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# Pi is IRRATIONAL: simplest proof on toughest test

Welcome to another Mathologer video. A couple of weeks ago I showed you an
animation of the first ever proof that pi is irrational by Johann Lambert who
published it in 1761. Really a huge milestone in the history of mathematics.
It came more than 2,000 years after the ancient Greeks first ran into those
annoying irrational numbers. After watching my video James Angel who is a
math teacher here in Australia told us about this really quite extraordinary
problem on the toughest Australian maths exam the New South Wales mathematics
extension 2 exam from 2003. The students who sit this exam are among the best in
this country. They are around 17 years old and this would be their last big
maths exam in school. Let's have a closer look. This is a three hour exam and it's
worth a total of 120 marks. The last set of problems on this exam on page 15
looks like this.
Looks pretty scary, right? It looks even scarier if you just focus on what's asked of the students at the
end. So it appears what they're supposed to do here is to prove that pi is
irrational. Really? And how much is this final problem on the hardest Australian
maths exam worth? Eight out of hundred and twenty that's less than seven
percent. Well, when you have a closer look you find that what is being asked there
is definitely quite demanding but not anywhere as insane as it looks at first
glance. What the students are asked to do is to fill in the details of a proof
which is essentially due to the famous French mathematician Charles Hermite.
That's the man. What I'd like to do today is to explain this proof to you. Well
actually today's goal is really quite ambitious. What I'll show you is a
streamlined version of this proof which on top of being completely
self-contained and accessible to high school students is probably also the
shortest and simplest proof for the irrationality of pi in existence. Now
unfortunately all the great attributes of this proof come at a price. Nothing's
free in life :) Unlike Lambert's long but easy-to-
motivate proof that I showed you before this short proof has been refined to
such an extent that it may very well seem to have fallen from the sky. Very
ingenious but the mathematical equivalent of black magic :) Since you're
watching this video you probably won't need this reminder but just in case, to
say that pi is irrational means that pi is not a rational number that it cannot
be written as a ratio of integers. Okay last chance to admire Hermite's hair before
things get serious. Like pretty much every irrationality proof the version of
Hermite's proof I'll show you is an ingenious proof by contradiction. It
works like this: we'll start by assuming that pi is actually a rational number.
So both u and V are positive integers. Based on this assumption we conjure up
an infinite sequence of positive numbers that gets closer and closer to zero.
But then when we have a really really close look it turns out that the numbers
in the sequence are actually all integers. But that's absolutely
impossible, positive integers cannot get arbitrarily close to zero. The closest
they can get is 1 away, right? And so our assumption that pi is a rational
number implies something that is false. But
since nothing false can follow from something true this means that our
assumption has to be false to start with and so since pi is not rational it
has to be irrational. Great! But now let's take the proof one step at a time.
Everybody, mathematical seatbelts on? Alright so we start by assuming that pi is
equal to u divided by v. Now let me show you this sequence of numbers that I just
talked about. So the zeroeth number in the sequence is just 2, the next number
is 4 times the denominator v squared. Then, in general, the nth term I_n is, whoa,
where did that come from? Looks pretty scary doesn't it. (Marty) Yeah, it's a little scary. (Burkard) Not for you, come
on. However, it's really just an area under a pretty simple curve. I'll explain
that in a moment, okay. So what do we have to do first? We have to convince
ourselves that all the numbers given by this scary formula are positive and that
they really get closer and closer to zero. Piece of cake :) The expression here
is called an integral. It simply represents the area between the x-axis
and the graph of this function here from x is equal to 0 to x is equal to pi. All
these graphs will roughly look the bump over there. So zeros at the end and
positive in between. Now what we need is a rough estimate of how big the area of
this bump is. That's easy. In between zero and pi the maximum value of x to the
power of n is, well, what? Well x cannot be bigger than pi
so the maximum value of x to the power of n is pi to the power of n, and
actually the same is true for the next factor. And sine can be at most 1.
