Welcome to another Mathologer video. Today's video is one I've been dreaming
about making for a long long time. Today I'd like to dazzle you with the
solutions of some of the most famous problems in the history of mathematics.
These problems had remained unsolved for more than 2,000 years after they were
first puzzled over in ancient Greece. The problems arose very naturally as part of
the Greeks mathematicians quest to determine the possible geometric
constructions when all were permitted is to draw lines and circles. That is what
is possible using just the most basic mathematical tools, the ruler and the
compass. For example, given a square, we could construct a new square of twice
the area just using our basic tools. However, unlike doubling a square like
this, doubling a cube turned out to be anything but easy.
Nobody was able to figure out whether, given a cube, we could construct a cube
of double the volume by just using ruler and compass. This is the first of those
ancient problems I'll be tackling today. Similarly, if you are given two lines
making some angle, then halving that angle with ruler and compass is easy. Let us
draw a couple of circles there and there is the bisector, but nobody could come up
with a ruler and compass construction that would trisect an arbitrary angle.
That's problem number two. Next, using ruler and compass, to construct
equilateral triangles, squares, regular pentagon's and regular hexagons is not a
problem at all. But, what about regular heptagons? Nobody had a clue for
thousands of years. That's problem number three. Finally, most famously, and what
turned out to be by far the hardest; Is it possible to square a circle? That is
given a circle, is it possible to use ruler and compass to construct a square
of exactly the same area? Well, as I said it took over two thousand years to
finally answer these questions. And what's the answer? The answer is: Stop
trying, you're wasting your time! In the 19th century it was proved that
it's impossible to use ruler and compass to double a cube, or to trisect an
arbitrary angle, or to construct a regular heptagon, or to square circle. But
who gets to see this impossibility proofs? Not many people, except a few pure
maths majors who may encounter them immersed in a course on Galois Theory.
Really heavy-duty stuff :) However, on close inspection it turns out that those
proofs only really require some of the semi heavy-duty parts of Galois theory.
Then if you are incredibly stubborn and you try really, really hard, it's possibly
possible to distill the essence of these proofs into one short and not too hard
YouTube video. I don't know about you, but I find the idea of making a video like
this super exciting. It's an opportunity to make a small but tangible
contribution to something that people have been struggling with for thousands
of years. And I hope that for you it's an opportunity to get to the core of some
classic and beautiful and hard mathematics, to get some insight into a
theory normally considered out of reach of mere mortals. Okay, here's the plan for
today. I'll first tell you exactly what it means to construct things with ruler
and compass, the precise rules of the game. Then, I'll go very carefully through
the proof that doubling the cube is impossible. Our argument that doubling
cube is impossible will be a proof by contradiction and will run like this: If
doubling the cube was possible, then we can use that to show that the cube root
of 2, a number closely related to our problem could be written as an
expression that only involves rational numbers and square roots.
However, as we'll also show, this would imply that cube root of 2 is in fact
equal to a rational number. Since this is definitely not the case,
doubling the cube cannot be possible either.
After the hard work of doubling the cube, I'll then tidy up, by sketching how this
wonderful proof by contradiction can be modified to also take care of trisecting
angles and constructing regular heptagons. And, finally, squaring the
circle can also be ticked off easily because we already did most of the heavy
lifting for the proof in a previous video on the transcendence of pi. As you
may already have guessed, this video is a challenging one.
So, as for my previous masterclass videos, this one is broken into levels of
enlightenment, six in all and each with its own mathematical superhero guide. Of
course, feel free to start skipping if it gets too tricky, and let me know how far
you make it in the comments. Okay, without further ado,
mathematical seatbelt on and on to level 1.
This video is all about the things that a ruler and compass cannot do. But first
let's ask what can they do? What can we construct this ruler and
compass? To start, let me begin drawing to just give you a feel for what's going on.
Okay? let's start with two points in the plane. Now draw the line through them.
Okay, draw a circle with this red point as it's
center through the other point. So, now we have a third point, the other spot
where the circle and line intersect. draw a circle with this red point as
center through this new green point. That gives us three new points of
intersection, and our construction is already getting pretty and also pretty
interesting. Now, draw a line connecting these two red points. Another
intersection. Clearly this point is the midpoint between our two starting points.
