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References:  

Koblitz, N. (2012). p-adic Numbers, p-adic Analysis, and Zeta-Functions (Vol. 58). Springer Science & Business Media.

Amazing intro to p-adic numbers here: https://youtu.be/3gyHKCDq1YA
Excellent series on p-adic numbers: https://youtu.be/VTtBDSWR1Ac
Great videos by James Tanton: @JamesTantonMath 

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Written by Derek Muller and Alex Kontorovich 
Edited by Trenton Oliver
Animated by Mike Radjabov, Ivy Tello, Fabio Albertelli and Jakub Misiek
Filmed by Derek Muller
Additional video/photos supplied by Getty Images & Pond5
Music from Epidemic Sound & Jonny Hyman
Produced by Derek Muller, Petr Lebedev, & Emily Zhang

Take the number 5 and square it, you get 25.

5 squared ends in a 5, 25 squared ends in 25, and 625 squared ends in 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.

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.

And if you square it, you'll get back that same number.

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?

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 1 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 tenadic numbers because they're written in base 10.

If you have 2 tenadic numbers, can you add them together?

Sure, you just go digit by digit from the right to the left, adding them up as usual.

Well, again, you can take any 2 tenatic numbers and multiply them out.

This works because the last digit of the answer depends only on the last digits of the tenadic 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 1 tenadic number ending in 857142857143 and multiply it by 7.

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 tenatic number must be equal to one-seventh.

What we have just found is that there are rational numbers, fractions, in the tenatic numbers without having to use the divided by symbol.

Say you want to find the tenatic number that equals a third.

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 units place?

Well, 3 times 7 is 21, so that gives us the 1 and then we carry the 2.

3 times 6 equals 18 plus 2 equals 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 0.

So an infinite string of 6s and 1 7 is equal to 1 third.

This looks similar to the infinite digits we're used to going off to the right of the decimal point, like .999999 repeating.

Well, I'll claim that it's exactly equal to 1.

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 1 to get 9 equals 9K.

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 9s went to the left of the decimal place.

Set it equal to say M and then multiply both sides by 10.

Now subtract this equation from the first 1 and we get 9 equals negative 9m, meaning m equals negative 1.

So this tenatic string of all nines is actually negative 1.

Now I know that seems weird so let's try adding 1 to it.

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 going to end up with a 1 all the way down on the left, but this never happens because the 9s go on forever.

All 9s plus 1 equals 0, therefore all 9s must be equal to negative 1.

This also means that all nines and then say a 3 equals negative 7.

What we have just discovered is that the 10 addicts contain negative numbers as well.

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 tenatic number, you could multiply by all nines, Or just perform these 2 steps.

First, take the nines complement, that is the difference between each digit and 9, and then add 1.

So if this number is 1 7th, then negative a 7th is 142857142856 plus 1.

And we can verify that this is indeed negative a seventh by adding it to positive 1 seventh and finding that these numbers annihilate each other to 0.

So to sum up, tenatic 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 1 big problem, and you can see it with the first tenatic number we found.

Remember, if you multiply this number by itself, you get back that same number.

And that's a problem which you can see if we move the n 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 kinetic 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.

Have you ever thought about why when faced with some complicated equation we move all the terms to 1 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 1 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.

And fundamentally, the reason is because we're working in base 10 and 10 is a composite number.

Say you wanna find 2 10 addict numbers that multiply to 0.

Then to start off, you know that the last digit must be 0.

So which 2 numbers could you multiply to get a 0 in the units place?

Well, you could multiply a 0 times any number, that's no problem, but you could also multiply say 5 times 4 and get 20, which gives you a 0 in the units place.

Then you can carry the 2 and find another 2 numbers that will give you a 0 in the tens 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, etc.

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 1, 0.

Now, how could you multiply 2 3-addict numbers to get 0?

Well, again, we can start by just looking at the last digit.

1 times 1 is 1, 2 times 1 is 2, and 2 times 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 to 0 is if 1 of the threeadic 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 1 of those numbers is itself 0.

This number means 1 times 3 to the 0 plus 2 times 3 plus 1 times 3 squared plus 1 times 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 negative 1 would be an infinite string of 2's.

If you add 1 then 2 plus 1 is 3 which in base 3 is 1 0.

So you leave the 0, carry the 1, and 2 plus 1 is again 1 0.

So you carry the 1, and you keep going on like that forever.

P-addicts have all the same properties as the 10-addicts, but in addition, you will never find a number that is its own square besides 0 and 1, nor will you find 1 non-zero number times another non-zero number being equal to 0.

And this is why professional mathematicians work with P addicts, where the P stands for prime, rather than say the 10 addicts.

