Finding and drawing equivalence class with a binary set defined on RxR In the question, it has been asked to find and draw the equivalence classes of the relation $∼$ on $(\Bbb R\times\Bbb R)\setminus\{(0, 0)\}$ which is defined as
$$(x_1, x_2) \sim (y_1, y_2)~\text{if}~(y_1, y_2) = \alpha(x_1, x_2)\text{, for some }\alpha \neq 0$$
On solving, I have obtained the equivalence class as
$$\begin{align}[(a,b)]&=\{(x,y)\in\Bbb R:(x,y)\sim(a,b)\}\\&
=\{(x,y)\in\Bbb R:(x,y)=\alpha(a,b)\}\end{align}$$
On solving this, I got $x=\alpha a$ and $y=\alpha b$
and $x-y=\alpha(a-b)$ which will correspond to the set of straight lines having slope $1$ and $y$-intercept $-\alpha(a-b)$ excluding the straight line passing.
But in the solution of this question, the equivalence class is given as the set of all straight lines passing through the origin(which is excluded).
I have just begun discrete math, and I am not getting any idea of the solution.
It would be really helpful if someone could help me with this.
 A: There is no problem with your algebra, but with which values are varying and which are constants.
Notice that $\alpha$ is a function of $x,y$, not some independent constant. We might do well to write it as $\alpha_{x,y}$ to emphasise this fact. This is the nuanced difference between your set, $$[(a,b)]=\{(x,y)\in\Bbb R:(x,y)=\alpha(a,b)\}$$ and the more formal $$[(a,b)]=\{(x,y)\in\Bbb R:\exists \alpha\in\mathbb{R}\setminus\{0\}:(x,y)=\alpha(a,b)\}$$
You have rearranged $x=\alpha_{x,y} a,y=\alpha_{x,y} b$ to $y=x-\alpha_{x,y} (a-b)$. Now we might see more clearly why this is not the equation of a straight line: $\alpha_{x,y}$ will vary as $x$ varies.
For instance, for some particular values $x,y$, it might be that $\alpha_{x,y}=1$ and $\alpha_{x+1,y}=3$. Here, the change that increasing $x$ by $1$ will have on $y$ is $+1-2(a-b)$, so it is not correct to say that the gradient is $1$.
Instead, we need to eliminate $\alpha_{x,y}$ from our equation. We can do that by approaching the algebra a bit differently:
$$\begin{align} &x=\alpha_{x,y} a,&y=\alpha_{x,y} b\\
&\alpha_{x,y}=x/a,&y=\alpha_{x,y} b\\
y&=\frac{b}{a}x \end{align}$$
Now we genuinely do have the equation of a straight line, with gradient $b/a$ and $y$-intercept $0$.
Varying $a,b$ can produce any possible real number: for instance, to achieve a straight line through the origin with gradient $r$, take $a=1,b=r$.
I'll leave you with this: there is one exceptional case, a straight line through the origin that does not have the form $y=rx$ for some $r\in\mathbb{R}$. What is the equation of that line, and what values of $a,b$ produce it?
