Welcome to the last Mathologer video of
the year. Almost everybody who watches these videos knows that root 2, pi and e
are irrational numbers, that these numbers cannot be written as ratios of
integers. Also a lot of you will be familiar with proofs of these facts in
the case of root 2 and e. On the other hand, my guess is that very few of you
will be able to say the same about everybody's favorite number pi.
In fact even many professional mathematicians have not seen a proof
that pi's irrational and they're really just taking someone else's word for it
in this respect. The main reason for this sorry state of affairs is that all the
known proofs for the irrationality of pi are very technical and for the most part
very unintuitive. My mission today is to make one of these proofs accessible to
you by maxing out what can be achieved with the sort of animated algebra that
I've been specializing in lately. The proof I have in mind was published in
1761 by the Swiss polymath Johann Lambert and that was actually the very
first proof of the irrationality of Pi. Why Lambert's prove and not any of the
others? Well, unlike all the other proofs it is quite easy to motivate Lambert's
approach and the maths in it was really calling out to me to be animated. I just
couldn't resist. Anyway let's get started. As a little warm-up let me show
you what is probably the simplest irrationality proof, a proof of the
irrationality of log 3. If log 3 was actually a fraction u/v that would
be the same as saying that if I take 10 to the power of u/v then I
get 3. But then raising both sides to the v's power we get this equation here. But
that is completely impossible because the left side is odd and the right side
is even. And so because our assumption that log 3 is rational leads to this false statement. And since nothing false can
logically follow from anything true we conclude that our assumption has to be
false itself. So log 3 is not a fraction. That means it is irrational.
Q.E.D. Neat, huh? It's neat. Most of you like puzzles and so here are some one-glance
puzzles for you: which of these logs is irrational? Leave your answers in the
comments, I'll grade them later ... just kidding! I really really, really hate
grading! (Marty) I like grading. (Burkard) You are very strange, Marty is very strange. So hopefully that was all fun
and really easy but now on to the big challenge, Lambert's proof that pi is
irrational. Here's a three step summary of the main ingredients. In step 1
Lambert proves a new miraculous formula for the circle function tangent. So tan x is equal to x over 1 minus x squared over 3 minus x squared over 5, and so on.
All the odd numbers in there, ok. In step 2 Lambert shows that if you plug
in any rational number for x other than 0, you always get an irrational number.
That's pretty amazing, isn't it? So tan of 1/2, tan of 3/4, all these numbers are
irrational. But now the third and final step is really, really easy. You all know
from school that tan of pi/4 that's tan or 45 degrees is equal to 1. But, of
course, the number 1 is not irrational and this implies that pi/4 cannot
be written as a fraction. In other words, pi/4 is irrational. But
if pi/4 is irrational then pi itself is 4 times an irrational number which is
also an irrational number. So pi is irrational. Now that's a really beautiful
argument, isn't it? But, as usual, the devil and the main part of the beauty of the
proof is in the details. And, of course, all the hardcore fans among you will ask
for my head if I don't fill in those details, right? (Burkard commenting on Marty giggling) He's just laughing! Well
I'll do that in a moment, promise, but first let me put Lambert's proof and
this formula in its historical context which will give you an idea of how
Lambert may actually have discovered his proof. Lambert was a contemporary of the
mathematical superstar Leonhard Euler. In fact, when Lambert came up with his proof
Lambert and Euler were colleagues at the Prussian Academy of Sciences in Berlin.
