means we're saved. Why? Well let's see. Let's just plug it in. There we go,

x^n goes to the front, all this stuff is another polynomial and we've

seen this thing before. This sort of expression on the left is an integer and

so D is an integer. And that's pretty much it. To recap, we assumed that e was

the solution of a cubic equation. We then wrote e cubed as an almost fraction. The

key was to make sure that the defining polynomial p(x) has a factor (x-3)^n.

In the same way, to get our expressions for e squared and e we

just may have to make sure that p(x) has factors (x-2)^n

and (x-1)^n. And so it seems we know exactly what to do.

We make p(x)=(x-1)^n times (x-2)^n

times (x-3)^n. Then we get an integer plus a tiny thing

equals 0. And it really looks like we've got our contradiction, yay :) (Marty and Michael cheering on).

Except it's not a contradiction, yet. After all, 0 is an integer and 0 is tiny,

so maybe the equation below amounts to

nothing more than 0 + 0 = 0. This

would not be much of a contradiction and that possibility is our very, very

annoying hiccup. How can we fix this? Although it's pretty annoying, it also

turns out to be pretty easy to fix. It turns out that by adjusting our

polynomial p slightly we can ensure that the integer bit in the left black box is

definitely not 0, and then we'll definitely have our contradiction. So how

can we make sure that the black box is not 0. Well, given that big A is the

common denominator of our fractions and big B, C and D are the numerators, the sum

naturally splits into this blue bit and this green bit, right? It now turns out

that to save our proof we simply have to raise the powers in p(x) from n to n+1

and then to choose the new exponent n+1 to be a large prime

number. Prime? Yes prime! The consequences of this

clever choice are that the green number will be divisible by n+1

and the blue number won't be divisible by this prime number and so the blue

will definitely not be a zero and the green cannot possibly cancel out the

blue. Now for the final tidying up, let's look at how the pink modification on the

top leads to the green and blue conclusions below. There's a fair amount

of fiddling but it's actually not too hard. We'll give you the basic idea here

and afterwards show you an animation and you can also check out transcendence

master Marty's detailed write-up linked in from the comments. Let's have another

look at the integral at the core of the numbers big A, B, C and D, that monster

there. Okay, the fact that the four numbers big A, B, C, D are integers hinges

on cancelling the 1/n! in front and this comes from the factor

x^n for big A and from the other polynomial factors for Big B, C

and D. Raising the exponents in our polynomial to n+1 has the effect of

giving big B, C and D an extra factor of n+1 which then translates into the

green bit being divisible by n+1. On the other hand, because we haven't raised

the exponent of the blue x^n, big A doesn't get this extra n+1

factor. Then cleverly choosing n+1 to be a big prime ensures that n+1

cannot sneak in some other way to be a factor of the blue number. Note that this

part of the proof also takes care of a related issue that until now we've let

hiccup under the radar. Division by 0. Blue not being 0 also guarantees that

our common denominator big A cannot be equal to 0, which is a good thing :)

Oh, and just in case you are now getting worried about little blue a, remember

that a long, long time ago at the beginning of our master plan, level 3. We

made sure that we can require little a to be non-zero and so we've definitely

arranged things so that the black box cannot sum to 0 and our proof, finally,

completely hiccup free is done. Oh my God :)

Now, before we proceed to our next level of enlightenment and pi I thought it

would be nice to have another close look at how everything we talked about in

this proof applies to one specific example, the equation down there. What I

want to show you is a fully animated version of our proof that this equation

is false. Of course for specific equations like this one you can very

easily check that they are not true just using a calculator, but the point is

to illustrate what works in general for all possible equations like this. Anyway

grab some popcorn sit back and enjoy the animation and the funky Mathologer

music, you've earned a rest.

(Music)

You should all congratulate yourselves, you've reached a truly high level of mathematical enlightenment. But not the

highest :) We've still got pi enlightenment waiting

for you. The thing with these transcendence proofs is each number is

pretty different usually needing different tricks and hmm are

we're really going to have to start again at square one?

Thankfully no. In 1882 the German mathematician Ferdinand von Lindemann

proved that pi is transcendental and the key was a simple but ingenious trick.

Lindemann figured out a way to have the proof for pi piggyback on Hermite's proof

for e. Now we're guessing you're pretty tired after all those levels of

enlightenment, so we'll just set up Lindemann's proof and leave most of the

details for you to ponder or to read in Marty's article. What was Lindemann's

trick? Well to prove pi is transcendental Lindemann set up the usual proof by

contradiction by assuming that pi is the solution to some polynomial equation. Now

how to relate pi to e? Well, let me see now. I think I vaguely recall we had an

equation in a previous video, once or twice, or maybe a zillion times. Yep

here's Euler's identity again and now all is clear, right? Nope :) Sure we have a

connection between e and pi, but so what? Lindermann's ingenious step was to show

that the supposed polynomial equation for pi can be combined with Euler's

identity to give a whole new equation involving powers of e. The blue exponents

are specially related to each other in a way we'll explain later. Lindermann was

then able to use Hermite's argument to prove this equation was impossible and

that gave the contradiction that showed that pi couldn't be the solution of a

polynomial equation. And now on to work. To start with we'll show how to get from

Euler's identity and the q equation for pi to Lindermann's a equation. That's

actually pretty easy. The first step is to note that if pi is the solution of

some non-trivial polynomial equation with integer coefficients then so is the

i pi in Euler's identity. That's yet another challenge for you, though a

pretty easy one if you're a little familiar with complex polynomials. Anyway

so big Q is a non-trivial polynomial with integer coefficients. As we did with

e let's just pretend that big Q is a cubic polynomial. Then apart from

i pi the polynomial big Q has two more zeros say s and t. Okay now comes Lindermann's

trick. Let's write Euler's identity like this. Just reshuffle things a bit, and

multiply by two factors that involve those extra zeros s and t. Okay, expanding

this new equation gives 1 plus a sum of powers of e equals zero. The gray

squares are the various sums of the three zeros i pi and s and t, Okay?

