Let "X = \\{1,2,3,4,5,6,7\\}" and "R = \\{(x,y)|x-y\\text{ is divisible by }3\\}" in "X". Let us show that "R" is an equivalence relation. Since "x-x=0" is divisible by 3 for any "x\\in X", we conclude that "(x,x)\\in R" for any "x\\in X", and hence "R" is a reflexive relation. If "x,y\\in X" and "(x,y)\\in R," then "x-y" is divisible by 3. It follows that "y-x=-(x-y)" is also divisible by 3, and hence "(y,x)\\in R."
We conclude that the relation "R" is symmetric. If "x,y,z\\in X" and "(x,y)\\in R,\\ (y,z)\\in R," then "x-y" is divisible by 3 and "y-z" is divisible by 3. It follows that "x\u2212z=(x\u2212y)+(y\u2212z)" is also divisible by 3, and hence "(x,z)\\in R". We conclude that the relation "R" is transitive. Consequently, "R" is an equivalence relation.
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