But what about 0.6 repeating? We can do the same thing, just replace the 9s with 6s.So 6.666 repeating equals 10k, subtract the top equation from the bottom one and we get 6 = 6k, so k = 1.So 0.666 repeating is equal to 1.
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 So if we set this equal to m, then we can multiply both sides by 3 and get 2 = 3m.Now, if we subtract this equation from the first one, we get 1 = -2m.So this 3-adic number of all 2s is equal to -1/2.Now, if we add 1 to this number, it’s gonna be a bit different from what happened in the 10-adics.So 2 + 1 is 3, carry the 1, 2 + 1 is 3, carry the 1, and all the way down the line, the digits become 0, and then the 1 will appear on the left-hand side.So this 3-adic number of all 2s + 1 is 1/2.
All 9s in a 10-adic number is equal to -1. Adding 1 to this number results in all 0s. To avoid this problem, a prime number base can be used, such as 3-adic numbers. For example, a 3-adic number of all 2s is equal to -1/2. Adding 1 to this number results in 1/2. 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 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 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. 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? - 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 $$x = x_0 + x_1 \times 3 + x_2 \times 3^2 + x_3 \times 3^3,$$ and $$y = y_0 + y_1 \times 3 + y_2 \times 3^2 + y_3 \times 3^3,$$ 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. $$36 \mod 10 \text{ or } 36 \mod 10$$ 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 $$x_0^2 + x_0^4 + x_0^8 = y_0^2.$$ And this will allow us to find the values of $x_0$ and $y_0$ that satisfy the equation mod 3. Now we know that $x_0$ can be either 0, 1, or 2, and $y_0$ can also be either 0, 1, or 2. If $x_0$ is 0, then so is $x_0^2$, $x_0^4$, and $x_0^8$. If $x_0$ is 1, then $x_0^2$ is 1, and so is $x_0^4$, and $x_0^8$. If $x_0$ is 2, then $x_0^2$ is 4. But remember, we’re working mod 3, and 4, mod 3 is just 1. To find $x_0^4$, we can just square $x_0^2$, so that also equals 1, and squaring again, $x_0^8 = 1$. Now we can sum up $x_0^2
- x_0^4 + x_0^8$ to find the left-hand side of the equation. If $x_0$ is 0, then the sum is equal to 0. If $x_0$ 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 $ But by taking a step back and looking at the bigger picture, we can see that these numbers can be used to solve real-world problems.
All of the terms higher than $x_1$ have a factor of 9 in them, so they’re all 0 mod 9. Expanding out the first term, we have 1 + 6$x_1$ + 9$x_1^2$. 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 + 12$x_1$ + 36$x_1^2$, but 36 is 9 x 4, so that’s 0 mod 9, and 12 mod 9 is 3. So we have 1 + 3$x_1$. The final term is just that squared, so 1 + 6$x_1$ + 9$x_1^2$. Again, the last part is 0. So on the left-hand side, we have 3 + 15$x_1$. 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 + 15$x_1$ 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 $x_1$ and $x_2$ 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 + 18$x_2$ + 81$x_2^2$, but 81 is 27 x 3, so that’s 0. The next term is the square of the first. So 256 + 576$x_2$ + 324$x_2^2$. 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 + 9$x_2$, and the last term is just that squared. So 169 + 234$x_2$ + 81$x_2^2$, which reduces to 7 + 18$x_2$. The right-hand side reduces to 0 + 81$y_2^2$, which is, again, 0. So we have 36 + 45$x_2$ equals 0, which mod 27 is the same as 9 + 18$x_2$ = 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.
But how do we make sense of this? What does this number equal? 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 $x^2$ is 1/4, $x^4$ So the absolute value of the product of two numbers should be the product of their absolute values.And it should satisfy the triangle inequality.So the absolute value of the sum of two numbers should be less than or equal to the sum of their absolute values.And the 3-adic absolute value satisfies all of these conditions.
Using the geometric series formula, we found that an infinite string of 1s in 3-adic notation is actually equal to -1/2. This is surprising since the ratio of each term is the previous 1 is 3, which should have resulted in the series blowing up to infinity. The key to understanding this is realizing that the geometry of the p-adics is 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 a growing tree, with 3 base cylinders representing the units 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, with coefficients multiplying higher powers of 3 making finer and finer adjustments. To determine the distance between two numbers, we 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. This is because the 3-adic absolute value satisfies the criteria of being non-negative, multiplicative, and satisfying the triangle inequality. 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. If you were inspired by the stories of Diophantus, Fermat, and Hensel, and you’re looking for more content like this, look no further than the math history course from this video sponsor, Brilliant.org.
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