But it's specifically an equivalence relation. In this case we have (a, b) ~ (c, d) if ad = cb. I had to prove that that made sense earlier because an equivalence relation has three properties of reflexivity (a~a), symmetry (if a~b, then b~a), and transitivity (if a~b and b~c, then a~c).
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But it's specifically an equivalence relation. In this case we have (a, b) ~ (c, d) if ad = cb. I had to prove that that made sense earlier because an equivalence relation has three properties of reflexivity (a~a), symmetry (if a~b, then b~a), and transitivity (if a~b and b~c, then a~c).
So yeah. It's weird...
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