Question

In the context of prime field GF(11), (i) What are the elements of this prime field? (ii) What is the additive inverse of 7 in this field? (iii) What is the multiplicative inverse of 7 in this field?

Answer 1

i) For GF(p) where p is the prime no, the elements are from 0....p-1. Thus, we have GF(11) , where 11 is prime no then elements will be GF(11) = {0,1,2,3,4,5,6,7,8,9,10}

ii)For additive inverse of 7 we have to make a table which looks like this( below in attached image). Add each row no with colomn no then mod them with given p(here it is 11). Now , look at the 7th row(here counting starts with 0) which is highlighted if you see the coloumn which have result as 0. That coloumn is additive inverse of 7. Therefore,[ 7+(some number) mod 11 = 0 ] = additive inverse of 7 in GF(11) = 4

iii)For multiplicative inverse of 7 we have to make a table which looks like this( below in attached image).Multiply each row no with colomn no then mod them with given p(here it is 11). Now , look at the 7th row(here counting starts with 0) which is highlighted if you see the coloumn which have result as 1. That coloumn is additive inverse of 7. Therefore,[ (7 * (some number) ) mod 11 = 1 ] = multiplicative inverse of 7 in GF(11) = 8

Hope you got the answer.