Proof: Is there a line in the xy plane that goes through only rational coordinates?

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.

Solution 1:

Two proofs.

First one based on cardinality. A line has the cardinality of the continuum like $\mathbb R$, while $\mathbb Q$ is countable.

Second one. A line has an equation $ax+by+c=0$ with $(a,b) \neq (0,0)$. If $a \neq 0$ then $(-\frac{b \pi +c}{a},\pi)$ belongs to the line and the second coordinate of that point is not rational. While if $a=0$, $(\pi,-\frac{c}{b})$ belongs to the line (thanks to immibis comment).

Solution 2:

Unless the line is vertical you could always pick an irrational $x$ (and consider the point $(x,y)$, on the line, which does not have all rational coordinates, regardless of whether $y$ happens to be rational or irrational). If the line is vertical, then you could pick any irrational $y$ and consider the point $(x,y)$. Assuming you work with straight lines. (Also, assuming you know that irrational numbers exist, e.g. $\sqrt{2}$, as from other comments and answers this seems to be part of your question. So, again, stated differently, if the line is not vertical then $(\sqrt{2},y)$ is on the line for some $y$ and $\sqrt{2}$ is irrational. If the line is vertical, then $(x,\sqrt{2})$ is on the line for some (unique) $x$, and this time the second coordinate is $\sqrt{2}$, irrational. It is also true that there are infinitely many points like that on the line, e.g. using $\frac pq\sqrt{2}$ or $\frac pq\pi$ and $\frac pq e$ in place of $\sqrt2$, they are all irrational whenever $\frac pq$ is rational, and many more, they form a dense set and you seem to make a comment related to that, but for the purposes of answering your question, just one point with (at least) one irrational coordinate suffices.)

Solution 3:

By a line, I take it to mean a the set of points satisfying $y=mx+b$.

If $b$ is irrational then let $x=0$. Assume $b$ is rational. If $m$ is irrational let $x=1$. Assume $m$ is rational. Let $x$ be your favorite irrational number.

Solution 4:

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.