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November 28th, 2017, 06:44 PM   #11
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Quote:
Originally Posted by SDK View Post
$(-1,0)$
I officially wish I'd never started with this
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November 28th, 2017, 08:59 PM   #12
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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.
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November 29th, 2017, 02:50 AM   #13
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Originally Posted by Maschke View Post
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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