Recent content by krispiekr3am

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) Prove that g(x) is irreducible over GF(2). can someone help me? my email is krispiekr3am@yahoo.com

Are the following polynomials irreducible over Z2? (a) x2 + x + 1 (b) x2 + 1 (c) x2 + x

(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 did this using excel 6^100mod13 is equal to 9? if its right, i think i did it right. 5^100mod13 is equal to 1?

Thank You So Much. That Really Helped. Thanks Everyone

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?

B Xor (b Ξ C)

Bc Xor B'c' =(b Xor C)' =b+c' =b' Xor C = Bc + B'c' = B=c? So The Answer Would Be B Xor (b = C) = 0??

A'b Xor Ab = B?

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

x^(a*b)? i don't quiet understand or how that help us.