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I just want to see if I am understanding this correctly.

I have to create an equivalence relation on 1,2,3 that has exactly two equivalence classes. So I came up with S = {(1,1), (2,2), (3,3), (1,2), (2,1)}.

So the equivalence classes would be [1] = {1, 2}, [2] = {2, 1}, [3] = {3} which works. [1] == [2] so theres one equivalence class and the other is [3]. There is exactly two equivalence classes. Is this correct?

Thanks

Xellic
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2 Answers2

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Yes, that's correct; your example works.

Brent Kerby
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Yes, it is of course the relation $R$ using the function $$ f(x)=-1.5x^2+6.5x-4 $$ and then defined by $$ aRb\iff f(b)-f(a)\text{ is even} $$

A maybe simpler, but actually quite similar one is given by $$ aRb\iff b-a\text{ is even} $$ which has classes $[1]=\{1,3\}$ and $[2]=\{2\}$.

String
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