Proving an Equivalence Relation on Real Numbers

[barbara]

I know that
1. To show the relation is reflexive, we need to show that for any x, using the definition of R, we have xRx. The definition of R means that we must have |x - x| is even.2. To show that R is symmetric, we would have to show that if xRy then yRx. In the context of the definition we would need to show that if |x - y| is even, then |y - x| is even.3. Show that R is transitive, we need to show that if |x-y| is even and |y-z| is even, then |x - z| is even. This part is probably the hardest to work through.

But what I can't do is look at this in the context of this specific example. I am totally lost trying to define the following relation on the set of real numbers

xRy if |x - y| is an even integer and how that R is an equivalence relation

 

[Deveno]

Any relation that is reflexive, symmetric and transitive is *by definition* an equivalence relation.

Ok, so suppose $|x - y|$ is even. What is an even integer? It is one divisible by $2$, so we can write:

$|x - y| = 2k$, where $k$ is an integer (we don't actually need to know *which* integer $k$ is, just that there is one).

Now, for $xRy \iff |x - y| = 2k$ (for some integer $k$), we ask ourselves: is it also true that for any such pair $(x,y)$ that:

$|y - x| = 2k'$ (we don't know that $k'$ is the same as $k$-it might be, it might not be).

Let's break it down into cases:

Case 1: $x > y$.

In this case,$ x - y > 0$, so $|x - y| = x - y$.

Now, when we look at $|y - x|$, we see that $y - x < 0$, so (by the definition of absolute value):

$|y - x| = -(y - x) = -y -(-x) = -y + x = x - y$. Since (by assumption of $xRy$) $|x - y| = 2k$, we see that:

$|y - x| = x - y = |x - y| = 2k$, as well, so this, too, is an even integer, just like $|x - y|$ is. That settles the case $x > y$.

Case 2: $x = y$. In this case, $|x - y| = |x - x| = |0| = 0$. This is, of course, an even integer, since $0 = 2\cdot 0$.

In this case, we also have $|y - x| = |y - y| = 0$, as well, so in this case, too, we see $R$ is symmetric.

Case 3: left to you.

A "faster" way to show that $R$ is symmetric is to observe that $|x - y| = |y - x|$ (Do you believe this? Can you prove it?).

Try to work through just this "symmetric" bit, when you're convinced in your heart-of-hearts that this relation really and truly IS symmetric, we'll move on to transitivity.