Skyrocket your IELTS band score by 1-2 points in under a month with our premium plan!
↔
- Take the number 5 and
square it, you get 25.
Now take 25 and square it, you get 625.
Square 625, and you get 390,625.
Do you see the pattern?
5 squared ends in a 5,
25 squared ends in 25, and
625 squared ends in 625.
So does this pattern continue?
Well, let's try squaring 390,625.
It doesn't quite end in itself,
but the last 5 digits match,
so it extends the pattern by a few places.
So let's try squaring
just that part, 90,625.
Well, that does end in itself,
and if we squared that whole number,
it also ends in itself,
and now we're up to 10 digits,
and you can keep doing this.
Squaring the part of the answer
that matches the previous number
and increasing the number of
digits they share in common,
it's as though we are
converging on a number,
but not in the usual sense of convergence.
This number will have infinite
digits, and if you square it,
you'll get back that same number.
The number is its own square.
Now, I bet you're thinking,
"Does it even make sense
to talk about numbers
that have infinite digits going off
to the left of the decimal point?
I mean, isn't that just infinity?"
In this video, my goal is to convince you
that such numbers do make sense.
They just belong to a number system
that works very differently
from the one we're used to,
and that allows these numbers
to solve problems that are impenetrable
using ordinary numbers,
which is why they are a fundamental tool
in cutting edge research
today in number theory,
algebraic geometry, and beyond.
So let's start by
looking at the properties
of the number system
that includes the number we just found.
We'll call these 10-adic numbers
because they're written in base 10.
If you have two 10-adic numbers,
can you add them together?
Sure, you just go digit by digit
from the right to the left,
adding them up as usual,
addition is not a problem.
What about multiplication?
Well, again, you can take
any two 10-adic numbers
and multiply them out.
This works because the
last digit of the answer
depends only on the last
digits of the 10-adic numbers,
and subsequent digits depend only
on the numbers to their right.
So it might take a lot of work,
but you can keep going for
as many digits as you like.
Let's take one 10-adic number ending
in 857142857143, and multiply it by 7.
So 7 times 3 is 21, carry the 2,
7 times 4 is 28 plus 2
makes 30, carry the 3,
7 times 1 is 7 plus 3
makes 10, carry the 1,
7 times 7 is 49 plus 1 is 50, carry the 5,
7 times 5 is 35 plus 5 is 40, carry the 4,
and you can keep going forever
and you'll find all
the other digits are 0.
So this number times 7 equals 1,
which means this 10-adic number
must be equal to 1/7th.
What we have just found
is that there are rational
numbers, fractions,
in the 10-adic numbers
without having to use
the divided by symbol.
Say you wanna find the 10-adic
number that equals 1/3.
How would you do it?
Well, let's imagine we have
an infinite string of digits,
and when we multiply them by 3, we get 1.
This implies that all the digits
to the left of the 1 must be 0.
So what do we multiply 3 by to get a 1
in the unit's place?
Well, 3 times 7 is 21,
so that gives us the 1,
and then we carry the 2.
Now what times 3 plus 2 gives us a 0? 6.
3 x 6 = 18 + 2 = 20.
So we have a 0 and we carry the 2.
Put another 6 there and
we get another 20 again.
So if we put a string of
6s all the way to the left,
they will all multiply to make a 0.
So an infinite string of 6s
and one 7 is equal to 1/3.
This looks similar to the
infinite digits we're used to
going off to the right
of the decimal point,
like 0.9999999 repeating.
What does this equal?
Well, I'll claim that
it's exactly equal to 1,
but how do we prove it?
Let's call this number k,
and then multiply both sides by 10.
So now we've got 9.999
repeating equals 10k.
Now, subtract the top
equation from the bottom one
to get 9 = 9k, so k = 1.
This is a fairly standard argument
for why 0.999 repeating
must be exactly equal to 1.
But what if instead of going
to the right of the decimal place,
the 9s went to the left
of the decimal place,
that is a 10-adic number of all 9s.
What does this number equal?
Well, we can do the same thing,
set it equal to say m,
and then multiply both sides by 10.
So we have 999999990 equals 10m.
Now, subtract this equation
from the first one,
and we get 9 = -9m, meaning m = -1.
So this 10-adic string
of all 9s is actually -1.
Now, I know that seems weird,
so let's try adding 1 to it.
Well, 9 + 1 is 10, carry the 1,
9 + 1 is 10, carry the 1,
and you just keep doing this
all the way down the line
and every digit becomes 0.
