Let $x\;\mathtt R\;y\iff xy = 4$ on the set of real numbers. What does it means to check reflexivity?

Question

I understand that something is reflexive if $\quad\forall a\in A : a\;\mathtt R\;a$.

But right now I'm confused as to what would be my $\mathbf a$ in this case.

For example, if I had: $(2,2)$ and $(4,1)$ would I be checking if $2\times 2=4$ in case one and $4\times 4=4$ in case two?

Or would I be checking if $2\times 4=4$?

Or would I say they're related because both $2\times2$ and $4\times 1=4$?

I'm just really confused as to what I would set equivalent to my $\mathbf a$ value and how I would check that it relates to itself.

Thanks for any help you can provide!

Answer 1

Hint:

Read the sentence "∀a∈A, aRa " aloud in your example:

"For every real number $a$, $a \times a = 4$."

If that's true then your relation is reflexive. If it's false, the relation is not.

Answer 2

Maybe it would be better to show you an example of a true equivalence relation first.

For instance, take $x\,\mathtt R\,y\iff x\equiv y\pmod{2\pi}\iff \exists k\in\mathbb Z\mid x = y +2k\pi$.

• reflexivity, for any real $x$, check if $x\,\mathtt R\,x$ ?

Since $x=x+0\times 2\pi$, then it is verified.

• symetry, for any reals $(x,y)$ check if $x\,\mathtt R\,y$ then $y\,\mathtt R\,x$ ?

Since $x=y+2k\pi$ also means $y=x+(-k)2\pi$ then it is verified.

• transitivity, for (x,y,z) reals, check if $x\,\mathtt R\,y$ and $y\,\mathtt R\,z$ implies $x\,\mathtt R\,z$ ?

Since $x = y +2k_1\pi = (z + 2k_2\pi) + 2k_1\pi= z+2(k_1+k_2)\pi$ then it is verified.

And our relation $\mathtt R$ is an equivalence relation.

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Let see other examples of only reflexivity :

• on $\mathbb N^2$, $x\,\mathtt R\,y\iff x,y$ have same parity.

$x$ has same parity than itself, so the relation is reflexive

• on $\mathbb R^2$, $x\,\mathtt R\,y\iff |x|=|y|$.

again it is obvious that $|x|=|x|$

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But the relation you proposed $x\,\mathtt R\,y\iff xy=4$ is not reflexive.

Because if we select a random real $x$, it is not true that $x^2=4$.

It is indeed true for $x=2$, but to be qualified of reflexive, the relation $\mathtt R$ should be verified for every real, and this is failing.

What you have when you say that $2\,\mathtt R\,2$ or $4\,\mathtt R\,1$, is that $2$ is in relation with $2$ by $\mathtt R$ and that $4$ is in relation with $1$ by $\mathtt R$, nothing more.

This relation is symetric since it is obvious that $xy=yx$.

But it is not transitive since $(x=1)\,\mathtt R\,(y=4)$ and $(y=4)\,\mathtt R\,(z=1)$ but $xz=1\neq 4$ so $x$ and $z$ are not in relation.

Finally your relation is not very interesting, because it doesn't have good properties, and maybe that's what is confusing you.

With a truly equivalent relation, like the ones I showed above, reflexivity should be something obvious, trivial like being equal to itself.

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But to conclude let see an example derived from yours, where the relation is an equivalence relation.

$x\,\mathtt R\,y\iff xy\ge 0$

$x^2$ is always positive so $x\,\mathtt R\,x$ and the relation is reflexive.

Symetry is obvious, since multiplication is commutative.

Finally $xy\ge$ and $yz\ge 0$ means $xyyz=(xz)(y^2)\ge 0$ but since $y^2$ always positive then $xz\ge 0$ and transitivity is verified.

So this time you could say, the reflexivity was a bit more interesting this time, it wasn't involving just being equal to itself. But look at the relation closer, $x\,\mathtt R\,y$ means $x,y$ have the same sign. So reflexivity is just the $sign(x)=sign(x)$, a triviality indeed.