MHB Proving an Equivalence Relation on Real Numbers

AI Thread Summary
The discussion focuses on proving that the relation R defined by xRy if |x - y| is an even integer is an equivalence relation on real numbers. To establish R as reflexive, it is shown that |x - x| equals zero, which is even. For symmetry, it is demonstrated that if |x - y| is even, then |y - x| is also even due to the properties of absolute values. The transitive property is acknowledged as more complex, requiring further exploration. Overall, the participants emphasize the need to verify reflexivity, symmetry, and transitivity to confirm R as an equivalence relation.
barbara
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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
 
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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.
 
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