I have to find the relation properties and the equivalence classes.

$$X = \mathbb{R}^{2}$$ $$(x,y) \sim (u,v) \Leftrightarrow x - y = u - > v$$

Showing the relation properties of the equivalence relation was easy I think:

Reflexive: $$(x,y) \sim (x,y) \Leftrightarrow x - y = x-y$$ Symmetric:

$(x, y) \sim (u, v)$. By definition, $x - y = u - v$. By symmetry of equality: $u - v = x - y$. By definition of $\sim$, we get that $(u, v) \sim (x, y)$.

Transitive: $$(x,y) \sim (u,v) \wedge (u,v) \sim (a,b)\Rightarrow (x,y) \sim (a,b)$$ $$x-y=u-v \wedge u-v=a-b \Rightarrow x-y = a-b$$ (by transitivity of equality)

But I'm not sure how I can show the equivalence classes. First I tried to make a set which has all the relation properties, so I can easily read the equivalence classes. $$ R= \left \{ [(x,y),(u,v)],[(u,v),(x,y)],[(x,y),(x,y)],[(u,v),(u,v)] \right \}$$

This should work. But now I'm not sure whether I should use the tuples like this because of R² or look at the variables separately. Also I completely ignored the right part of the equivalence (x - y = u - v). Also I don't know whether I should even use the variables, or work with real numbers.

The equivalence classes read from above would be: $$[(x,y)] = \left \{ (x,y),(u,v) \right \}$$ $$[(u,v)] = \left \{ (x,y),(u,v) \right \}$$

They would also apply to this sentence: $$a \sim b \Rightarrow [a] = [b]$$

Is this right or wrong?

Another equivalence classes approach:

$(x,y)$ equivalent to $(0,0)$
Pairs of $(x,y)$ where $x-y = 0-0$, i.e. $x=y$

$(x,y)$ equivalent to $(0,1)$
Pairs of $(x,y)$ where $x-y = 0-1$, i.e. $x=y-1$

$(x,y)$ equivalent to $(1,0)$
Pairs of $(x,y)$ where $x-y = 1-0$, i.e. $x=y+1$

$(x,y)$ equivalent to $(1,1)$
Pairs of $(x,y)$ where $x-y = 1-1$, i.e. $x=y$

But what does that tell me and how do I write it down formally?


You were sort of OK up until you looked for equivalence classes.

Let me fixc up your "SYMMETRY" proof. You need to show that IF $(x, y) \sim (u, v)$, THEN $(u, v) \sim (x, y)$. Your "proof" says something about what happens if both of these are true. Here's a repaired version:

Suppose that $(x, y) \sim (u, v)$. Then by definition, $x - y = u - v$. By symmetry of equality, we have $u - v = x - y$. And then by the definition of $\sim$, we get that $(u, v) \sim (x, y)$.

Do you see the difference? Can you fix up the transitivity "proof" as well? Hint: start by assuming that $(x, y) \sim (u, v)$ and $(u, v) \sim (a, b)$, and find a sequence of statements that leads to the conclusion that $(x, y) \sim (a, b)$. The best route? Write out what each of these means without the "$\sim$" and see whether you can fill in the (few) gaps.

On to finding the classes:

Let's look for things equivalent to $(0,0)$. Those are ordered pairs $(x, y)$ where $x - y = 0 - 0$, i.e., where $x = y$. So this equivalence class is a straight line of slope 1 through the origin.

Can you do the same thing for the point $(0, 1)$? Once you've done that, you'll see a pattern, I expect.

As a general hint, when you're facing things with notions you don't quite understand, making them as specific as possible (e.g., trying to work out the $(0,0)$ example rather than some generic $(x, y)$ example) can often help a lot.

  • $\begingroup$ Check out my improved transitivity proof in my edited answer, is it better? $\endgroup$ – Thursday Next Mar 20 '15 at 17:40
  • $\begingroup$ Also I added my approach for your $(0,0)$ hint. $\endgroup$ – Thursday Next Mar 20 '15 at 17:45
  • $\begingroup$ Your improved transitivity proof is better, but you might want to justify your "=>" by saying "by transitivity of equality" (if you want to be really pedantic). For the classes: the class I wrote down contained all points on a diagonal line (with slope 1) through the origin. Each of your classes consists of all points on a different diagonal (with slope 1) line through some other point. Indeed, EACH equiv class is exactly the set of points on a diagonal line (with slope 1). To make this more concrete, you could say "each class is a line with slope 1 through $(0, a)$ for some $a \in \mathbb R$. $\endgroup$ – John Hughes Mar 20 '15 at 18:01
  • $\begingroup$ This does make sense, but how could I write that with a bracket and set notation? "$[x]$ = {a line with slope 1 through $(0,a)$ for some $a∈\mathbb{R}$}" Because we never used the word slope and that concept for equivalent classes. $\endgroup$ – Thursday Next Mar 20 '15 at 18:38

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