Can't determine if given relation is equivalence relation Definition of relation  ~    
$(a,b)$ ~ $(c,d)$ $\iff$ $bc^2=da^2$,
where $(a,b),(c,d)$ are from $\mathbb{R}\times\mathbb{R}$ and $(a,b),(c,d)$ are different from $(0,0)$ 
First of all, I wonder if R can be equivalence relation if it is not defined on $\mathbb{R} \times \mathbb{R}$($(0,0)$ is not included ), because I will not be able to find classes that divide the whole $\mathbb{R} \times \mathbb{R}$.
If I can observe $\mathbb{R} \times \mathbb{R}$ without $(0,0)$, then I should see if this relation is reflexive, symmetric and transitive.
Clearly, this relation is reflexive, but I have troubles with determining if it is symmetric and transitive. 
( Because of elements like $(a,0)$ or $(0,b)$ and dividing with zero )
Also, if it is, I can't imagine what classes are.
Anything you can think of will be helpful.
Thank you in advance.
 A: Reflexive: 
We need to prove that $(a,b)R(a,b)$, let us check if this is true. We have $$ba^2 = ba^2 \implies (a,b)R(a,b).$$
So our relation is reflexive.
Symmetric: We need to prove that if we have $(a,b)R(c,d)$ then this implies $(c,d)R(a,b)$. We have $$bc^2 = da^2 \implies da^2 = bc^2 \equiv (c,d)R(a,b).$$
So our relation is symmetric. 
Transitive: We need to show that if we have $(a,b)R(c,d)$ and $(c,d)R(e,f)$ then we have $(a,b)R(e,f)$. We do so below: We need to show $be^2 = da^2$
$$bc^2 = da^2 \text{ and } de^2 = fc^2 \implies c^2 = \frac{de^2}{f}$$
Substituting the latter into the first equation, we get $$b \cdot \frac{de^2}{f}  = da^2 \iff be^2 = fa^2 \equiv (a,b)R(e,f).$$ 
To complete the proof of transitivity, you need to consider the case $f=0$, look in the comment section for how to do so. 
Since our relation is reflexive, symmetric and transitive, it must be an equivalence relation and as you can see, there was no need to divide by $0$ or any other funky stuff. You can split the set $\mathbb{R} \times \mathbb{R}$ not including $(0,0)$ into equivalence classes.
Note: I used $R$ instead of ~ for the relation. 
A: We can either do this using division, or not using division.  Using division is quicker, but not doing so has implicit theoretical value (since we can use this relation to more generally define division).

Method 1: We handle the relation in two cases
If all elements are non-zero, we note that
$$
(x_1,x_2) \sim (y_1,y_2) \iff \frac{x_1}{x_2^2} = \frac{y_1}{y_2^2}
\iff \frac{x_2^2}{x_1} = \frac{y_2^2}{y_1}
$$
That is, if we define $f:(x_1,x_2) \mapsto \frac{x_1}{x_2^2}$, then we can define our relation on $(\Bbb R \setminus \{0\})\times (\Bbb R \setminus \{0\})$ by either


*

*$(x_1,x_2) \sim (y_1,y_2) \iff f(x_1,x_2) = f(y_1,y_2)$

*$(x_1,x_2) \sim (y_1,y_2) \iff g(x_1,x_2) = g(y_1,y_2)$


From either of these definitions, we can see that $\sim$ is an equivalence relation simply by noting that $f$ (or $g$) is a function.
If we look at the set of elements $(0,x)$ with $x \neq 0$, we can apply definition $1$ to see that for $y_1,y_2$ not both zero, $(0,1) \sim (y_1,y_2) \iff y_1 = 0$.  So, the set $\{(0,x): x \neq 0\} = f^{-1}(0)$ is an equivalence class under $\sim$.
Similarly, we can use definition $2$ to see that $\{(x,0):x \neq 0\} = g^{-1}(0)$ is an equivalence class under $\sim$.
We can conclude that $\sim$ divides $\Bbb R^2 \setminus (0,0)$ into equivalence classes.  It follows that $\sim$ is an equivalence relation.

