Let \(\displaystyle{A}={\left\lbrace{0},{1},{2},{3},{4},{5},{6},{7},{8},{9},{10}\right\rbrace}{\quad\text{and}\quad}\) \(\displaystyle{R}={\left\lbrace{r}{\left({x}_{{{1}}}{y}\right)}:{7}\right)}{\left({2}{x}+{5}{y}\right)}\) where \(\displaystyle{x}_{{{1}}}{y}\epsilon{A}\rbrace\) Phase that

David Rodgers

David Rodgers

Answered question

2022-03-23

Let A={0,1,2,3,4,5,6,7,8,9,10}and
R={r(x1y):7)(2x+5y) where x1yϵA}
Phase that R is an equivalence relation on A!

Answer & Explanation

Cecilia Nolan

Cecilia Nolan

Beginner2022-03-24Added 13 answers

Step 1
Given: A={0,1,2,3,4,5,6,7,8,9,10}
R={(x1y):7)(2x+5y) where x1yϵA}
We have to prave that R is an wquivalence
Now we day r is an equivalence velation on
A:if
a) reflexiving: for all aϵA, a R a
b) symmetre: for all a,bϵA if a R b
then b R a
c) transitibity:foe all a,b,cϵA, if
a R b and b R c then a R c
Step 2
Now when (x1x),then 2x+5x=7x,xϵA
which is always divide by 7
So R is reflexive
Now suppose x R y ie 72x+5y,xyϵA
then we have that
2y+6x=7(x+y)(2x+5y)
then 7(2y+5x)(y1x)ϵR
Step 3
So (x1y)ϵR implies (y1x)ϵR
Thefore R is transitive
Now suppose x R y and y R z where x1y1zϵA
Then (x1y)ϵR,(y1z)ϵR
Then 7|2x+5y,7|2y+5z
ie 2x+5y=7k and 2y+5x=7h for some h,kϵ
Now
2x+5z=(7k5y)+(7h2y)
=7k+7h7y
=7(k+hy)
Since k+hyϵZ
Thefore 7|2x+5z (x1z)ϵR
So,(x1y)ϵR,(y1z)ϵR(x1z)ϵR,x1y1zϵA
ie r is transitive 
thefore r is an equivalence relation on A

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