Kind: captions
Language: en
take the number five and square it you
get 25 now take 25 and square it you get
625 Square 625 and you get
39,6 125 do you see the pattern 5 squar
ends in a five 25 squar ends in 25 and
625 squar ends in
625 so does this pattern continue well
let's try squaring
39,6
125 it doesn't quite end in itself but
the last five digits match so it extends
the pattern by a few places so let's try
squaring just that part
9,625 that does end in itself and if we
square 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 adct numbers
because they're written in base 10 if
you have two 10 addict 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 addict numbers and
multiply them out this works because the
last digit of the answer depends only on
the last digits of the 10 addict 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 adct number ending in
8571 42 8571 43 and multiply it by 7 so
7 * 3 is 21 carry the 2 7 * 4 is 28 + 2
makes 30 carry the 3 7 * 1 is 7 + 3
makes 10 carry the 1 7 * 7 is 49 + 1 is
50 carry the 5 7 * 5 is 35 + 5 is 40
carry the four and you can keep going
forever and you'll find all the other
digits are zero so this number times 7
equals 1 which means this 10 adct number
must be equal to
17th what we have just found is that
there are rational numbers fractions in
the T addic Numbers without having to
use the divid by
symbol say you want to find the 10
addict number that equals a thir how
would you do it well let's imagine we
have an infinite string of digits and
when we multiply them by three we get
one this implies that all the digits to
the left of the one must be zero so what
do we multiply three by to get a one in
the units Place well 3 * 7 is 21 so that
gives us the 1 and then we carry the two
now what * 3 + 2 gives us a zero 6 3 * 6
= 18 + 2 equals 20 so we have a zero and
we carry the two put another six there
and we get another 20 again so if we put
a string of sixes all the way to the
left they will all multiply to make zero
so an infinite string of sixes and 1 7
is equal
to3 this looks similar to the infinite
digits we're used to going off to the
right of the decimal point like
9999999 repeating what does this equal
well I'll claim that it's it's exactly
equal to one but how do we prove it
let's call this number K and then
multiply both sides by 10 so now we've
got 99.999 repeating equals
10K now subtract the top equation from
the bottom one to get 9 = 9 K so k equal
1 this is a fairly standard argument for
why 999 repeating must be exactly equal
to 1 but what if instead of going to the
right of the decimal place the nines
went to the left of the decimal place
that is a 10 addict number of all Nines
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
99999999 0 equal 10 m now subtract this
equation from the first one and we get 9
= -9 M meaning m =
-1 so this 10 adct string of all nines
is actually -1 now I know that seems
weird so let's try adding one 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 zero I know it seems like
at some point you're going to end up
with a one all the way down on the left
but this never happens because the nines
go on forever all 9es + 1 equal 0
therefore all 9es must be equal to -1
this also means that all 9es and then
say a 3 equals -7 what we have just
discovered is that the 10 adct 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 addict number you
could multiply by all nine
or just perform these two steps first
take the n's complement that is the
difference between each digit and nine
and then add one so if this number is
17th then negative a 7th is
2856 + 1 and we can verify that this is
indeed negative a 7th by adding it to
positive 17th and finding that these
numbers annihilated each other to
zero so to sum up 10 addict 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 addict
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
you can see if we move the N to the left
hand side and Factor it well then we
have n * n -1 = 0 the numbers 0 or 1
would satisfy this equation but our ten
adct number is not 0 or one you can even
verify by multiplying it out this number
n * nus1 really does work out to zero
this breaks one 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 zero
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 zero if
several terms multiplied together equal
zero then you know at least one of those
terms must be zero 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
addicts 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 * 2 say you want to find two 10
addict numbers that multiply to zero
then to start off you know that the last
digit must be zero so which two numbers
could you multiply to get a zero in the
units Place well you could multiply zero
times any number that's no problem but
you could also multiply say five * 4 and
get 20 which gives you a zero in the
unit's place then you can carry the two
and find another two numbers that will
give you a zero in the 10's place and
you can keep building the numbers from
there so that all the digits work out to
zero 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 Etc as an example let's
create a three addict number so this is
a base three number with infinite digits
to the left of the decimal point in the
three atcts the only digits we can use
are 0 1 and 2 because three is the same
thing as 1 0 now how could you multiply
two three addict numbers to get zero
well again we can start by just looking
at the last digit 1 * 1 is 1 2 * 1 is 2
and 2 * 2 is 4 which in base 3 is 1 one
so the only way to get a zero is if one
of those three addict digits itself is
zero and it's the same for all the
digits going to the left the only way
they multiply to zero is if one of the
three adct numbers is itself entirely
zero and this works for any Prime base
and restores the useful property that
the product of several numbers will only
be zero if one of those numbers is
itself
zero here is a random three adct number
this number means 1 * 3 0 + 2 * * 3 + 1
* 3^ 2 + 1 * 3 cubed and so on so you
can think of a three adct number as an
infinite expansion in powers of three
the three adic integer that equals -1
would be an infinite string of twos if
