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. So what does this number equal?Well, just like before, we can set it equal to say m and then multiply both sides by 3.So we have 22222222 equals 3m.Now, subtract this equation from the first one, and we get 1 = -2m, meaning m = -1/2.So this 3-adic string of all 2s is actually -1/2.Now, if we try adding 1 to it, we get 0, and if we try adding 2, we get 1, so this works out exactly as we’d expect it to.Plus, these numbers can be multiplied and squared, and they will never give us 0 as a result, so we can solve equations using the 3-adics.
The 10-adic number of all 9s is actually equal to -1. We can verify this by adding 1 to it, and all the digits become 0. We can also find the negative of any 10-adic number by taking the 9s complement and then adding 1.
Negative numbers are included in the 10-adics, and subtraction can be done by adding the negative of the number. However, a problem arises when we try to factor the first 10-adic number we found, as it is its own square. This is because 10 is a composite number, and not a prime.
To avoid this problem, we can use a prime number base instead of base 10, such as 2, 3, 5, 7, etc. As an example, the 3-adic number of all 2s is equal to -1/2. When we add 1 to it, we get 0, and when we add 2, we get 1. This works out exactly as we’d expect it to, and equations can be solved using the 3-adics. Now, how could you multiply two 3-adic numbers to get 0? Well, the only way to get a 0 is if one of the 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. 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.
P-adics, such as 3-adic numbers, 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. This is why professional mathematicians work with p-adics rather than 10-adics. P-adics were even involved in cracking one of the most legendary math problems of all time - Fermat’s Last Theorem. 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, finding 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. 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 mod 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 again, we’re working But this number is actually just a way of writing 1/2 + 1/4 + 1/16 + 1/256 and so on.
All of the terms higher than $x_1$ have a factor of 9 in them, so they’re all 0 mod 9. So we’re left with this expression:
$$ 1 + 6x_1 + 9x_1^2 $$
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 $1 + 12x_1 + 36x_1^2$, 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 $3 + 15x_1 = 0$. 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. So the absolute value of a product should be the product of the absolute values.And it should be sub-additive, so the absolute value of a sum should be less than or equal to the sum of the absolute values.
The geometric series formula was used to find that an infinite string of 1s in 3-adic notation is actually -1/2, despite the ratio of each term being 3. This goes against common sense, as the series should have blown up to infinity. However, the key idea is that the geometry of the p-adics is different from that of the real numbers, and they do not exist on a number line. One way to visualize them is with a growing tree with 3 base cylinders representing the unit’s digit, and above each cylinder is a trio of shorter cylinders corresponding to the 3s digit. This reflects the relative contributions each successive cylinder makes to the value of the 3-adic number, as the coefficients multiplying higher powers of 3 make finer and finer adjustments. To determine the distance between two numbers, we look at the lowest level of the tree where they disagree, and if they differ in the unit’s place, we say their separation is 1. However, 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. This notion of size fits the criteria for an absolute value, as it is non-negative, multiplicative, and sub-additive. 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 $y$. 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.
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. They inspire us to find new connections and even make discoveries ourselves.
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