1. 10110000 2. [1,0,0,1,1,1,0,0] + [1,0,0,1,1,1,0,0] mod 2; 3. Other than the fact that it should say cases 1 and 4 rather than cases 1 and 2, the proof seems to be correct. 4. The associativity property 5. 01010101

1. 10110000 2. [1,0,0,1,1,1,0,0,]+[1,0,0,1,1,1,0,0]; 3. The proof is more than likely correct since you have shown that 1 and 4 work, but what about 2 and 3. Would that not show a contradiction? 4. the associative property 5. 10101010

1. 10110001 2. 00000000

1. 10110001 2. [1,0,0,1,1,1,0,0]+[1,1,1,1,1,1,1,1] mod 2; =[0,0,0,0,0,0,0,0] =00000000 3. Yes, it is. 4. #2 from definition of a group: for all a,b,c in G a*(b*c)=(a*b)*c associative property 5. [1,1,1,1,1,1,1,1]+[0,1,0,1,0,1,0,1] mod 2; =[1,0,1,0,1,0,1,0]

1) 10110001 2) [1,0,0,1,1,1,0,0]+[1,0,0,1,1,1,0,0] mod 2; = [0,0,0,0,0,0,0,0] 3) The proof states that case 1 and case 2 satisfy a_i+b_i mod 2=0, but case 3 and case 4 do not satisfy a_i+b_i mod 2=0. Therefor it would not be correct to to say that a_i=b_i for all i. If the proof stated that this was true because of case 1 and case 4 I think that it would be ok. 4) #2 for the definition of group under * ( associative) 5) 01010101

1. 10110000 2. [1,0,0,1,1,1,0,0]+[1,0,0,1,1,1,0,0] mod 2 3. yes 4. #2 for all a,b,cEG a*(b*c)=(a*b)*c (assoc) 5. p= 10101010

1) 10110000 2) [10011100]+[10011100]mod 2; 00000000 3)Yes with the exception on you say 2 instead of 4 at then end. 4)Assocative 5)10101010

1.[1,0,1,1,0,0,0,0]=10110000 2.[1,0,0,1,1,1,0,0]+[1,0,0,1,1,1,0,0]mod2; 3.yes 4.Associative group property 5.[1,0,1,0,1,0,1,0]

Question 1: (1 mod 2,2 mod 2,1 mod 2,1 mod 2,0 mod 2,2 mod 2,2 mod 2,0 mod 2)-----(1,0,1,1,0,0,0,0)------10110000 Question 2: [1,0,0,1,1,1,0,0] + [1,0,0,1,1,1,0,0] mod 2; [0,0,0,0,0,0,0,0] Question 3: The proof is not correct due to the fact that it states that cases 1 and 2 satisfy the given information. The actual fact is that cases 1 and 4 satisfy a_i + b_i mod 2=0. Question 4: Property two satisfies the given for k_i and p in the given group. Question 5: [1,1,1,1,1,1,1,1] + [0,1,0,1,0,1,0,1] mod 2; gives you the password of [1,0,1,0,1,0,1,0]