Define a relation $ \sim$ on $\mathbb{R}^2 \setminus (0,0)$ by $(a,b)\sim (c,d)$ if there is some real number $x$ with $a=xc$ and $b=xd$. I need to prove the relation is an equivalence relation and determine the equivalence classes.

Here's what I have started.

Reflexive: Let $(a,b) \epsilon \sim$, then $a=1\cdot a$ and $b=1\cdot b$. Thus $(a,b)\sim(a,b)$ and $\sim$ is reflexive.

Can I have a nudge to finish the rest?

Even more important though...can I some deeper intuition to their relation? Help with understanding that will help me determine the equivalence classes on my own.

  • $\begingroup$ Viewing $(a,b)$ as a vector from $(0,0)$ to $(a,b)$ then the relation says $V\sim U\iff V = xU$ is a scaling of $U$ by some real $x$ (necessarily $x\neq 0$ since $V\neq (0,0)$ by hypothesis). That this is an equivalence relation is equivalent to the fact that the scalars $\Bbb R\backslash0$ contain $1$ (reflexive) and are closed under inverses (symmetric) and multiplication (transitive) i.e. the scalars $\Bbb R\backslash0$ form a group. $\endgroup$ – Bill Dubuque Aug 31 '17 at 17:41

Symmetry comes from the fact that you can divide by $x$ since neither $a$ or $b$ are $0$, and transitivity comes from considering the product of $x_1$ and $x_2$.

For what the equivalence classes are, think of $\mathbb{R}^2$ as the Euclidean plane, and imagine lines through the origin.

  • $\begingroup$ If $x = 0$ then we have $a = b = 0$, but we've forbidden that from the choice of set we're working on. $\endgroup$ – Duncan Ramage Aug 31 '17 at 17:24
  • $\begingroup$ Then for symmetry I can say: If $(a,b),(c,d))\epsilon\sim$, then $ a=xc, b=xd$ and so $c=x^{-1}a, d=x^{-1}b$. Thus $\sim$ is symmetric. $\endgroup$ – PerpetualStudent Aug 31 '17 at 17:51
  • $\begingroup$ Exactly. You will probably also want to include a line about why $x^{-1}$ exists. $\endgroup$ – Duncan Ramage Aug 31 '17 at 17:53
  • $\begingroup$ Transitivity: If $(a,b), (c,d), (e,f) \epsilon \sim$, then $a=xc, b=xd$ $c=xe, d=xf$ Thus, $a=x^{2}e, b=x^{2}f$. Therefore, $\sim$ is transitive. $\endgroup$ – PerpetualStudent Aug 31 '17 at 18:01
  • $\begingroup$ Yes, thanks. Since $x \epsilon \mathbb{R}, x^{-1} \epsilon \mathbb{R}$ $\endgroup$ – PerpetualStudent Aug 31 '17 at 18:03

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