At that time infinite nested fractions like the one discovered by Lambert were
are very much an in thing among mathematicians. These continued fractions as they are usually called were made popular by Euler who had succeeded in
proving all sorts of amazing results using them. For example, he showed that
any number that can be written as an especially simple infinite fraction in
which all the numerators are equal to 1 and all the denominators are positive
integers, any number like this is irrational. For example this infinite fraction here with the simple periodic pattern of
denominators 12121212 must be irrational. In fact, this fraction turns out to be root 3 which is not too
difficult to show and which you can attempt in the comments. Should they attempt it? (Marty) I thinks so. (Mathologer) I think so, too, it's a nice one. So Euler's result leads to a very pretty proof that root 3 is irrational. Even better, all Euler used the same irrationality of
infinite fractions to give a first proof that the number e is irrational and he did
this by demonstrating that e can be written as this amazing infinite
fraction here with the numbers in the denominators continuing in the infinite
regular pattern 2 1 1 4 1 1 6 1 1 8 1 1 10 1 1, and so on. I already talked
about all the stuff in another video so check that one out too, that's an
order! (Marty) Jawohl. Okay, anyway, Lambert was undoubtedly
aware of Euler's work when he came up with a great idea to turn tan X into
an infinite fraction. He starts out with the basic fact that tan x is equal to
sin x divided by cos x. Lambert then uses another well-known fact that the
functions sin x and cos x can be written as these pretty infinite sums.
Now I've already inflicted these sums upon you many many times in other videos
so here I just run with them, hope you are okay with. (Marty) What if they are not okay with it? (Mathologer) Let's see what
happens. Anyway this means that tan x is equal to this beautiful monster here.
Okay now we have to transform this monster ratio into Lambert's infinite
fraction for tan x. In Lambert's manuscript this transformation is an absolute
nightmare. So ready for the nightmare? Well hopefully the animations, which only
took me 10 million hours to create, JUST FOR YOU, will make it all a lot easier to
follow. In any case, now is a really, really good time to fasten your
mathematical seat belts, promise. First, let's write all those powers of x
as products like this, very pretty isn't it? Pull out a common factor in the
numerators. Okay, now we're dealing with a product of x and a fraction. At this
point I have to remind you of a little fraction trick that we'll be using over
and over to build the infinite fraction. The product over there is the same as
the quotient of x and the reciprocal of the fraction. Just think about it for a
second. You're ok with that, great! Now, the
quotient at the bottom we can also write like this big nested fraction here. All
good. Of course this is simple junior school algebra but we'll be using this
trick over and over with the green and yellow expressions both monster sums so
I thought I better make absolutely sure that we're on the same page. Now let's unleash the reciprocal trick for the first time. Just watch it. Here we've got our product
and abracadabra this real magic as far as I'm concerned. So we're off to a good
start, great. Now if this 1 here was not there,
we could move the common factor x squared out in front of the fraction at
the bottom and repeat the reciprocal trick, right? But since the 1 is there we'll first get rid of it using another fraction
trick I call it the add-and-subtract-the same-stuff-from-the-numerator trick. (Marty) Catchy title :)
Okay so let's just launch into it. For the moment don't worry too much about following every detail, okay. We add and subtract the green denominator to and
from the yellow numerator. Whoa, looks scary, but really isn't. (Marty) Yes, it is. (Mathologer) Isn't! Just think
about it, we've just added and subtracted the same thing to and from the numerator
that does not change the value of the numerator. Now swap the top two sums and do some housekeeping. First in the pink box pull the minus into the brackets.
Okay, now 1-1 in the pink is zero, so we're about to get rid of the
blue 1, as planned. Also, for later, note that we get rid of
the blue 1 using the other 1 here which is the constant of the sum in the
denominator, the green sum. Anyway we just got rid of the 1 we wanted to get
rid of. It's not clear how this will help but just wait for it. Now some quick
routine housekeeping on autopilot. Okay that's minus two thirds put it
together, okay. Do it again and again again. At this point we spot a pattern
and can confidently say, etc. Now the blue boxes contain the same stuff and so the
red fraction is equal to blue / blue plus pink / blue but blue / blue is 1
and so we get 1 plus pink / blue. Now again some easy housekeeping in the numerator. Pull out the minus, there you go,
pull out the common factor there, cancel the common factors that I've highlighted
here and just, generally, clean up a bit. Very nice, now we are ready to unleash
the reciprocal trick for the second time. just watch it - magic again. Magic, magic
magic, magic. Yes, now rinse and repeat well perhaps there was a little too much
for you to handle the next rinse on your own so lets us go through the whole
thing one more time real quick as before we'll get rid of the constant in the
numerator of the red fraction using the constant in the denominator now the blue
is equal to three times the pink and so this time we add and subtract three
times the denominator to and from the numerator. Okay, now just follow our nose
again, there we go. Okay and this is the point where the numbers at the front get
cancelled again. Alright now just some housekeeping again and we're back to
this picture here. Our red fraction at the bottom is now three times blue /
blue + pink / blue but three times blue / blue is three and so we get three plus
that. Clean up again really quick. This time I promise quick. quick, quick, quick,
good. And we are ready for the reciprocal trick again.