you add one then 2 + 1 is three which in
base three is 1 Zer so you leave the
zero carry the one and 2 + 1 is again
one Zer so you carry the one and you
keep going on like that
forever pics have the same properties as
the 10 addicts but in addition you will
never find a number that is its own
Square besides 0 and one nor will you
find one non-zero number times another
nonzero number being equal to zero and
this is why professional mathematicians
work with P addicts where the P stands
for Prime rather than say the 10 addicts
pics 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
[Music]
in 1637 Pier de FMA was reading the book
arithmetica by the ancient Greek
mathematician diaphanus diaphanus was
interested in the solutions to
polinomial equations phrased in
geometric terms like the Pythagorean
theorem for a right triangle x^2 + y^2 =
z ^2 the set of solutions to this
equation in the real numbers is pretty
easy to find it's just an infinite cone
but diaphanus wanted to find solutions
that were whole numbers or fractions
like 345 and
51213 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 diaphanus
discussion of the Pythagorean theorem
FMA writes a statement that will go down
in history as one of the most infamous
of all time the equation x n + y n = z
the N has no Solutions in integers 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
for Ma'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 pics and these provide a
systematic method for solving other
problems in diaphanus 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 uh algebra so if
we set the side of the first Square to
be X then its area is x squ this is the
side length of the second Square which
therefore has an area of x 4 this is
then the side length of the third Square
which has an area of x 8 and we want
these three areas x^2 + x 4 + x 8 to add
to make a new Square so let's call its
area y^2 so x^2 + x 4 + x 8 =
y^2 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 < tk3 in fact we can make a plot of
all the real number solutions to this
equation but diaphanus 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 and we're totally stuck like what
do you do you know you have this cliff
and you're looking for anywhere to to to
grab a hold of and there's nothing in
sight 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 three the solutions would
take the form of x = x + X1 * 3 + X2 *
3^ 2 + X3 * 3 cubed and so on and Y
would also be a similar expansion in
powers of three 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 two so we know the
units digit of our base 3 number is two
next divide 17 by 9 the next higher
power of three and you get a remainder
of 8 subtract off the two we found
before and you have 6 which is 2 * 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 eight
we've already accounted for and you have
nine so the n's digit is 1 so 17 in base
three is 1
22 what we're doing here is a form of
modular arithmetic in modular arithmetic
numbers reset back to zero once they
reach a certain value called the modulus
the hours on a clock work kind of like
this with a modulus of 12 the hours
increase up to 11: but then 12:00 is the
same thing as
0:00 if it's 10: in the morning say what
time will it be in 4 hours you could say
14 but usually we'd say 2:00 p.m.
because 2 is 14 modulo 12 it's two more
than a multiple of 12 so so in other
words modular arithmetic is only about
finding the remainder 36 modulo 10 or 36
mod 10 is six and 25 mod 5 is zero mod 3
means your clock only has three numbers
on it zero uh 1 and
two and if you multiply two by two you
get four and four is the same thing as
4:00 is the same thing as 1:00 on a
3-hour
clock 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 zero if we're
working mod 3 so we're left with x^ 2 +
x 4 plus x to the 8 equals y^ 2ar and
this will allow us to find the values of
xot and Y that satisfy the equation mod
3 now we know that X notot can be either
0 1 or 2 and Y KN can also be either 0 1
or 2 if x is zero then so is x^2 x 4 and
X to 8 if x is 1 then X2 is 1 and so is
x 4 and x 8 if x is 2 then x^2 is 4 but
remember we're working mod 3 and 4 mod 3
is just 1 to find x 4 we can just Square
Square x^2 so that also equals 1 and
squaring again x 8 = 1 now we can sum up
x + x 4 + x 8 to find the left hand side
of the equation if x is0 then the sum is
equal to 0 if x is 1 or two the sum is
three but again we're working mod 3 so
three is the same as zero now let's
calculate y^2 if Y is 0 y^2 is 0 if Y is
1 y^2 is 1 and if Y is 2 then y^2 is 4
but again 4 mod 3 is 1 now since for all
values of X the left hand side of the
equation is zero the only value of y
that satisfies the equation is y equals
0 but X can be 0 1 or two so we have
three potential solutions that satisfy
our equation mod 3 we've got 0 0 1 0 and
two 0 now we shouldn't be surprised to
find 0 0 as a solution since xal 0 and Y
equals 0 does satisfy the equation but
squares of zero size don't really count
as solutions to dian's geometric problem
so let's try to expand on one of the
other Solutions I'll pick 1 0 this means
xal 1 and y equals 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 terms
higher than X1 have a factor of N9 in
them so they're all zero mod 9 so we're
left with this expression expanding out
the first term we have 1 + 6 X1 + 9
X12 but again since we're working mod 9
the last part is zero the next term is
just the first term squared so that
equals 1 + 12 X1 + 36 X1 2 but 36 is 9 *
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 + 6 X1 + 9 X1 2 again the last part
is zero so on the left hand side we have
3 + 15 X1 and on the right hand side we
have 0 + 9 y1 2 which is also 0 because
it contains a factor of 9 so 3 + 15 X1
equal 0 now remember since we're working
mod 9 the zero 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 zero 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^ of 3 or higher contain a factor of
27 so they're zero leaving only this
expression but but we know X knot and X1
are equal to 1 but right now we've
learned nothing about y1 because when we
Square uh we anything can happen if we
find a good solution in X it'll come
with a good solution in y so we can
