Chapter 1: Question about Finite Field Division about programmingbitcoin

[smh69]

I'll give it a shot jachiang, mostly to help clear my own mind, let me know if this makes sense.

b^-1 is just 1/b being multiplied to both sides of the equation b^(p-1) = 1,
effectively giving us (b^(p-1))/b = 1/b, which is exactly what we want... except we still cannot divide.

But since we can multiply, and pow(), we're just going to multiply b by itself one less time. If b^(p-1) gives us 1, then b^(p-2) will give u 1/b, our multiplicative inverse, which we can use without dividing.

certainly for large primes, we'll never have a negative exponent to deal with, no real division.

Im totally lost on what to say about field of prime=2 {0,1}, we'd get 0^0 = 1 and 1^0 = 1. same goes for prime=3 {0,1,2} as x^1=x, however even in these cases, an element multiplied by it's inverse gets us 1 mod the prime, maybe thats good enough to define division??? x(

[jimmysong]

We're defining b^-1 more than anything else. Division doesn't have meaning until you say it's the inverse of multiplication. To figure out the value analytically, you end up having to use Fermat's Little Theorem and get the b^(p-2)