An example of a straight line passing through the origin and no other point in $\mathbb{Q}$ Consider all the straight lines passing through the origin $A=\{ f(x) : f(x)=ax$, $a\in R\}$
Do a function $g\in A$ exists such that $g(x)\notin \mathbb{Q}$ for every $x\in R$
My intuition says yes, since $A$~$[0, 2\pi]$ (every line is defined by it's angle) and $|[0,2\pi]|=\aleph$ and $|\mathbb{Q}^2 - \{0,0\}|=\aleph_0$ 
But why would that imply such a line exists? it only proves exists infinitely more point lines than rational numbers, but each line also goes through $\aleph$ points.
Is there an example of such a line? (in other words, a number $x$ such that $yx$ is irrational for every $y$)
 A: No, this is not possible. In fact far from avoiding rationals, every element of $A$ a bijection from $\mathbb{R} \to \mathbb{R} $ as for any $y \in \mathbb{R}$, $a\cdot y/a = y$ 
A: 1) Continuity.  If the line ever achieves a value $g(x_0) = a$ and a value $g(x_1) = b$, then for any $c$ between $a$ and $b$ there is an $x_2$ between $x_0$ and $x_1$ so that $g(x_2) = c$.
Now between any $a$ and $b$ there will an infinite number of rational numbers.  The line can not "jump" over them.  It must "go through" them.
2) Direct solution.  Let $q \in \mathbb Q$.  Let $g(x) = a*x; a\ne 0$ then $f(\frac qa) = a\frac qa = q$.
3) D'oh:  $0 \in \mathbb Q$.  $g(0) = 0 \in \mathbb Q$.
4) Now as to your angle observation.  Let $\alpha \in [0, 2\pi)$ and let the line determined by the angle by $\{(r*\cos \alpha, r*\sin \alpha)\}$ the question is.  Is there any an $\alpha$ so that $r*\sin \alpha$ is never rational?  The answer is ... no.  As in 2) above, we simply let $r = \frac q{\sin \alpha}$ for some $q \in \mathbb Q$ (assume $\alpha \ne 0, \pi, 2\pi$)  [This is the same as assuming $a \ne 0$.]
====
BUT all said and done, a better question is let $\{(x, g(x))\}$ be a graph of a line.  Is it possible that for all $x \ne 0$ that $(x, g(x)) \not \in \mathbb Q \times \mathbb Q$?  
In other words is it possible for $g(x) \not \in \mathbb Q$ for all $x \in \mathbb Q; x \ne 0$?
The answer to THAT question is !YES!.  Indeed if $a \not \in \mathbb Q$ and $x = q \in \mathbb Q$ then $g(x) = a*q \not \in \mathbb Q$.  So for instance $g(x) = \pi x$ or $h(x) = \sqrt{2} x$ will never be rational for any non-zero rational $x$.