I know it seems like at some point,
you're gonna end up with a 1
all the way down on the left,
but this never happens
because the 9s go on forever.
All 9s + 1 = 0,
therefore all 9s must be equal to -1.
This also means that all 9s
and then say a 3 equals -7.
What we have just discovered
is that the 10-adics
contain negative numbers as well.
You don't need a negative sign
by the structure of these numbers alone,
negatives are included.
To do subtraction,
you just add the negative of that number.
To find the negative
of any 10-adic number,
you could multiply by all 9s
or just perform these 2 steps.
First, take the 9s complement,
that is the difference
between each digit and 9, and then add 1.
So if this number is 1/7, then -1/7
is 142857142856 + 1,
and we can verify that this is indeed -1/7
by adding it to positive 1/7,
and finding that these numbers
annihilate each other to 0.
So to sum up, 10-adic numbers
can be added, subtracted,
multiplied, and they work
exactly as you'd expect.
Plus they contain fractions
and negative numbers
without having to use additional symbols.
There is just one big problem,
and you can see it with the
first 10-adic number we found.
Remember, if you multiply
this number by itself,
you get back that same number.
This number is its own square,
and that's a problem which you can see
if we move the end to the
left-hand side and factor it.
Well, then we have n
times n minus 1 equals 0.
The numbers 0 or 1 would
satisfy this equation,
but our 10-adic number is not 0 or 1.
You can even verify by multiplying it out.
This number n times n minus
1 really does work out to 0.
This breaks 1 of the tools
mathematicians rely on
to solve equations.
I mean, have you ever thought
about why when faced with
some complicated equation,
we move all the terms to one side,
set them equal to 0, and then factor them?
Well, I certainly haven't
before making this video,
but there is a good reason,
and it comes down to the
special property of 0.
If several terms multiplied
together equal 0,
then you know at least one
of those terms must be 0,
and this allows us
to break down complicated
higher order equations
into a set of smaller,
simpler equations, and solve.
But this won't work with the 10-adics,
and fundamentally, the reason
is because we're working
in base 10, and 10 is a composite number,
it's not a prime, it's 5 x 2.
Say you wanna find two 10-adic numbers
that multiply to 0,
then to start off, you know
that the last digit must be 0.
So which two numbers could
you multiply to get a 0
in the unit's place?
Well, you could multiply
a 0 times any number,
that's no problem,
but you could also multiply
say 5 x 4 and get 20,
which gives you a 0 in the unit's place.
Then you can carry the 2
and find another two numbers
that will give you a 0 in the ten's place,
and you can keep building
the numbers from there,
so that all the digits work out to 0.
There is a way to avoid this,
and that is to use a prime
number base instead of base 10.
It could be any prime like
2, 3, 5, 7, et cetera.
As an example, let's
create a 3-adic number.
So this is a base 3 number
with infinite digits
to the left of the decimal point.
In the 3-adics,
the only digits we can use are 0, 1, and 2
because 3 is the same thing as 10.
Now, how could you multiply
2, 3-adic numbers to get 0?
Well, again, we can start by just looking
at the last digit, 1 x 1 is 1, 2 x 1 is 2,
and 2 x 2 is 4, which, in base 3, is 11.
So the only way to get a 0 is
if 1 of those 3-adic digits itself is 0,
and it's the same for all
the digits going to the left.
The only way they multiply the 0 is
if one of the 3-adic numbers
is itself entirely 0.
And this works for any prime base
and restores the useful property
that the product of several
numbers will only be 0
if one of those numbers is itself 0.
Here is a random 3-adic
number, this number
means 1 x 3 to the 0 + 2 x 3
+ 1 x 3 squared + 1 x 3 cubed,
and so on.
So you can think of a 3-adic number
as an infinite expansion in powers of 3.
The 3-adic integer that equals -1
would be an infinite string of 2s.
If you add 1, then 2 + 1 is
3, which in base 3 is 10.
So you leave the 0, carry the 1,
and 2 + 1 is again 10.
So you carry the 1,
and you keep going on like that forever.
P-adics have all the same
properties as the 10-adics,
but in addition, you
will never find a number
that is its own square besides 0 and 1
nor will you find one non-0 number
times another non-0
number being equal to 0.
And this is why
professional mathematicians
work with p-adics where
the p stands for prime
rather than say the 10-adics.
P-adics are the real tool.