And, our second line is clearly the perpendicular bisector of our starting
points. Plus, these three red points there are the corners of an equilateral
triangle. Pretty impressive, by just drawing a
couple of lines and circles you get all this good stuff. What else is there? Let's
draw another, smaller, circle, like so, okay. Two more points of intersection which,
together with the original two points, from the corners of a square. Pretty easy
right? A regular hexagon is also hiding just around the corner. Let's go and find
that one too.
All child's play. But what about regular pentagon's? That's not so obvious. Let me
show you how you can do this, starting with this part of the picture. There
just that one. First, we locate the point halfways between these two points. Now
you all remember how to do that, right? So, I'll skip the details. Nothing very
five-ish yet. But just watch.
Pretty cool right? But maybe I cheated? Is that really a perfect regular pentagon?
It's definitely is not obvious, but yep, it is.
And, your first real challenge today is to work through the details and nail
down the proof. Okay, so having played with ruler and compass a bit we are
ready to declare the rules of our game. Here we go. At any stage of the game, we
have before us a collection of points, lines and circles. Right? Then there are
two ways we can proceed. First way: choose any two points and draw the line through
them. As well as adding the new line to our collection, it also adds all the
intersection points of this new line with the previous lines and circles. So
there you go. All of them go in and there's a second way and the second way is to
again start by picking any two points. And then you draw a circle centered at
one of the points and passing through the second point. The new circle is then
added to the collection, as are all the new intersection points. So, those are the
rules of our game. Pretty easy right? Now, what about those special problems,
doubling the cube, and so forth?
Okay, you have a cube and your aim is to double its volume using just ruler and
compass. What this means in our game is that you start with two points the side
length of our cube apart and your aim is to construct, in a finite number of ruler
and compass steps, two points that are the side lengths of the doubled cube
apart. For the purposes of our game we can assume that our original cube has
side length one, and so also volume of one, and the volume of the double cube would have
to be 2 of course, right? And so it's side lengths would have to be the cube root
of 2. Got it? Starting with two points a distance 1
apart, our aim is to construct two points that are a distant cube root of 2 apart.
Similarly, a circle of radius 1 has area pi. So constructing a square of the same
area means the square should have side lengths of root pi. As I already showed
you, constructing squares is not a problem.
So, in our game "squaring the circle"' amounts to a beginning with two point a
distance 1 apart, and using them to construct two new points a distance root
pi apart. In exactly the same way, the other two problems I showed you at the
beginning boil down two starting again with two points a distance 1 apart and,
using these points, to construct two specific new lengths. In fact, let's go all
Cartesian and pin down things a little bit further. Let's make our two beginning
points the origin and (1,0) the point one unit to the right along the x-axis. Now there's a
little ruler and compass magic that I won't go into, but which helps simplify
our problems. It turns out that if we can construct a segment of a certain length
somewhere, then we can always translate and rotate using ruler and compass to
obtain a segment of the same length starting at the origin and lying along
the positive x-axis. That means our cube doubling problem boils down to using a
ruler and compass to locate the point cube root of two along the x-axis.
Similarly, squaring the circle boils down to locating root pi on the x-axis. And
there are two more mysterious numbers which arise from our other two problems.
Those two guys. Of course, faced with the task of constructing numbers such as the
cube root of 2 and root pi, it's natural to just stare helplessly
into space. Which is pretty much all that happened for 2000 years. So let's look at
it another way. Let's ask what numbers can we construct?
Well, obviously, all the integers can be constructed like this. Then, as we've seen,
we can construct midpoints. These are the integer multiples of 1/2. There are more
midpoints here. This gives all the integer multiples of 1/4, and so on.
What else? Lots! It turns out that once you've
constructed two numbers A and B, you can also construct their sum, their difference,
their product and a quotient by executing four easy constructions. For
example, to construct A+B, you just need to transfer a distance which,
remember, is no problem with ruler and compass. So there we go, just push it
over and that's A+B there. And B-A is just as easy. Okay let's have
a look. Let's push this over. That's B-A. Next, to construct A times B, we
also need to be able to construct perpendiculars, which we've already seen
and to draw parallel lines, which is also straightforward with ruler and compass.
Ignoring the fiddly details, here is the key idea of this product construction.
Pretty nice, huh? And, constructing A/B is similarly easy.
