So, it comes in two parts:

a. Prove that $\sim$ is an equivalence relation on ℝ².

b. Give a geometric description of the partition of ℝ² formed by the equivalence classes.

For a, I have to prove that $\sim$ is reflexive, transitive, and symmetric. So, for reflexive, can I say, let $a$ be in ℝ². Then, $a \sim a$ because $a+a = a+a$?

For symmetric, can I say: assume $x,y,w,z$ are in ℝ² and that $x+y~w+z$. Then, by definition $x+y=w+z$. So, if we multiply by $1$, we get $y+x=z+w$?

I'm not sure how to set up transitive.

Also, can someone help me with the partitions as well?


1 Answer 1


IMHO the best way to begin an equivalence relation question is to see if you can informally convince yourself that the relation is an equivalence. If you can describe the meaning of the relation using the word "same" then you can be reasonably sure that it is an equivalence relation. For example:

  • "being parallel" is an equivalence relation: two lines are parallel if they have the same direction;
  • congruence modulo $5$ is an equivalence relation: two integers are congruent modulo $5$ if they have the same remainder after dividing by $5$;
  • in your relation, two pairs are related if their elements have the same sum.

Then you should do the question formally by showing that the relation is reflexive, symmetric and transitive.

Your proof for reflexivity is not good because you are adding pairs of numbers, whereas the definition of the relation says that you should add the components of the pairs. Try this.

Let $(x,y)\in\Bbb R^2$. Then obviously $x+y=x+y$, and by definition this means $(x,y)\sim(x,y)$. So $\sim$ is reflexive.

For transitivity, try to fill in the gaps in the following proof.

Let $(x,y),\,(w,z),\,(u,v)$ be in $\Bbb R^2$.
Suppose that $(x,y)\sim(w,z)$ and $(w,z)\sim(u,v)$.
By definition this means that $x+y=w+z$ and. . .
Therefore. . .
So $(x,y)\sim(u,v)$.
Therefore $\sim$ is transitive.

The equivalence class of a pair $(a,b)$ is by definition $$\{\,(x,y)\in\Bbb R^2\mid(x,y)\sim(a,b)\,\}\ ,$$ that is, $$\{\,(x,y)\in\Bbb R^2\mid x+y=a+b\,\}\ .$$ If $a,b$ are fixed numbers, what does this set look like geometrically? If you vary $a$ and/or $b$, can you describe all the equivalence classes you get?

Good luck!

  • $\begingroup$ And how about my symmetry proof? Is that okay? $\endgroup$
    – JCMcRae
    Commented Nov 3, 2014 at 3:10
  • $\begingroup$ Sorry but no, not really. First you said that $x,y,w,z$ are in $\Bbb R^2$: but $\Bbb R^2$ does not contain numbers, it contains pairs of numbers. Then you said $x+y\sim w+z$, which cannot be right since $\sim$ relates pairs of numbers, not numbers. You ended up proving that $y+x=z+w$ which by definition means $(y,x)\sim(z,w)$. But you didn't want to prove this, you wanted to prove $(w,z)\sim(x,y)$. Try modelling a symmetry proof on the transitivity proof I gave you. $\endgroup$
    – David
    Commented Nov 3, 2014 at 3:15
  • $\begingroup$ Okay, so how about this one: Let (x,y) and (w,z) be in ℝ², and (x,y)~(w,z). Then, by definition, x+y = w+z. If we multiply by one, we get w+z = x+y. Therefore, (w,z)~(x,y) and therefore, ~ is symmetric. $\endgroup$
    – JCMcRae
    Commented Nov 3, 2014 at 3:28
  • $\begingroup$ Much better, well done! I would just have one very minor criticism: getting from $x+y=w+z$ to $w+z=x+y$ is not multiplying by $1$, it is simply a basic property of equality. Actually, you are using the well-known fact that equality is a symmetric relation in order to prove that $\sim$ is also a symmetric relation. $\endgroup$
    – David
    Commented Nov 3, 2014 at 3:36
  • $\begingroup$ Okay, thanks so much! $\endgroup$
    – JCMcRae
    Commented Nov 3, 2014 at 3:45

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