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