please answer to all of the questions and write neatly Other

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] }