So this means that for every possible value x the whole function is definitely
at most as large as this product here which is equal to this. And this means
that the bump is located underneath the horizontal at this height here. But then
the area of the bump is less than the area of this rectangle and so width
times height. That's this guy here and this means that if we replace the
integral which is the area of the bump by this rectangle area we get something
bigger than the term I_n that we are pondering here. Okay let's do that. Clean
up a bit and let's put these three numbers on the
number line. Now it's not hard to show that the number on the right will tend
to zero as we crank up n. But then, since I_n is always sandwiched between this
number and zero it has to tend to zero as well, right? So, squeeze, squeeze, squeeze, the
thing in the middle has to go to zero, too. Actually, a bit of a puzzle: why don't you
try to show that this number actually goes to zero as n goes to infinity.
Here's a hint. Call this number here x and then show
that this expression goes to zero for all possible choices of x. Now, second
hint check out the previous video around the 29 minutes 20 seconds mark. As usual
leave your solutions in the comments. Okay so in terms of our master plan
we're up to here. We just got to know our special sequence of positive numbers a
little bit and convinced ourselves that it really goes to zero. Now so far we've
not made use of our assumption that pi is a fraction we'll do that in the next
step to conclude that all the terms of our sequence are also integers which
then leads to the contradiction we're chasing.
Okay to indicate the kind of argument we'll make let's take a quick look at
the Fibonacci numbers. Well you all know this but the sequence starts out with
a 0 and a 1 which are usually called F_0 and F_1. Then the next term F_2 is 0 plus
1 equals 1. Next is 1 plus 1 is 2, then 2 plus 1 is 3, and so on. Now this whole
sequence is pinned down by the first two terms F_0 and F_1 together with a
general rule that the nth Fibonacci number F_n is equal to the sum of the
previous two Fibonacci numbers. In maths lingo a rule like this is called a
recurrence relation. The point I'd like to make is that it's completely clear
from this recurrence relation and the fact that the seed numbers F_0 and F_1 are
integers that all the terms of the Fibonacci sequence are integers, right? An
integer plus an integer is always an integer. So if we left the recurrence
relation in place but changed the seed integers to other integers the terms of
the new sequence would still be all integers, right? Ok now it turns out that
the sequence of numbers were interested in for our pi calculation is also pinned
down by its first two terms and the recurrence relation like this. Ok, ok this
recurrence relation is a bit more complicated than the one for the
Fibonacci sequence. Also remember that u and v are the numerator and denominator of
our pi fraction. Anyhow it should also be clear at a glance that if the pink and
green numbers are both integers then the orange number they combine into has to
be an integer as well. And this means that if we can show that the
recurrence relation up there is really true and if we can show that the seed
numbers I_0 and I_1 are the integers that we indicated in the blue, then all
the terms of our sequence are automatically integers. This then
establishes the contradiction and we'll be done, we'll have conclusively proved
that pi is irrational. What a super duper, amazing,
fantastic, nifty argument, don't you think?Well I think he agrees. (Marty) I agree. (Burkard) He agrees.
Everything so far was pretty easy but now we really have to calculate the
numbers in our sequence precisely and for that we need to unleash the power of...
calculus. To check that the blue numbers are correct we just just have to
evaluate the integral. First for n is equal to 0 which gives this guy
here. Of course, if you know a little bit of integral calculus you see at a glance
that this is really equal to 2. But just in case this is new to you I hope
that I make your day by pointing out that this just says that the area under
one of the waves of the sine function is exactly 2. I still remember being
really intrigued by this the first time I came across this fact. Also we're
talking about pi here but there seems to be no circle in sight. Well in this proof
the circle is represented by the sine function and it is two of its nicest
properties that are at the core of the recurrence relation, as we shall see. How
about the second blue number. Hmm this is also not hard if you know a little bit
of calculus. Otherwise just go with the flow for the moment. For the rest of you
simplify on auto-pilot. It looks roughly like this here, right, you don't even
think about this stuff. And now you probably recognize the integrals in the
boxes as standard examples of integrals that can be evaluated with the technique
of integration by parts. And if you do integration by parts this is what you
get. This and that and everything cancels out and we are done. Okay as the
last part of the puzzle it remains to prove that our recurrence relation is
true for all n starting with n = 2. Similar to the calculation of I_1
this is done by applying integration by parts. Not hard in spirit but fiddly
enough that the students actually get 4 out of the total of 8 marks for getting
this part right. I actually now have to show you the details of this calculation
because without those details it's just not
clear why the integral is built the way it is and what role sine our proxy for the
circle really plays in this proof. The calculation itself is completely routine
if you're familiar with integration by parts. Basically just algebra with some
calculus facts thrown in. If you're not familiar with any of this, you should
still be able to understand most of it and it's only going to take two or three
minutes anyway. As a reward for those of you who make it to the end of the proof
I've prepared something extra nice for the end of the video. Okay now what
integration by parts allows us to do is to transform one integral into a
different integral in the hope that the new integral is easier to evaluate. Let
me show you using one instance of the integral we are dealing with here. To be
able to apply integration by parts you have to split the function into two
factors like this. Now here's the integration by parts formula. So the
first factor in the original integral turns into its derivative in the second
integral and the second factor turns into one of its anti derivatives. Alright let's
have a close look at our special integral. As you can see, the orange f(x)
vanishes at x=0 and x=pi. This means that all those
non-integral terms down there vanish, too. Gone! The pink on the left is sine which
when integrated gives minus cosine. The two minuses cancel out.
