Let $X = \{1, 2, 3, 4, 5\}.$ Define a relation $R = \{(x, y)\mid x, y \text{ even}; x, y \in X\}.$ Show that $R$ is an equivalence relation on X

Question

Can you help me with this problem? I have a problem demonstrating it. I know that it needs to be reflexive, symmetric, and transitive. I just don't understand when they said even, does it mean $R=\{(2,2),(2,4),(4,2),(4,4)\}?$ If yes how can I demonstrate it for example reflexive will be: for all x belongs to $X$ $(x,x)R(x,x)\rightarrow (x,x)=(x,x) ?$

Answer 1

First of all: $x-y$ even does not mean $x, y$ are even. If we took $x=1, y=3$, then $x-y=-2, y-x=2$ and both are even. The reflexivity comes from this thing quite easily, as $0$ is even. Symmetry is a quick thing to check, and transitivity is the only tricky part, but can be done by writing $x-y=2m, y-z=2n$ for some integers $m, n$, and then seeing what that gives for $x-z$.

Answer 2

I just don't understand when they said even,

Yeah, I don't either. But given any option, I'd take it to mean $R= \{(x,y)| x, y$ are both even$\}$.

does it mean R={(2,2),(2,4),(4,2),(4,4)}?

That's the only real thing that would make sense.

If yes how can I demonstrate it for example reflexive will be: for all x belongs to X (x,x)R(x,x)→(x,x)=(x,x)?

You can't. Because it wouldn't be true.

$1 \not R 1$ because $(1,1) \not \in R$. Nor is $(3,3)$ or $(5,5)\in R$.

So the relation just isn't reflexive.

I suspect the book has a typo and meant.

$R = \{(x,y): x, y$ are both even or are both odd$\}$. That would be an equivalence relation with exactly to equivalet classes.

Class 1: $ \{1,3,5\}$ as all elements in the class are equivalent to each other, and Class 2: $\{2,4\}$ as all elements in the class are equivalent to each other. Every element in $X$ is in one of the classes and the cases are disjoint and every element of $X$ belongs to exactly one class.

(That's not the usual definition of equivalence relation, but it is an... ahem, equivalent, definition. It gets down to what equivalence means and why the text relation obviously isn't. In the texts case there is only one class, the class of evens: $\{2,4\}$ and $1,3,5$ aren't in any class.

....

Anyway it looks like you are on top of everything and have a good grasp on everything... except.....

The relation is on $X$ and contains ordered pairs of $X$ but we say $xRy$ which is the same thing as $(x,y) \in R$ or if $R$ is an equivalence relationship (which it may not be) then we can say "$x$ is equivalent to $y$" or notation $x \equiv y$.

We wouldn't write, as you did when you wrote "$(x,x)R(x,x)\rightarrow (x,x)=(x,x)$", about the pairs being related. Either the elements are related, OR the ordered pairs are in $R$. But not that the ordered pairs are related.