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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.

mrp
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Jimmy Hoang Huy Nguyen
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  • 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. – P Vanchinathan Feb 11 '15 at 03:14
  • 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. – Jimmy Hoang Huy Nguyen Feb 11 '15 at 03:28
  • This question is about a very similar equivalence relation but on a different set: http://math.stackexchange.com/questions/297748/showing-that-r-is-an-equivalence-relation-on-x-times-x – Martin Sleziak Feb 11 '15 at 11:04
  • And if it helps you can also have a look at a similar equivalence relation defined using addition instead of multiplication: http://math.stackexchange.com/questions/21256/equivalence-relation-a-b-r-c-d-leftrightarrow-a-d-b-c – Martin Sleziak Feb 11 '15 at 11:09
  • And I should also mention that a more general construction is field of fractions. – Martin Sleziak Feb 11 '15 at 11:10
  • BTW are you sure it is supposed to be $\mathbb Z\times\mathbb Z$ and not $\mathbb Z\times(\mathbb Z\setminus{0}$? (This is what I would expect since you are asked: Do you recognize the arithmetic system you just constructed?) – Martin Sleziak Feb 11 '15 at 11:12
  • Okay, so I have come to where (x,y) = x/y and (x',y') = x'/y' and when you multiply the two you get xx'/yy'. And if you do the same thing for the equivalence classes you get bar(x,y) * bar(x'y') = bar(xx'/yy'). I understand the intuition behind the statement now, but I don't know what I need to show in order to prove that it's well defined. – Jimmy Hoang Huy Nguyen Feb 11 '15 at 21:01

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