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 November 14th, 2017, 01:12 PM #1 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Straight line that doesn't touch any rational Does exist a straight line (in $\mathbb{R}^2$) that touches only irrational numbers?
November 14th, 2017, 01:36 PM   #2
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Quote:
 Originally Posted by Berker Does exist a straight line (in $\mathbb{R}^2$) that touches only irrational numbers?
Do you mean that it only passes through points both of whose coordinates are irrational? Or only through points at least one of which is irrational?

Consider first lines through the origin, which are described by functions of the form $y = mx$.

If $x = \frac{1}{m}$ then the line passes through the point $(\frac{1}{m}, 1)$. So clearly every line through the origin passes through at least one point with a rational coordinate.

This generalizes to arbitrary lines. If $f(x) = mx + b$ then

$f(\frac{-b}{m}) = 0$ so that regardless of what $m$ and $b$ are, some point on the graph has $0$ as its second coordinate.

There's a related question about what happens when we confine this question to the integer lattice: the set of points in the plane whose coordinates are both integers. In this case lines through the origin having rational slope must eventually hit some point in the lattice; and a line with irrational slope never does. In other words if the plane is a forest and the points are trees, we can find a straight-line path through the forest that never hits a tree. Any line with irrational slope will do.

Finally there's yet another way to interpret this. A rational point is defined as a point where both coordinates are rational. Is it possible to have a line that passes only through non-rational points? Yes. $f(x) = \pi x$ is one such. If $x$ is rational, $f(x)$ is irrational. And the only way to make $f(x)$ rational is for $x$ to be irrational, for example $\frac{1}{\pi}$, $\frac{2}{\pi}$, etc.

Last edited by Maschke; November 14th, 2017 at 01:50 PM.

November 14th, 2017, 05:07 PM   #3
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Quote:
 Originally Posted by Maschke So clearly every line through the origin passes through at least one point with a rational coordinate.
$(0,0)$ is perhaps a more obvious example.

Quote:
 Originally Posted by Maschke regardless of what $m$ and $b$ are, some point on the graph has $0$ as its second coordinate.
Unless $m=0$.

Every non-vertical line intersects every line of the form $x=c$ where $c \in \mathbb R$. For any rational value of $c$, we therefore have a rational coordinate. A similar argument deals with non-horizontal lines and the combination covers all possibilities.

The line $x+y=\pi$ never has both coordinates being rational as this would require $\pi$ to be rational because the rationals are closed.

November 14th, 2017, 05:10 PM   #4
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Quote:
 Originally Posted by Berker Does exist a straight line (in $\mathbb{R}^2$) that touches only irrational numbers?
x-coordinate, y-coordinate or both? If both, no.

November 14th, 2017, 05:22 PM   #5
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Quote:
 Originally Posted by v8archie $(0,0)$ is perhaps a more obvious example.
Good point, missing the obvious on my part.

November 14th, 2017, 06:05 PM   #6
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Quote:
 Originally Posted by Maschke Yes. $f(x) = \pi x$ is one such. If $x$ is rational, $f(x)$ is irrational. And the only way to make $f(x)$ rational is for $x$ to be irrational, for example $\frac{1}{\pi}$, $\frac{2}{\pi}$, etc.
$(0,0)$. Cough.

November 14th, 2017, 09:20 PM   #7
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Quote:
 Originally Posted by v8archie $(0,0)$. Cough.
LOL. My bad. Again. I think $\pi x + \pi$ works. Can't pass through a rational point.

Last edited by Maschke; November 14th, 2017 at 09:37 PM.

 November 14th, 2017, 09:37 PM #8 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Thanks for your replies. So if I have a numerical sequence of rational numbers anddivergent (${r_n}$), the set ${(x,y) \in \mathbb{R}^2 : y= r_n x}$ doesn't cover all the plane?
November 14th, 2017, 09:45 PM   #9
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Quote:
 Originally Posted by Berker Thanks for your replies. So if I have a numerical sequence of rational numbers anddivergent (${r_n}$), the set ${(x,y) \in \mathbb{R}^2 : y= r_n x}$ doesn't cover all the plane?
For $x = 1$ you can only hit points of the form $(1, r_n$). All of those are rational points. That holds for any rational $x$.

For any $x$ in general there are only countably many $y$ values. You're missing most of the points on the vertical line above every $x$.

Last edited by Maschke; November 14th, 2017 at 09:52 PM.

November 28th, 2017, 07:02 PM   #10
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Quote:
 Originally Posted by Maschke LOL. My bad. Again. I think $\pi x + \pi$ works. Can't pass through a rational point.
$(-1,0)$

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