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We have this relation on $\mathbb{R}^* \times \mathbb{R}^*$ defined by: $(x,y) \sim (a,b)$ if $xa > 0$ and $yb>0$.

I have proven it is an equivalence relation, and I know the equivalence classes should be four: the set of all $x$ and $a$ positive, $y$ and $b$ negative; $x$, $a$, $y$ and $b$ positive; $x$ and $a$ negative, $y$ and $b$ positive; $x, a, y$ and $b$ negative.

But I'm failing to give a mathematical description of these classes, or justify it. Any help? Thank you!

Gibbs
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bluemuse
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    What you've written is a "mathematical description". Words can make acceptable mathematics. As an instructor I would gladly accept this answer. – Ethan Bolker Jun 08 '18 at 19:57

2 Answers2

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Hint: $(1,1); (1,-1), (-1,1); (-1,-1)$ are in different classes. And every element is equivalent to one of these elements.

Tsemo Aristide
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  • So I can describe them by partitioning the cartesian plane into 4 parts. But how do I describe that as an equivalence class? – bluemuse Jun 08 '18 at 19:44
  • As @TsemoAristide mentioned, you have the class containing $(1,1)$.

    $$[(1,1)] = \left{ (x,y) \in \mathbb{R}^* \times \mathbb{R}^*: x>0, y>0 \right}$$

    Can you do the rest?

     – SlipEternal Jun 08 '18 at 19:46
  • The element $(x,y)$ is equivalent to $(1,1)$ if $x>0,y>0$ it is equivalent to $(1,-1)$ if $x>0,y<0$ it is equivalent to $(-1,1)$ if $x<0,y>0$ it is equivalent to $(-1,-1)$ if $x<0,y<0$. – Tsemo Aristide Jun 08 '18 at 19:47
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Geometrically, $x, a$ have the same sign and $y, b$ also have the same sign means the points $(x,y)$ and $(a,b)$ are in the same quadrant.

Bernard
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