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  
Senior Member Joined: Aug 2012 Posts: 2,099 Thanks: 604  Quote:
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 straightline 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 nonrational 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  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra  Quote:
Quote:
Every nonvertical 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 nonhorizontal 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 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond  
November 14th, 2017, 05:22 PM  #5 
Senior Member Joined: Aug 2012 Posts: 2,099 Thanks: 604  
November 14th, 2017, 06:05 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra  
November 14th, 2017, 09:20 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,099 Thanks: 604  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  
Senior Member Joined: Aug 2012 Posts: 2,099 Thanks: 604  Quote:
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 
Senior Member Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics  

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