Compiled Articles.
Article No. 13: posted by Alan Marion on Sun, Mar. 26, 2000, 15:26
Subject: Question Answers 
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
Article No. 14: posted by Anna Wright on Sun, Mar. 26, 2000, 16:03
Subject: Question Answers 
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
Article No. 19: [Branch from no. 14] posted by Melissa Shoaf on Mon, Mar. 27, 2000, 09:45
Subject: re: Question Answers 
1.  10110001



2.  00000000
Article No. 21: [Branch from no. 19] posted by Melissa Shoaf on Mon, Mar. 27, 2000, 17:13
Subject: re: Question Answers 
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]
Article No. 15: posted by Larien Chilton on Sun, Mar. 26, 2000, 22:49
Subject: Answers to Maple Demo 
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
Article No. 16: posted by Michael McEllen on Sun, Mar. 26, 2000, 23:10
Subject: homework 
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
Article No. 17: posted by Mackenzie Inman on Sun, Mar. 26, 2000, 23:48
Subject: Maple Demom Answers 
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
Article No. 18: posted by Brandon Brooks on Mon, Mar. 27, 2000, 08:49
Subject: Maple Demos 
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]
Article No. 20: posted by Tracie Wallace on Mon, Mar. 27, 2000, 09:55
Subject: Answers to Questions 
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]