Problem 3 Explain why a5 amod5 for any positive integer a Hi

Problem 3: Explain why a^5 a(mod5) for any positive integer a. (Hint: think about the congruence classes [a]5).

Solution

Let take set Z. On that set you have the relation

R={(x,y)xy(mod5)}

Another way to say this is that xx and yy are equivalent (xy) if xy is divisible by 5.

Now for an integer nn you have the equivalence class (congruence class).

This class is defined as all those x for which xn. So for example if n=2, you get

           [2]₅

={xZx2}={xZx2(mod 5)}

={xZ:5x2}={…,8,3,2,7,12,…}

in this case i took a=2 it is real number and positive

         congruence mod 5\" means; thus [n]5=n+5

                                                             a+5 is positive integer.