Faantaaaastic!! Okay so we can add, subtract, multiply and divide... Karl was here >:) ...
constructible numbers and, with the integers already constructed, this means
that all quotients of integers are constructible. In other words, we can
construct all rational numbers with ruler and compass. Anything else? Well, yes,
we can definitely construct some irrational numbers. For example, root 2 is
the diagonal of a unit square which is easy to construct. In fact, with a little
Pythagorassing, it's easy to construct the square root of any positive integer.
But there's more. I'll now show you how, taking any positive number
we've already constructed, we can always construct that number's square root.
Suppose we've already constructed the number A down there. First add 1 to A to
get the green number here, using ruler and compass of course. Now half the green
number, draw this circle, draw a perpendicular through the blue point.
Then the pink segment has length root A. A pretty easy recipe to follow. I'll leave
it as homework for you and your friend Pythagoras to check out that this
actually works. But, just for fun, let's calculate root 2 this way. That should be
root 2 but root 2 is also the length of the diagonal of a unit square. There's
the square, well let's check. Yep that looks about right which means it must be
true. Well of course not but as you will show in your homework it is right. RIGHT?
RIGHT, you will show this! So, it turns out we can construct a lot of numbers.
Starting with the numbers 0 and 1, we can add, subtract, multiply, divide and square
root lengths any finite number of times. And here then are just a few examples of
numbers we can construct. That's a ton of constructible numbers, but are there even
more? What do you think? The answer to this question is... NO!
These nested square rooty creature's based upon the integers are the only kind of
numbers we can construct with ruler and compass and it's actually very easy to
see that this is the case. Suppose you have two points with square rooty
coordinates. Then the equation of the line through those points will have square
rooty coefficients. Pretty obvious that they are square rooty right?
And, a circle with square rooty center and passing through a square rooty point
will have a square rooty radius, and so will also have an equation with square
rooty coefficients. There that's the equation. (as well as Descartes). :)
And then how do we get new points? By intersecting lines with lines, lines with
circles and circles with circles. That corresponds to solving pairs of linear
and quadratic equations. And the solutions of those systems of equations
are of course all square rooty again. And so, to summarise. Starting with our two
numbers 0 and 1 on the x-axis, we can use ruler and compass to produce any square
rooty coordinates we want, but only square rooty coordinates.
Nothing else. So you're back for more. (laugh) Very good!! So, now we know exactly what type of
of numbers we can construct, those square rooty monsters. But what about the cube
root of 2? Can that number be written as a mess of square roots? It seems unlikely
but how do we go about proving it? Well, some of those square rooty monsters
are more monstrous than others, right? So, let's start easy consider some of the
less scary monsters first and work our way up.
Hopefully, we'll detect a pattern which will suggest a plan of attack. So, what
are the very simplest square rooty numbers? Well, there would be the
fractions, with no roots at all. So, let's first ask whether the cube root of 2 can
be written as a fraction. That is, our base level question is `Is cube root of 2
a rational number? And, as all regular Mathologerers would know the answer is
a big NO. The cube root of 2 is not a rational number. I recently did a whole
video about how you can quickly prove the irrationality of this and a bunch of
other rooty numbers using the amazing rational root theorem. So, let's take the
irrationality of the cube root of 2 as given. We'll also take it as known that
squared of 2, the square root of 3, the square root of 5, and so on, are all irrational. Okay?
Not a big deal. Okay, now, as a warm-up exercise, let's think about all square
rooty numbers of this form there, where both a and B are rational numbers.
There's absolutely nothing special about 7
just that it's a rational number whose square root is irrational. We could have
used lots of other numbers there. Now, what we'd like to prove is that the cube
root of two cannot be written as one of those root seveny numbers over there. That
will be progress, right? Little, tiny, special progress but
progress. Now, if you're worried that this tiny progress will be too tiny
don't. Later there will be a humungous plot revelation that will more than
compensate for the incy wincyness here. Anyway, to prove our special case, we set
up the usual proof by contradiction. So, we start by assuming the cube root of
2 is equal to one of these special numbers. In other words, we are assuming
that our number is the solution of this equation here. Cubing the cube root of 2
gives 2, right? In a moment we'll show that the assumption that a plus b root
7 solves the equation implies that the conjugate of this number a minus b
root 7 does so as well. However, ignoring complex number
possibilities, this equation has one and just one solution, the cube root of 2
which means that that our plus solution and our minus solution must be the same and
b must therefore be 0. But that would mean that the root 7 stuff is irrelevant and
that the rational number a itself would solve our equation. In other words, the
conclusion at the end of this chain of consequences is that cube root of 2 would
have to be rational which, as you agreed, YOU AGREED (pointing finger) is nonsense.
