1 Billion is Tiny in an Alternate Universe: Introduction to p-adic Numbers

limit is the analogue of i; is the square root of -1. This is extremely different from the real

numbers

. In real

numbers

, there is no square root of -1.  But in the 5-

adic

numbers there is. We can show that this limit is the square root of -1 by factoring x^5-x. Since 5-

adic

numbers admit division, the only way for this product to be 0 is for one of the factors to be 0. The first three factors are non-zero, so it must be the last one. The existence of p-adic limits like this is quite common. Let's mix things up by looking at a different type of sequence: the Fibonacci number sequence.

Let's choose p to be 2 and look at the (2^n)th Fibonacci numbers. Fibonacci numbers (2^n) do not converge 2-adically. But if we take

alternate

rows, we get convergence. It's not obvious at all, but in fact these two limits are the two square roots of -3/5. Be careful not to fall into the trap of thinking that these are complex numbers. These are not complex numbers! These are 2-adic integers that, when you square them, multiply them by 5, and then add 3, give 0. Since 2-adic numbers contain square roots of -3/5, what about the square roots of 1? In 2-adic numbers, -1 is the number whose digits are all 1, since when we add 1 to this number we get 0.

So what would a 2-adic integer whose square is -1 look like? There are two possibilities for its last digit: 0 or 1. If it is 0, then 0*0 is 0, then we get a 0 in the product where we want a 1. So the last digit cannot be 0. If the last digit is 1, so 1*1 is 1, which is what we want. So the last digit is 1. What about the second to last digit? If it is 0, then 1*0 + 0*1 is 0, which is not what we want. So it can't be 0. If it's 1, then 1*1 + 1*1 is 2, so we write 0 and carry the 1. This is still not what we want, so the penultimate digit can also be It won't be 1.

We have tried both possibilities for the penultimate digit and neither worked, so there is no square root of -1 in 2-adic numbers. What happens if p=3? Do 3-adic numbers contain square roots of -1? Not there either. But 5-adic numbers contain square roots of -1. We saw this before; one of the square roots is the 5-adic limit of 2^(5^n). Try this on your own... Choose a prime number p and see if you can construct a p-adic integer, from right to left, whose square is -1. If you do this for a set of prime numbers, you can probably guess a characterization of the prime numbers for which this is possible.

Now we have completed our properties table. So what are p-adic numbers for? Well, here's an application: When is n^2+7 divisible by a large power of 2? In other words, when is it highly divisible by 2? When we enter the first n values, the highest power of 2 we see is 2^5. When do we get a value divisible by 2^6? Well, not up to the value 11, and it is actually divisible by 2^7. For 2^8, we have to go further. For 2^9, we have to go up to 75. And the next record occurs at 181. And after that, it's not until 16203. So what are these numbers? What is a more direct way to find them?

If we're trying to make something highly divisible by 2, then we want it to have a lot of zeros at the end of its base 2 representation. This means we want n^2+7 to be close to 0 in 2-adic numbers. Converting this approximation into an equation and solving it, we see that we want n to be close to a square root of -7, 2-adic, if in fact the 2-adic numbers contain square roots of -7, which is not guaranteed. But it turns out that it is, and we can calculate approximations to a square root of -7 just as we could calculate approximations to solutions of an equation in real numbers: using Newton's method.

We set the function that takes x to x minus f(x) over the derivative of f, where f(x) is x^2+7. Then we choose an initial value for x and iterate. For example, we could choose x to be 1; then the function outputs -3. Then we plug in -3 and the function outputs -1/3. Then we plug in -1/3, the function outputs 31/3, and so on. In real numbers, this sequence does not converge. After all, it supposedly approximates a solution of x^2=-7, and the real numbers do not contain square roots of -7. But, thanks to the 2-adic numbers that support division, these rational numbers are also 2-adic numbers, and this sequence converges 2-adic.

By truncating one of these 2-acid integers, we obtain an integer n. And sure enough, n^2+7 is highly divisible by 2. Isn't it beautiful? So p-adic numbers are worthy competitors of real numbers. They have many of the main properties that real numbers have, and unlike real numbers, they allow us to extend rational numbers to a larger number system in a way that reflects properties of integers that are related to divisibility by the cousin p. From there, they allow us to use powerful calculus concepts, such as limits, continuity, and derivatives, to obtain information about whole numbers that we simply cannot obtain any other way.