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Problem: We say that a point (the Cartesian plane) is rational if both its coordinates are rational numbers. There are lines of the plane, not parallel to any axis that does not intersect any rational point?

My answer: Yes, there are. For example, consider $ y = x + \pi $ and show that it does not intersect any rational point. Each point on the line is given by $(x, x + \pi)$ and divided the two cases: 1) x is irrational -> point has not both coordinates rational.

2) x is rational, then $x + \pi$ is irrational because the sum of a rational and an irrational ($\pi$) generates a number irrazinale (hence the point has not both coordinates rational. So this line does not intersect any rational point. cvd

Is that okay?

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in title is always irrational? – user52188 Jan 20 '13 at 19:53
cases: 1) x is irrational -> point has both coordinates rational? – user52188 Jan 20 '13 at 19:55
I corrected now – Agenog Jan 20 '13 at 19:57
any line with inclination irrational do not intersect rational points – user52188 Jan 20 '13 at 19:59
@Edgar: $y = \pi x$ intersects $(0,0)$ – Hurkyl Jan 20 '13 at 20:03

EDIT: Thanks all for the feedback. The initial versions of this question was hard to interpret.

ORIGINAL: Your question is a little confusing since you seem to be talking about two separate questions at the same time. Let's break it down:

Definition: a point in $\mathbb{R}^2$ is rational if both $x$ and $y$ are rational; otherwise it's called irrational. A line is irrational if all points on that line are irrational.

The question (I think) you are asking is if there are any lines that aren't parallel to an axis that are irrational?

(Any line of the form $y = \alpha$ or $x = \alpha$ for any irrational $\alpha$ are irrational, but they are parallel to the axis)

Your example $y = x + \pi$ does not work, because in your proof you are assuming that the sum of two irrational points is always irrational. This is false; if $x = -\pi$ then $x + \pi = 0$ which is rational.

I claim that the line $y = \sqrt{2} + x\sqrt{3}$ is irrational. To show this, we need to prove that if $x = \frac{a}{b}$ is rational, then $\sqrt{2} + x\sqrt{3}$ is irrational. We'll prove by contradiction: suppose $\sqrt{2} + \frac{a}{b}\sqrt{3} = \frac{c}{d}$, where $a$, $b$, $c$, $d$ are integers. Then

$$ db\sqrt{2} + da\sqrt{3} = bc $$

However, if we square the whole equation we get

$$ 2(db)^2 + 3(da)^2 + d^2ab\sqrt{6} = (bc)^2 $$

which implies that

$$ \sqrt{6} = \frac{(bc)^2 - 2(db)^2 - 3(da)^2}{d^2ab} $$

is rational (remember, $a$, $b$, $c$, and $d$ are integers), a contradiction.

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Where is he assuming that the sum of two irrationals is irrational? If $x$ is irrational, the point $(x,y)$ is already irrational, so I don't see any problem with $y=x+\pi$. – Brusko651 Jan 20 '13 at 20:37
"your example $y=x+\pi$ does not work..." But it does work, as put in the question. If $x$ is irrational then $(x,y)$ is an irrational point (not both coordinates rational), and again if $x$ is rational then $y=x+\pi$ is irrational and again not both coordinates rational. The OP's proof does not use that the sum of two irrationals is irrational. – coffeemath Jan 20 '13 at 20:37
He does not assume that the sum of two irrationals is irrational. – David Mitra Jan 20 '13 at 20:37
I didn't assumed that the sum of two irrational is irrational!!! – Agenog Jan 20 '13 at 22:03

Your answer $y=x+\pi$ seems fine to me and your proof is also good. I don't see what could be wrong about it.

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