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.
