   Chapter 1.7, Problem 9E

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# In Exercises 6 − 10 , a relation R is defined on the set Z of all integers, In each case, prove that R is an equivalence relation. Find the distinct equivalence classes of R and list at least four members of each. x R y if and only if 3 x − 10 y is a multiple of 7 .

To determine

The proof of the statement,” The relation defined by xRy if and only if 3x10y is a multiple of 7 on the set of all integers Z is an equivalence relation” and also find the distinct equivalence classes of R with at least four members of each.

Explanation

Formula Used:

A relation R on a nonempty set A is an equivalence relation if the following conditions are satisfied for arbitrary x, y, and z in A:

1. xRx for all xA. (Reflexive Property)

2. If xRy, then yRx. (Symmetric Property)

3. If xRy and yRz, then xRz. (Transitive Property)

If R is an equivalence relation on the nonempty set A, then for each aA, the set [a]={xA|xRa} is the equivalence class containing a.

Explanation:

Consider the relation xRy if and only if 3x10y is a multiple of 7 defined on Z.

1. xRx, since 3x10x=7x.

2. xRy3x10y=7kforsomekZx=7k3+10y3forsomekZ

3y10x=3y10(7k3+10y3)=3y70k3100y3=70k391y3=7(10k3+13y3)

Simplify further.

3y10x=7(10k3+7y3+2y)=7(x+2y)

So, xRyyRx.

3. xRyandyRz3x10y=7kand3y10z=7mforsomek,mZ3x=7k+10yandy=7m3+10z3forsomek,mZ

Here, 3x=7k+10(7m3+10z3)=7k+70m3+100z3

Now, 3x10z=7k+70m3+100z310z=7k+70m3+70z3=7k+7(10m3+10z3)=7k+7(m+7m3+10z3)

Simplify further

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