Whatever. A few of these exponents may be zero and of course e to the 0 is 1.

So we can combine all these 1s at the front giving something like this, ok? Then

the a at the front is a positive integer and the remaining exponents are nonzero

complex numbers. This is Linderman's magic equation and,

believe it or not, we're well on our way to proving pi is transcendental.

Well, maybe you can believe that we're well on our way to proving pi is transcendental

since Lindermann's equation looks very similar to the starting equation for our

e proof that one here. What's different? Well little b, c and d are all equal to 1

and in Lindemann's equation which if anything should make things easier, right?

But, of course, the nice exponents 1, 2 and 3 have been replaced by the nasty complex

numbers beta, gamma and delta. That of course could be a much bigger deal, right?

Hmm. But, amazingly, despite all those scary complex demons floating around we

can pretty much mimic our e proof with the exponents beta,

gamma and delta playing the roles of 1, 2 and 3 and we can prove that

Linderman's equation is also impossible which, as I've shown you, proves that pi

is transcendental. So we're definitely getting there, right? So to begin, as with

the e proof, we write those powers of e like this: big A, B, C, D and the small numbers

are defined with integrals pretty much exactly as before. For example, for e to

the delta the expressions for big D, small and big A should look very

familiar. And so should the polynomial p inside those integrals. As in the

e proof the exponent n+1 will be a big prime number. However, apart from beta,

gamma and delta, there are two other obvious differences. First, because of the

complex numbers floating around we write z everywhere instead of x.

That may make the integrals puzzling but don't worry about their

precise meaning for now. Second, the polynomial p has a final clunky factor.

The number k in that factor is the constant term in the polynomial big Q for i pi.

Then something amazing happens. The polynomial p is apparently full of

complex demons, right? However because of the symmetric way p is defined, once we

expand, it turns out that p has ordinary everyday integer coefficients. Nothing

really nasty complex anywhere to be seen. What this means is that we can think of

the integral for big A just as we did in the e proof. And, just as in the e

proof it follows that big A is an integer and is not divisible by the prime n+1.

That's a very promising start, don't you think? Yeah, very promising.

Let's now fast-forward to the critical point of the contradiction argument

where we had to show this. The blue bit is now under control, right? We already

know that we can ensure that big A is not divisible by the big prime n+1

and the little a from Lindermann's equation is a positive integer. So if the

prime n+1 is chosen bigger than little a, then definitely the blue bit

will not be divisible by n+1. What about the green bit? In Lindermann's

equation the little b, c and d are all equal to 1, but now we have to finally

face some very real complex demons. Unlike our blue A, the numbers B, C and D

in the green are usually not even real numbers, let alone integers.This is

hardly surprising, given that each of these numbers is obtained by integrating

from some complex number to infinity. For us this means integrating along an

infinite red path in the complex plane. This looks very, very bad. However, the

symmetry of the polynomial p again comes to the rescue.

Together with an infamous complex demon slayer known as Cauchy's theorem. And

this is the super, super, super, super amazing bit. Though B and C and D will be

nasty complex numbers, the whole green sum B+C+D is an integer, and

then, just as in the e proof the sum will be divisible by n+1. And then, just

as in the proof for e we have our contradiction. Truly great stuff. Of

course, as advertised this was just a quick flyover of the pi proof and

there's still a few :) details to fill in. But, what we've given you here is the

honest structure of the proof and if you've made it this far you should now

enjoy going through Marty's detailed and really nice write-up. And that's about it.

You just experienced the most ambitious and most "insaniest" Mathologer

video ever. All our colleagues told us that it was crazy to try to Mathologerise these

transcendence proofs. Maybe they were right. We hope not

since all up this video took me and Marty a couple hundred hours to make. But

you can decide. Hopefully it was worth it and now many, many people out there can really appreciate what it means for e

and pi to be transcendental. Oh and I really should mention that the

transcendence proofs for this video are based on proofs by the great David

Hilbert and refinements of Hilbert's proof by Steinberg and Redheffer. I've

linked to the papers in the comments. And that's it. Except, of course, there is

still level 7. Quadratic reciprocity? No. Riemann

hypothesis? Nope. No? No. So what's next? Definitely not

transcendence again, no. Irrationality? No, no. Not ever again? Not

ever again. We've done it. I just noticed. You know at the very

beginning of Mathologer we did a video on 0.999.... = 1.

Mm-hmm It's now got ten thousand comments. I

think it's time to do a follow up. You're gonna try and convince them, it's really

equal to 1. Yep. Convince the unconvincable. Yep I'm gonna do it. It's gonna be harder

than proving that e is transcendental but I'm gonna try it again. That's a challenge.