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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
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, fröhliche Weihnachten and I'll see you in
the new year :)