They have been used in the work
of over a dozen recent Fields Medalists.
They were even involved in cracking one
of the most legendary
math problems of all time.
In 1637, Pierre de Fermat
was reading the book "Arithmetica"
by the ancient Greek
mathematician Diophantus.
Diophantus was interested in the solutions
to polynomial equations
phrased in geometric terms
like the Pythagorean theorem.
For a right triangle, x squared
+ y squared = z squared.
The set of solutions to this equation
in the real numbers is
pretty easy to find.
It's just an infinite cone.
But Diophantus wanted to find solutions
that were whole numbers or fractions
like 3, 4, 5 and 5, 12, 13,
and he wasn't the first.
Here is an ancient Babylonian clay tablet
from about 2000 BC with a huge list
of these Pythagorean triples.
By the way, this list predates Pythagoras
by more than a millennium.
Right next to Diophantus' discussion
of the Pythagorean theorem,
Fermat writes a statement
that will go down in history
as one of the most infamous of all time.
The equation x to the n
+ y to the n = z to the n
has no solutions in integer
for any n greater than 2.
I have a truly marvelous
proof of this fact,
but it's too long to be
contained in the margin.
Fermat's last theorem,
as this became known,
would go unproven for 358 years.
In fact, to solve it,
new numbers had to be
invented, the p-adics,
and these provide a systematic method
for solving other problems
in Diophantus' "Arithmetica."
For example, find three squares
whose areas add to create a bigger square
and the area of the first
square is the side length
of the second square,
and the area of the second
square is the side length
of the third square.
- He's really giving
the the first instance
of algebra many, many
centuries before, al-jabr.
- [Derek] So if we set the side
of the first square to be x,
then its area is x squared.
This is the side length
of the second square
which therefore has an area of x to the 4.
This is then the side
length of the third square
which has an area of x to the 8.
And we want these three areas,
x squared + x to the 4 + x to the 8
to add to make a new square.
So let's call its area y squared.
So x squared + x to the 4
+ x to the 8 = y squared.
Now, it's not hard to find
solutions to this equation
in the real numbers.
For example, set x equal to 1,
and you find y is root 3.
In fact, we can make a plot of
all the real number solutions
to this equation,
but Diophantus wasn't
interested in real solutions.
He wanted rational solutions.
Solutions that are whole
numbers or fractions.
These are much harder to find.
I mean, where would you even start?
I mean, where would you even start?
- Like what do you do?
You have this cliff
and you're looking for
anywhere to grab a hold of.
- Well, in the late 1800s,
a mathematician named Kurt Hensel
tried to find solutions
to equations like this one
in the form of an expansion
of increasing powers of primes.
So working with the prime 3,
the solutions would take
the form of x = x not
+ x1 times 3 + x2 times 3
squared + x3 times 3 cubed,
and so on,
and y would also be a similar
expansion in powers of 3.
Each of the coefficients
would be either 0, 1, or 2.
Now imagine inserting these expressions
into our equation for x and y,
and you can see it's going
to get messy real fast,
but there is a way to simplify things.
Say you wanted to write 17 in base 3.
Well, one way to do it
is to divide 17 by 3
and find the remainder,
which in this case is 2.
So we know the unit's digit
of our base 3 number is 2.
Next, divide 17 by 9, the
next higher power of 3,
and you get a remainder of 8.
Subtract off the 2 we found before
and you have 6 which is 2 x 3.
So we know the second to last digit is 2.
Next, divide 17 by 27, and
you get a remainder of 17.
Subtract off the 8 we've
already accounted for
and you have 9, so the 9's digit is 1.
So 17 in base 3 is 122.
What we're doing here is a
form of modular arithmetic.
In modular arithmetic,
numbers reset back to 0 once
they reach a certain
value called the modulus.
The hours on a clock work kinda like this
with a modulus of 12.
The hours increase up to 11,
but then 12:00 is the same thing as 00:00.
If it's 10 in the morning,
say what time will it be in four hours?
You could say 14, but
usually, we'd say 2:00 PM
because 2 is 14 modulo 12,
it's two more than a multiple of 12.
So in other words,
modular arithmetic is only
about finding the remainder.
36 modulo 10 or 36 mod 10
is 6, and 25 mod 5 is 0.
- Mod 3 means your clock
only has three numbers on it,
0, 1, and 2,
and if you multiply 2 by 2, you get a 4,
and 4 is the same thing as 04:00,
the same thing as 1:00 on a 3-hour clock.