And therefore we've arrived at the contradiction you're chasing. So, again,
the assumption that cube root of 2 is of the form a plus b root 7 implies a
contradiction which proves that the cube root of 2 is not of this form. But, of
course, to make the proof complete, I still have to fill in that crucial step
of the argument. I have to show that if a plus b root 7 solves this equation,
then a minus b root 7 does so too. Okay, you're ready?
Then here we go. So when in doubt calculate, right? So moving the 2 to the
left, plugging our route 7 thing in and
carefully expanding we get, let's see, ... all of this mess. Okay, but since we assumed a
and b are rational numbers, the blue and the yellow numbers must be rational,
right? But then much more than being rational,
the yellow number must be exactly 0. Why? Because if it weren't, we could solve
for root 7 like this. That would mean root 7 is a quotient of rational
numbers and so would also be rational. We already know that this is not the case,
so the yellow number is definitely equal to 0. But with the whole left side
being equal to zero that means that the blue number has to be equal to zero, too.
And now everything's clear, right? Well maybe not but we're basically done. The
key observation is that the blue expression only contains an even power
of the number b and both terms in the yellow part have b to an odd power. So
what is the trick? Let's replace b with minus b. What changes? Well, the blue
number is left unchanged since the one minus sign gets squared away. And because
of the odd powers the yellow becomes negative, like this. So there, the
minuses goes out, that's it. So, the overall effect of replacing b by minus b
in this left side of our equation was simply to turn the plus sign in the
middle into a minus. Overall that's the only change, right? But since both blue and
yellow are zero the overall expression with the minus is still equal to 0.
How does that help? Remember the original blue and yellow
expression was obtained by plugging our original root seveny number into x^3-2.
But that means we get the new blue and yellow expression by replacing
b in our number by minus b and then again plugging this new number into
x^3-2. But since the new blue and yellow expression is also equal to 0
this means that our new number is also a solution of our cubic equation. Tada!!!! Well
maybe you'll have to go through this stuff again but that's basically it. Once
again, the assumption that our original number a plus b root 7 solves x cubed
minus 2 is equal to 0 implies that it's conjugate, a minus b root 7 solves
this equation, as well. And this then implies the contradiction and we
conclude that cube root of 2 cannot possibly be of the form a plus b root 7. Great!!!
So you're still here?! Fantastic!!! Now I promised you a big plot revelation. We're
getting there. It turns out that we can iterate our a
plus b root 7 proof by contradiction to conclude that the cube root of 2 is
not equal to any square rooty number. To make the iteration work we need to
isolate the essential ingredients of our proof. It turns out that these ingredients
all have to do with the rational numbers which served as a foundation for what
we're doing. There are four ingredients. First, we know that cube root of 2 is an
irrational number. Second, we needed the property that
adding, subtracting, multiplying, and dividing of rational numbers always
gives rational numbers. Right? we need this to be able to conclude that the
blue and yellow numbers being made up of rational numbers are rational themselves.