Okay let's calculate the derivative of the orange bit and fill it in. So
this new integral is not all that different from the one we started with.
Just a cos instead of a sine in the pink, the exponents here are down by 1
from 5 and a bit of junk in front. Now integrating by parts again has all
those non integral bits in the formula vanish again, cos turns back into sine
and the rest is just this here. Well, okay, yes this looks a little bit like a mess
but it turns out that a little easy algebra cleans it up very nicely. So
basically I mean it's just the pen doing this, you don't even have to think,
easy, easy, easy, done. Now this mini mess splits up into
two integrals like this. Of course, the term I_5 which is what we are really
interested in is this guy here's. So multiply this factor here. Now back on
algebra autopilot, alright, it's all standard stuff: simplifying here, kind of
massaging a bit so that everything looks more like the bit at the top and at this
stage you can see these bits are actually I_3 and I_4 and that's almost it.
I actually think it's really cool that the first time we need to use our
assumption that pi is the fraction u divided by v is in this final step here.
So let's do it. Things cancel out, great. Now it turns out that for the general
formula we just have to replace the 5s by ns. So let's just do that. Phew,
okay, okay I have to take a bow here, take a bow here. That was quite something
again. So anyway what a great final maths problem to face in school and
congratulations to all the students who managed to ace it.
It's really quite marvellous how everything in this proof fits together but
then you may ask yourself how could anybody come up with anything like this.
This proof really seems to have fallen from the sky. Well, to answer this
question is almost worth another video all by itself. If you're interested I've
linked to some articles that give some details. Fascinating stuff and well worth
checking out. As a bit of a crazy challenge for you, can you think of a way
in which our recurrence relation up there can be turned into an infinite
fraction like the one in Lambert's proof that I showed you earlier. Hint: divide
by I_(n-1). So now we know that numbers like pi and e are irrational but in
the grand scheme of things we still know very little when it comes to
the irrationality of numbers. For example, nobody knows whether the numbers pi+e
or pi x e are irrational. Having said that there is this absolutely bizarre
result which says that at least one of these two numbers has to be irrational,
with no indication whatsoever as to which. How irrational is that? I
absolutely love this sort of thing. Actually my friend Marty who is giggling
there behind the camera pointed out to me that Hermite's proof can actually be
refined to show that pi is not only irrational but that pi can also not be
of this form here with u, v and w integers. If you ponder this a bit and think
quadratic formula you'll see that this means that pi is not the solution of any
quadratic equation with integer coefficients. And with this insight an
ingenious one-glance proof of the bizarre result I just mentioned is just
one step away. So let's finish with this proof. Okay what have we got up there?
Okay we've got x minus pi times x minus e on the left that gives the special
quadratic polynomial on the right which features those two mystery numberi
pi + e and pi x e. Now let's assume that both these numbers are rational.
Then multiplying through with the denominators b and d at the top
gives this, which means that this quadratic polynomial with integer
coefficients on the right has a zero at pi. But our souped-up irrationality proof
shows that this is not the case and so we conclude that our assumption that
both pi + e and pi x e are rational numbers has to be wrong. At
least one of these numbers has to be irrational but that's all we know for
sure, we don't know which. Strange, strange, strange. And that's all for today but
before you switch channels if you know of any other really interesting and
challenging questions on school exams could you
please let me know in the comments or send me an email.