compute the inverse of 550 in GF (1997). Notice GF (p), Zp, or Ip I covered before consisting of all integers 0, 1, .., p -1 modulo p are the same thing with different names. Can we compute the inverse of 550 in Z 1995 ? Why?
(a) Is (Z4, +) a group? Is (Z4, +, *) a ring? Explain.
(b) Is Z4 a field, in other words, does every integer in Z4 have a multiplicative inverse?
(c) Generate the addition table and multiplication table of GF(4).
can someone help me. i am clueless?
i follow the notes from my class.
the teacher created a chart of
all possible number n, 6^n, 6^n%13.
6^100 = (6^10)^10?
and knowing that 6^10=4
4^10mod13 =9
therefore 6^100mod13=9
n 6^n 6^n % 13
1 6 6
2 36...
i think i did the 1st problem wrong...
here is the simplied solution
A'B xor BC xor AB xor B'C'
becomes
A'B xor AB xor BC xor B'C'
A'B xor AB = B (we know that A A' =1)
BC xor B'C'
= (B xor C')(C xor B')
B xor (B xor C')(C xor B')
the answer to 5x = 1 (mod 13) is x = 8 (mod 13)
and the answer to 13x = 2 (mod 23) is x = 9 (mod 23)
and the answer to 37x = 5 (mod 13) is x = 4 (mod 13)
TO approve this other problem
AB XOR A'B XOR A'B' XOR B'A
B(A XOR A') XOR B'(A' XOR A)
we know that A XOR A' =1
B XOR B'
and from one of the XOR property, X XOR X' = 1
THEREFORE B EQUALY 1?
THE SOLUTION WOULD BE 1?
let me know if that right?
x^a*x^b
=x^(a+b)
?
can you guys give me the answer and i can work backward? because i can try many ways and will not know its the correct answer or not.
thanks