TITLE: Proof: Is there a line in the xy plane that goes through only rational coordinates?
QUESTION [11 upvotes]: Question: Is there a line in the XY plane that has all rational coordinates. Prove your answer. 
Idea: There is most certainly not. I believe it can be shown that between any 2 rational points that there is at least one irrational coordinate. Therefore, there can not be a line that contains only rational points. The issue is that I am not sure how to show this. Any ideas? I am also open to any other ideas of how to do this. Thanks.
Note: this is for an intro to proofs study guide. So, I would prefer not to use advanced theorems.

REPLY [2 votes]: Well no.
We can nitpick.  Is it assumed your plane  is the usual two dimension $\mathbb R \times \mathbb R$ plane?  We can argue that if your "universe" is just the rational numbers then $\mathbb Q \times \mathbb Q$ is also a plane but this "universe" is defined to only have rationals so lines will only have rational coordinates.
But that's me being pedantic and trying to intimidate.
If your "universe" is the real numbers then:
In general a line has the formula $y = mx + b$ (there is one type of exception).  And we know irrational numbers exists.  So for an irrational $z$ the point $(z, mz + b)$ exist and this is not a rational coordinate as $z$ is not rational.
The exception is vertical lines.  These are of the form $x = c$.  But these lines pass through every value of $y$.  (Actually, all non-horizontal lines pass through all values of $y$ too.  Vertical lines pass through all $y$, horizontal lines pass through all $x$, and all other lines pass through all $x$ and all $y$).  So this will have the point $(c, z)$ which is not a rational coordinate.