November 28th, 2017, 06:44 PM  #11 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  
November 28th, 2017, 08:59 PM  #12 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740 
Ok, I think I've got this. $y = \sqrt{2} x + \sqrt{3}$ passes through no rational point, because $\sqrt{2}$ and $\sqrt{3}$ are linearly independent over the rationals. That is, if $\frac{p}{q} = \frac{r}{s} \sqrt{2} + \sqrt{3}$ then $p = \frac{qr}{s} \sqrt{2} + q \sqrt{3}$ and $\frac{qr}{s} \sqrt{2} + q \sqrt{3}  p = 0$, which is impossible. https://math.stackexchange.com/quest...tovermathbbq It's interesting that using transcendentals doesn't work. For example it's unknown whether $\pi + e$ is rational. Nobody knows how to answer these types of questions. My apologies to the group for posting several times without thinking. Last edited by Maschke; November 28th, 2017 at 09:12 PM. 
November 29th, 2017, 02:50 AM  #13 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra  I gave the example $x+y=\pi$ earlier. It's guaranteed to have at least one irrational coordinate in every point $(x,y)$ by the closure of the rationals. It works with any irrational number as the constant term and any pair of rational coefficients for the variables.


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 13th, 2017 11:55 PM 
a curve to a straight line  waynebutt  Algebra  9  April 11th, 2016 05:32 AM 
straight line  outsos  Algebra  7  December 23rd, 2010 07:15 PM 
another straight line question  Kiranpreet  Algebra  3  August 10th, 2008 07:45 AM 
straight line Q  Kiranpreet  Algebra  2  August 9th, 2008 09:29 AM 