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Consider the following relation on $\mathbb{Z} \times \mathbb{Z}$: $(x, y)\sim (x', y')$ iff $xy' = x'y$.

(a) Prove that it is an equivalence relation.

(b) Consider the set of equivalence classes $\mathbb{Z}\times \mathbb{Z}/\sim$, and denote the equivalence class of $(x,y)$ by $\overline{(x,y)}$. Define the following binary operations between equivalence classes:

  1. $\overline{(x,y)} \cdot \overline{(x',y')} = \overline{(xx', yy')}$;

  2. $\overline{(x,y)} + \overline{(x', y')} = (xy' + x'y, yy')$.

Prove that these operations are well defined, that is does not depend on a choice of a representative. Do you recognize the arithmetic system $\mathbb{Z} \times \mathbb{Z}/ \sim$ you just constructed?

My attempt:

(a) Prove that $\sim$ is an equivalence relation. Proof: We need to show that $\sim$ has the properties of reflexivity, symmetry, and transitivity.

  1. Reflexivity: We need to show that $(x,y) \sim (x,y)$. Suppose $x' = x$ and $y' = y$, then $(x,y)\sim (x',y') \Rightarrow xy' = x'y \Rightarrow xy = xy \Rightarrow (x,y) \sim (x,y)$.

  2. Symmetry: We need to show that $(x',y') \sim (x,y)$ given that $(x,y) \sim (x',y')$. Suppose $(x,y) \sim (x',y') \Rightarrow xy' = x'y \Rightarrow x'y = xy' \Rightarrow (x',y') \sim (x,y)$.

  3. Transitivity: We need to show that $(x,y) \sim (x'',y'')$ given that $(x,y) \sim (x',y')$ and $(x',y') \sim (x'',y'')$. Suppose $(x,y) \sim (x',y')$ and $(x',y') \sim (x'',y'') \Rightarrow xy' = x'y$ and $x'y'' = x''y' \Rightarrow x' = xy'/y \Rightarrow xy'y''/y = x''y \Rightarrow xy'y'' = x''y'y \Rightarrow xy'' = x''y$. Therefore $(x,y) \sim (x'',y'')$.

(b) If I want to prove that something is well defined, generally I would use $x'$ and $y'$ and try to show that the representative doesn't matter. Would I do something like:

Since $\overline{(x,y)} \cdot \overline{(x',y')} = \overline{(xx',yy')}$, then $\overline{(x,y)} \cdot \overline{(x'',y'')} = \overline{(xx'',yy'')}$. If I rearrange both equations so that they equal $\overline{(x,y)}$ and then set both equations equal to each other, then I would get something like:

$\overline{(xx',yy')}/\overline{(x',y')} = \overline{(xx'',y'')}/\overline{(x'',y'')}$

and then multiply both sides by their denominator, I would get:

$\overline{(xx',yy')} \cdot \overline{(x'',y'')} = \overline{(x',y')} \cdot \overline{(xx'',y'')}$.

I have no idea what I'm doing.

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  • $\begingroup$ Think of $(x,y)$ as the rational number $\frac xy$, then you will get an idea of what you are doing and everything will fall in place. $\endgroup$ – P Vanchinathan Feb 11 '15 at 3:14
  • $\begingroup$ So would the equivalence class for (x,y) for every element a in Z except for 0 such that x' = ax and y' = ay. Since the equivalent relation requires that xy' = x'y => x/y = x'/y' which just means that they're in a ratio with each other. $\endgroup$ – Jimmy Hoang Huy Nguyen Feb 11 '15 at 3:28
  • $\begingroup$ This question is about a very similar equivalence relation but on a different set: math.stackexchange.com/questions/297748/… $\endgroup$ – Martin Sleziak Feb 11 '15 at 11:04
  • $\begingroup$ And if it helps you can also have a look at a similar equivalence relation defined using addition instead of multiplication: math.stackexchange.com/questions/21256/… $\endgroup$ – Martin Sleziak Feb 11 '15 at 11:09
  • $\begingroup$ And I should also mention that a more general construction is field of fractions. $\endgroup$ – Martin Sleziak Feb 11 '15 at 11:10

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