- [Derek] What's
great about this approach
is it allows us to work
out the coefficients
in our expansion one at a time
by first solving the equation mod 3,
and then mod 9, and
then mod 27, and so on.
So first, let's try to
solve the equation mod 3.
And since all the higher
terms are divisible by 3,
they're all 0 if we're working mod 3.
So we're left with x not
squared + x not to the 4th
+ x not to the 8 = y not squared.
And this will allow us to find the values
of x not and y not that
satisfy the equation mod 3.
Now we know that x not
can be either 0, 1, or 2,
and y not can also be either 0, 1, or 2.
If x is 0, then so is x squared,
x to the 4, and x to the 8.
If x is 1, then x squared is 1,
and so is x to the 4, and x to the 8.
If x is 2, then x squared is 4.
But remember, we're working mod 3,
and 4, mod 3 is just 1.
To find x to the 4, we
can just square x squared,
so that also equals 1,
and squaring again, x to the 8 = 1.
Now we can sum up x squared
+ x to the 4 + x to the 8
to find the left-hand
side of the equation.
If x is 0, then the sum is equal to 0.
If x is 1 or 2, the sum is 3.
But again, we're working mod 3,
so 3 is the same as 0.
Now, let's calculate y squared.
If y is 0, y squared is 0.
If y is 1, y squared is 1.
And if y is 2, then y squared is 4,
but again, 4 mod 3 is 1.
Now, since for all values of x,
the left-hand side of the equation is 0,
the only value of y
that satisfies the equation is y = 0,
but x can be 0, 1, or 2.
So we have three potential solutions
that satisfy our equation mod 3,
we've got 0, 0, 1, 0, and 2, 0.
Now we shouldn't be surprised to find 0, 0
as a solution since x = 0 and y = 0
does satisfy the equation,
but squares of 0 size don't really count
as solutions to Diophantus'
geometric problem.
So let's try to expand on
one of the other solutions.
I'll pick 1, 0.
This means x not = 1 and y not = 0
satisfies the equation mod 3.
Now let's try to find x1,
and we'll do this by
solving the equation mod 9.
All of the terms higher than x1
have a factor of 9 in them,
so they're all 0 mod 9.
So we're left with this expression.
Expanding out the first term,
we have 1 + 6x1 + 9x1 squared.
But again, since we're working mod 9,
the last part is 0.
The next term is just
the first term squared.
So that equals 1 + 12x1 + 36x1 squared,
but 36 is 9 x 4, so that's
0 mod 9, and 12 mod 9 is 3.
So we have 1 + 3x1.
The final term is just that squared,
so 1 + 6x1 + 9x1 squared.
Again, the last part is 0.
So on the left-hand
side, we have 3 + 15x1.
And on the right-hand side,
we have 0 + 9y1 squared,
which is also 0 because
it contains a factor of 9.
So 3 + 15x1 equals 0.
Now remember, since we're working mod 9,
the 0 on the right-hand side
represents any multiple of 9.
So in this case, if x1 = 1,
then 3 + 15 = 18, which is
a multiple of 9, so it's 0.
So x1 = 1 is a solution to the equation.
Let's find one more term of the expansion
by solving the equation mod 27.
Again, all the terms with
3 raised to the power of 3
or higher contain a factor of 27.
So there's 0, leaving
only this expression,
but we know x not and x1 are equal to 1.
- But right now, we've
learned nothing about y1
because when we square,
anything can happen.
If we find a good solution in x,
it'll come with a good solution in y.
- [Derek] So we can simplify to this.
Expanding again, we get
16 + 18x2 + 81x squared,
but 81 is 27 x 3, so that's 0.
The next term is the square of the first.
So 256 + 576x2 + 324x2 squared.
But 324 is 27 x 12, so that's 0.
And since we're working mod 27,
we can simplify this down.
576 is nine more than a multiple of 27,
and 256 is 13 more than a multiple of 27.
So we're left with 13 + 9x2,
and the last term is just that squared.
So 169 + 234x2 + 81x2 squared,
which reduces to 7 + 18x2.
The right-hand side
reduces to 0 + 81y squared,
which is, again, 0.
So we have 36 + 45x2 equals 0,
which mod 27 is the same as 9 + 18x2 = 0.
So x2 must be equal to 1, 9 + 18 is 27,
which mod 27 is 0.
So what we've discovered is
the first three coefficients
in our expansion are all 1.
