# Prove that any point I pick on the plane will lie on a line defined by integer coordinates and $(0,0)$?

I was playing around with the idea of infinity, and I thought of this question. It may or may not have a solution, but nonetheless it is interesting.

I start at the origin of a cartesian plane $(0,0)$. Then, I draw a line to each and every lattice point. $~$i.e. I draw a line from $(0,0)$ to all $(x,y)$ for $x,y \in \mathbb Z$.

It might look something like this, except there would obviously be a lot more lines:

The question is to prove (or disprove) that any point $(p,q)$ for $p,q \in \mathbb R$ I pick will lie on a line.

For example, I'll pick the point $(2.5,-7)$. That lies on the line created by joining $(0,0)$ with $(5,-14)$. It is simple to prove rational numbers, as you just multiply until you get a whole number.

I am more interested in the irrationals. Say I pick the point $(\pi, \sqrt 2)$. Because the lattice points go on forever, intuitively, there must be a line which the point $(\pi, \sqrt 2)$ lies on. But I have no way to find it. Is this some sort of a paradox?

• Your lines all have rational slope. That means you can only hit $(x,y)$ if $\frac yx\in \mathbb Q$.In particular, $(\pi, \sqrt 2)$ isn't on any of your lines.
– lulu
Commented Jan 13, 2018 at 1:27
• Congratulations; you've discovered for yourself the fundamental difference between rational and irrational numbers! (I'm not being sarcastic, lest there be any doubt—that's very well done.) Commented Jan 13, 2018 at 3:55

Suppose I have the point $(\pi, \sqrt{2})$, and there were a line $y = mx$ that it lies on. That line would need slope $m = \frac{\sqrt{2}}{\pi}$. What lattice point can we choose with that kind of slope? We can't choose any, since, for all lattice points $(a,b)\in\mathbb{Z}^2$, the resulting slope is a rational number $\frac{b}{a}\in\mathbb{Q}$, and $m$ is irrational.
You might have tried a simpler proof, where "the lattice points go on forever", by noticing that the line $y = mx$ maps the real line onto the plane. The number of lines can be enumerated; the space of lines has the same cardinality as $\mathbb{R}^1\times\mathbb{N}^2$, which has equal cardinality as $\mathbb{R}^2$. Surely this means they're the same space, like a space-filling curve!