November 28th, 2017, 06:44 PM  #11 
Senior Member Joined: Aug 2012 Posts: 2,045 Thanks: 584  
November 28th, 2017, 08:59 PM  #12 
Senior Member Joined: Aug 2012 Posts: 2,045 Thanks: 584 
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: 7,445 Thanks: 2499 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.


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line, rational, straight, touch 
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