The Strange Number System Where Infinity Is Tiny

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. We have just found that there are rational numbers, fractions, in the 10-adic numbers without having to use the divided by symbol. What we have just found is that 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. 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. So if we take all 0s, this is the same thing as 0.If we take all 1s, this is the same thing as 1.And if we take all 2s, this is the same thing as 2.Now, let’s take all 1s and add 1 to it.So this is 1 + 1, which is 2, carry the 1, 1 + 1 is 2, carry the 1, and every digit just becomes 0.So all 1s + 1 = 0.Now, let’s try multiplying this by itself.So 1 times 1 is 1, carry the 1, 1 times 1 is 1, carry the 1, and every digit just becomes 0.So all 1s times all 1s is 0.Now, this is the same thing as our 10-adic number, but it works because 3 is a prime number, and the only two numbers that multiply to 0 in base 3 are 0 and 0.So this means that if we use a prime number base, then we can still factor equations, and this is why mathematicians always use prime number bases when working with infinite numbers, so that they can still use the same tools they’re used to.

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 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 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 But with the help of the p-adics, we can find them.

Now, how could you multiply two 3-adic numbers to get 0? Well, again, we can start by just looking at the last digit, 1 x 1 = 1, 2 x 1 = 2, and 2 x 2 = 4, which, in base 3, is 11. So the only way to get a 0 is if one 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 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 3^0 + 2 x 3 + 1 x 3^2 + 1 x 3^3, 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 = 3, which in base 3 is 10. So you leave the 0, carry the 1, and 2 + 1 = 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^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 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^n + y^n = z^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-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 first instance of algebra many, many centuries before, al-jabr. So if we set the side of the first square to be x, then its area is x^2. 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 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 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 = x₀ + x₁ $\times$ 3 + x₂ $\times$ 3² + x₃ $\times$ 3³, 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 $\times$ 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, 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.

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 x₀² + x₀⁴ + x₀⁸ = y₀². And this will allow us to find the values of x₀ and y₀ that satisfy the equation mod 3.

Now we know that x₀ can be either 0, 1, or 2, and y₀ can also be either 0, 1, or 2. If x is 0, then so is x², x⁴, and x⁸. If x is 1, then x² is 1, and so is x⁴, and x⁸. If x is 2, then x² 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, and squaring again, x⁸ = 1. Now we can sum up x² + x⁴ + x⁸ 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 this number is actually a real number, and it’s equal to -1/2.

All of the terms higher than $x_1$ have a factor of 9 in them, so they’re all 0 mod 9. We’re left with the expression $1 + 6x_1 + 9x_1^2$. Since we’re working mod 9, the last part is 0. The next term is just the first term squared, so that equals $1 + 12x_1 + 36x_1^2$, but 36 is 9 x 4, so that’s 0 mod 9, and 12 mod 9 is 3. So we have $1 + 3x_1$, and the final term is just that squared, so $1 + 6x_1 + 9x_1^2$. Again, the last part is 0. So on the left-hand side, we have $3 + 15x_1$, and on the right-hand side, we have 0 + 9$y_1^2$, which is also 0 because it contains a factor of 9. So $3 + 15x_1 = 0$. 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 $x_1 = 1$, then $3 + 15 = 18$, which is a multiple of 9, so it’s 0. So $x_1 = 1$ 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 $x_1$ and $x_2$ are equal to 1. We can simplify to this. Expanding again, we get $16 + 18x_2 + 81x_2^2$, but 81 is 27 x 3, so that’s 0. The next term is the square of the first. So $256 + 576x_2 + 324x_2^2$. 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 $13 + 9x_2$, and the last term is just that squared. So $169 + 234x_2 + 81x_2^2$, which reduces to $7 + 18x_2$. The right-hand side reduces to 0 + 81$y_2^2$, which is, again, 0. So we have $36 + 45x_2 = 0$, which mod 27 is the same as $9 + 18x_2 = 0$. So $x_2$ must be equal to 1, $9 + 18 = 27$, 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. So if we have two numbers, x and y, then the absolute value of x times y should be the absolute value of x times the absolute value of y.And it should be consistent with the distance between two numbers.So if we have two numbers, x and y, the absolute value of x minus y should be less than or equal to the distance between x and y.

We used the geometric series formula to find that an infinite string of 1s in 3-adic notation is actually -1/2, even though the ratio of each term is 3. This result was surprising as it seemed like the series should have blown up to infinity. The key to understanding this lies in the geometry of the p-adics, which do not exist on a number line. Instead, they are represented as an infinite triply branching tree, which looks like a Sierpinski gasket. Each 3-adic number is represented by a stack of infinite cylinders that get shorter and narrower as they go up, and this reflects the relative contributions each successive cylinder makes to the value of the 3-adic number.

In the world of p-adics, what we’re used to thinking of as big is small, and vice versa. If two numbers differ in the unit’s place, we say their separation is 1, and if they differ in the 27th’s place, we say they differ not by 27, but by 1/27. This means that the distance between two numbers is determined by the lowest level of the tower where they disagree.

To make sense of this, we have to open our mind to other notions of size. When we do this, we find that this new world fits all the criteria for an absolute value: it is non-negative, multiplicative, and consistent with the distance between two numbers. If I take $x$ times $y$ and multiply those two, that should be the same as the absolute value of $x$ times the absolute value of $y$. But if I 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. 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.

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. 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.

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