Is binary-relation $\left\{\left(a,b\right)\mid a,b\in\mathbb{N}\wedge a,b \text{ are even numbers}\right\}$ reflexive?

Question

I'm a novice in set theory and I'm not clear about reflexive relation.

My question is the title.

Is binary-relation $R:=\left\{\left(a,b\right)\mid a,b\in\mathbb{N}\wedge a,b \text{ are even numbers}\right\}$ reflexive?

The relation set $R$ is the set of all positive ordered pair of even numbers, and it satisfies $\alpha R\alpha$.

But $\alpha$ does not demonstrates all elements in $\mathbb{N}$ where the definition of reflexive is $\left(\forall a\in \mathrm{A}\right)\left(aRa\right)$.

Is this means for all $a$ in $\mathrm{A}$ or in $R$?

Added OK. I think I've understood. For checking ->

In the case above, $R$ is not reflexive binary-relation on $\mathbb{N}$. But it is reflexive on $R$. Is this correct?

Answer 1

Recall that $R$ is reflexive on the set $A$ if and only if for every $a\in A$, the pair $(a,a)$ is in $R$.

This means that the same relation can be reflexive on one set, but not on another. For example $\{(0,0)\}$ is reflexive on $\{0\}$ but not on $\{0,1\}$.

So the question depends on who is your $A$. If $A$ is the set of even numbers, then the answer is yes. If it's not then the answer is no, because there will be some $x\in A$ which is not even, and therefore $(x,x)\notin R$.

Answer 2

$\alpha$ is a "stand in" to denote for any (arbitrary) $a \in A$, $(aRa)$. The relation must be reflexive for every $a$ in $A$, no exclusions.

That is, since its value is arbitrary, we treat it like a variable which can take on the value of any element in $A$: no exclusions. And so it means exactly what is meant by $\forall a \in A (aRa)$.

Can you see how your relation is not reflexive? Put $a = 1 \in A$. Note that $a =1 \in \mathbb N$, but $(1, 1)\notin R$, since it is not an ordered pair of even numbers.

Answer 3

It means for all $a\in A$, so this relation is not reflexive: $1\in\Bbb N$, but $\langle 1,1\rangle\notin R$.