And in fact, if you kept going
with modulus 81, 243, and so on,
you would find that all
of the coefficients are 1.
So the number that solves
Diophantus' equation
about the squares is
actually a 3-adic number
where all the digits are 1s.
But how do we make sense of this?
What does this number equal?
- And of course, this number
makes no sense at all,
at least not as a real number.
- [Derek] Well, remember
that this is just another way
of writing 1 x 3 to the 0 + 1 x 3
+ 1 x 3 squared + 1 x 3 cubed, and so on.
So each term is just 3
times the term before it.
This is a geometric series.
And to find the sum of an
infinite geometric series,
you can use the equation 1/1 - lambda
where lambda is the ratio of
one term to the previous one.
So in this case, it's 3.
Now, I know for this to work,
lambda is meant to be strictly less than 1
because otherwise, the terms keep growing
and the sum doesn't converge,
it just diverges to infinity.
I promise I'll come back to this,
but for now, let's just
say that lambda = 3
and see what happens.
Well, then we have 1/1 - 3, which is -1/2.
So if we believe this formula,
then x = -1/2 should be a
solution to our original equation.
If we sub it in,
we get x squared is
1/4, x to the 4 is 1/16,
and x to the 8 is 1/256.
Let's put all of these
over a common denominator.
So a quarter becomes
64/256, now 16th is 16/256.
And if we add them all
together, we get 81/256,
which is indeed the area of a square
with sides of length 9/16ths.
We have found a rational solution
to Diophantus' sum of squares problem.
The first square has sides of length 1/2,
the second has sides of length 1/4,
the third has sides of length 1/16,
and all three squares
together make a square
with sides of length 9/16ths.
To find this solution,
we used new seemingly
absurd numbers, p-adics,
infinite digits going off to
the left of the decimal point,
implying an infinite expansion
of increasing powers of 3.
Then we used the geometric series formula
to find that an infinite string
of 1s in 3-adic notation is actually -1/2.
This works, even though
the ratio of each term,
the previous 1 is 3.
So by common sense, the series
shouldn't have converged,
it should have blown up to infinity.
So the real question is how did this work?
Well, the key idea is that the geometry
of the p-adics is totally different
from that of the real numbers.
In fact, they don't exist
on a number line at all.
One way to visualize
them is with something
like a growing tree.
For the 3-adic integers
we've been working with,
3 base cylinders
represent the unit's digit
or x not being 0, 1, or 2.
Above each cylinder is a
trio of shorter cylinders
corresponding to the 3s digit or x1.
And we continue in this way forever
making an infinite triply branching tree.
Looking down from above,
it looks like a Sierpinski gasket.
Each 3-adic number is represented here
as a stack of infinite cylinders
that get shorter and
narrower as they go up.
And this actually
reflects the relative contributions
each successive cylinder
makes to the value of the 3-adic number.
Contrary to what you'd expect,
the coefficients multiplying
higher powers of 3
actually make finer and finer adjustments.
547
00:26:18,300 --> 00:26:20,790
So when we calculated
successive coefficients
to solve Diophantus' squares problem,
we were actually zooming
in more and more accurately
on the solution.
- This is the feeling of slowly zooming in
on the value of a number.
- [Derek] Normally, we
think of the size of a number
as being determined approximately at least
by how many digits it has to
the left of the decimal point.
But here, all the numbers
have infinitely many digits.
So people realize that
to determine the distance
between two numbers,
we need to look at the
lowest level of the tower
where they disagree.
If two numbers differ in the unit's place,
we say their separation is 1.
But if they differ in the 27th's place,
we say they differ not by 27, but by 1/27.
In the world of p-adics,
what we're used to thinking
of as big is small,
and vice versa.
- If we have a number, let's say, s1,
which will be some sequence
like 2 x 1 + 1 x 3 + 0 x 3 squared
+ 1 x 3 cubed + 2 x 3
to the 4th, and so on
we'll be close to another number
with a similar expansion
except at the, let's say,
3 to the third place,
I change a digit,
and then doesn't matter
what I do after that.
The digits can be all the
same or all different.
These two numbers will
say that their distance,
their 3-adic distance is roughly of size,
the first place where they go wrong.
So this'll be 3 to the -3.
- Am I supposed to think of the 3s
as kind of like, the bigger they are,
the smaller the numbers
that they're multiplying, or something?
- Exactly.