Method 2: Division free approach
Reflexivity is easy.  For any $a,b$,
$$
a^2b = a^2b \implies (a,b) \sim (a,b)
$$
Symmetry is slightly more difficult:
$$
(a,b) \sim (c,d) \implies  bc^2 = da^2 \implies\\
da^2 = bc^2 \implies
(c,d)\sim(a,b)
$$
Transitivity is a bit tough:
$$
(a,b) \sim (c,d) \text{ and } (c,d) \sim (e,f) \implies\\
bc^2 = da^2 \text{ and } de^2 = fc^2\\
$$
From there, if $c$ and $d$ are non-zero, we have
$$
(da^2)(fc^2) = (bc^2)(de^2) \implies\\
c^2d[fa^2] = c^2d[be^2] \implies\\
fa^2 = be^2
$$
Suppose now that $c = 0$ and $d \neq 0$.  We can deduce that $a = 0$ and $e = 0$. We conclude that
$$
fa^2 = be^2 = 0
$$
Similarly, suppose that $d = 0$ and $c \neq 0$. We can deduce that $b = 0$ and $f = 0$.  Conclude that
$$
fa^2 = be^2 = 0
$$
So, $\sim$ is transitive.
A: It is obvious that the relation
$$(x,y)\sim(x',y')\quad\Leftrightarrow\quad y\>x'^2=y'\>x^2$$
on $\dot {\mathbb R}^2$ is reflexive and symmetric. The transitivity will follow from the following discussion which at the same time describes the resulting equivalence classes.
If $(x,y)\sim(x',y')$ and $x=0$ then by assumption $y\ne0$, whence $x'=0$. Conversely, all pairs $(0,y)$, $y\ne0$, are equivalent.
By the foregoing, if $(x,y)\sim(x',y')$ with $x\ne0$ then $x'\ne0$ as well, and 
$${y\over x^2}={y'\over x'^2}=c$$
for some $c\in{\mathbb R}$. Conversely, all pairs $(x,y)\in\dot {\mathbb R}^2$ lying on the same parabola (resp. line) $y=cx^2$ are equivalent.
A: If $f\colon X\to Y$ is a function then the relation ${\sim} \subseteq X\times X$ given by $x\sim x'\iff f(x)=f(x')$ is immediately verified to be an equivalence relation (and for any equivalence relation the canonical map $X\to X/{\sim}$ has this property). We have $X=\mathbb R^2\setminus\{(0,0)\}$ and try
$$\begin{align}f\colon\mathbb R^2\setminus\{(0,0)\}&\to \mathbb R\cup\{\infty\}\\(x,y)\ &\mapsto \begin{cases}\frac{x^2}{y}&\text{if }y\ne 0\\\infty&\text{if }y=0\end{cases}\end{align} $$
where $\infty$ is just an arbitrary symbol, an objetc that is $\notin\mathbb R$. (One could more formally let $\infty=\mathbb R$ because then certainly $\infty \notin \mathbb R$, but that may be even more confusing than the notion of "new symbol").
a) $(x,y)\sim (x'y')\implies f(x,y)=f(x'y')$
We are given that $yx'^2=y'x^2$. There are two cases: $y=0$ or $y\ne0$. In the first case we have $x\ne 0$ and hence find $y'=\frac{yx'^2}{x^2}=0$, so $f(x,y)=\infty=f(x',y')$ as desired.
In the second case we have $x'^2=\frac{y'x^2}{y}$. Then $y'=0$ would imply $x'=0$, which is absurd. Therefore $y'\ne0$ and we obtain $f(x',y')=\frac{x'^2}{y'}=\frac{x^2}y=f(x,y)$, again as desired. $_\square$
b) $f(x,y)=f(x',y')\implies (x,y)\sim (x'y')$
Again there are two cases: $f(x,y)\in \mathbb R$ or $f(x,y)=\infty$.
In the first case $yx'^2=yy'f(x'y')=yy'f(x,y)=y'x^2$.  In the second case, $y=y'=0$ so that $yx'^2=0=y'x^2$. $_\square$
A: Unless otherwise stated, assume that all $x$-coordinates are nonzero. Also note that this answer is mostly for fun; I wanted to visualize the equivalence classes!
If we say that $(x_1, y_1) \sim (x_2, y_2)$ if and only if $y_1x_2^2 =  y_2x_1^2$, then (assuming neither $x_1$ nor $x_2$ is $0$) dividing both sides by $x_1^2x_2^2$ we have
$$\frac{y_1}{x_1^2} = \frac{y_2}{x_2^2}.$$
Now, choose a fixed $x_1$ and $y_1$, and let $k =\frac{y_1}{x_1^2}$. Any pair $(x, y)$ related to $(x_1, y_1)$ satisfies $\frac{y}{x^2} = k = \frac{y_1}{x_1^2}$, so that $y = kx^2$. In other words, pairs $(x_1, y_1)$ and $(x_2, y_2)$ with nonzero $x$-coordinates are related if their $y$-coordinates are proportional to the square of their $x$-coordinates, with the same constant of proportionality. 
For a fixed constant of proportionality $k$, we have then that points satisfying the equation $y = kx^2$ are related. For points with nonzero $x$-coordinates, that the relation is an equivalence relation follows from the properties of equality (namely, that equality is an equivalence relation). 
Note that all points whose only nonzero coordinate is $y$ are related (as they correspond to $k = 0$). I'll let you think about what to do with points whose $x$-coordinate is $0$, and what all they're related to.
Thus, (most of) the equivalence classes can be visualized as the graphs of parabolas $$y = kx^2,$$ with one equivalence class for each choice of a real number $k$. Since each of these parabolas go through $(0, 0)$, it should be clear why we needed to remove it from the set under consideration: Every other pair would have been related to it!
