Question

8.) Consider the integers Z. Dene the relation on Z by x y if and only
if 7j(y + 6x). Prove:
a.) The relation is an equivalence relation.
b.) Find the equivalence class of 0 and prove that it is a subgroup of Z
with the usual addition operator on the integers.

8.) Consider the integers Z. Define the relation ~ on Z by x ~ y if and only if 7)(y + 6x). Prove: a.) The relation is an equivalence relation. b.) Find the equivalence class of 0 and prove that it is a subgroup of Z with the usual addition operator on the integers.

Answer 1

③ The relation mon ze defined by any iff 7164+6x). a hef ny € Z and y+60=7k., Now, at het net. Then and since 71 (+62). Tre 717X , HXEZ. This shown that ~ is reflenir... any holdis, Then. 71 (Y+62) and this implies that there exist an integer keit n+by =*+6(7K-Sw) +7(64-5) > 71 Entby) , [since 6K -5K is an integer > yon. Thus any yon , nytt. Hence a is symmetric D leth, J, Z EZ and nny Then 7 (y +6n) and 71 (Z+by) and there fore there exists K2 sot integers 4+603 Z+by =.7k2 from this we have 7+6076,-6) = 732 7 Z+61 = 7(*2-669-60) > 7/2+6n f since Karbk, mon å integer > nne holde. Therefore my and Y~Z) no Z holda, tn, 7, z E ZI. Hence ~ is transitive: confequently the relations is an relation and yuz holds . k, and +60=7kg and an equivalence

6 The o le co] sub group of Z. equivalence class of o OJ = 7867 1.0 onas = {REZI 7/1} = 77 o We now prove that give 77 is a Here 7Z ů a non empty subset of liet a, b € 77. Then a=fki, 6= 7k2 for some ku kz ih Z now, a+b = a+b = 7(ky+ki) E 77 Thus a, b EFZ => a+b € 7 O at Alho a = 7k, 774 → -Q=-7k, = 76K) E 77 80 a E77 a E77 [ Here inverse of a s-a] Therefore by definition of Rubgroup, colie 77 is a subgroup of Z