In the 3-adics, we want
numbers to be close
when they agree up to large powers of 3,
so that if they agree
up to 10 3-adic places,
then this has a distance
like 3 to the -10.
- And it turns out that if
you do this crazy thing,
swap what we think of as big and small,
then all the other laws of mathematics
work out in the usual way.
This is why the geometric sum worked out,
even though we thought it
would blow up to infinity.
- You have to open your mind
to other notions of size,
and when you do, this
whole other world appears,
and it's a very useful world,
just like negative numbers became useful,
just like square roots of
negative numbers became useful.
- This feels even crazier
than negative numbers or
square roots of negatives.
- Just 'cause it's less familiar.
- [Derek] And you can
prove this notion of size
fits the criteria you would
want for an absolute value.
- What do we want an absolute value to be?
It should be non-negative.
So the absolute value of any number, x,
should be non-negative,
and the absolute value should be 0
if and only if that number is itself 0.
So that's called positive definite.
So it should be multiplicative.
If I take x times y and
I multiply those two,
that should be the same
as the absolute value
of x times the absolute value of y.
And I want one more property,
which is that if you add x and y,
should that be the same
as the absolute value
of x plus the absolute value of Y? No.
But it should be, at most, the sum, right?
This is the triangle inequality.
So we have multiplicativity,
positive definites, and
the triangle inequality.
You give me only these
three very abstract things
and I can prove that your function is,
in fact, the usual absolute value
or one of these p-adic absolute values
or the thing that gives 0 to
0 and 1 to everything else.
So that's the trivial absolute value.
These are the only games you can play
on the rational numbers that
give you absolute values
that behave the way we want them to.
- [Derek] This geometry
makes the p-adics much more disconnected
when compared to the real numbers.
And this is actually useful
for finding rational solutions
to an equation.
There are many fewer p-adic solutions
in a neighborhood of a rational solution.
If we tried the same strategy
of solving Diophantus'
squares problem digit by digit
in the real numbers, it
would be doomed to fail
because there are simply
too many real solutions all
over the place.
They get in the way.
In a groundbreaking
pair of papers in 1995,
one by Andrew Wiles,
and another by Wiles and Richard Taylor,
they finally proved Fermat's last theorem,
but the proof could not possibly
have been the one Fermat
alluded to in the margin
because it made heavy
use of p-adic numbers.
- Wiles' proof of Fermat's last theorem
used the prime 3, and then at some point,
he got stuck with the prime 3,
and he had to switch to the prime 5.
And this is literally called
the three, five trick.
There was something that worked
for the prime 3 most of the time,
but sometimes it didn't work,
and when it didn't work for the prime 3,
it did work for the prime 5.
Each prime gives you a completely
unrelated number system,
just like these number
systems are unrelated
to the real numbers.
- There is a great quote I like
by a Japanese mathematician, Kazuya Kato.
He says, "Real numbers are like the sun
and the p-adics are like the stars.
The sun blocks out the
stars during the day,
and humans are asleep at night
and don't see the stars,
even though they are just as important."
Well, I hope this video has revealed
at least a glimpse of those stars to you.
The discovery of p-adic
numbers is a great reminder
of just how much we have yet
to discover in mathematics,
not to mention science, computer science,
and just about every technical field.
They inspire us to find new connections
and even make discoveries ourselves.
If you were inspired by the stories
of Diophantus, Fermat, and Hensel,
and you're looking for
more content like this,
look no further than
the math history course
from this video sponsor, brilliant.org
Brilliant's course
introduces you to the minds
behind some of the most
fascinating discoveries
in mathematics, while giving
you a deeper understanding
of the concepts they pioneered,
all of which enable our
modern technologies.
Brilliant actually has thousands
of bite-sized lessons on everything
from math and science to data programming
and computer science,
and beyond inspiring you
to write your own story of discovery,
the most amazing thing about Brilliant
is that you get hands-on
with the building blocks of innovation.
Brilliant's latest
course, Thinking in Code,
is a perfect example designed
with beginners in mind.
It's got everything you need
to start thinking like a programmer.
You can try out everything
Brilliant has to offer free
for 30 days.
Just go to brilliant.org/veritasium
And for viewers of this video,
Brilliant are offering 20% off
an annual premium subscription
for the first 200 people to sign up.
Use that same link,
which I will put down in the description.
So I wanna thank Brilliant
for sponsoring Veritasium,
and I wanna thank you for watching.
Please play the YouTube video first