# The PROOF: e and pi are transcendental

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.