please answer to all of the questions and write neatly. Other exercises: 1], but that there can be other ways of writing down representatives of the congruence classes, Here are some more examples of (a) Show that z/sz {I-2, (-11, 10, 11, 12) and z/6z f -2l, l-1l, lol (1), (2,13l). (b) Show that (with some terrible notation) Ln/2l 1] where, lrJ means the \"floor\" of r, i the biggest integer m that is S r. .e. 3s), but s not equal to (d) Show that Z/7z (e) Show that there is no a E z such that z/4z {I01 Ullao), lail, la l Solution So for Z/5Z = {-[5-1]/2, -[5/2]+1,-[5/2]+2, -[5/2]+3,[5/2] } ={-2,-1,0,1,2} Z/6Z = {-[6-1]/2, -[6/2]+1,-[6/2]+2, -[6/2]+3,-[6/2]+4,-[6/2]+5,[6/2] } ={-2,-1,0,1,2,3} Now 20 % 5 = 1 21 % 5 = 2 22 % 5 = 4 23 % 5 = 3 24 % 5 = 1 25 % 5 = 2 As we can see repetition occurs. Therefore it is cyclic. Now 30 % 5 = 1 31 % 5 = 3 32 % 5 = 4 33 % 5 = 2 34 % 5 = 1 35 % 5 = 3 As we can see repetition occurs. Therefore it is cyclic. 0 % 7 = 0 30 % 7 = 1 31 % 7 = 3 32 % 7 = 2 33 % 7 = 6 34 % 7 = 4 35 % 7 = 5 36 % 7 = 1 37 % 7 = 3 As repetition get started. Therefore Z7 is cyclic for power of three. 20 % 7 = 1 21 % 7 = 2 22 % 7 = 4 23 % 7 = 1 24 % 7 = 2 25 % 7 = 4 26 % 7 = 1 From above we can say that Z/7Z is not {[0]} union {[20][21][22] [23] [24][25]} a0 % 4 = 1 a1 % 4 = 2 a2 % 4 = 0 a3 % 4 = 2 a4 % 4 = 0 From this we can say Z/4Z is not equal to {[0]} union {[a0][a1][a2] } Let say Let ‘a’ is odd no. so it can give only reminder 1 or 3 not 2 That is why we can say Z/4Z is not equal to {[0]} union {[a0][a1][a2] }.

1. please answer to all of the questions and write neatly. Other exercises: 1], but that there can be
other ways of writing down representatives of the congruence classes, Here are some more
examples of (a) Show that z/sz {I-2, (-11, 10, 11, 12) and z/6z f -2l, l-1l, lol (1), (2,13l). (b)
Show that (with some terrible notation) Ln/2l 1] where, lrJ means the "floor" of r, i the biggest
integer m that is S r. .e. 3s), but s not equal to (d) Show that Z/7z (e) Show that there is no a E z
such that z/4z {I01 Ullao), lail, la l
Solution
So for Z/5Z = {-[5-1]/2, -[5/2]+1,-[5/2]+2, -[5/2]+3,[5/2] }
={-2,-1,0,1,2}
Z/6Z = {-[6-1]/2, -[6/2]+1,-[6/2]+2, -[6/2]+3,-[6/2]+4,-[6/2]+5,[6/2] }
={-2,-1,0,1,2,3}
Now 20 % 5 = 1
21 % 5 = 2
22 % 5 = 4
23 % 5 = 3
24 % 5 = 1
25 % 5 = 2
As we can see repetition occurs. Therefore it is cyclic.
Now 30 % 5 = 1
31 % 5 = 3
32 % 5 = 4
33 % 5 = 2
34 % 5 = 1
35 % 5 = 3
As we can see repetition occurs. Therefore it is cyclic.
0 % 7 = 0
30 % 7 = 1
31 % 7 = 3
32 % 7 = 2
33 % 7 = 6
34 % 7 = 4
35 % 7 = 5

2. 36 % 7 = 1
37 % 7 = 3
As repetition get started. Therefore Z7 is cyclic for power of three.
20 % 7 = 1
21 % 7 = 2
22 % 7 = 4
23 % 7 = 1
24 % 7 = 2
25 % 7 = 4
26 % 7 = 1
From above we can say that Z/7Z is not {[0]} union {[20][21][22] [23] [24][25]}
a0 % 4 = 1
a1 % 4 = 2
a2 % 4 = 0
a3 % 4 = 2
a4 % 4 = 0
From this we can say Z/4Z is not equal to {[0]} union {[a0][a1][a2] }
Let say
Let ‘a’ is odd no. so it can give only reminder 1 or 3 not 2
That is why we can say Z/4Z is not equal to {[0]} union {[a0][a1][a2] }