# Prove that E is an equivalence relation

Let $\mathbb{Q}^* = \mathbb{Q}\setminus\{0\}$ and let $E$ be the binary relation on $\mathbb{R}$ defined by $$r E s \Longleftrightarrow \text{there exists } q\in\mathbb{Q}^* \text{ such that } qr=s.$$

Prove that $E$ is an equivalence relation on $\mathbb{R}$.

I understand that I have to show that the relation is reflexive, symmetric, and transitive.

To show reflexivity, consider $x E x$. Then $xq=x$ and $\exists q \in \mathbb{Q}^*$, namely $q=\frac{1}{1}$ which holds.

EDIT: Now getting stuck on proving reflexivity.

• $1$ is a perfectly respectable element of $\mathbb{Q}^*$, which is the set of nonzero rationals. – Ethan Bolker Dec 13 '17 at 21:22
• Well, actually $1 \in \Bbb Q^*$. – Crostul Dec 13 '17 at 21:22
• What do you mean $1\notin \Bbb Q^*$? $1$ is a rational number different than zero, is it not? You might know the rational number $1$ by a different notation: $\frac{1}{1}$... Just because $1$ happens to be an integer and a natural number does not preclude it from also being a rational number. – JMoravitz Dec 13 '17 at 21:22
• As an aside, you should find that it is indeed an equivalence relation. There will be uncountably infinitely many equivalence classes, but they will include the class containing only zero, the nonzero rational numbers, the nonzero rational multiples of $\pi$, the nonzero rational multiples of $\sqrt{2}$, the nonzero rational multiples of $\sqrt{3}$ and so on – JMoravitz Dec 13 '17 at 21:30
• You will make use of the following two facts: The multiplicative inverse of a nonzero rational number is again a nonzero rational number. The product of two nonzero rational numbers is again a nonzero rational number. Feel free to prove these first if you haven't seen these properties before (they should be a short one-line proof each). – JMoravitz Dec 13 '17 at 21:40

## Setup and major hints:

Reflexivity:

You seem to have already figured this one out:

Suppose $x\in \Bbb R$. We wish to show that $xEx$. Indeed, since $1\in\Bbb Q^*$ and since $1\cdot x = x$ we have shown that $x$ is a rational multiple of itself, hence $xEx$ is true.

Symmetry:

Suppose that $x$ and $y$ are (not necessarily distinct) real numbers such that $xEy$. We wish to prove that it follows that $yEx$.

Since $xEy$ this means by definition of $E$ that...

... there is some nonzero rational number $q$ such that $q\cdot x = y$

Now, since nonzero rational numbers all have multiplicative inverses which are again nonzero rational numbers, we notice that:

$q^{-1}\cdot y =$ _______ ... which implies that...

... which means that $yEx$ is also true.

Transitivity

Suppose that $x,y,z$ are (not necessarily distinct) real numbers such that $xEy$ and that $yEz$. We wish to prove that it follows that $xEz$.

Since $xEy$ and $yEz$ this means by the definition of $E$ that...

there is some nonzero rational number $q$ and some nonzero rational number $r$ such that...

Now, since the product of two nonzero rational numbers is again a nonzero rational number we find that:

$z=\dots$

... which implies that $xEz$ is also true.

• Cannot thank you enough for this. I am going to soak it in and get back to you. – 1011011010010100011 Dec 13 '17 at 21:54
• Does it suffice to say..... $qx=y, qy=z \Leftrightarrow y=\frac{z}{q}$. Hence, $qx=\frac{z}{q} \Leftrightarrow qqx=z$. Thus, $xEz$ holds true. – 1011011010010100011 Dec 13 '17 at 22:25
• @sgerbhctim close, however these might be different values of $q$. There is no reason to think that the ratio between $x$ and $y$ must be the same as the ratio between $y$ and $z$. Use different letters. I would also phrase this differently and without division (though that may simply be a result of habit, division should be avoided like the plague when doing problems involving only natural numbers or only integers, but in this case it shouldn't be a problem). – JMoravitz Dec 13 '17 at 22:28