Proof for number of rational ordered pairs on a line

Question

It is given that the function $y=ax+b,\; a \neq 0$ has an ordered pair $(x,y)=( \sqrt{2}, 0)$. Prove that $y=ax+b$ does not have two or more rational ordered pairs.

From the above I know that $-\frac ba= \sqrt2$.

Answer 1

Assume that in fact there are two rational points $(q,r)$ and $(s,t)$. Then the slope of this line is just

$$a={t-r\over s-q}$$

which is also a rational. However, in that case

$$0=\sqrt 2\cdot a+b$$

so $b=-a\sqrt 2$

But then $r=aq-a\sqrt 2$ is what we conclude, and we know this is impossible, since $q-r/a=\sqrt 2$ would say $\sqrt 2$ is rational. We conclude that the two hypotheses "the line has two rational points" and "the point $(\sqrt 2,0)$ is on the line" are incompatible, and cannot both be true.