Prove that $E$ is an equivalence relation, where $E$ is given by $qr=s$ for some $q\in\mathbb Q^*$

Question

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 $\mathbb{R}$.

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

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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.

Answer 1

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.