Answer to Question #145983 in Abstract Algebra for Sourav Mondal

Question #145983
Define a relation R on the set of integers Z
by R= {(n, n+ 3k) I k belongs to Z}. Show that R is an equivalence relation. Also find all distinct
equivalence classes.
1
Expert's answer
2020-11-24T05:26:48-0500

By given relation, aRbaRb if b=a+3k,kZb = a+3k, k \in \Z .

Reflexivea=a+3(0)a = a+3 (0)     aRa\implies aRa . So RR is an reflexive relation.

Symmetric: Let aRb    b=a+3kaRb \implies b = a+3k where kk is an integer.

    a=b3k=b+3(k)=b+3k1\implies a = b - 3k = b + 3(-k) = b + 3k_1 and k1=kZk_1 = -k \in \Z     bRa\implies bRa.

So. RR is an symmetric relation.

Transitive: Let aRb,bRcaRb, bRc     b=a+3k1,c=b+3k2\implies b = a+3k_1 , c =b +3k_2

    c=a+3k1+3k2=a+3(k1+k2)=a+3k\implies c = a+3k_1 +3k_2 = a+3(k_1+k_2) = a+3k where k=k1+k2Zk = k_1+k_2 \in \Z.

Hence, aRcaRc . So, RR is an transitive relation.

Thus, RR is an equivalence relation.


Equivalence class of 0 = {3k:kZ}={_ _ _,6,3,0,3,6,_ _ _}\{3k: k\in \Z\} = \{ \_ \ \_ \ \_, -6, -3, 0, 3 , 6, \_ \ \_ \ \_ \}

Equivalence class of 1 = {3k+1:kZ}={_ _ _,5,2,1,4,7,_ _ _}\{3k+1 : k\in \Z\} = \{ \_ \ \_ \ \_, -5, -2, 1 , 4 , 7, \_ \ \_ \ \_ \}

Equivalence class of 2 = {3k+2:kZ}={_ _ _,4,1,2,5,8,_ _ _}\{3k+2 : k\in \Z\} = \{ \_ \ \_ \ \_, -4, -1, 2 , 5 , 8, \_ \ \_ \ \_ \}

These three distinct equivalence classes are possible.



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