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I'm reflecting on something rather abstract, but which gave me great insight and satisfaction about my comprehension of algebra.

I'm wondering if a line, (consider here in $ \mathbb{R}^2 $ Euclidean Space), represents in fact a coordinate-wise equivalence relation (given by parametric representation of the line) between ANY two distinct points in the plane?

Hint:

Let $ f: X \subset \mathbb{R} \to Y \subset \mathbb{R} $ be the equation for a line in two dimensional Euclidean Space, so that the line denoted by $(L)$ will be nothing but the graph of the equation ie. set of points $(X, f(X))$. Then, fix the point $u = (u_x,u_y) \in (L)$ for the value $x=1$ on the line: Consequently, $ u = (u_x,u_y) = (1, f(1)) $ would be coordinates of this point.

Let's define now the relation we intend to call $ R $ between any two points of this two dimensional $\mathbb{R}^2$ Euclidean Space:

  • $ \forall (a,b) \in X^2 $ and pairwise distinct, we have, for some $ \lambda $ real:

$$ a R b \implies [ a_x = \lambda b_x \implies f(a_x) = \lambda f(b_x) ] $$

which is true on any Archimidean Field. Note that, by analogy to the notation of my initial hypothesis, $a=(a_x,a_y)$ and $b=(b_x,b_y)$ represent two distinct points on the line $(L)$.

So, we can define a line $(C)$ as the set: $ \{ (a,f(a)) | aRu \} $, or just $ \{ \dot{f(1)} \} $.

NOTE: it's just a matter of point of view, but i need validation and comment about it.

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  • $\begingroup$ What are $X$ and $Y$? What is a unity point? What does $a_x$ and $b_x$ mean? I think these and the overall question needs to be clarified. $\endgroup$ – Pratyush Sarkar Dec 29 '18 at 1:09
  • $\begingroup$ Appearantly, as I said it is in $ \mathbb{R}^2 $, one cannot have doubt about $X$ and $Y$ being any subset of $ \mathbb{R}^2 $. And, if you'd understood the idea, you'd see that we want to fix a point, and define every other points on the line regards to that one. You can pick any point for that, but not the $(0,0)$ not to risk any definition issues, so I picked up the point $u=(1,f(1))$. you should have seen too. That is one of points you could have hepled me to adjust though. $\endgroup$ – freehumorist Dec 29 '18 at 13:11
  • $\begingroup$ if you intend to generalize the situation by questioning possible natures $X$ and $Y$ could have (without accepting my restricted suppositions), your help is wanted too. But I tell you, you can't. This is valid for lines as it seems... But the reason why I'm sharing this idea on the forum is that I'm not good enough at Complex Analysis. $\endgroup$ – freehumorist Dec 29 '18 at 13:17
  • $\begingroup$ If $X \subset \mathbb R^2$, then what does $f(1)$ mean? $\endgroup$ – Pratyush Sarkar Dec 29 '18 at 17:17
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    $\begingroup$ No problem! This makes more sense. $\endgroup$ – Pratyush Sarkar Dec 30 '18 at 3:59
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Sure, you can define a line through the origin in $\mathbb R^2$ as a set as you explained and as you recognized we can use lines in $\mathbb R^2$ to put a equivalence relation.

One change I would make though in your explanation is that instead of linear functions $f: \mathbb R \to \mathbb R$ defined by $f(x) = mx$ for some slope $m \in \mathbb R$, I would use instead $f: \mathbb R \to \mathbb R^2$ defined by $f(t) = vt$ for some direction vector $v \in \mathbb R^2 \setminus \{(0, 0)\}$ (think of the latter as a parametrized line). The reason the latter is better is because we no longer have to think of one coordinate as a function of the other and so we can include vertical lines (and really any kind of curves in $\mathbb R^2$) as well.

Purely as a set, we can define lines with direction $v \in \mathbb R^2 \setminus \{(0, 0)\}$ as $$ \ell_v = \{vt \in \mathbb R^2: t \in \mathbb R\}. $$ Here is something you might be interested in related to your observations: we can go further and talk about the set of lines in $\mathbb R^2$ defined as $$ L = \{\ell_v \subset \mathbb R^2: v \in \mathbb R^2 \setminus \{(0, 0)\}\}. $$ Since directions determine the line, an equivalent definition is $$ L = (\mathbb R^2 \setminus \{(0, 0)\})/\sim $$ where we quotient using the equivalence relation $\sim$ on $\mathbb R^2 \setminus \{(0, 0)\}$ defined by $v \sim w$ if $v = \lambda w$ for some $\lambda \in \mathbb R \setminus \{0\}$. Sets with more structure is usually more interesting than without and often times there are natural structures. In this case it's natural to want to talk about when two lines are "close" (we would like to say that if the slopes/directions are very similar with minimal error then the lines themselves are close). In this case we can put a "topology" on $L$ and this is usually called $\mathbb R P^1$ (whose points are lines in $\mathbb R^2$). This space is actually in fact just a circle (where again, each point on the circle represents a line in $\mathbb R^2$). To see this recall that each line has an associated slope $m \in \mathbb R$ but as $m \to +\infty$ the line becomes vertical and similarly as $m \to -\infty$ the line becomes vertical. Let's geometrically add points $+\infty$ and $-\infty$ which represent "infinite and negative infinite slopes". But by our observation, if we want these slopes to represent lines then we have to join $+\infty$ and $-\infty$ to a single point, and let's call it $\infty$. Geometrically we can visualize it as the real line $\mathbb R$ and then taking the two ends and connecting them to form a circle. See https://en.wikipedia.org/wiki/Real_projective_line to see a picture and more information.

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  • $\begingroup$ Thank you very much. Your insight is helpful, and corresponds exactly to the kind of ouverture I wanted to have on the notions. $\endgroup$ – freehumorist Dec 30 '18 at 6:10
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    $\begingroup$ Glad this helped. $\endgroup$ – Pratyush Sarkar Dec 30 '18 at 6:30

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