I am looking at constructible points in abstract algebra, particularly in $\mathbb{C}$. Alongside a proof of a theorem, I came across this expression which I cannot work out how it's been derived. It just comes out as follows, but some notations;

$L(z_1,z_2)$ represents the straight line that connects points $z_1,z_2 \in \mathbb{C}$. $\mathbb{Q}^{Py}$ represents the Pyhtagorean Closure of $\mathbb{Q}$.

The theorem is stated as follows

A point $z \in \mathbb{C}$ is constructible if and only if $z \in \mathbb{Q}^{Py}$.

Omitting the bits I understood(The proof uses induction), the specific part I don't understand is,

Say $z$ lies on a "line meeting another line", namely $\{z\} \in L(z_1,z_2) \cap L(z_3,z_4)$ where the lines are distinct. Then $\exists a,b \in \mathbb{R}$ such that



Well, in all honesty, it looks familiar; points on a line that goes through $2$ distinct points. I think the above expressions come from rather elementary facts and ideas but as much as I am embarrassed to say this, I can't see how it gets derived. Considering the Complex plane as $\mathbb{R}^2$, taking $x,y$ coordinates, I thought I could find $a$ or $b$ wrp to $z_j=x_j+iy_j$ but I cannot.

Would someone show me how that part is derived?


You have the right idea, you should be considering it as $\mathbb{R}^2$. The line $L(z_1,z_2)$ can be defined as the image of the straight-line path between its endpoints:

$$ f : [0,1] \rightarrow \mathbb{C} \\ f(t) = z_1 + t(z_2 - z_1) \\ = tz_2 + (1-t)z_1 $$

So any point on the line can be represented as $f(t)$ for some $t$. The step you're looking at is just choosing $a$ and $b$ to be the appropriate values for the point being considered, in terms of the two different lines it is on.

  • $\begingroup$ I replaced the idea with vectors and I think I got it; I arrived at the same form as yours up there, thanks for the help! $\endgroup$ – John Trail Feb 5 '16 at 18:39

Rewrite $z = a z_1 + (1 − a) z_2$ as:

$$ a ( z - z_1) + (1 - a) (z - z_2) = 0 $$

Since $a \in \mathbb R$ this means that $z - z_1$ and $z - z_2$ have a real ratio, which happens iff $z$ is on the $z_1 z_2$ line.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.