simplify to this expanding again we get
16 + 18 X2 + 81 x^2 but 81 is 27 * 3 so
that's zero the next term is the square
of the first so 256 + 576 X2 + 324
x22 but 324 is 27 * 12 so that's zero
and since we're working mod 27 we can
simplify this down 576 is 9 more than a
multiple of 27 and 256 is 13 more than a
multiple of 27 so we're left with 13 + 9
X2 and the last term is just that
squared so 169 + 234 X2 + 81 X2 2 which
reduces 2 7 + 8
X2 the right hand side reduces to 0 + 81
y^2 which is again 0 so we have 36 + 45
X2 = 0 which mod 27 is the same as 9 +
18 X2 equals 0 so X2 must be equal to 1
9 + 18 is 27 which mod 27 is zero so
what we've discovered is the first three
coefficients in our expansion are all
one and in fact if you kept going with
modulus 81 243 and so on you would find
that all of the coefficients are one so
the number that solves dian's equation
about the squares is actually a three
adct number where all the digits are
ones 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 well
remember that this is just another way
of writing one time * 3 0 + 1 * 3 + 1 *
3^ 2 + 1 * 3 cubed and so on so each
term is just three 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/ 1us
Lambda where Lambda is the ratio of one
term to the previous one so in this case
it's three now I know for this to work
Lambda is meant to be strictly less than
one 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 equals 3 and see what
happens well then we have 1/ 1 - 3 which
is half so if we believe this formula
then x = a half should be a solution to
our original equation if we sub it in we
get x^2 is a/4 x 4 is a 16th and x 8 is
1 over
256 let's put all of these over a common
denominator so a quarter becomes 64 over
256 uh 16th is 16 over 256 and if we add
them all together we get 81 over
256 which is indeed the area of a square
with sides of length
916 we have found a rational solution to
diaphanes sum of squares problem the
first square has sides of length 1/2 the
second has sides of length a/ quar the
third has sides of length 16th and all
three squares together make a square
with sides of length
916 to find this solution we used new
seemingly absurd numbers pics infinite
digits going off to the left of the
decimal point implying an infinite
expansion of increasing powers of three
then we Ed the geometric series formula
to find that infinite string of ones in
three addict notation is actually
negative a half this works even though
the ratio of each term to the previous
one is three 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 pics 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 three attic integers we've been
working with three base cylinders
represent the units digit or X knot
being 01 or two above each cylinder is a
trio of shorter cylinders corresponding
to the threes digit or X1 and we
continue in this way forever making an
infinite triple branching tree looking
down from above it looks like a serinsky
gasket each three atct 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 three attic number contrary to
what you'd expect the coefficients
multiplying higher powers of three
actually make finer and finer
adjustments so when we calculated
successive coefficients to solve
diaphanus squares problem we were
actually zooming in more and more
accurately on the solution the this is
the feeling of slowly zooming in on the
value of a number 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
one
but if they differ in the 27's place we
say they differ not by 27 but by 1 over
27 in the world of pics 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 *
1 plus 1 * 3 + 0 * 3^ 2 + 1 1 * 3 cubed
+ 2 * 3 4th and so on will be close to
another number with a similar expansion
except at the let's say 3 to the thir
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 three attic distance is
is roughly of size the first place where
they go wrong so this will be 3 to the
three am I supposed to think of the
threes as kind of like the bigger they
are the smaller the numbers that they're
multiplying or something exactly in the
three addicts we want numbers to be
close when they agree up to large powers
of three so that if they agree up to 10
three addict places then this is has a
distance like 3 to Theus 10 it turns out
that if you do this crazy thing swap big
and small all the other laws of
mathematics work 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 cuz it's less familiar and you
could improve 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 um
non- negative so the absolute value of
any number X should be non- negative and
the absolute value should be zero if and
only if that number is itself zero 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 uh 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 that your function is in
fact the usual absolute value or one of
these ptic absolute values or the thing
that gives zero to zero and one to
everything else so that's the tri
absolute value these are the only games
you can play on the rational numbers
that give that give you absolute values
that behave the way we we want them to
this geometry makes the pics 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 patic
Solutions in a neighborhood of a
rational solution if we tried the same
strategy of solving dianas squares
problem digit by digit in the real
numbers it would be Doom 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 ws and another by
ws and Richard Taylor they finally
proved forma's Last Theorem but the
proof could not possibly have been the
one Forma alluded to in the margin
because it made heavy use of ptic
numbers W's uh proof of fas theorem used
the prime three 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 35 trick there was
something that worked for the prime
three 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 pics are like the stars the sun
blocks out the stars during the day 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 ptic 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
diaphanus FMA and Hensel and you're
looking for more content like this look
no further than the math history course
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