My Math Forum Straight line that doesn't touch any rational

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November 28th, 2017, 06:44 PM   #11
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Quote:
 Originally Posted by SDK $(-1,0)$
I officially wish I'd never started with this

 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...t-over-mathbbq 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
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Quote:
 Originally Posted by Maschke 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.
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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