My discrete math class has this problem: we need to find the equivalence classes for the equivalence relation over the set of $\mathcal{N}$ onto $\mathcal{N}$ defined as follows: $$xRy \iff (x-3.3)(y-3.6)>0$$ I thought that there's only 1 class: $\{\mathcal{N}-{0,1,2,3}\}$ that is all natural numbers greater than 3. However the solution says that there's another equivalance class: $\{0,1,2,3\}$. But how can this be? $$1R2 \to (1-3.3)(2-3,6) <0$$ so clearly such input renders the output to be below zero hence it can't be part of the relation.
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How is it less than zero? Both terms are negative, so the product $(1-3.3)(2-3.6)$ is greater than zero. For greater clarification, that product is $3.68$. This is an equivalence relation because there is no natural number between $3.3$ and $3.6$. – Sarvesh Ravichandran Iyer Sep 10 '16 at 10:21
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you're right I was confused with the product. Thanks! – Yos Sep 10 '16 at 10:33
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You are welcome. Top mathematicians make this mistake, so nobody's new to it. – Sarvesh Ravichandran Iyer Sep 10 '16 at 10:37
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$$(1-3.3)(2-3.6)\color{red}{>}0$$ or more generally $$x\cdot y>0\iff (x>0\ \land y>0)\ \lor \ (x<0\ \land y<0)$$
b00n heT
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