My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Thanks Tree4Thanks
  • 3 Post By Maschke
  • 1 Post By Maschke
Reply
 
LinkBack Thread Tools Display Modes
November 14th, 2017, 01:12 PM   #1
Newbie
 
Joined: Oct 2017
From: Italy

Posts: 13
Thanks: 0

Straight line that doesn't touch any rational

Does exist a straight line (in $\mathbb{R}^2$) that touches only irrational numbers?
Berker is offline  
 
November 14th, 2017, 01:36 PM   #2
Senior Member
 
Joined: Aug 2012

Posts: 1,628
Thanks: 413

Quote:
Originally Posted by Berker View Post
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.
Thanks from topsquark, Joppy and JeffM1

Last edited by Maschke; November 14th, 2017 at 01:50 PM.
Maschke is online now  
November 14th, 2017, 05:07 PM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Maschke View Post
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 View Post
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.
v8archie is offline  
November 14th, 2017, 05:10 PM   #4
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,641
Thanks: 959

Math Focus: Elementary mathematics and beyond
Quote:
Originally Posted by Berker View Post
Does exist a straight line (in $\mathbb{R}^2$) that touches only irrational numbers?
x-coordinate, y-coordinate or both? If both, no.
greg1313 is offline  
November 14th, 2017, 05:22 PM   #5
Senior Member
 
Joined: Aug 2012

Posts: 1,628
Thanks: 413

Quote:
Originally Posted by v8archie View Post
$(0,0)$ is perhaps a more obvious example.
Good point, missing the obvious on my part.
Maschke is online now  
November 14th, 2017, 06:05 PM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Maschke View Post
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.
v8archie is offline  
November 14th, 2017, 09:20 PM   #7
Senior Member
 
Joined: Aug 2012

Posts: 1,628
Thanks: 413

Quote:
Originally Posted by v8archie View Post
$(0,0)$. 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.
Maschke is online now  
November 14th, 2017, 09:37 PM   #8
Newbie
 
Joined: Oct 2017
From: Italy

Posts: 13
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?
Berker is offline  
November 14th, 2017, 09:45 PM   #9
Senior Member
 
Joined: Aug 2012

Posts: 1,628
Thanks: 413

Quote:
Originally Posted by Berker View Post
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 $x$.
Thanks from Berker

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

  My Math Forum > College Math Forum > Real Analysis

Tags
line, rational, straight, touch



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Straight Line workings jl1974 Algebra 7 September 14th, 2017 12:55 AM
a curve to a straight line waynebutt Algebra 9 April 11th, 2016 06:32 AM
straight line outsos Algebra 7 December 23rd, 2010 08:15 PM
another straight line question Kiranpreet Algebra 3 August 10th, 2008 08:45 AM
straight line Q Kiranpreet Algebra 2 August 9th, 2008 10:29 AM





Copyright © 2017 My Math Forum. All rights reserved.