Is $x \sim y \Leftrightarrow x,y$ are even an Equivalence relation?

Question

The following instruction defineds an Equivalence relation on the set of natural numbers.

$x \sim y \Leftrightarrow x,y$ are even

My idea:

Reflexivity: $x \sim x \Leftrightarrow x,x$ is even

Symmetry: $x \sim y \Leftrightarrow x,y$ are even

Transitivity: $x \sim y \Leftrightarrow x,y$ are even, $y \sim z \Leftrightarrow y,z$ are even $\Rightarrow x \sim z \Leftrightarrow x,z$ are even

So my result is, that it is an Equivalence relation, but the solution says it is not, but doesn't say why. Where is my mistake?

Answer 1

Reflexivity means that no equivalence class is empty, since the equivalence class of an element $x$ always contains at least one element: $x$ itself. As Marc pointed out in his comment, reflexivity fails, indeed if $x$ is odd then the set of $y$ such that $x \sim y$ is empty, thus $\sim$ is not an equivalence relation.

Answer 2

According to the question the relation is defined on $\Bbb N$ in which case reflexivity does not work.

If $\mathscr R$ is an equivalence relation on the set $A$ then $\mathscr R$ is a reflexive relation iff $a \mathscr aR \;\; \forall a \in A$. I'm sure according to the definition of the relation above in your exercise the relation is not reflexive since it would not work for every element in the given set. If my assumption is correct your counter example is $1 \not \sim 1$ and hence this is not an equivalence relation.

Answer 3

If $\sim=\left\{ \left(x,y\right)\in\mathbb{N}^{2}\mid x\; and\; y\; even\right\} $ then $\left(1,1\right)\notin\sim$ showing that $\sim$ is not reflexive.

Answer 4

For the reflexivity: is $(1,1)$ an element of the relation?