We also needed this ingredient when we argued that the ratio of two rational numbers is
rational. The second ingredient the property of being "closed" under the four
arithmetic operations, is described by saying that the set of rational numbers
is a subfield of the field of all real numbers. It tells us that the rationals
from a self-contained universe of numbers. Okay, the third important
ingredient was the fact that 7, inside the square root is a rational
number. Right? We needed 7 to be rational because at various spots we use
that squaring the square root brought us back to a rational number. For example,
when we expanded right at the beginning. Let's do it again! So there expand, expand,
expand. That 7 there is the result of squaring the square root and we need
that to be rational. The last ingredient of our proof was the
fact that although 7 is a rational number, it's square root is not. We use
this in two spots. Okay, to summarise, our magic ingredients are: the cube root of
two is not a rational number, rational numbers form a subfield,
7 is rational, square of 7 is not. And from all that follows that cube root
of 2 is not one of the red numbers. Okay, now it turns out
that the red numbers themselves are also a subfield, that is, if you add, subtract,
multiply or divide two root seveny numbers, then you again get a root seveny
number, there like this. All straightforward except for the division which
involves a conjugate trick that some of you may remember from school. And you can
worry about division by zero, if you're the worrying type. :D Challenge for you is
to fill in the details in the comments. Ok, ingredients again: cube root of 2 is
not a rational number, rational numbers are a subfield, 7 is rational, square root of 7 is
not. From this it follows that cube root of 2 is not one of the red numbers. Now
the red numbers themselves are subfield, and now things are supposed to repeat. So
what comes next? Well something like this! So what we need
is a number, that is red, but whose square root is not. Okay and I'll just give you
one okay? So 1 plus root 7 is such a number and root 1 plus root 7 is not a
red number. And now with exactly the same arguments as before we can conclude that,
all these blue numbers are not candidates for anything that can be
equal to cube root 2. And now things repeat again because the blue numbers
can also be seen to be a subfield, with exactly the same argument as before. And
now? Well, repeat so we need a line here so we need a number that is blue but
whose square is not. And there are lots of possibilities so let me just give you one again.
86 divided by 5 is such a number and now with exactly the same arguments as before
it follows that cube root of 2 is not one of the pink numbers and we can go on
At this stage, perhaps you're feeling a little "rooted", as we say in Australia. :D
Time to catch our breath and take stock. The field property of the subworlds of
numbers that we've been looking at shows that the final pink numbers are exactly
the numbers that can be constructed by combining the rational numbers and these
three rooty expressions with the usual arithmetic: adding, subtracting,
multiplying and dividing. For example, this complicated rooty
number is one of the pink numbers. Same for this one. And this one. And, because
they are all pink numbers none of them can equal cube root of 2. Of course, any
individual pink number can be ruled out using a calculator. But the point is that
we've ruled out ALL the pink numbers of that subfield. And, we can go to bigger
and bigger subfields. Given any square rooty number, we can start from the
rational numbers and by iterating the extension process, we can construct, in a
finite number of steps, a subfield that contains that particular square rooty
number. And, so that square rooty number, ANY square rooty number, cannot be cube
root of 2. And that's it. That completes the proof that cube root of 2 is not a
square rooty number, and therefore that the cube cannot be doubled with ruler
and compass. Very hard work, but also very very nice. Don't you think?
Now before going to the next level, there's just a little tidying up to do.
Remember that we knew or we assumed or you simply trusted me that root 1 plus root
7 is not part of the a plus b root 7 subfield and that route 86 divided by 5 is not part of the last subfield, and so on. but even if root 1 plus root 7 was part
But even if 1 plus root 7 was part of the a plus b root 7 subfield that would have been a problem. Why? Because
that supposedly new subfield would be just the same as the previous subfield. So
there would just be no new numbers to worry about,
that would just mean that to get to this point we would need one less step of the
extension process. It's the same for all the square roots that go into building
our subfield extensions. If the subfield is enlarged, we know how to handle it and
if not then nothing is changed. All that matters is that whenever our subfield is
enlarged we are guaranteed that the new subfield still cannot contain cube root
of 2. And this really finishes the proof that no square rooty number is equal
to the cube root of 2. In turn this shows that it is impossible to double a cube
just using ruler and compass.
What an ingenious argument, don't you think? But there were four problems that
I promised to solve in this video. One down after quite a fight and still three
to go. How long a video is this going to be? Well, not that much longer. It turns out
that proving the impossibility of trisecting angles and constructing a
regular heptagon with ruler and compass can be taken care of in a very similar
fashion. Just like doubling the cube boiled down to showing that the solution
of the cubic equation x^3-2=0 cannot be a square rooty
expression, showing the impossibility of trisecting and heptagoning reduces to
showing that the roots of two other cubic equations cannot be square rooty
expressions. First trisecting angles. Of course, some angles can be trisected. The
angle 90 degrees, for example, is easy to trisect. But the point is that not all
angles can be trisected and to show this we just have to prove that one
particular angle cannot be trisected and we've chosen 60 degrees as our
victim. Here's a quick run-through of how you
show that ruler and compass cannot be used for constructing 20 degree angles.
