We can make the same argument for the 10-adic numbers.So if we have an infinite string of 6s and one 7, then multiplying both sides by 10, we get 9.999 repeating equals 10k.Subtract the top equation from the bottom one and we get 9 = 9k, so k = 1, and that’s why 0.6 repeating is equal to 1/3.
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 So all 3-adic numbers are either 0, 1, or 2 repeated infinitely to the left of the decimal point.So what number do we get if all these digits are 0? Well, we can do the same thing we did with the 10-adics, set it equal to m, multiply both sides by 3, and then subtract the equation from itself to find m.So 0.000… equals 3m, and subtracting this equation from itself gives us 0 = 0, meaning m = 0.So this 3-adic string of all 0s is actually 0.Now, let’s try adding 1 to it.Well, 0 + 1 is 1, carry the 1, 0 + 1 is 1, carry the 1, and you just keep doing this all the way down the line and every digit becomes 1.And, you don’t get a 1 all the way down on the left because the 0s go on forever.So all 0s + 1 = 1, therefore all 0s must be equal to 0.This also means that all 0s and then say a 2 equals 2.Now, let’s try multiplying this 3-adic number by itself.So 0 times 0 is 0, carry the 0, 0 times 0 is 0, carry the 0, and you just keep doing this all the way down the line and every digit is still 0.So all 0s times all 0s is 0, and this works out perfectly.We can even verify by multiplying it out.This number 0 times 0 really does work out to 0.This is a special property of 0, and this is why prime number bases are so important. But with the help of p-adics, we can actually solve this problem.
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 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 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 + 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 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. 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 × 3 + x2 × 32 + x3 × 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 × 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 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
- x But in the end, we were able to make sense of them and to find a rational solution to Diophantus’ equation.
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 $1 + 6x_1 + 9x_1^2$. But 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 $1 + 3x_1$. 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 + 9y_1^2, which is also 0 because it contains a factor of 9. So $3 + 15x_1$ 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 $x_1 = 1$, 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 $x_0$ and $x_1$ are equal to 1. But right now, we’ve learned nothing about $y_1$ because when we square, 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 $16 + 18x_2 + 81x_2^2$, 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, $9 + 18$ 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. So the absolute value of x times y should be equal to the absolute value of x times the absolute value of y.And it should be subadditive.So the absolute value of x plus y should be less than or equal to the absolute value of x plus the absolute value of y.And this is exactly what the 3-adic absolute value does.- So by thinking of numbers in this new way, we can actually use the geometric series formula to find that an infinite string of 1s in 3-adic notation is actually -1/2.
By using the geometric series formula, we found that an infinite string of 1s in 3-adic notation is actually -1/2, even though the ratio of each term, the previous 1 is 3, which would lead one to believe the series should have converged to infinity. The key to understanding this is to realize that the geometry of the p-adics is totally different from that of the real numbers, and they don’t exist on a number line at all. One way to visualize them is with something like a growing tree, with 3 base cylinders representing 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 this continues on forever. Looking down from above, it looks like a Sierpinski gasket, and each 3-adic number is represented as a stack of infinite cylinders that get shorter and narrower as they go up. This reflects the relative contributions each successive cylinder makes to the value of the 3-adic number, with coefficients multiplying higher powers of 3 making finer and finer adjustments.
In the world of p-adics, what is normally thought of as big is actually small, and vice versa. To determine the distance between two numbers, one must look at the lowest level of the tower where they disagree. If two numbers differ in the unit’s place, their separation is 1, but if they differ in the 27th’s place, their separation is 1/27. This is why the geometric sum worked out, even though it was expected to blow up to infinity. By thinking of numbers in this new way, the 3-adic absolute value satisfies the criteria of being non-negative, multiplicative, and subadditive, and the geometric series formula can be used to calculate the value of an infinite string of 1s in 3-adic notation. If I take $x$ times $y$ and multiply them, that should be the same as the absolute value of $x$ times the absolute value of $y$. And if I want one more property, which is that 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, which 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. 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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