So let's watch it again: magic. magic :) And what's next? Well we need to kill the
1/3 with the 1 / 3 times 5 and since five times the pink is equal to the blue
after the next reciprocal trick we get the five out in front, and so on, pretty
neat, huh? Phew, that was a hundred slides since I told you to fasten your seat belts :) Anyway all these manipulations are definitely quite
complicated but in the end also very natural and something that a lot of you
would have ended up doing if I had challenged you with
the input output of this manipulation. Especially those of you who know how to
divide power series should have experienced an eerie feeling of deja
vu while you were watching me. Maybe one of you can explain the connection in
the comments. Anyway all this suggests to Lambert and to us that somehow this
equality should hold. Well we're not going to get anywhere with making sense
of this formula unless I finally tell you how mathematicians actually assign a
value to the infinite fraction. To motivate our rule for doing this let's
chop off the fraction at the first denominator. Okay, so we end up with a
finite expression which evaluates to the function x. Let's graph both tan x and
x, here we go. As you can see, the blue function x approximates the orange tan x very well close to 0. Let's chop the infinite fraction at the second
denominator and plot the new function. Amazing right?
The new partial fraction approximates tan spectacularly well. Let's keep going:
third partial fraction, fourth partial fraction, fifths, and so on. Overall these
partial fractions converge to tan x and it is in this sense that tan x really appears
to be equal to the infinite fraction. However here's a little bit of bad news.
Although Lambert's calculation starts with a valid formula for tan x and ends up
with this infinite fraction and despite what our graphing suggests, we cannot be
absolutely sure that the equal sign between tan x and the infinite fraction is
really justified. What's the problem? Well let's just quickly switch back to the
point where we said "and so on". Here you see at any stage of the calculation
there is one of these infinite monster fractions at the bottom. When we say, etc.
we simply discard these monsters and effectively hope for the best. Well so
what's really still needed at this point is a bullet proof proof
that we really do have equality between tan x and the infinite fraction in the
sense that I just described. This part of the proof involves looking very
carefully at all those chopped off partial fractions. Lambert actually does
that. The calculations are again quite laborious but in the end pretty much as
straightforward as the ones that spit out the infinite fraction in the first
place. I'll skip those calculations here but link to an article that has the
full details and if you ask me really nicely I can also put a supplementary
video on Mathologer 2 where I nut out some more details. Anyway for our
purposes I hereby declare Lambert's tan formula under control. Okay, hands up if you're still here! Okay, fantastic,
you've all earned the Mathologer algebra merit badge but there's a little bit
more work to be done and another real highlight to look forward to: an infinite
descent into mathematical hell. Okay, so let's plug in a nonzero fraction u/v for x and let's see how Lambert concludes that the tangent of this
fraction is irrational. That then finishes the proof
remember the infinite fractions like the one for root 3 or for e I that I showed you
earlier were of a special simple type, all the numerators were 1 and all the
denominators began with positive integers and pluses and being of this
special type Euler had shown that these infinite fractions always represent
irrational numbers. Lambert's infinite fractions are obviously not so simple so
Euler's result does not automatically apply to prove the irrationality of
Lambert's fractions. This may seem like just the technicality but in fact there
are infinite fractions that amount to rational numbers. For example, as a puzzle
for you, you can try to show that the infinite fraction down there equals 1. As
a harder puzzle try to come up with infinitely many infinite fractions that
are equal to 1. Anyway to get somewhere with the
irrationality proof, let's see how close we can massage Lambert's infinite
fraction to Euler's special form. Getting rid of the fractions in the numerators
is easy. We'll get rid of them one at a time. This animation of the
transformation should be fairly self-explanatory. Okay now let's show why, as long as both
u and v are positive integers the infinite fraction is always equal to an
irrational number. That will then finish Lambert's proof. Now, apart from the very
first numerator u, all the other numerators are equal to u squared, all
right. On the other hand, the denominators get larger and larger. This means that
from some point on the denominators are all going to be a lot larger than the
numerators. Now there's a general theorem which says that any infinite fraction of
this form is irrational. However, waving that magic wand now to
finish off would be a bit unsatisfying don't you think. Wouldn't be good, right?