They have been used in the work of over a dozen recent Fields Medalists.

They were even involved in cracking 1 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 plus y squared equals 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.

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 1 of the most infamous of all time.

The equation x to the n plus y to the n equals z to 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.

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-addix.

And these provide a systematic method for solving other problems in Diophantus's Arithmetica.

For example, find 3 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 first instance of algebra, many many centuries before algebra.

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 3 areas, x squared plus x to the 4 plus x to the 8, to add to make a new square.

So x squared plus x to the 4 plus x to the 8 equals 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.

You know, 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 1 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 equals x naught plus x 1 times 3 plus x 2 times 3 squared plus x 3 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.

Well 1 way to do it is to divide 17 by 3 and find the remainder, which in this case is 2.

So we know the units 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 times 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.

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 kind of like this with a modulus of 12.

The hours increase up to 11, but then 12 o'clock is the same thing as 0 o'clock.

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 p.m.

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 3 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 01:00 on a 3 hour o'clock.

What's great about this approach is it allows us to work out the coefficients in our expansion 1 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 plus x not to the fourth plus x not to the 8 equals y not squared.

And this will allow us to find the values of x-naught and y-naught that satisfy the equation mod 3.

Now we know that x-naught can be either 0, 1, or 2, and y-naught 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 x⁴, we can just square x², so that also equals 1.

Now we can sum up x² plus x⁴ plus x⁸ to find the left-hand side of the equation.

But again, we're working mod 3, so 3 is the same as 0.

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 equals 0 but x can be 0 1 or 2 So we have 3 potential solutions that satisfy our equation mod 3.

Now we shouldn't be surprised to find 0, 0 as a solution, since x equals 0 and y equals 0 does satisfy the equation.

But squares of 0 size don't really count as solutions to Diophantus's geometric problem So let's try to expand on 1 of the other solutions I'll pick 1 0 this means x naught equals 1 and y naught equals 0 satisfies the equation mod 3.

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.

Expanding out the first term, we have 1 plus 6x1 plus 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 plus 12 x 1 plus 36 x 1 squared but 36 is 9 times 4 so that's 0 mod 9 and 12 mod 9 is 3 so we have 1 plus 3 x 1.

The final term is just that squared, so 1 plus 6x1 plus 9x1 squared.

Again, the last part is 0, so on the left-hand side we have 3 plus 15x1.

And on the right-hand side we have 0 plus 9y1 squared, which is also 0 because it contains a factor of 9.

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 equals 1, then 3 plus 15 equals 18, which is a multiple of 9, so it's 0.

So x1 equals 1 is a solution to the equation.

Let's find 1 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 they're 0, leaving only this expression.

But we know x0 and x1 are equal to 1, So we can simplify to this.

Expanding again, we get 16 plus 18x2 plus 81x2, but 81 is 27 times 3, so that's 0.

The next term is the square of the first, so 256 plus 576x2 plus 324x2 squared.

But 324 is 27 times 12, so that's 0 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 plus 9 x2 and the last term is just that squared so 169 plus 234 x2 plus 81 x2 squared which reduces to 7 plus 18 x2 the right-hand side reduces to 0 plus 81 y squared which is again 0.

So we have 36 plus 45x2 equals 0, which mod 27 is the same as 9 plus 18x2 equals 0.

So what we've discovered is the first 3 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.

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 1 times 3 to the 0, plus 1 times 3, plus 1 times 3 squared, plus 1 times 3 cubed, and so on.

So each term is just 3 times the term before it.

And to find the sum of an infinite geometric series, you can use the equation 1 over 1 minus lambda, where lambda is the ratio of 1 term to the previous 1.

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 equals 3 and see what happens.

Well then we have 1 over 1 minus 3, which is negative a half.

So if we believe this formula, then x equals negative a half should be a solution to our original equation.

If we sub it in, we get x squared is a quarter, x to the 4 is a sixteenth, and x to the 8 is 1 over 256.

Let's put all of these over a common denominator.

So a quarter becomes 64 over 256, sixteenth 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 9 16ths.

We have found a rational solution to Diophantus' sum of squares problem.

The first square has sides of length 1 half, the second has sides of length 1 quarter, the third has sides of length 1 sixteenth, and all 3 squares together make a square with sides of length 9 sixteenths.

To find this solution, we used new seemingly absurd numbers, p-addicts, 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 ones in 3-addict notation is actually negative a half.

This works even though the ratio of each term to the previous 1 is 3, so by common sense the series shouldn't have converged, it should have blown up to infinity.

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Elite Mathematicians Use Numbers Differently From The Rest of Us