Wikipedia article on Finite Field Arithmetic says addition and subtraction are equivalent operations a+b=a^b and a-b=a^b.

I'm trying to implement multiplication and division in GF(256) by using the identity a*b=antilog(log(a)+log(b)) mod 255 and a/b =anitlog(log(a)-log(b)) mod 255

Does this mean that the addition and subtraction in the multiplication and division identities above are equal to a^b?

Wouldn't this give you the same result for multiplication and division?

## finite field question

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: finite field question

You seem to be assuming that log in this context is a function from GF(2

The multiplicative group of GF(2

^{n}) to GF(2^{n}), but it is not.The multiplicative group of GF(2

^{n}) is cyclic of order 2^{n}-1. If g is a generator of that group, then you are multiplying two elements a=g^{e}and b=g^{f}as a*b = g^{e}g^{f}= g^{e+f}= c. The addition of these exponents is not done in the field GF(2^n) itself but in (an additive version of) this cyclic group. So log is a function from GF(2^{n}) to Z_(2^{n}-1).### Who is online

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