That then proves the impossibility of trisecting the constructible angle 60
degrees. And, you guessed it, it will be a proof by a contradiction. So, let's
assume that starting with our two points it was possible to construct a 20 degree
angle. Then we can transfer the angle to the origin like this and then
constructing this perpendicular also gives us the cosine of 20 degrees. Time
to dust off your high school trig. Remember your double angle formulas? Well
there are also triple angle formulas and the formula for the cosine is this.
Substituting 20 degrees for theta we get this. The cosine of 60 degrees is
one-half and this means that the cosine of 20 degrees is a solution of this
cubic equation here.
Now we can use the rational root theorem from two videos ago to prove in the
blink of an eye that this equation has no rational roots. And because we are again
dealing with a cubic equation things proceed very much along the lines of the
cube root of 2 proof. In particular, assuming that a solution of this
equation is contained in a square rooty extension field forces the conclusion
that there is also a solution in the smaller subfield and this then gives the
usual contradiction. Challenge for you: fill in the details. One slight
difference you have to deal with: the cubic for the trisection has three real
solutions rather than just one. For those who get lost or are just plain exhausted
I can understand. I'll also provide some links to a write-up. And on to
heptagons. If it was possible to construct a regular heptagon with ruler and
compass, then starting with the usual two points it would be possible to transfer
the heptagon here. This would mean that it was possibly to construct the number
cosine 360/7 degrees. There that creature. And then, it's fiddly
but it can be shown, that this cosine is a solution to this cubic equation there.
And it's straight sailing from there. Again I'd say first try to fill in the
details yourself in the comments and if you get desperate follow the links in
Finally, finally, finally what about squaring a circle and what about using
ruler and compass the the number root Pi.
Well if root Pi was constructible, then it's square the number Pi would also be
constructible. However, it turns out that all square rooty numbers are algebraic,
that is, all square rooty numbers and in fact all rooty numbers are
solutions of polynomial equations with integer coefficients. But Pi is not as we
showed you in a previous masterclass Mathologer video. Pi is a transcendental
number, it is not a solution of such a polynomial equation.
This proves that Pi cannot be constructed and that's all there is to
it. But to nail down the last little bit of the proof, how does one prove that at all
square rooty numbers are algebraic. It turns out to be pretty easy. The idea is
to start with a square rooty expression, set it equal to x and then unravel the
resulting equation, successively isolating and squaring away the square
roots. Eventually all that is left are integers and powers of x. To give you
some intuition I'll finish this video with an animation of constructing such
an equation for one of our previous scary square rooty expressions. But before
the pretty ending, first let me finish with a few thoughts on impossibilities.
Today's video was a tricky tour through some difficult mathematics. It easily
took me 200 hours to put it together and still, even after me and Marty there
behind the camera have agonized and reagonzed over every detail of the
video, it's tough going and I don't expect that everybody who watches this will get
everything in a first viewing. Don't feel bad if you don't, just give it another go
and remember it took mathematicians a couple thousand years to sort out these
ideas and it took that long for a reason. And, so, if it takes you a few viewings
and maybe a question or two in the comments
that's perfectly fine. But there's one more thing. Every
year I get at least a few people writing to me with the news that after devoting
anything from a few seconds to half a lifetime of study they have managed to
square the circle or have achieved some other impossibility. Of course, they are
all wrong. It's important to realize that if you change the rules of our game just
a tiny, tiny little bit, it's then no problem at all to square the circle, and
so on. And this has been known since the days of Euclid. The plan is actually to
make another video about this fascinating topic of rule changing in
the near future. Let's see. Anyway, it's very easy to
misinterpret the rules of the Greek game and end up with false solutions of those
ancient problems, especially if you base your understanding on a bare-bones
exposition of the rules such as the one I have given in this video. Just like
those people who write to me, there have been many thousands of people who've
made this mistake over the years and whole books have been written about
their pointless endeavours. To make absolutely sure that you really, truly
understand the rules and most definitely before you embark on a pointless quest
of squaring the circle, I recommend you very carefully study those rules and the
common mistakes people make interpreting them. The wiki page on ruler and compass
constructions is a great starting point. Ok enough of that. To finish off in a
much prettier style here's the construction of a polynomial equation
with integer coefficients that has this square rooty expression as a solution. Ok,
let's call this guy, oh I don't know, how about x. Now you can watch the polynomial
materialize. And that's all for today.