- 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.
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2023-06-09