So let's see if we can convince ourselves of this directly. Just so that
we can keep things within a small window let's suppose that 5 times v is at
least 2 larger than u squared. So, from this point on the denominators are all
at least 2 larger than the numerators. Okay, alright now it's a bit fiddly but
with this extra plus 2 wiggle room in the denominators it's now not too hard
to show that just like u squared over five v is a positive number less than 1,
this whole infinite fraction here which I've highlighted in blue is also equal
to a positive number that is smaller than 1. And the same is true for this
infinite fraction here and for the following ones. What we'll show is that
this positive and less than one property of all these nested infinite fractions
implies that the first blue infinite fraction is an irrational number which
then implies that the number we are really interested in namely this one
here is irrational - Why? Because three times an integer minus an irrational
number is irrational, a positive integer divided by an irrational number is
irrational, and so on, Okay, so why is the blue infinite fraction irrational. Well
let's assume that it was actually equal to the fraction B over A. Remember the
infinite fraction is special in that it is positive and less than 1. But that
means that B is less than A, right? Let's note this down somewhere. Now let's
consider the size of the next infinite fraction. To do this we just solve it.
Okay here we go, solve, solve, solve, magic magic, magic and all the bits in the numerator
are integers and this means that the whole numerator is an integer, let's call
it C. But the yellow number was also special in that it is positive and less
than one and that means that C is less than B. Let's also note that down and now
we repeat and solve for the next infinite fraction which will generate a
positive integer D less than C, there we go. And now repeating this infinitely
often gives us an infinite sequence of ever-decreasing positive integers and now for the punchline: There is no such thing as an
infinite sequence of ever-decreasing positive integers. Why? Well suppose A was 1,000. Then we could count down for a while but not forever.
Eventually we'd have to go below zero and the same is true no matter how large
an A we start with. We can only descend finitely many steps, each ending on
positive integers, until we hit zero. And so, because our assumption that the blue
infinite fraction is rational leads to this impossible infinite descent of
positive integers, we conclude that our assumption has to be false itself that
is the blue stuff cannot be a fraction, and so must be irrational.
What an absolutely beautiful argument to finish of things, don't you think? Lovely. Anyway, this was the last piece of our puzzle so I hope you enjoyed my take
on Lambert's beautiful proof. As usual let me know in the comments what did and
did not work for you. If you made it all the way to here you earned yourself a
special Mathologer seal of approval. Merry Christmas, fröhliche Weihnachten and I'll see you in the new year :)
I just love how indirect the last part of the proof is. It's cool to see two completely different questions (Is pi irrational? Can there be an infinite sequence of decreasing positive integers?) end up being related.
I love Mathologer videos, besides being really informative his presentation really makes me smile.
Very cool! Never realized transforming continued fraction could be so suitable for animating.
There is a mistake at about the 7 minute mark where “x/(12)” should be “x2/(12)”.
If only my typo did the same!