Prove $(x, y)R(s, t)$ iff $2(x - s) = (y - t)$ defines an equivalence relation

Question

Define the relation $R$ on the set of all ordered pairs of real numbers as follows: $(x, y)R(s, t)$ iff $2(x - s) = (y - t)$.

Prove that $R$ is an equivalence relation.

Find the equivalence class of the point (1, 1).

The only ideas I know of how to solve this include proving the relation has reflexive, symmetric, and transitive properties. Then I would prove $2(x - s) = 2(s - x)$, so it is reflexive; $2(x - s) = (y - t) \Rightarrow (y - t) = 2(x - s)$, and then doing something similar for antisymmetry.

However I am not so sure if this is too simple and I would like to know if I am on the right track.

Answer 1

To prove that $R$ is an equivalence relation, you must show (as you said) that it is symmetry, reflexive, and transitive. But, it seems to me that you are trying to deal with real numbers, rather than pairs of real numbers, as in the relation definition.

Let's start with reflexive. We must show that for any $(x,y)$, we have $(x,y)R(x,y)$. By definition, this is true if and only if $2(x-x)=y-y$, which clearly holds. So, the relation is reflexive.

For symmetry, you must show that whenever $(x,y)R(s,t)$, we also have $(s,t)R(x,y)$. Can you prove this using the definition of $R$?

Lastly, for transitivity, you must show that whenever it is true that both $(x,y)R(s,t)$ and $(s,t)R(p,q)$, we have $(x,y)R(p,q)$. In other words, assuming $2(x-s)=y-t$ and $2(s-p)=t-q$, you must prove that $2(x-p)=y-q$. It may help to note that $$ 2(x-p)=2(x-s)+2(s-p). $$

Answer 2

As you said you have to prove: Reflexive, symmetric and transitive.

Reflexive: WTS that $(x,y)R(x,y)$, so you check that $2(x-x)=(y-y)$, which is true as $0=0$.

Symmetric: WTS that given $(x,y)R(s,t)$ then you have $(s,t)R(x,y)$, so unravel the definition of the last one: $2(s-x)=(t-y)$ is this true given that $2(x-s)=(y-t)$?

For transitivity you have to assume $(a,b)R(c,d)$ and $(c,d)R(e,f)$, and you want to prove that $(a,b)R(e,f)$, this one you should try. Unravel what does it mean for $(a,b)R(e,f)$ to hold and use the two relations you know they are true.

Answer 3

As other answers point out, there are perfectly straightforward approaches to this problem by working directly from the definition of $R$. In this case, however, there is also a more visual approach. For $\langle x,y\rangle,\langle s,t\rangle\in\Bbb R\times\Bbb R$ we have $\langle x,y\rangle\mathbin{R}\langle s,t\rangle$ if and only if $2(x-s)=y-t$, which in turn is equivalent to $2x-y=2s-t$. Thus, we could just as well have defined $R$ by saying that

$$\langle x,y\rangle\mathbin{R}\langle s,t\rangle\quad\text{if and only if}\quad 2x-y=2s-t\;.$$

Let $\alpha$ be any real number. If $x,y,s,t\in\Bbb R$ satisfy $2x-y=\alpha$ and $2s-t=\alpha$, then automatically $2x-y=2s-t$ and $\langle x,y\rangle\mathbin{R}\langle s,t\rangle$. What are the pairs $\langle s,t\rangle\in\Bbb R\times\Bbb R$ such that $2s-t=\alpha$? They’re exactly the points on the line $2x-y=\alpha$ or, if you prefer, $y=2x-\alpha$. Each possible value of $\alpha$ gives such a line, and all of the points on any one of those lines are in the relation $R$ to one another. Conversely, if $\langle x,y\rangle\mathbin{R}\langle s,t\rangle$, then $2x-y=2s-t=\alpha$ for some $\alpha\in\Bbb R$, and $\langle x,y\rangle$ and $\langle s,t\rangle$ both lie on the line $y=2x-\alpha$.

Note that all of these lines have the same slope, $2$, so they must be parallel: each point of the plane lies on exactly one of them.

With this geometric model in mind, it’s easy to check that $R$ is an equivalence relation whose equivalence classes are precisely the lines $y=2x-\alpha$ for all possible real values of $\alpha$.